Introduction to Quantum Mechanics
Welcome to the fascinating world of quantum mechanics, a realm where classical physics breaks down and the rules governing nature become fundamentally altered. This introductory lecture is designed to clarify the necessity of quantum mechanics, provide historical context, and illustrate pivotal experiments that led to the development of this revolutionary theory.
The Need for Quantum Mechanics
Historical Context
Distance yourself to the year 1900: physicists believed that by applying classical mechanics, they could describe the universe's intricate workings. However, the advent of electricity changed everything. By the turn of the century, scientific advancements had reached an all-time high, aligning with various quotes from experts like Joseph-Louis Laplace, who asserted that perfect knowledge of the state of the universe predicts its future.
Yet, as experimental anomalies began to surface, it was clear that classical physics couldn’t explain them. Unexplainable phenomena like the black body spectrum, photoelectric effect, and atomic spectrum revealed deep flaws in the classical understanding of light and matter interactions, representing the seeds of quantum theory.
Historical Key Experiments
- Black Body Radiation:
- Classical predictions failed at high frequencies, leading to the ultraviolet catastrophe.
- Photoelectric Effect:
- Classical wave theory of light could not explain ejected electrons. Einstein proposed the photon to solve this, earning him the Nobel Prize.
- Atomic Spectra:
- Lines observed in atomic emissions could not be reconciled with classical models.
Key Concepts of Quantum Mechanics
Wave Function
A wave function, often denoted \( \psi \, \text{or } \Psi \, ext{in quantum mechanics, conveys the system's state, providing probabilities rather than certainties.}
- The key characteristics of wave functions are:
- They are complex functions of position and time.
- |
Operators
Operators are used to extract physical information from the wave function. For example:
- Position Operator (x): Simply multiplying the wave function by position.
- Momentum Operator (p): Given by \(-i\hbar\frac{\partial}{\partial x} ).
The Schrödinger Equation
The foundational equation of quantum mechanics, relating the wave function's time evolution to the energy and potential of the system:
- Time-dependent Schrödinger Equation: $$i\hbar \frac{\partial \Psi}{\partial t} = -\frac{\hbar^2}{2m} \frac{\partial^2 \Psi}{\partial x^2} + V(x)\Psi$$
- Time-independent Schrödinger Equation: $$\hat{H}\Psi = E\Psi$$
Quantum Systems: Types & Examples
The Particle in a Box
Consider a one-dimensional particle confined in a box with infinite potential walls. The wave function of this particle has defined boundary conditions:
- Solutions are quantized: Energies are proportional to the square of integers.
- Wavefunctions form standing waves:
- $$\Psi_n(x) = \sqrt{\frac{2}{L}} \sin\left(\frac{n\pi x}{L}\right)$$
Quantum Harmonic Oscillator
This model describes a particle attached to a spring, leading to quantized energy levels:
- Energy Levels:
$$E_n = \left(n + \frac{1}{2}\right)\hbar\omega,$$ where \(n = 0, 1, 2, \ldots \) - Wavefunctions:
$$\Psi_n(x) = \sqrt{\frac{m\omega}{\pi\hbar}} e^{-\frac{m\omega}{2\hbar} x^2} H_n\left(\sqrt{\frac{m\omega}{\hbar}} x\right),$$ where (H_n) are Hermite polynomials.
The Uncertainty Principle
The famous uncertainty principle, encapsulated as (\Delta x \Delta p \geq \frac{\hbar}{2}), illustrates the limit of precision in measuring certain pairs of observable properties:
- Position and momentum cannot both be measured precisely at the same time.
- Wider implications on energy and time as well with (\Delta E \Delta t \geq \frac{\hbar}{2}).
Conclusion
Quantum mechanics presents a paradigm shift from classical physics, chronicling how light and matter behave on atomic and subatomic scales. Through understanding key concepts like the wave function and uncertainty principles, we have only scratched the surface of a deep and mystical field of physics. Despite its counterintuitive nature, quantum mechanics continues to be crucial in explaining the intricate fabric of our universe.
welcome to quantum mechanics my name is brent carlson since this is the first lecture on
quantum mechanics um we ought to have some sort of an introduction and what i want to do to
introduce quantum mechanics is to explain first of all why it's necessary and second of all to put it in
historical context to well i'll show one of the most famous photographs in all of physics
that really gives you a feel for the brain power that went into the construction of
this theory and hopefully we'll put it in some historical context as well so you can
understand where it fits in the broader philosophy of science but the the main goal of this lecture is
about the need for quantum mechanics which i really ought to just have called why do we need quantum mechanics
uh this subject has a reputation for being a little bit annoying so why do we
bother with it well first off for some historical context
imagine yourself back in 1900 turn of the century science has really advanced a lot we
have electricity we have all this fabulous stuff that electricity can do
and even almost 100 years before that physicists thought they had things
figured out there's a famous quote from laplace given for one instant an intelligence which could comprehend all
the forces by which nature is animated and the respective position of the beings which compose it nothing would be
uncertain in the future as the past would be present to its eyes now
maybe you think intelligence which can comprehend all the forces of nature is a bit of a stretch
and maybe such a being which can know all the respective positions of everything in the universe is a bit of a
stretch as well but the feeling at the time was that if you could do that you would know
everything if you had perfect knowledge of the present you could predict the future
and of course you can infer what happened in the past and everything is connected by one
unbroken chain of causality now in 1903 albert michaelson another famous quote from that time period
said the more important fundamental laws and facts of physical science have all been
discovered our future discoveries must be looked for in the sixth place of decimals
now this sounds rather audacious this is 1903 and he thought that the only thing that we had left to nail down was the
part in a million level precision well to be fair to him he wasn't talking about never discovering new fundamental
laws of physics he was talking about really astonishing discoveries like the discovery of uranus on the basis of
orbital perturbations of neptune never having seen the planet uranus before they figured out that it
had to exist just by looking at things that they had seen that's pretty impressive
and michaelson was really on to something precision measurements are really really useful especially today
but back in 1903 it wasn't quite so simple and michaelson probably regretted that remark for the rest of his life
the attitude that i want you guys to take when you approach quantum mechanics though is not this
sort of 1900s notion that everything is predicted it comes from shakespeare horatio says one
oh day and night but this is wondrous strange to which hamlet replies one of the most
famous lines in all of shakespeare and therefore as a stranger give it welcome there are more things in heaven and
earth horatio than are dreamt of in your philosophy so that's the attitude i want you guys
to take when you approach quantum mechanics it is wondrous strange
and we should give it welcome there are some things in quantum mechanics that are deeply non-intuitive
but if you approach them with an open mind quantum mechanics is a fascinating
subject there's a lot of really fun stuff that goes on now to move on to the necessity for
quantum mechanics there were some dark clouds on the horizon even at the early 20th century
michelson wasn't quite having a big enough picture in his mind when he said that everything was down to the sixth
place of decimals the dark clouds on the horizon at least according to kelvin here
were a couple of unexplainable experiments one the black body spectrum now a black
body you can just think of as a hot object and a hot object like for example the
coils on an electrical stove when they get hot will glow and the question is what color do they
glow do they glow red they go blue what is the distribution of radiation that is emitted by a hot object
another difficult to explain experiment is the photoelectric effect if you have some light
and it strikes a material electrons will be ejected from the surface
and as we'll discuss in a minute the properties of this experiment do not fit
what we think we know about or at least what physicists thought they knew about the physics of light in the
physics of electrons at the turn of the 20th century the final difficult experiment to
explain is bright line spectra for example if i have a flame coming from say a bunsen burner
and i put a chunk of something perhaps sodium in that flame
it will emit a very particular set of frequencies that looks absolutely nothing like a black body
we'll talk about all these experiments in general or in a little bit more detail in a minute or two but
just looking at these experiments now these are all experiments that are very difficult to explain
knowing what we knew at the turn of the 20th century about classical physics they're also also experiments that
involve light and matter so we're really getting down to the
details of what stuff is really made of and how it interacts with the things around it
so these are some pretty fundamental notions and that's where quantum mechanics really got its start
so let's pick apart these experiments in a little more detail the black body spectrum as i mentioned
you can think of as the light that's emitted just by a hot object
and while hot objects have some temperature associated with them let's call that t
the plot here on the right is showing very qualitatively i'll just call it the intensity of the
light emitted as a function of the wavelength of that light
so short wavelengths high energy long wavelengths low energy now if you look at t equals 3 500 kelvin
curve here it has a long tail to long wavelengths and it cuts off pretty quickly as you go
to short wavelength so it doesn't emit very much high energy light whereas if you have a much hotter object
5500 kelvin it emits a lot more high energy light the red curve here is much higher than the black curve
now if you try to explain this knowing what early 20th century physicists knew
about radiation and about electrons and about atoms and how they could possibly emit
light you get a prediction and it works wonderfully well up until about here at which point it
blows up to infinity um infinities are bad in physics this is the the rayleigh jeans law and
it works wonderfully well for long wavelengths but does not work at all for short wavelengths that's called the
ultraviolet catastrophe if you've heard that term on the other end of things
if you look at what happens down here well it's not so much a prediction but an observation
but there's a nice formula that fits here so on one side we have a prediction
that works well on one end but doesn't work on the other and on the other hand we have a sort of
empirical formula called veen's law that works really well at the short wavelengths but well also blows up to
infinity at the long wavelengths both of these blowing up things are a problem the question is how do you get something
that explains both of them this is the essence of the the black body spectrum and how it was difficult
to interpret in the context of classical physics the next experiment i mentioned is the
photoelectric effect this is sort of the opposite
problem it's not how a material emits light it's how light interacts with the material
so you have light coming in and the experiment is usually done like this
you have your chunk of material typically a metal and when light hits it electrons are
ejected from the surface hence the electric part of the photoelectric effect
and you do all this in a vacuum and the electrons are then allowed to go across a gap to some other material
another chunk of metal where they strike this metal and the experiment is usually done like
this you connect it up to a battery so you have your material on one side
and your material on the other you have light hitting one of these materials and ejecting electrons
and you tune the voltage on this battery such that your electrons when they're ejected never quite make it
so the electric field produced by this voltage is opposing the motion of the electrons
when that voltage is just high enough to stop the motion of the electrons keep them from completely making it all the
way across we'll call that the stopping voltage now
it turns out that what classical e m predicts as i mentioned doesn't match what
actually happens in reality but let's think about what does classical e m predict here
well classical electricity and magnetism says that
electromagnetic waves here have electric fields and magnetic fields associated with them
and these are propagating waves if i increase the intensity
of the electromagnetic wave that means the magnitude of the electric field
involved in the electromagnetic wave is going to increase and if i'm an electron
sitting in that electric field the energy i acquire is going to increase
that means the stop is going to increase because i'll have to have more voltage to stop a higher
energy electron as would be produced by higher intensity beam of light the other parameter of this incoming
light is its frequency so we can think about varying the frequency if i increase the frequency i have more
intense light now that doesn't say anything about the string sorry if i increase the frequency
i don't necessarily have more intense light the electric field magnitude
is going to be the same which means the energy and the stopping voltage
will also be the same now it turns out what actually happens in reality
does not match this at all in reality when the intensity increases the energy
which i should really write as v stop the stopping voltage necessary doesn't change
and when i increase the frequency the voltage necessary to stop those electrons increases
so this is sort of exactly the opposite what's going on here that's the puzzle in explaining the
photoelectric effect just to briefly check your understanding consider these plots of stopping voltage
as a function of the parameters of the incident light and check off which you think shows the
classical prediction for the photoelectric effect the third experiment that i mentioned is
bright line spectra and as i mentioned this is what happens if you take a flame
or some other means of heating a material like the bar of sodium i mentioned
earlier this will emit light
and uh in this case the spectrum of light from red to blue of sodium
looks like this oh actually i'm sorry that's not sodium that's mercury
uh the these are four different elements hydrogen mercury neon
and xenon and instead of getting a broad continuous distribution like you would
from a black body under these circumstances where you're talking about gases you get these very
bright regions it's the spectrum instead of looking like a smooth curve like this
looks like spikes those bright lines are extraordinarily
difficult to explain with classical physics and this is really the straw that broke the camel's back broke
classical physics is back that really kicked off quantum mechanics how do you explain this
this is that famous photograph that i mentioned this is really the group of people who first built quantum mechanics
now i mentioned three key experiments the
black body spectrum this guy figured that out this is plonk the photoelectric effect
this guy who i hope needs no introduction this is einstein that out
this is the paper that won einstein the nobel prize and
as far as the brightline spectra of atoms it took a much longer time to figure out
how all of that fit together and it took a much larger group of people but they all happen to be present
in this photograph there's this guy and this guy
and these two guys and this guy this photograph is famous because
these guys worked out quantum mechanics but that's not the only these aren't the only famous people in
this photograph you know this lady as well this is marie curie this is lorenz
which if you studied special relativity you know einstein used the lorentz transformations
pretty much everyone in this photograph is a name that you know i went through and looked up who these people were
these were all of the names that i recognized which doesn't mean that the people whose names i didn't recognize
weren't also excellent scientists for example ctr wilson here one of my personal favorites inventor of the cloud
chamber this is the brain trust that gave birth to quantum mechanics and it was quite a
brain trust you had some of the most brilliant minds of the century working on some of the
most difficult problems of the century and what's astonishing is they didn't really like what they found they
discovered explanations that made astonishingly accurate predictions but throughout the history you keep seeing
them disagreeing like no that can't possibly be right not necessarily because
the predictions were wrong or they thought there was a mistake somewhere but because they just disliked the
nature of what they were doing they were upending their view of reality einstein in particular really disliked quantum
mechanics to the day that he died just because it was so counter-intuitive and so with that introduction
to a counter-intuitive subject i'd like to remind you again of that shakespeare quote
there are more things in heaven and earth horatio than are dreamt of in your philosophy
uh try to keep an open mind and hopefully we'll have some fun at this
knowing that quantum mechanics has something to do with explaining the interactions of light and matter for
instance in the context of the photoelectric effect or
black body radiation or bright line spectra of atoms and molecules one might be led to the question of when
is quantum mechanics actually relevant the domain of quantum mechanics is unfortunately not a particularly simple
question when does it apply well on the one hand you have classical physics
and on the other hand you have quantum physics and
the boundary between them is not really all that clear on the classical side you have things that are certain
whereas on the quantum side you have things that are uncertain what that means in the context of
physics is that on the classical side things are predictable they may be chaotic and difficult to
predict but in principle they can be predicted well on the quantum side things are
predictable too but with a caveat in the classical side you determine
everything basically every property of the system can be
known with perfect precision whereas in quantum mechanics what you predict are probabilities
and learning to work with probabilities is going to be the first step to getting comfortable with quantum mechanics
the boundary between these two realms when the uncertain and probabilistic effects of quantum mechanics start to
become relevant is really a dividing line between things that are large
and things that are small and that's not a particularly precise way of stating things
doing things more mathematically quantum mechanics applies for instance when angular momentum
l is on the scale of planck's constant or the reduced flux constant h bar now
h bar is the fundamental scale of quantum mechanics and it appears not only in the context of angular momentum
planck's constant has units of angular momentum so if your angular momentum is of order planck's constant or smaller
you're in the domain of quantum mechanics we'll learn more about uncertainty
principles later as well but uncertainties in this context have to do with
products of uncertainties for instance the uncertainty in the momentum of a particle times the
uncertainty in the position of the particle this if it's comparable to planck's
constant is also going to give you the realm of quantum mechanics energy and time also have an uncertainty
relation again approximately equal to planck's constant most fundamentally
the classical action when you get into more advanced studies of classical mechanics you'll learn
about a quantity called the action which has to do with the path the system takes as it evolves in space and time
if the action of the system is of order planck's constant then you're in the quantum mechanical
domain now klonk's constant is a really small number it's 1.05
times 10 to the negative 34 kilogram meters squared per second times 10 to the negative 34 is a small
number so if we have really small numbers then
we're in the domain of quantum mechanics in practice these guys are the most useful
whereas this is the most fundamental but we're more interested in useful things than we are in fundamental things
after all for example the electron in the hydrogen atom now
you know from looking at the bright line spectra that this should be in the domain of quantum mechanics
but how can we tell well to use one of the uncertainty principles
as a calculation consider the energy the energy
of an electron in a hydrogen atom is you know let's say
about 10 electron volts if we say that's p squared over 2m using the classical kinetic energy
relation between momentum and kinetic energy that tells us
that the momentum p is going to be about 1.7 times 10 to the
minus 24th kilogram meter square sorry kilogram where'd it go where's my eraser
kilogram meters per second now this suggests that the momentum of
the electron is you know non-zero but if the hydrogen atom itself is not moving we know the average
momentum of the electron is zero so if the momentum of the electron is going to be zero with
still some momentum being given to the electron this is more the uncertainty in
the electron momentum than the electron momentum itself the next quantity if we're looking at
the uncertainty relation between momentum and position is we need to know the size of or the uncertainty in the
position of the electron which has to do with the size of the atom now the size of the atom
that's about 0.1 nanometers which if you don't remember the conversion from nanometers is 10 to the
minus 10th meters so let's treat this as delta x our uncertainty in position
because we don't really know where the electron is within the atom so this is a reasonable guess at the uncertainty
now if we calculate these two things together delta p delta x you get something
i should say this is approximate because this is very approximate 1.7 times 10 to the negative 34th
and if you plug through the units it's kilogram meters squared per second this
is about equal to h-bar so this tells us that quantum mechanics is definitely important here
we have to do some quantum in order to understand this system as an example
of another small object that might have quantum mechanics relevant to it this is one that we would
actually have to do a calculation i don't know intuitively whether a speck of dust in a light breeze is in the
realm of quantum mechanics or classical physics now
i went online and looked up some numbers for a speck of dust let's say the mass is about 10 to the minus sixth
kilograms a microgram uh has a velocity in this light breeze
of let's say one meter per second and let me make myself some more space
here um the size of this speck of dust
is going to be about 10 to the minus 5 meters
so these are the basic parameters of this speck of dust in a light breeze now we can do some calculations with
this for instance momentum well
in order to understand quantum mechanics there's some basic vocabulary that needs to that i need to go over so let's talk
about the key concepts in quantum mechanics thankfully there are only a few there's
really only three and the first is the wave function the wave function is and always has been
written as psi the greek letter my handwriting gets a little lazy sometimes and it'll end up just looking
like this but technically it's supposed to look something like that
details are important provided you recognize the symbol psi
is a function of position potentially in three dimensions x y and z
and time and the key facts here is that psi is a complex function which means that while x y z and t here
are real numbers psi evaluated at a particular point in space will potentially be a complex number
with both real and imaginary part what is subtle about the wave function and we'll talk about this in great
detail later is that it while it represents the state of the system
it doesn't tell you with any certainty what the observable properties of the system are it really only gives you
probabilities so for instance if i have
coordinate system something like this where say this is position in the x direction
psi with both real and imaginary parts might look something like this this could be the real part of psi
and this could be say the complex or the imaginary part of psi
what is physically meaningful is the squared magnitude of psi which might look something like this
in this particular case and that is related to the probability of finding the particle at a particular
point in space as i said we'll talk about this later but the key facts that you need to know
about the wave function is that it's complex and it describes the state of the system but not with certainty
the next key concept in quantum mechanics is that of an operator
now operators are what
connect psi to observable quantities that is one thing operators can do
just a bit of notation usually we use hats for operators for instance x hat
or p-hat our operators that you'll encounter shortly operators
act on psi so if you want to apply for instance the x-hat operator to psi you would write x
hat psi as if this were something that were as it appears on the left of psi the
assumption is that x acts on psi if i write psi x hat doesn't necessarily mean that x
hat acts on psi you assume operators act on whatever lies to the right likewise of course p hat psi
now we'll talk about this in more detail later but x hat
the operator can be thought of as just multiplying by x
so if i have psi as a function of x x hat psi is just going to be x times psi of x
so if psi was a polynomial you could multiply x by that polynomial the
the p operator p hat is another example is a little bit more complicated this is
just an example now and technically this is the momentum operator but we'll talk more about that later it's equal to
minus h bar times the derivative with respect to x so this is again something that
needs a function needs the wave function to actually give you anything meaningful
now the important thing to note about the operators is that they don't give you the observable quantities either
but in quantum mechanics you can't really say the momentum of the
wave function for instance p hat psi is not
and i'll put this in quotes because you won't hear this phrase very often momentum
of psi it's the momentum operator acting on psi and that's not the same thing as the
momentum of psi the final key concept in quantum mechanics is the schrodinger
equation and this is really the big equation so i'll write it big
i h bar partial derivative of psi
with respect to time is equal to h hat that's an operator
acting on psi now h hat here is the hamiltonian
which you can think of as the energy operator so the property of the physical system
that h is associated with is the energy of the system and the energy of the system
can be thought of as a kinetic energy so we can write a kinetic energy operator plus a potential energy
operator together acting on psi and it turns out the kinetic energy operator can be written down this is
going to end up looking like minus h bar squared over 2m
partial derivative of psi with respect to sorry second partial derivative of side with respect to
position plus and then the potential energy operator is going to look like the
potential energy as a function of position just multiplied by psi so this is the schrodinger equation
typically you'll be working with it in this form so i h bar times the partial derivative
with respect to time is related to the partial derivative with respect to space and then multiply multiplied by some
function the basic quantum mechanics that we're going to learn in this course mostly
revolves around solving this function and interpreting the results so to put these in a bit of a roadmap
we have operators we have the schrodinger equation and we have the wave function
now operators act on the wave function and operators are
used in the schrodinger equation now the wave function that actually describes the state of the system is
going to be the solution to the schrodinger equation now i mentioned operators acting on the
wave function what they give you when they act on the wave function is some property of the system
some observable perhaps and the other key fact that i mentioned so far is that the wave function doesn't
describe the system perfectly it only gives you probabilities so that's our overall concept map
to put this in the context of the course outline the probabilities are really the key feature of quantum mechanics and
we're going to start this course with the discussion of probabilities we'll talk about the wave function after
that and how the wave function is related to those probabilities and
we'll end up talking about operators and how those operators and the wave functions together give you
probabilities associated with observable quantities that will lead us into a discussion of
the schrodinger equation which will be most of the course really the bulk of the material before the
first exam will be considered with very concerned with various examples a solution to the schrodinger equation
under various circumstances this is really the main meat of quantum mechanics in the beginning
after that we'll do some formalism and what that means is we'll learn about some advanced mathematical tools that
make keeping track of all the details of how all of this fits together a lot more straightforward
and then we'll finish up the course by doing some applications so those are our key concepts and a
general road map through the course hopefully now you have the basic vocabulary necessary to understand
phrases like the momentum operator acts on the wave function or the solution to the schrodinger equation describes the
state of the system and that sort of thing don't worry too much if these concepts
haven't quite clicked in order to really understand quantum mechanics you have to get experience
with them these are not things that you really have any intuition for based on anything you've seen in physics so far
so bear with me and this will all make sense in the end i promise
complex numbers or numbers involving conceptually you can think about it as the square root of negative one
i are essential to understanding quantum mechanics since some of the most
fundamental concepts in quantum mechanics for instance the wave function are expressed in terms of complex
numbers complex analysis is also one of the most beautiful subjects in all of mathematics
but unfortunately in this course i don't have the time to go into the details lucky you perhaps
here's what i think you absolutely need to know to understand quantum mechanics from the perspective of complex analysis
first of all there's basic definition i squared is equal to negative 1 which you can think of also as i equals the square
root of negative 1. a in general a complex number z then can be written as a the sum of a
purely real part x and a purely imaginary part i times y
note in this expression z is complex x and y are real where i times y is purely imaginary
the terms purely real or purely imaginary in the context of this expression like
this x plus i y something is purely real if y is zero something is purely imaginary if x is zero
as far as some notation for extracting the real and imaginary parts typically mathematicians will use this funny
calligraphic font to indicate the real part of x plus iy or the imaginary part of x plus iy and
that just pulls out x and y note that both of these are real numbers when you pull out the
imaginary part you get x and y you don't get i y for instance another one of the most beautiful
results in mathematics is e to the i pi plus one equals zero
this formula kind of astonished me when i first encountered it but it is a logical extension of this
more general formula that e raised to a purely imaginary power i y is equal to the cosine of y plus i times
the sine of y this can be shown in a variety of ways in particular involving the taylor
series if you know the taylor series for the exponential the taylor series for cosine of y and the taylor series for
sine of y you can show quite readily that the taylor series for complex exponential is the taylor series of
cosine plus the taylor series of sine and while that might not necessarily constitute a rigorous proof
it's really quite fun if you get the chance to go through it at any rate the trigonometric functions
here cosine and sine should should be suggestive and there is a
geometric interpretation of complex numbers that we'll come back to in a minute
but for now know that while we have rectangular forms like this x plus i y where x and y the
nomenclature there is chosen on purpose you can also express this in terms of r e to the i theta where you have now a
radius and an angle the angle here by the way is going to be the
arc tangent of y over x and we'll see why that is in uh in a moment when we talk about the geometric
interpretation but given these rectangular and polar forms
of complex numbers what do the basic operations look like how do we manipulate these things
well addition and subtraction in rectangular form is straightforward if we have two complex numbers a plus ib
plus and we want to add to that the second complex number c plus id we just add the
real parts a and c and we add the imaginary parts b and d this is just like adding in any other
sort of algebraic expression multiplication is a little bit more complicated you have to distribute
and you distribute in the usual sort of draw smiley face kind of way a times c and b times d are going to end
up together in the real part the reason for that is well a times c a and c both being real numbers a times c will be
real whereas ib times id both being purely complex numbers
you'll end up with b times d times i squared and i squared is minus 1. so you just
end up with minus bd which is what we see here otherwise the complex part is perhaps a
little more easy to understand you have i times b times c and you have a times i times d both of which end up with plus
signs in the complex part division in this case
is like rationalizing the denominator except instead of involving radicals you have complex numbers
if i have some number a plus i b divided by c plus id i can simplify this by both multiplying
and dividing by c minus id note the sign change in the denominator here c plus id is then
prompting me to multiply by c minus id over c minus id now when you do the distribution there
for instance let's just do it in the denominator c plus
id times c minus id my top eyebrows here of the smiley face c squared
minus sorry c squared times id c squared plus now id
times minus id which is well i'll just write it out i times minus id
which is going to be d squared times i times minus i so i squared times minus one and i
squared is minus one so i have minus one times minus one which is just one so i can ignore that
i've just got d squared so what i end up with in the denominator is just c squared plus d
squared what i end up with in the numerator well that's the same sort of multiplication thing that we just
discussed so the simplified form of this has no complex part in the denominator
which helps keep things a little simple and a little easier to interpret now in polar form addition and
subtraction while they're complicated under most circumstances if you have two complex numbers given in polar form it's
easiest just to convert to rectangular form and add them together there multiplication and division though in
polar form have very nice expressions q e to the i theta times r e to the i phi
well these are just all real numbers multiplying together and then i can use the rules regarding multiplication of
exponentials meaning if i have two things like e to the i theta and e to the iv i can just add
the exponents together it's like taking x squared times x to the fourth and getting x to the sixth
but q are e to the i theta plus v so that was easy we didn't have to do any distribution at all
the key factor is that you add the angles together in the case of division it's also quite
easy you simply divide the radii q over r and instead of adding you subtract the angles
so polar complex numbers expressed in polar form
are much easier to manipulate in multiplication and division while complex numbers represented in
rectangular form are much easier to manipulate for addition and subtraction taking the magnitude of complex number
usually we'll write that as something like z if z is a complex number just using the same notation for
absolute value of a real number usually is expressed in terms of the complex conjugate the complex conjugate
notationally speaking is usually written by whatever complex number you have here in
this case x plus iy with a star after it and what that signifies is you flip the sign
on the complex part on the imaginary part x plus iy becomes x minus iy the squared magnitude then which is
always going to be a real and positive number this
absolute value squared notation is what you get for multiplying a number by its complex conjugate and that's what
we saw earlier with c plus id say i take the complex conjugate of c plus id and multiply it by c plus i d
well the complex conjugate of c plus id is c minus id times
c plus id and doing the distribution like we did when we calculated the denominator when
we were simplifying the division of complex numbers in rectangular form just gave us c squared
plus d squared this should be suggestive if you have something like
x plus i y that's really messy x plus i y and i want to know the squared absolute
magnitude thinking about this as a position in cartesian space
should make this formula c squared plus d squared in this case just make uh make a little more sense
you can also of course write that in terms of real and imaginary parts but let's do an example
if w is 3 plus 4i and z is -1 plus 2i first of all let's find w plus z well w
plus z is three plus four i plus minus one plus two i that's straightforward if you can keep
track of your terms 3 minus 1 is going to be our real part so that's 2 and 4i plus 2i which is plus 6i is going
to be our complex part sorry our imaginary part now w times z
3 plus 4 i times minus 1 plus 2i for this we have to distribute
like usual so from our top eyebrow terms here we've got three
times minus one which is minus three and four i times 2i both positive so i
have 4 times 2 which is 8 and i times i which is minus 1 minus 8.
then for my imaginary part the i guess the mouth and the chin if you want to think about it that way i
have 4i times minus 1 minus 4 with the i out front will just be minus 4 inside the parentheses here
and 3 times 2i is going to give me 6i plus 6 inside the end result you get here is 8 or
minus 8 minus 3 is minus 11 and minus 4 plus 6 is going to be
2. so i get minus 11 plus 2i for my multiplication here i guess i'm going to circle that answer i should
circle this answer as well now slightly more complicatedly w over z w is three plus four i
and z is minus one plus two i and you know when you want to simplify an expression like this you multiply by
the complex conjugate of the denominator divided by the complex conjugate of the denominator so minus 1 minus 2i divided
by -1 minus 2i and if we continue the
same sort of distribution i'll do the numerator first same sort of multiplication we just did
here only the signs will be flipped a little bit we'll end up with minus three plus eight instead of minus three minus
eight and for the complex sorry for the imaginary part we'll end up with minus 4
minus 6 instead of minus 4 plus 6 and you can work out the details of that distribution on your own if you want
the denominator is not terribly complicated since we know we're taking the absolute magnitude of a complex
number by multiplying a complex number by its complex conjugate we can just write this out as the square
of the real part 1 plus the square of the imaginary part
minus 2 which squared is 4. so if i continue this final step
this is going to be 5 this is going to be minus 10 i and our denominator here is just going to be
5. so in the end what i'll end up with is going to be 1
minus 2 i so it actually ended up being pretty simple in this case now for the absolute magnitude of w
3 plus 4 i you can think of this as w times w star
square root you can think of this as the square root of the real part of w plus the imaginary
part of w sorry square root of the squared of the real real part plus the square of
the imaginary part which is perhaps a little easier to work with in this case so you don't have to
distribute out complex numbers in that in that way real part is three imaginary part is
four so we end up with the square root of three squared plus four squared
which is five now this was all in rectangular form let me
move this stuff out of the way a little bit and let's do it again at least a subset
of it in polar form in polar form w
three plus four i we know the magnitude of w that's five so that's going to be our radius 5
and our e to the i theta where theta is like i said the arctan since complex numbers are so important
to quantum mechanics let's do a few more examples in this case i'm going to demonstrate how to manipulate complex
numbers in a more general way not so much just doing examples with numbers first example simplify this expression
you have two complex numbers multiplied in the numerator and then a division
first of all the first thing to simplify is this multiplication you have x plus iy times
ic this is pretty easy it's a simple sort of distribution
we're going to have x times ic that's going to be a complex part so i'm going to write that down a
little bit to the right i x c and then we're going to have i y times i c which is going to be minus
y c that's going to be real we also have a real part in the numerator from d here so i'm going to write this as d minus y
c plus i c that's the
result of multiplying this out that's then going to be divided by f plus i g
now in order to simplify this we have a complex number in the denominator you know you need to
multiply by the complex conjugate and divide by the complex conjugate so f minus i g
divided by f minus ig now expanding this out is a little bit messier
but fundamentally you've seen this sort of thing before
you have real part real part an imaginary part imaginary part in the numerator
and then you're going to have imaginary part real part and real part imaginary part
and what you're going to end up with from this first term you get f times d minus yc
from the second term you have minus ig times ixc which is going to give you xcg
we have a minus i times an i which is going to give us a plus incidentally if you're having trouble
figuring out something like minus i times i think about it in the geometric
interpretation this is i in the complex plane this is minus i in the complex plane
so i have one angle going up one angle going down if i'm multiplying them together i'm adding the angles together
so i essentially go up and back down and i just end up with 1 equals i times minus i
otherwise you can keep track of i squared equals minus 1s and just count up your minus signs
this then is the real part
suppose i should write that in green unless my fonts get too confusing excuse me
so that's the real part the imaginary part then is what you get from these terms here
i'm going to write an i out front and now we have x c times f so x c f with an i from here
and then we have d minus y yc times ig which i'll just write as g
d minus yc in the denominator we're now multiplying a number by its
complex conjugate you know what to do there f squared plus g squared
this is just the magnitude of this complex number sorry squared magnitude
now this doesn't necessarily look more simple than what we started with but this is effectively fully simplified you
could further distribute this and distribute this but it's not really going to help you very much
the thing to notice about this is that the denominator is purely real
we've also separated out the real part of the numerator and the imaginary part
of the numerator my handwriting is getting messier as i go
imaginary part of the numerator so we can look at this numerator now and
say ah this is the complex number real part imaginary part and then it's just divided by this real number which
effectively is just a scaling it's it's a relatively simple thing to do to divide by a real number
as a second example consider solving this equation for x now this is the same expression that we had
in the last problem only now we're solving it for equal to zero
so from the last page i'm going to borrow that first simplification step we did distributing
this through we had d minus y c for the real part
plus i x c for the imaginary part and that was divided by f plus i g if we're setting this equal to zero
the nice part about dealing with complex expressions like this is that 0 treated as a complex number is 0 plus
0 i it has a real part and an imaginary part as well it's just kind of trivial
and in order for this complex number to be equal to zero the real part must be zero and the imaginary part must be zero
so we can think of this as d minus y c plus i x c this has to equal zero and this has to
equal 0 separately so we effectively have two equations here not just 1 which is nice we have d
minus yc equals 0 and xc equals 0 which unless c equals 0 just means x equals zero
that's the only way that this equation can hold is if x equals zero
the key factor is to keep in mind that the in order for two complex numbers to be
equal both the real parts and the imaginary parts have to be equal as a slightly more involved example
consider finding this the cubed roots of one now you know one cubed is one that's a
good place to start we'll see that fall out of the algebra pretty quickly what we're trying to do is solve the
equation z cubed equals one
which you can think of as x plus i y where x and y are real numbers
cubed equals one now if we expand out this cubic you get
x cubed plus three x squared times i y plus 3 x times i y squared
plus i y cubed and this is going to have to equal 1.
excuse me equal 1. now
looking at these expressions here we have an i y here we have an i y squared this is going to give me an i squared
which is going to be a minus sign and here i have an i y cubed this is going to give me an i cubed which is
going to be minus i so i have two complex parts and two real parts
so i'm going to rewrite that x cubed and then now a minus sign from the i squared
3 x y squared plus pulling an i out front the imaginary part then is going to come
from this 3x squared y and this y cubed so i've got a 3 x squared y here and then a minus y cubed minus coming from
the i squared and this is also going to have to equal 1.
now in order for this complex number to equal this complex number both the real parts and the imaginary parts have to be
equal so let's write those two separate equations x cubed minus three x y
squared equals the real part of this is the real part of the left hand side has to equal
the real part of the right hand side one and the imaginary part of the left hand side three x squared y minus y cubed has
to equal the imaginary part of the right hand side zero so those are our two equations
this one in particular is pretty easy to work with um we can simplify this
this is you know we can factor a y out this is y times three x squared minus y squared
equals zero one possible solution then is going to come from this
you know you have a product like this equals zero either this is equal to zero or this is equal to zero and saying y
equals to zero is rather straightforward so let's say y equals zero and let's substitute that into this
expression that's going to give us x cubed equals 1
which might look a lot like the equation we started with z cubed equals 1 but it's subtly different because z is a
general complex number whereas our assumption in starting the problem this way is that x is a purely real number
so a purely real number which when cubed gives you 1 that means x equals 1. so x equals one y equals zero that's one
of our solutions z equals one plus zero i or just zero z equals
one now we could have told me that right off the bat z
z cubed equals one well z one possible solution is that z equals one since one cubed is one
the other thing we can do here is we can say three x squared minus y squared
is equal to zero this means that i'll just cheat a little bit and
simplify this 3x squared equals y squared now i can substitute
this in this y squared into this expression as well
and what you end up with is x cubed minus 3x and then y squared was equal to 3x squared so 3x squared is going to go
in there that has to equal 1. now let's move up here what does that leave us with that says
x cubed minus nine x cubed equals one so minus
eight x cubed equals one this means x again being a purely real
number is equal to minus one-half minus one-half times minus one-half times minus one-half
times eight times minus one is equal to one you can check that pretty easily now
where does that leave us where do they go that leaves us substituting this back in to this
expression which tells us that three x squared
equals y squared x equals minus one half so three minus one half squared equals y
squared which tells you that y equals plus or minus the square root of three fourths
if you finish your solution so now we have two solutions for y here coming from one value for x and that
gives us our other two solutions to this cubic we have a cubic equation we would expect there to be three solutions
especially when we're working with complex numbers like this and this is our other solution
z equals minus one half plus or minus the square root of three fourths
i so those are our three solutions now
finding the cubed roots of one to be these complex numbers is not necessarily particularly instructive
however there's a nice geometric interpretation the cubed roots of unity like this
the nth roots of unity doesn't have to be a cubed root all lie on a circle of radius 1 in the
complex plane and if you check the complex magnitude of this number the complex magnitude of
this number you will find that it is indeed unity to check your understanding of this
slightly simpler problem is to find the square roots of i um put another way you've got z some
generic complex number here equals to x squared plus x plus i y quantity squared if that's going to
equal y we'll expand this out solve for x and y in the two equations that will result
from setting real and imaginary parts equal to each other same as with the cubed roots of one
the square roots of i will also fall on a circle of radius one in the complex plane
so those are a few examples of how complex numbers can actually be manipulated
in particular finding the roots of unity there are better formulas for that than the approach that we took here
but i feel this was hopefully instructive if probability is at the heart of
quantum mechanics what does that actually mean well the fundamental source of
probability in quantum mechanics is the wave function psi psi tells you everything that you can in
principle know about the state of the system but it doesn't tell you everything with perfect precision
how that actually gives rise to probability distributions in observable quantities like position or energy or
momentum is something that we'll talk more about later but from the most basic perspective
psi can be thought of as related to a probability distribution
but let's take a step back and talk about probabilistic measurements in general first
if i have some space let's say it's position space
say this is the floor of a lab and i have a ball that is
somewhere on in the floor somewhere on the floor i can measure the position of that ball
maybe i measure the ball to be there on the floor if i prepare the experiment in exactly
the same way attempting to put the ball in the same position on the floor and measure the position of the ball again i
won't always get the same answer because of perhaps some imprecision in my measurements or some
imprecision in how i'm reproducing the system so i might make a second measurement
there or a third measurement there um if i repeat this experiment many
times i'll get a variety of measurements at a variety of locations and maybe they cluster in certain
regions or maybe they're very unlikely in other regions but this distribution of measurements we
can describe that mathematically with the probability distribution uh probability distribution for instance
i could plot p of x here and p of x tells you roughly how many or how likely you are to make a
measurement so i would expect p of x as a function to be larger here where there's a lot of measurements and 0 here
where there's no measurements and relatively small here where there's few measurements so p of x might look
something like this so the height of p of x here tells us how likely we are to make a measurement
in a given location this concept of a probability distribution is intimately related to
the wave function so the most simple way that you can think of probability in quantum
mechanics is to think of the wave function psi of x now psi of x you know is a complex
function and a complex number can never really be observable what would it mean for
example to measure a position of say two plus three i
meters this isn't something that's going to occur in the physical universe
but the fundamental interpretation of quantum mechanics that
most that your book in this book in particular that most uh physicists think of is the interpretation that psy
in the context of a probability distribution the absolute magnitude of psi squared
is related to the probability of finding the particle described by psi
so if the squared magnitude of psi is large at a particular location that means it is likely that the
particle will be found at that location now the squared magnitude here means that we're not that we have to
say well we have to take the squared magnitude of psi we can't just take psi itself
so for instance in the context of the plot that i just made on the last page if this is x
and our y axis here is
psi psi has real and imaginary parts so the real part of psi might look something like this
and the imaginary part might look something like this and the squared magnitude
would look something like well what you can imagine the square magnitude of that function looking like
you can think of the squared magnitude of size the probability distribution let me move this up a little bit give
myself some more space the squared magnitude of psi then can be thought of as a probability
distribution in the likelihood of finding the particle at a particular location like i
said now what does that mean mathematically mathematically suppose you had two positions
a and b and you wanted to know what the probability of finding the particle between a and b was
given a probability distribution you can find that by integrating the probability distribution
so the probability that the particle is between a
and b is given by the integral from a to b of
the squared absolute magnitude of psi dx you can think of this as a definition
you can think of this as an interpretation but fundamentally this is what the
physical meaning of the wave function is it is related to the probability distribution of position
associated with this particular state of the system now what does that actually mean
and that's a bit of a complicated question it's very difficult to answer suppose i have
a wave function which i'm just going to write as the square plot is the square of magnitude
of psi now suppose it looks something like this now that means i'm perhaps likely to
measure the position of the particle somewhere in the middle here so suppose
wrong color so suppose i do that suppose i measure the position of the
particle here so i've made a measurement now
messy handwriting i've made a measurement and i've observed the particle b here
what does that mean in the context of the wave function now everything that i can possibly know about the particle has
to be encapsulated in the wave function so after the measurement when i know the particle is here you can
think of the wave function as looking something like this it's not going to
be infinitely narrow because there might be some uncertainty the width of this is related to the precision of the
measurement but the wave function before the measurement was broad like this and the
wave function after the measurement is narrow what actually happened here what about the measurement caused this to
happen this is one of the deep issues in quantum mechanics that is quite
difficult to interpret so what do we make of this well
one thing that you could think just intuitively is that well this probability distribution wasn't really
all the information that was there really the particle was there let's say this is point c
one interpretation is that the particle really was at c all along
that means that this distribution reflects ignorance on our part as physicists not fundamental uncertainty
in the physical system this turns out to not be true and you can show mathematically and in
experiments that this is not the case the main interpretation that physicists use is to say that this wave function
psi here also shown here collapses
now that's a strange term collapses but it's hard to think of it any other
way suppose you were concerned with the wavefunction's value here before the measurement it's non-zero
whereas after the measurement it's zero so this decrease in the wave function
out here is a well it's reasonable to call that a collapse
what that wave function collapse means is subject to some debate and there are
other interpretations one interpretation that i'll mention very briefly but we won't really discuss
very much is the many worlds interpretation and that's that when you make a measurement like this
the universe splits so it's not that the wave function all of a sudden decreases here it's that for
us in our tiny little chunk of the universe the wave function is now
this and there's another universe somewhere else where the wave function is this because
the particle is observed to be here don't worry too much about that but the interpretation issues in quantum
mechanics are really fascinating once you start to get into them you can think about this as the universe
splitting into oh sorry
splits the universe you can think about this as the universe splitting into many little subuniverses where the
probability of uh observable where the particle is observed at a variety of locations
one location per universe really this question of how measurements take place is really fundamental
but hopefully this explains a little bit of where probability comes from in quantum
mechanics the wave function itself can be thought of as a probability distribution
for position measurements and unfortunately the measurement process is not something that's
particularly easy to understand but that's the fundamental origin of probability in quantum mechanics
to check your understanding here is a simple question about probability distributions and how to interpret them
variance and standard deviation are properties of a probability distribution that are related to the uncertainty
since uncertainty is such an important concept in quantum mechanics we need to know how to quantify how uncertainty
results from probability distributions so let's talk about the variance and the standard deviation
these questions are related to the shape of a probability distribution so if i have
a set of coordinates let's say this is the x-axis and i'm going to be plotting then
the probability density function as a function of x
probability distributions come in lots of shapes and sizes you can have probability distributions
that look like this probability distributions that look like this you can even have probability
distributions that look like this or probability distributions that look like this
and these are all different the narrow peak here
versus the broad distribution here the distribution with multiple peaks or
multiple modes in this case it has two modes so we call this distribution bimodal
or multimodal and then this distribution which is asymmetric has a long tail in the positive direction and
a short tail in the negative direction we would say this distribution is skewed so distributions have lots of different
shapes and if what we're interested in is the uncertainty you can think about that roughly as the width of the
distribution for instance if i'm drawing random numbers from the orange distribution the narrow one here
they'll come over roughly this range whereas if i'm drawing from the blue distribution
they'll come over roughly this range so if this were say the probability density for
position say this is the squared magnitude of the wave function for a particle
i know where the particle represented by the orange distribution is much more accurately
than the particle represented by the blue distribution so this concept of width
of a distribution and the uncertainty in the position for instance
are closely related the broadness is related to the uncertainty uh this is fundamental to quantum
mechanics so how do we quantify it in statistics the the broadness of a distribution is
called the variance variance is a way of measuring the broadness of a distribution for example
so suppose this is my distribution the mean of my distribution is going to
fall roughly in the middle here let's say that's the expected value of x if this is the x-axis
now if i draw a random number from this distribution i won't always get the expected value suppose i get a value
here if i'm interested in the typical deviation of this value from the mean
that will tell me something about how broad this distribution is so let's define this displacement here
to be delta x delta x is going to be equal to x minus the
expected value of x and first of all you might think well if i'm looking for the typical values of
delta x let's just try the expected value of delta x well what is that
unfortunately the expected value of x doesn't really work for this purpose because delta x is positive if you're on
this side of the mean and negative if you're on this side of the mean so the expected value of delta x
is zero sometimes it's positive sometimes it's negative and they end up cancelling out
now if you're interested in only positive numbers the next guess you might come up with is let's use
not delta x but let's use the absolute value of delta x what is that well absolute values are difficult to
work with since you have to keep track of whether a number is positive or negative and keep flipping signs if it's
negative so this turns out to just be kind of painful
what is this what statisticians and physicists do in the end then is instead of taking the absolute value of a number
just to uh make it positive we square it so you calculate the expected value of the squared deviation sort of
the mean squared deviation this has a name in statistics it's written as sigma squared and it's called
the variance to do an example let's do a discrete example
suppose i have two probability distributions all with equally likely outcomes say the
outcomes of one distribution are one two and 3 while the outcomes for the second
distribution are 0 2 and 4. photographically these numbers are more closely spaced than these numbers
so i would expect the broadness of this distribution to be larger than the broadness of this distribution
you can calculate this out by calculating the mean squared deviation so first of all we need to know the mean
expected value of x is 2 in this case and also in this case knowing the expected value of x you can
calculate the deviations so let's say delta x here is going to be
-1 0 and 1 are the possible deviations from the mean for this probability distribution
whereas in this case it's -2 0 and 2. then we can calculate the delta x squareds that are possible
and you get 1 0 and 1 for this distribution and
4 0 and 4. for this distribution
now when you calculate the mean of these squared deviations in this case the expected
value of the squared deviation is two thirds whereas in this case
the expected value of the squared deviation is eight thirds
so indeed we did get a larger number for the variance in this distribution so you can think of that as the
definition this is not the easiest way of calculating the variance though
it's actually much easier to calculate the variance as an expected value of a squared quantity and an expected and
minus the square of the expected value of the quantity itself so the mean of the square minus the square of the mean
if that helps you to remember it you can see how this results fairly easily by plugging through some
basic algebra so given our definition the expected value of delta x squared we're calculating an expected value so
suppose we have a continuous distribution now the continuous distribution expected value
has an integral in it so we're going to have the integral of delta x squared
times rho of x dx now delta x squared we can we know what
delta x is delta x is x minus the expected value of x so we can plug that in here
and we're going to get the integral of x minus expected value of x squared times rho of x dx
i can expand this out and i'll get integral of x squared minus 2 x expected value of x
plus expected value of x quantity squared rho of x dx
and now i'm going to split this integral up into three separate pieces first piece integral of x squared rho of
x dx second piece integral of 2 x expected value of x
rho of x dx and third piece
integral of expected value of x squared rho of x dx
now this integral you recognize right away this is the expected value of x squared
this integral i can pull this out front since this is a constant this is just a number this is
the expected value so this integral is going to become 2 i can pull the 2 out of course as well
2 times the expected value of x and then what's left is the integral of x rho of x dx which is just the expected
value of x this integral again this is a constant so i can pull it out front
and when i do that i end up with just the integral of rho of x dx and we know the integral of rho of x dx
over the entire domain i should specify that this is the integral from minus infinity to infinity
now all of these are integrals from minus infinity to infinity
the integral of minus infinity to infinity of rho of x dx is 1. so this after i pull the expected the
expected value of x quantity squared out is just going to be the expected value of x quantity squared
so this is expected value of x squared this is well i can simplify this as well this is
the expected value of x quantity squared as well so i'm going to erase that and say squared there
so i have this minus twice this plus this and in the end that gives you
expected value of x squared minus the expected value of x squared
so mean of the square minus the square of the mean
to check your understanding of how to use this formula i'd like you to complete the following table now i'll
give you a head start on this if your probability distribution is given by 1 2 4 5 and 8
all equally likely you can calculate the mean
now once you know the mean you can calculate the deviations x minus the mean which i'd like you to
fill in here then square that quantity and fill it in here and take the mean of that square
deviation same as what we did when we talked about the variance as the mean squared deviation
then taking the other approach i'd like you to calculate the squares of all of the x's and calculate
the mean square you know the mean you know the mean square
you can calculate this quantity mean of the square minus the square of the mean and you should get something
that equals the mean squared deviation that's about it for variance but just to say
a little bit more about this variance is not the end of the story
it turns out there's well there's more i mentioned the distributions that we
were talking about earlier on the first slide here keep forgetting to turn my ruler off the
distributions that look like this versus distributions that look like this this is a question of symmetry
and the mathematical name for this is skew or skewness
there's also distributions that look like this versus distributions
that look like this and this is what mathematically this is called kurtosis
which kind of sounds like a disease or perhaps a villain from a comic book kurtosis has to do with the relative
weights of things near the peak versus things in the tails now mathematically speaking you know the
variance sorry let me go back a little further you know the mean
that was related to the integral of x rho of x dx
we also just learned about the variance which was related to the integral of x squared
rho of x dx it turns out the skewness is related to the integral of x cubed
row of x dx and the kurtosis is related to the
integral of x to the fourth row of x dx at least those are common ways of
measuring skewness and kurtosis these are not exact formulas for skewness and kurtosis nor is this an
exact formula for the variance of course so i'm taking some liberties with the math
but you can imagine well what happens if you take the integral of x to the fifth row of x dx
you could keep going and you would keep getting properties of the probability distribution
that are relevant to its shape now you won't hear very much about skewness and kurtosis in physics but i
thought you should know that this field does sort of continue on for the purposes of quantum mechanics
what you need to know is that variance is related to the uncertainty and we will be doing lots of calculations of
variance on the basis of probability distributions derived from wave functions in this class
we talked a little bit about the probabilistic interpretation of the wave function psi
that's one of the really remarkable aspects of quantum mechanics that there are probabilities rolled up in your
description of the physical state we also talked a fair amount about probability itself and one of the things
we learned was that probabilities had to be normalized meaning the total sum of all of the probable outcomes the
probabilities of all of the outcomes in a probability distribution has to equal 1.
that has some implications for the wave function especially in the context of the schrodinger equation so let's talk
about that in a little more detail normalization in the context of a probability distribution
just means that the integral from minus infinity to infinity of rho of x dx is equal to
1. you can think about that as the sort of extreme case of the probability that say
x is between a and b being given by the prob the integral
from a to b of row of x dx in the context of the wave function
that that statement becomes the probability that the particle
is between a and b is given by the integral from a to b of
the squared magnitude of psi of x integrated between a and b so this is the same sort of statement
you're integrating from a to b and in the case of the probability density you have just the probability density in the
case of the wave function you have the squared absolute magnitude of the wave function this is our probabilistic
interpretation we're may sort of making an analogy between psi squared magnitude and a probability
density this normalization condition then has to also hold for psi if the squared
magnitude of psi is going to is going to be treated as a probability density so integral from minus infinity to
infinity of squared absolute magnitude of psi dx
has to equal 1. this is necessary for our statistical interpretation of
the wave function this brings up an interesting question though
because not just any function can be a probability distribution therefore
this normalization condition treating size of probability density means there are some conditions on what
sorts of functions are allowed to be wave functions this is a question of normalizability
suppose for instance i had a couple of functions that i was interested in say one of those functions looks sort of
like this keeps on rising as it goes to infinity
if i wanted to consider the squared magnitude of this function
this is our possible psi this is our possible psi squared sorry about the messy there
this function since it's going to you know it's it's continuing to
increase as x increases both in the negative direction and in the positive direction its squared magnitude is going
to look something like this i can do a little better there sorry if i tried to say calculate the integral
from minus infinity to infinity of this function i've got a lot of area out here
from say 3 to infinity where the wave function is positive this
would go to infinity therefore what that means is that this function is not
normalizable not all functions can be normalized if i drew a different function for
example something that looked maybe something like this its squared magnitude might look
something like this there is a finite amount of area here so if i integrated the squared magnitude
of the blue curve i would get something finite what that means
is that whatever this function is i could multiply or divide it by a constant such
that this area was equal to one i could take this function and convert it into something such that the integral
from minus infinity to infinity of the squared magnitude of psi equaled one and it obeyed our sort of statistical
constraint on the probability distribution in order for this to be possible psi has
to have this property and the mathematical way of stating it is that psi
must be square integrable and all this means is that the integral
from minus infinity to infinity of the squared magnitude of psi is finite
you don't get zero you don't get infinity in order for this square integrability
to hold for example though you need a slightly weaker condition that
psi goes to zero as x goes to either plus or minus
infinity it's not possible to have a function that
stays non-zero or goes to infinity itself as x goes to infinity and still have things be integrable
like i said if this holds if this integral here is finite you can convert any function into
something that is normalized by just multiplying or dividing by a constant
is that possible though in the schrodinger equation does multiplying or dividing by a
constant do anything well the schrodinger equation here you can just glance at it and see that
multiplying and dividing by a constant doesn't do anything the short injury equation is i
h bar partial derivative with respect to time of psi equals minus h bar squared over 2m
second derivative of psi with respect to position plus the potential times psi
now if i made the substitution psi went to
some multiple or some constant a multiplied by psi you can see what would happen here i
would have psi times a here i would have psi times a and here i would have psi times a
so i would have an a here an a here and an a here so i could divide through this entire equation by a and all of those
a's would disappear and i would just get the original schrodinger equation back what that means is that if
psi solves the schrodinger equation a psi
does 2. i'll just say a psi works now this is only if a is a constant does not depend on time does not depend
on space if a depended on time i would not be able to divide it out of this partial derivative because the
partial derivative would act on on that a same goes for if a was a function of
space if a was a function of space i wouldn't be able to divide it out of this partial derivative with respect to
x so this only holds if a is a constant that means that i might run into some
problems with time evolution i can choose a constant and i can multiply psi by that constant
such that psi is properly normalized at say time t equals zero but will that hold for future times
it's a question of normalization and time evolution what we're really interested in here is
the integral from minus infinity to infinity of
psi of x and time squared dx
if this is going to always be equal to 1 supposing it's equal to 1 at some initial time
what we really want to know is what the time derivative of this is if the time derivative of this is equal
to zero then we'll know that whatever the normalization of this is it will hold throughout the evolution
of the well throughout the evolution of the wave function now i'm going to make a little bit of
simplifying notation here and i'm going to drop the integral limits since it takes a
while to write and we're going to mult or sorry we're going to manipulate this expression
a little bit we're going to use the schrodinger equation we're going to use the rules of complex numbers
i'm going to use the rules of differential calculus i'm going to get something that will
show that indeed this does hold so let's step through that manipulations of the schrodinger equation like this
are a little tricky to follow so i'm going to go slowly and if it seems like i'm being extra
pedantic please bear with me some of the details are important so the first thing that we're going to
do pretty much the only thing that we can do with this equation is we're going to exchange the order of
integration and differentiation instead of differentiating with respect to time the integral with respect to x
we're going to integrate with respect to x
of the time derivative of this psi of x and t
quantity squared basically i've just pushed the derivative inside the integral
now notationally speaking i'm going to move some stuff around here give myself a little more room
and notationally oops didn't mean to change the colors notationally speaking here
the d dt became a partial derivative with respect to time the total derivative d by dt
is now a partial what the notation is keeping track of here
is just the fact that this is a function only of time since you've integrated over x and you've substituted
in limits whereas this is a function of both space and time
so whereas this derivative is acting on something that's only a function of time i can write it as a simple d by dt
a total derivative in this case since what the derivative is acting on as a function of both
position and time i have to treat this as a partial derivative now so
the next thing that we're going to do aside from after pushing this derivative inside and converting it to a partial
derivative is rewrite this squared absolute magnitude of psi as psi star times psi
now the squared absolute magnitude of a complex number is equal to the complex number times its
complex conjugate it's just simple complex analysis rules there
so what we've got is the integral of the partial derivative with respect to time of psi star times psi
integral dx now we have a time derivative applied to a product we can apply the product rule
from differential calculus what we end up with is the integral of the partial derivative with respect to
time of psi star times psi plus psi star
partial derivative of psi with respect to time that's integrated dx
now what i'm going to do is i'm going to notice these partial derivatives with respect to time
and i'm going to ask you to bear with me for a minute while i make a little more space
it's probably a bad sign if i'm running out of space on a computer where i have effectively infinite space
but bear with me the partial derivatives with respect to time
appear in the schrodinger equation i h bar d by dt of psi
equals minus h bar squared over 2m partial derivative second partial derivative of psi with
respect to position plus potential times psi these
are the time derivatives that i'm interested in i can use the schrodinger equation to
substitute in say the right hand side for these time derivatives both for psi star and for psi
so first i'm going to manipulate this by dividing through by i h bar which gives me
d partial psi partial time equals i h bar over 2m
second partial of psi with respect to x minus
where did it go i v over h bar psi
so that can be substituted in here i also need to know something for the complex conjugate of psi so i'm going to
take the complex conjugate of this entire equation what that looks like is partial
derivative of psi star with respect to time now i'm taking the complex conjugate of
this so i have a complex part here the sign of that needs to be flipped and i have a complex number here
that needs to be complex conjugated since the complex conjugate of a product is the product of the complex conjugates
what that means is this is going to become minus i h bar over 2 m d squared psi
star dx squared sorry i forgot the squared there
my plus i v over h bar
psi so i've just gone through and changed the signs on all of the imaginary parts
of all these numbers psi became psi star i became minus i minus i became i this can be substituted in for that
what you get when you make that substitution this equation isn't really getting simpler is it it's getting
longer what you get is the integral of something i'll put an open square
brackets at the beginning here i've got this equation minus i h bar over 2m
second partial derivative of psi star partial x squared plus i
v over h bar psi star that's multiplied by psi
from here so i've just substituted in this expression for this
now the next part i have plus psi star and whatever i'm going to substitute in from
this which is what i get from this version of the schrodinger equation here i h bar over 2m
second partial derivative of psi with respect to x minus
i v over h bar psi close
parentheses close square brackets and i'm integrating dx now this doesn't look particularly
simple but if you notice what we've got here this
term if i distributed this psi in would have i v over h bar psi star times psi this term
if i distributed this psi star in would have an i v over h bar psi star and psi this term has a plus sign this term has
a minus sign so these terms actually cancel out what we're left with then to rewrite
things both of the terms that remain have this minus i h bar over 2m out front
so we're going to have equals to i h bar over 2m and here
i have a minus second partial derivative of psi star with respect to x
times psi and here i have plus psi star times the corresponding second
partial of psi with respect to x
and this is integrated dx is that all right yes now what i'd like you to notice here
is that we've got d by dx and we've got an integral dx
we don't have any time anymore so we're making progress and we're actually almost done
where where did we get so far we started with the time derivative of this
effective total probability which should have been equal to one if
which would be equal to one if this were proper probability distribution but we're just considered with the time
evolution since we know that we whatever
psi is we can multiply it by some constant to make it properly normalized at a particular time now we're
interested in the time evolution we're looking at the time derivative of this
and we've gone to this expression which has complex conjugates of psi and second partial derivatives
with respect to x now what i'd like you to do and this is a check your understanding
question is think about why this statement is true
this is the partial derivative with respect to x of psi star d psi dx
minus d psi star dx so sorry i'm saying d i should be
saying partial these are partial derivatives this is true and it's up to you to
figure out why but since this is true what we're left with is we have our i h bar over 2m
an integral over minus infinity to infinity of this expression partial with respect to
x of psi star partial psi partial x minus
partial psi star partial x psi we're integrating dx now
and this is nice because we're integrating dx of a derivative of something with
respect to x so that's easy fundamental theorem of
calculus we end up with i h bar over 2m
psi star partial psi partial x minus partial psi star
partial x psi evaluated at the limits of our integral which are minus infinity to infinity
now if psi is going to be normalizable we know something about the value of psi at negative and positive infinity
if psi is normalizable psi has to go to zero as x goes to negative and positive
infinity what that means is that when i plug in the infinity here
psi star d psi dx d psi e x and psi they're all everything here is going to be 0.
so when i enter in my limits i'm just going to get 0 and 0. so the bottom line here after all of
this manipulation is that this is equal to 0. what that means
is that the integral from negative infinity to infinity of the squared absolute magnitude of psi
as a function of both x and time is equal to a constant put another way
time evolution does not affect
normalization what that means is that i can take my candidate wave
function not normalized integrate it find out what i would have
to multiply or divided by to make it normalized and if i'm successful i have my
normalized wave function i don't need to worry about how the system evolves in time the
schrodinger equation does not affect the normalization so this is that check your understanding
question i mentioned the following statement was that crucial step
in the derivation and i want you to show that this is true explain why in your own words
now to do an example here normalize this wave function
what that means is that we're going to have to find a constant and i've already put the constant in the wave function a
such that the integral from minus infinity to infinity of the squared absolute
magnitude of psi of x in this case i've left the time dependence out is equal to
1. and same as in the last problem the first thing we're going to do is substitute the squared absolute
magnitude of psi for psi star times psi the other thing i'm going to do before i get started is notice that my
wavefunction is zero if the absolute value of x is greater than one meaning for x
above one or below negative one so instead of integrating from minus infinity to infinity here i'm just going
to focus on the part where psi is non-zero and integrate from -1 to 1. integral from minus 1 to 1 of
psi star which is going to be a e to the ix is going to become e to the
minus ix and 1 minus x squared is still going to be 1 minus x squared
now i have a complex conjugated a because part of the assumption about normalization constants like this is
usually that you can choose them to be purely real so i'm not going to worry about taking
the complex conjugate of a just to make my life a little easier psi
well that's just right here a e to the ix 1 minus x squared i'm integrating dx this is psi star this is psi integral dx
from -1 to 1 should be equal to 1. so let's do this we end up with a squared times the integral from -1 to 1
of e to the minus ix and e to the ix what's e to the minus ix times e to the
ix well thinking about this in terms of the geometric interpretation we have e
to the ix which is cosine theta plus i sine theta you can think about that as being
somewhere on the unit circle at an angle theta minus i x or minus i theta would just be
in the exact opposite direction so when i multiply them together i'm going to get something
that has the product of the magnitudes the magnitudes are both one and it's purely real
you can see that also by looking at just the the rules for multiplying exponentials like this
e to the minus ix times e to the plus ix is e to the minus ix plus ix or e to the zero which is one
so i can cancel these out and what i'm left with is 1 minus x squared quantity squared dx
plugging through the algebra a little further a squared integral minus 1 to 1 of 1 minus 2x squared
plus x to the fourth dx you can do this integral equals a
squared 2 sorry
x minus two thirds x cubed plus
x to the fifth over five we know in quantum mechanics that all of the information about the physical
system is encapsulated in the wave function psi psi then ought to be related to
physical quantities for like like example for example position velocity and momentum of the particle
we know a little bit about the position we know how to calculate things like the expected value of the position
and we know how to calculate the probability that the particles within a particular range of positions
but what about other dynamical variables like velocity or momentum the connection with velocity and
momentum brings us to the point where we really have to talk about operators operators are one of our
fundamental concepts in quantum mechanics and they connect the wave function with physical
quantities but let's take a step back first and think about what it means for a quantum system to move
um the position of the particle we know say
the integral from a to b of the squared magnitude of the wave function dx gives us the probability that the
particle is between a and b and we know that the expected position is given by a similar expression the
integral from minus infinity to infinity of psi star of x times x times psi of x dx
now these expressions are related you know by the fact that the squared magnitude of psi is the probability
density function describing position and this is really just the calculation of the expected value of x
given that probability density function now what if i want to know what the motion of the particle is
one way to consider this is suppose i have a box
and if i know the particle is say here at time t equals zero what can quantum
mechanics tell me about where the particle is later physically speaking you could wait
until say t equals one second and then measure the position of the particle
maybe it would be here you could then wait a little longer and measure the particle again
maybe at that point it would be here that say t equals two seconds or if i wait a little bit longer
and measure the particle yet again at say t equals three seconds maybe the particle would be up here
now does that mean that the particle followed a path that looked something like this no we know that the position
of the particle is not something that we can observe at any given time with impunity because
of the way the observation process affects the wave function back when we talked about measurement we
talked about having a wave function that looks something like this a probability density that looks
something like that and then after we measure the problem measure the position of the particle the probability density
has changed if we say measure the particle to be here the new wave function has to accommodate that new
probability density function the fact that measurement affects the system like this
means that we really can't imagine repeatedly measuring the position of a particle in
the same system what we really need is an ensemble
that's the technical term for what we need and what what an ensemble means in this
context is that you have many identically prepared systems now if i had many identically prepared
systems i could measure the position over and over and over and over again once per system if i have you know 100
systems i could measure this measure the position 100 times and that would give me a pretty good feel for what the
probability density for position measurements is at the particular time when i'm making those measurements
if i wanted to know about the motion of the particle i could do that again except instead of taking my 100
measurements all at the same time i would take them at slightly different times
so instead of this being the same system this would be these would all be excuse me these would all be different
systems that have been allowed to evolve for different amounts of time
and as such the motion of the particle isn't going to end up looking something like that
it's going to end up looking like some sort of probabilistic motion of the wave function in space
what we're really interested in here sorry i should make a note of that many i'm sorry
single measurement per system this notion of averaging over many identically prepared systems is
important in quantum mechanics because of this effect that measurement has on the
system so what we're interested in now in the context of something like motion
is well can we predict this can we predict where the particle is likely to be as a function of time
and yes we can and what i'd like to do to talk about that is to consider
a quantum mechanical calculation that we can actually do the time derivative
of the expected value of position this time derivative tells us how the
center of the probability distribution if you want to think about it that way how the center of the wave function
moves with time so this
time derivative d by dt of
the expected value of x that's d by dt of let's just write out the
expected value of x integral from minus infinity to infinity of x times
psi star of x psi of x where this is the probability density
function that described given by the wave function and this is x
we're integrating dx now if you remember when we talked about normalization whether the normalization
of the wave function changed as the wave function evolved in time we're going to do the same sort of calculation with
this we're going to do some calculus with this expression we're going to apply the schrodinger equation
but as before the first thing we're going to do is move this derivative inside the equation this is a total time
derivative of something that's a function of in principal position and time i should write these as functions
of x and t and
what you get when you push that in is as before the integral or the
total derivative becomes a partial derivative since x is just the coordinate x in
these contexts of functions of both space and time the total time derivative will not
affect the coordinate x even when it comes becomes a partial derivative so what we'll end up with is x times the
partial time derivative of psi star
psi integral dx i'm not going to write the integral from minus infinity to infinity
here just to save myself some time now if you remember
this expression the integral or sorry not the not the full integral
just the partial time derivative of psi star psi that was what we worked with in the
lecture on normalization so if we apply the result from the electron normalization and it's equation
126 yes in the book if we apply that you can simplify this down
a lot right off the bat and what you end up with is i h bar over 2 m
times this integral x and then what we substitute in the equation 126 is gives an expression for
this highlighted part here in orange and what you get is the partial derivative with respect to x
of psi star partial of psi with respect to
x minus partial of psi star with respect to x times psi
integral still with respect to dx of course now if we
look at this equation we're making the same sort of progress we made when we did the normalization derivation
we had time derivatives here now we have only space derivatives and we have only space derivatives in an integral over
space so this is definitely progress now we can start thinking about what we can do
with integration by parts the first integration by parts i'm going to do has
the non-differential part just being x and the differential part being dv
is equal to you know i'm not going to have space to write this here i'm going to move stuff around a little
bit so the differential part is dv
is the partial derivative well what's left of this equation the partial derivative with respect to x of psi star
d psi dx minus d psi
dx psi sorry d psi star dx psi
and then there's the dx from the integral sorry i'm running out of space this
differential part here is just this part of the equation now i can take this derivative dudx in
my integration by parts procedure d u equals dx and
dv here is easy to integrate because this is a derivative
so when i integrate the derivative there i'll just end up with v equals psi star d psi
dx minus d psi star dx
psi now when i actually apply those that integration by part
the boundary term here with the without the integral in it is going to involve these two
so i'm going to have x times psi star partial psi partial x minus
partial psi star partial x psi
and that's going to be evaluated between minus infinity and infinity the limits on my integral
the integral part which comes in with the minus sign is going to be composed of these bottom two terms
integral of psi star partial psi partial x minus
partial psi star partial x psi and it's integral dx
from minus infinity to infinity now what's nice oh you know i forgot something here what
did i forget my leading constants i still have this i h bar over 2m out there
i h bar over 2m is multiplied by this entire expression now the boundary terms here vanish
boundary terms in integration by parts and quantum mechanics will often vanish because if you're evaluating something
at say infinity psi has to go to zero at infinity so this term is going to vanish psi star
has to go to zero at infinity so this is going to vanish so even though x is going to infinity psi is going to zero
and if you dig into the mathematics of quantum mechanics you can show convincingly that the limit as x times
psi goes to infinity is going to be zero so this boundary term vanishes both at infinity and at minus infinity
and all we're left with is this yes all you're left with is that
so i'll write that over i h bar over 2m times the integral of
psi star partial psi partial x minus partial psi star
partial x psi integral dx i'm actually going to
split that up into two separate integrals so i'll stick another integral sign in
here and i'll put a dx there and i'll put parentheses around everything so my leading constant gets
multiplied in properly and now i'm going to apply integration by parts again but this time just to the
second integral here so here we're going to say u is equal to psi
and dv is equal to again using the fact that when we do this integral if we can
integrate a derivative that potentially simplifies things so this is going to be partial psi star partial x dx
so when we derivative take the derivative of this we're going to get d u is equal to partial psi
partial x and when we integrate this we're going to get v equals
psi star now when we do the integration when we write
down the answer from this integration by parts the boundary term here psi star times psi
is going to vanish again because we're evaluating it at a region where both psi star and psi
well vanish so the boundary term vanishes and
you notice i have a minus sign here when we do the integration by parts the integral term has a minus sign in it
here so we're going to have the partial psi with respect to x and psi star with a minus sign coming from the
integration by parts and a minus sign coming from the leading term here so we're going to end up with a plus
sign there so we get a minus from the integral part
um what that means though is that i have psi star and partial psi partial x in my integration by parts i end up with
partial psi partial x and size star it's the same the fact that i had a minus and another
minus means i get a plus so i have two identical terms here the result of this then is i h bar
over m i'm adding a half and a half and getting one basically times
the integral of psi star partial psi partial x
dx and this is going to be something that i'm going
to call now the expectation of the velocity vector velocity operator this is the sort of thing that you get
out of operators in quantum mechanics you end up with expressions like this
and this i'm sort of equating just by analogy with the expectation of a velocity operator this is not really a
probability distribution anymore at least not obviously we started with the probability distribution due to psi the
absolute magnitude of psi squared and we end up with the
partial derivative on one of the size so it's not obvious that this is a probability distribution anymore and
well it's the probability distribution in velocity and it's giving you the expected velocity
in some sense in a quantum mechanical sense so this is really a more general sort of
thing we have the velocity operator
the expectation of the velocity operator oh and operator wise i will try to put hats on things
i will probably forget i don't have that much attention to detail when i'm making lectures like this
the hat notation means operator if you see something that you really serve as an operator but it doesn't have a hat
that's probably just because i made a mistake but this expression for the expectation
of the velocity operator is the one we just derived minus i h bar over m times the integral of
psi star partial derivative of psi with respect to x integral dx
now it's customary to talk about momentum instead of velocity momentum has more meaning
because it's a conserved quantity under you know most physics so we can talk about the momentum
operator the expectation of the momentum operator and i'm going to write this momentum
operator expression in a slightly more suggestive way the integral of psi star times something
in parentheses here which is minus i h bar partial derivative with respect to x i'm
going to close the parentheses there put a psi after it and a dx for the integral
it had the same sort of expression for the position operator we were just writing that as the expected value of
position without the hat earlier but that's going to be the integral of psi star what goes in the parenthesis
now is just x psi dx
so this you recognize is the expectation of the variable x uh subject to the probability
distribution given by psi star times psi this is slightly more subtle you have psi star and psi which looks like a
probability distribution but what you have in the parentheses now is very obviously an operator that does
something it does more than just multiply by x it multiplies by minus i h bar and takes the derivative of psi
operators in general do that we can write them as say x hat
equals x times where there's very obviously something that has to go after the x in order for
it to be considered an operator or we can say the same for v hat it's minus i h bar over m times the
partial derivative with respect to x where there obviously has to be something that goes here
likewise for momentum um minus i h bar partial derivative with
respect to x something has to go there another example of an operator is the kinetic energy operator usually that's
written as t and that's minus h bar squared over 2m
you can think of it as the momentum operator squared it's got a second derivative
with respect to x and again there very obviously has to be something that goes there the operator
acts on the wave function that's what i said back when i talked about the fundamental concepts of quantum
mechanics and this is what it means for the operator to act on the wave function the operator itself is not meaningful
it's only meaningful in the context when it's acting on away function in general
in general the expectation value of some has an introduction to the uncertainty
principle we're going to talk about waves and how waves are related to each other
we'll get into a little bit of the context of fourier analysis which is something we'll come back to later
but the overall context of this lecture is the uncertainty principle and the uncertainty principle is one of the key
results from quantum mechanics and it's related to what we discussed earlier in the context of the boundary between
classical physics and quantum physics quantum mechanics has these inherent uncertainties that
are built into the equations built into this state built into the nature of reality
that we really can't surmount and the uncertainty principle is one way in which those or is the mathematical
description uh it's those relationships that i gave you earlier delta p delta x is greater
than about equal to h bar over 2. i think i just said greater than about equal to h bar
earlier we'll do things a little more mathematically here and it turns out there's a factor of 2 there
to start off though conceptually think about position and wavelength
and this really is now in the context of a wave so say i had
a coordinate system here something like this and if i had some wave
with a very specific wavelength you can just think about it as a sinusoid if i asked you to measure the wavelength
of this wave you could take a ruler and you could plop it down there
and say okay well how many inches are there from peak to peak
or from zero crossing to zero crossing or if you really wanted to you could get a tape measure
and measure many wavelengths one two three four wavelengths in this case
that would allow you to very accurately determine what the wavelength was if on the other hand
the wave looked more like this give you another coordinate system here the wave looks something like this
you wouldn't be able to measure the wavelength very accurately you could as usual put your ruler down
on top of the wave for instance and count up the number of inches or centimeters from one side to the other
but that's just one wavelength it's not nearly as accurate as say measuring four wavelengths or ten wavelengths or a
hundred wavelengths you can think of some limiting cases suppose you had a wave
with many many many many many oscillations it looks like i'm crossing out the wave
underneath there so i'm going to erase this in a moment but if you had a wave with many wavelengths and you could
measure the total length of many wavelengths you would have a very precise measurement of the wavelength of
the wave the opposite is the case here you only have one wavelength you can't really
measure the wavelength very accurately what you can do however is measure the position very accurately here i can say
pretty certainly the wave is there you know plus or minus a very short spread in position
the other hand here i cannot measure the position of this wave accurately at all you know if this thing continues i can't
really say where the wave is it's not really a sensical question to ask where is this wave this wave is everywhere
these are the sorts of built-in uncertainties that you get out of quantum mechanics where is the wave the
wave is everywhere it's a wave it doesn't have a local position it turns out if you get into the
mathematics of fourier analysis that there is a relationship between the spread of wavelengths and the spread of
positions if you have a series of waves of all different wavelengths and they're added up
the spread in the wavelength will is related to the spread in positions of
the sun and we'll talk more about fourier analysis later but for now just realize
that this product is always going to be greater than or equal to about one wavelength is
something with units of inverse length and link when the position of course is something with units of length
so the dimensions of this equation are sort of a guideline wavelength and position have this sort
of relationship and this comes from fourier analysis so how do these waves come into quantum
mechanics well waves in quantum mechanics really first got their start with
louis de bruy i always thought his name was pronounced de broglie but it's uh well he's french so there's all sorts of
weird pronunciations in french is my best guess at how it would probably be pronounced
de voy proposed that matter could travel in waves as well and he did this with a
interesting argument on the basis of three fundamental equations that had just recently been
discovered when he was doing his analysis this was in his phd thesis by the way
e equals m c squared
you all know that equation you all hopefully also know this equation e equals h f
planck's constant times the frequency of a beam of light is the energy associated with a quanta of light
this was another one of einstein's contributions and it has to do with his explanation of
the photoelectric effect the final equation that de bruy was working with was c
c equals f lambda the speed of light is equal to the frequency of the light
times the wavelength of the light and this is really not true just for light this is true for any wave phenomenon
the speed the frequency and the wavelength are related now if these expressions are both equal
to waves or are both equal to energy then i ought to be able to say m c squared equals h f
and this expression tells me something about f it tells me that
f equals c over lambda so i can substitute this expression in here and get m c squared equals h c
over lambda now i can cancel out one of the c's
and i'm left with m c equals h over lambda now what the voice said was
this this is like momentum so i'm going to write this equation as p
equals h over lambda and then i'm going to wave my hands extraordinarily vigorously
and say while this equation is only true for light and this equation is only true for waves this is also
true for matter how actually this
happened in the context of quantum mechanics in the early historical development of
quantum mechanics is de broglie noticed that the spectrum of the hydrogen atom this
bright line spectra that we were talking about where a hydrogen atom emits light of only very specific wavelengths
intensity as a function of wavelength looks something like this but that could be explained if he
assumed that the electrons were traveling around the nucleus of the hydrogen atom as waves and that only an
integer number of waves would fit the one that i just drew here didn't end up back where it started so
that wouldn't work if you had a wavelength that looked something like this going
around say three full times in a circle that that would potentially count for these
allowed emission energies that was quite a deep insight and it was one of the things
that really kicked off quantum mechanics at the beginning the bottom line here for our purpose is
that we're talking about waves and we're talking about matter waves so that uncertainty relation or the
relationship between the spreads of wavelengths and the spreads in positions that i mentioned in the context of
fourier analysis will also potentially hold for matter
and that gets us into the position momentum uncertainty relation the wave momentum relationship we just
derived on the last slide was p equals h over lambda this tells you that the momentum and the
wavelength are related from two slides ago we were talking about waves and
whether or not you could say exactly where a wave was we had a relationship that was something like delta lambda the
spread in wavelengths times the spread and positions of the wave is always greater than about equal to one
combining these relationships together in quantum mechanics and this is not something that i'm doing rigorously now
i'm just waving my hands gives you delta p
delta x is always greater than about equal to h bar
over two and this is the correct mathematical expression of the heisenberg uncertainty
principle that we'll talk more about and derive more formally in chapter three
but for now just realize that the position of a wave the position of a particle
are on certain quantities and the uncertainties are related by this
which in one perspective results from consideration of adding many waves together in the context of fourier
analysis which is something we'll talk about later as well extended through the use of or the
interpretation of matter as also a wave phenomenon to check your understanding
here are four possible wave packets and i would like to rank i would like you to rank them in two different ways one
according to the uncertainties in their positions and two according to the uncertainties in their momentum
so if you consider say wave b to have a very certain position you would rank that one highest in terms of the
certainty of its position perhaps you think wave b has a very low uncertainty in position you would put it
on the other end of the scale i'm looking for something like the uncertainty if b is greater than the
uncertainty of a is greater than the uncertainty of d is greater than the uncertainty of c for both position and
momentum the last comment i want to make in this lecture is on energy time uncertainty
this was the other equation i gave you when i was talking about the boundary between classical physics and quantum
physics we had delta p delta x is greater than or equal to h bar over 2 and now
we also had excuse me for a moment here delta e delta t
greater than about equal to h bar over two same sort of uncertainty relation except now we're talking about spreads
in energy and spreads in time i'd like to make an analogy between these two equations
delta p and delta x delta p according to deploy is related to the wavelength
which is sort of a spatial frequency it's a the frequency of the wave
in space delta x of course is just well i'll just say that's a space
and these are related according to this equation in the context of energy and time we
have the same sort of thing delta t well that's pretty clear that's time and delta e
well that then therefore by analogy here has to have something to do with the
frequency of the wave now in time and that's simple that's just the
frequency the fact that these are also related by an uncertainty principle
tells you that there's something about energy and frequency and time
and this is something that we'll talk about in more detail in the next lecture when we start digging into the
schrodinger equation now the time dependent sure on your equation and deriving the time
independent schrodinger equation which will give us the relationship exactly but for now position and momentum
energy and time we're all talking are both talking about sort of wave phenomenon except in the
context of position and momentum you're talking about wavelength frequency of the wave
in space whereas energy and time you're talking about the frequency of the wave in time
how quickly it oscillates that's about all the uncertainty principle as i've said is something that
we'll treat in much more detail in chapter three but for now the uncertainty principle is
important because you have these equations and these are fundamental properties
of the universe if you want to think of them that way and there's something that we're going to be
working with as a way of checking the validity of quantum mechanics
throughout the rest of the next throughout chapter two um that's all for now you just need to
conceptually understand how these wave lengths and positions or frequencies and times are interrelated
the last few lectures have been all about the wave function psi
since psi is such an important concept in quantum mechanics really the first entire chapter of the
textbook is devoted to the wavefunction and all of its various properties since we've reached the end of chapter
one now now is a good opportunity to go and review the key concepts of quantum mechanics in particular the wave
function and how it is related to the rest of quantum mechanics the key concepts as i stated them
earlier were operators the schrodinger equation and the wave function operators are used in the schrodinger
equation and act on the wave function your friend and mine psi
what we haven't really talked about a lot yet is how to determine the wave function and the wave function is
determined as solutions to the schrodinger equation that's what chapter 2 is all about
solving the schrodinger equation for various circumstances the key concepts that we've talked about
so far operators and the wave function conspire together to give you observable quantities
things like position or momentum or say the kinetic energy of a particle
but they don't give us these properties with certainty in particular the wave function really only gives us
probabilities and these probabilities don't give us really any certainty about what will
happen uncertainty is one of the key concepts that we have to work with
in quantum mechanics so let's take each of these concepts in turn and talk about them in a little
more detail since now we have some actual results that we can use some mathematics we can put more meat on
this concept map than just simply the concept map first the wave function
the wave function psi does not tell us anything with a with certainty
and it's a good thing too because psi as a function of position and time is complex
it's not a real number and it's hard to imagine what it would mean to actually observe a real number
so the wave function is already on somewhat suspect ground here but it has a meaningful connection to
probability distributions if we more or less define the squared modulus the absolute
magnitude of the wave function to be equal to a probability distribution
this is the probability distribution for what well it's the probability distribution for outcomes of
measurements of position for instance you can think about this as a probability distribution for where
you're likely to find the particle should you go looking for it this interpretation as a probability
distribution requires the wave function to be normalized namely
that if i integrate the squared magnitude of the wave function over the entire space that i'm
interested in i have to get one this means that if i look hard enough
for the particle everywhere i have to find it somewhere the probability distributions as i
mentioned earlier don't tell you anything with certainty in particular there is a good deal of uncertainty
which we express as a standard deviation or variance for instance if i'm interested in the standard deviation of
the uncertainty or standard deviation of the position excuse me
it's most easy to express as the variance which is the square of the standard deviation
and the square of this standard deviation or the variance is equal to the expectation value of the square
of the position minus the square of the expectation value of the position and we'll talk about expectation values
in a moment expectation values are calculated using expressions with operators
that look a lot like these sorts of integrals
in fact i can re-express this as the expectation of the square in terms of a probability distribution is just
the x squared times multiply multiplied by the probability distribution with respect to x
integrated overall space this is the expectation of x squared
i can add to that or subtract from that sorry the square of the expectation of x which
has a very similar form and that gives us our variance so our wave function which is complex gives us
probability distributions which can be used to calculate expectation values and uncertainties
this probabilistic interpretation of quantum mechanics gets us into some trouble pretty quickly i'm going to move
this up now give myself some more space namely with the concept of wave function
collapse now collapse bothers a lot of people and it should this is really a
philosophical problem with quantum mechanics that we don't really have a good interpretation of what quantum
mechanics really means for the nature of reality but the collapse of the wave function is more or less a necessary
consequence of the interpretation of the wave function as a probability distribution
if i have some states some space some coordinate system and i plot on this coordinate system
the squared magnitude of psi this is related to our probability distribution with respect to position
if i then measure the position of the particle what i'm going to get is
say i measure the particle to be here now if i measure the position of the
particle again immediately i should get a number that's not too different than the number that i just
got this is just sort of to make sure that if i repeat a measurement it's consistent with itself that i don't have
particles jumping around truly randomly if i know the position i know the position that's a reasonable assumption
what that means is that the new probability distribution for the position of the particle after the
measurement is very sharply peaked about the position of the measurement
if this transition from a wave function for instance that has support here to a
wavefunction that has no support here did not happen instantaneously it's
imaginable that if i tried to measure the particle's position twice in very rapid succession that i would have one
particle measured here and another particle measured here does that really mean i have one
particle or do i have two particles these particles could be separated by quite a large distance in space and my
measurements could be not separated by very much in time so i might be getting into problems with special relativity in
the speed of light and these sorts of considerations are what leads to the copenhagen
interpretation of quantum mechanics which centers on this idea of wave functions as probability distributions
and wave function collapse as part of the measurement process now i mentioned operators in the context
of expectation values operators are our second major concept in quantum mechanics
what about operators in the wave function well
operators let's just write a general operator as q hat hats usually signify operators operators always act on
something you can never really have an operator in isolation and what the operators act on
is usually the wave function we have a couple of operators that we've encountered namely
the position operator x hat which is defined as x times and what's it multiplied by well it's multiplied by
the wave function we also have the momentum operator p hat and that's equal to minus i h bar times
the partial derivative with respect to x of what well of the wave function we also have the kinetic energy which
i'll write as k e hat you could also write it as t hat that
operator is equal to minus h bar squared over 2m times the second derivative with respect to position
of what well of the wave function and finally we have
h hat the hamiltonian which is an expression of the total energy
in the wave function it's a combination of the kinetic energy operator here which you can see first of all as
p squared we have a second derivative with respect to position and minus h bar squared this is just p squared divided
by 2m p squared over 2m is a classical kinetic energy the analogy is reasonably clear there
you add a potential energy term in here and you get the hamiltonian now expectation values of operators like
this are calculated as integrals
the expectation value of q for instance is the integral of psi star times q acting on psi
overall space this bears a striking resemblance to our expression for instance for the
expectation of the position which was the integral of just x times rho of x where rho of x
is now given by the absolute magnitude of psi squared which is given by psi star times psi
now basically the pattern here is you take your operator and you sandwich it
between psi star and psi and you can think about this position as being sandwiched between psi star and
psi as well because we're just multiplying by it doesn't really matter where i put it in the expression
the sandwich between psi star and psi of the operator is more significant when you have operators with derivatives in
them but i'm getting a little long-winded about this
perhaps suffice it to say that operators in the wave function allow us to calculate meaningful
physical quantities like x the expectation of position this is more or less where we would expect to
find the particle or the expectation of p and i should be
putting hats on these since technically they're operators the expectation of p is more or less the expected value of
the momentum the sort of sorts of momentum momenta that the system can have
or the expectation value of h the typical energy the system has and all of these are tied together in
the context of uncertainty for instance if i wanted to calculate the uncertainty in the momentum
i can do that with the same sort of machinery we used when we were talking about probability
that i calculate the expectation of p squared and i subtract the expectation of p
squared so the expectation of the square minus the square of the expectations is
directly related to the uncertainty so that's a little bit about operators and a little bit about the wave function
and a little bit about how they're used operators acting on the wave function calculating expectations in the context
of the wave function being treated as a probability distribution now
where are we all going with this we're going towards the schrodinger equation in the schrodinger equation to write it
out is i h bar partial derivative with respect to time of the wave function and that's equal to
minus h bar squared over 2m second partial derivative with respect to position of the wave
function plus some potential function function of x
times the wave function now the wave function psi here i've left it off as a function of position and
time so this is really the granddaddy of them all this is the equation that we will be
working with throughout chapter two we will be writing this equation for various scenarios and solving it and
describing the properties of the solutions so hopefully now you have a reasonable
understanding of the wave function and the schrode and enough understanding of operators to understand what to do with
the wave function the sorts of questions you can ask of the wave function are things like what sorts of energy does
this system have how big is the spread in momenta where am i likely to find the particle if i went looking for it
but all of that relies on having the wave function and you get the wave function by solving the schrodinger
equation so that's where we're going with this and that's all of the material for
chapter one and without further ado moving on to the next lecture we'll start solving the schrodinger equation
we're going to move now in to actually solving the schrodinger equation this is really the main meat of quantum
mechanics and in order to start tackling the schrodinger equation we need to know a
little bit about how equations like the schrodinger equation are solved in general
one of those solution techniques is separation of variables and that's the solution technique that we're going to
be applying repeatedly to the schrodinger equation first of all though let's talk a little
bit about ordinary and partial differential equations the schrodinger equation is a partial differential
equation which means it's a good deal more difficult than an ordinary differential
equation but what does that actually mean first of all
let's talk about ordinary differential equations what an ordinary differential equation
tells you is how specific coordinates change with time at least that's most applications so you have something like
x as a function of time y as a function of time sorry not y is a function of x y is a function of time
z as a function of time for example the position of a projectile moving through the air
could be determined by three functions x y and z if you're only working in two dimensions
for instance let me drop the z but we might have a velocity as well say v x of t and v y of t
these four coordinates position in two dimensions and velocity in two dimensions fully specifies the state of
a projectile moving in two dimensions what an ordinary differential equation might look like to govern the motion of
this projectile would be something like the following dx dt
is vx dy dt is vy
nothing terribly shocking there the position coordinates change at a rate of change given by the velocity
well the velocity change velocities change dv x dt is given by let's say minus k
v x and d v y d t
is minus k v y sorry k v subscript y now k v y minus g
this tells you that um well where i got these equations this is
a effectively damped frictional motion in the plane uh xy where gravity is pulling you down
so in the absence of any velocity gravity leads to an acceleration in the negative y direction
and the rest of this system evolves accordingly what that tells you though in the end is
the trajectory of the particle if you launch it as a function of time tick tick tick
tick tick tick tick tick tick as the projectile moves through the air in say x y space
partial differential equations on the other hand pdes
you have several independent variables so where an ordinary differential equation we only had time
and everything was a function of time in a partial differential equation what you're trying to solve for will have
several independent variables for example the electric field
the vector electric field in particular as a function of x y and z the electric field
has a value both a magnitude and a direction at every point in space so x y and z potentially vary over the entire
universe now you know how
excuse me you know a few equations that pertain to the electric field that maybe you could
use to solve to determine what the electric field is one of these is gauss's law which we usually give an
integral form the electric field the integral of the electric field dotted with an area
vector over a closed surface is equal to the charge enclosed by that surface over epsilon not
now hopefully you also know there is a differential form for gauss's law and it usually is written like this
this upside down delta is read as del so you can say this is del dot e and this is a vector differential operator
i'm going to skip the details of this because this is all electromagnetism and if you go on to take advanced
electromagnetism courses you will learn about this in excruciating detail perhaps suffice to say here that most of
the time when we're trying to solve equations like this we don't work with the electric field we work with the
potential let's call that v and this system of equations here if you treat
the electric field as minus the gradient of the potential gives you
this equation or this equation gives you the laplace equation del squared v
equals rho over epsilon naught what that actually writes out to if you
go through all the vector algebra is the second derivative of v with respect to x plus the second derivative of v with
respect to y plus the second derivative of v with respect to z and i've left off all my squares in the denominator here
is equal to rho over epsilon naught this is a partial differential equation
and if we had some machinery for solving partial differential equations we would be able to determine the potential at
every point in space and that would then allow us to determine the electric field at every
point in space this is just an example hopefully you're familiar with some of the terms i'm
using here the main solution technique that is used for partial differential equations is
separation of variables and separation of variables is fundamentally a guess
suppose we want to find some function in the case of electromagnetism it's the
potential x y and z the potential is a function of x y and z let's make a guess that v of x
y and z can be written as x of x
times y of y times z of z so instead of having one function of
three variables we have the product of three functions of one variable each does this guess work well it's
astonishing how often this guess actually does work this is a very restrictive sort of thing
but under many realistic circumstances this actually tells you a lot about the solution
for example the wave equation the wave equation
is what you get mathematically if you think about say having a string stretched between two solid objects
now under those circumstances if you zoom in on if you say pluck the string
you know it's going to vibrate up and down mathematically speaking if you zoom in
on a portion of that string say it looks like this you know the center of this string is
going to be accelerating downwards and the reason it's going to accelerate downwards is because there is tension in
the string and the tension force pulls that direction on that side and that direction on that side so it's being
pulled to the right and pulled to the left and the net force then ends up being in the downward direction
if the string curved the other direction you would have effectively a net force
pulling up into the right and a net force pulling up into a force pulling up into
the right a force pulling up into the left and your net force would be up this tells you about forces in terms of
curvatures and that thought leads directly to the wave equation the acceleration
as a result of the force is related to the curvature of the string and how we express that
mathematically is with derivatives the acceleration is the second
derivative of the position so if we have the position of this string is
u as a function of position and time then the acceleration of the string at a given point and at a given time is going
to be equal to some constant traditionally written c squared times the curvature which is the
second derivative of u with respect to x again u being a function of position and
time so this is the weight equation i should probably put a box around this
because the wave equation shows up a lot in physics this is an important one to know
but let's proceed with separation of variables u
as a function of position and time
is going to be x a function of not time x a function of position
and t a function of time so capital x and capital t are functions
of a single variable each and their product is what we're guessing reproduce reproduce the behavior of u
so if i substitute this u into this equation what i end up with is the second
derivative of x of x t of t with respect to time
equals c squared times the second derivative of x of x t of t
with respect to position so this hasn't really gotten this
anywhere yet but what you notice here is we have derivatives with respect to time and
then we have this function of position since these are partial derivatives they're derivatives taken with
everything else other than the variable that you're concerned with held constant which means this
part here which is only a function of position can be treated as a constant and taken outside of the derivative
the same sort of thing happens here we have second derivatives partial second derivatives with respect to
position and here we have only a function of time effectively a constant for this partial
derivative which means we can pull things out and what we've got then is capital x i'm
going to drop the parentheses x because you know capital x is a function of lowercase x
so you've got big x second partial derivative with respect
to time of big t equals c squared big t
second partial derivative of big x with respect to x that's nice
because you can see we're starting to actually be able to pull x and t out here the next step is to divide both
sides of this equation by x t by basically dividing through by u in order for this to work we need to
know that our solution is non-trivial meaning if x and t are everywhere zero dividing through by this will do bad
things to this equation but what you're left with after you divide by this
is one over t second partial of t big t with respect to little t
and c squared one over big x second partial of big x with respect to little x
this is fully separated what that means is that the left hand side here is a
function only of t
the right hand side is a function only of x
that's very interesting suppose i write this
function of t as say f of t this then this part let's call that g of x
i have two different functions of t and x normally you would say oh i have f of t
and i have g of x and i know what those forms are i could in principle solve for t as a
function of x but that isn't what you're going to do and the reason that's not the case is
that this is a partial differential equation both x and t are independent variables
all of this analysis in order for separation of variables to work must hold at every point in space at every x
and at every time so suppose this relationship held for a certain value of t for a certain value
of x i ought to be able to change x and have the have the relationship still hold
so if i change x without changing t the left-hand side of the equation isn't changing
if changing x led to a change in g of x then my relationship wouldn't hold
anymore so effectively what this means is that g of x is a constant in order for this relationship to hold both f of
t and g of x have to be constant essentially what this is saying in the context of the partial differential
equation is that if we look at the x part here when i change the position
any change in the second derivative of the position function is mimicked by this one over x such that the overall
function ends up being a constant that's nice because that means i actually have two
separate equations f of t is a constant and g of x is a constant what these equations actually look like
this was my f of x this is my g or f of t and this is my g of x
that constant which i've called a here and the notation is arbitrary though you can in principle save yourself some time
by thinking ahead and figuring out what might be a reasonable value for a what's especially nice about these is
that this equation is now only an ordinary differential equation
since t is big t is only a function of little t we just have a function of a single variable we only have a single
variable here we don't need to worry about what variables are being held constant what variables aren't being
held constant so we can write this as total derivative with d instead of uh partial derivative with
the partial derivative symbol so we've reduced our partial differential equation into
two ordinary differential equations this is wonderful and we can write we can rearrange these
things to make them a little more recognizable you've got d squared t
dt squared equals a t and c squared
d squared big x d little x squared equals a times big x multiplying through by big t in this equation and big x in
this equation and these are equations that you should know how to solve
if not you can go back to your ordinary differential equations books and
solution to ordinary differential equations like this are very commonly studied
in this case we're taking the second derivative of something and we're getting the something back
with a constant out front anytime you take the derivative of something and get itself
or itself times a constant you should think exponentials and in this case the solution is t
equals e to the square root of a
times time if you take the second derivative of this you'll get two square roots of a
factors that come down a time times e to the root a t which is just big t
you can in principle also have a normalization constant out front and you end up with the same sort of
thing for x big x is going to be
e to the square root of a over c x
with again in principle a normalization constant out front what that means is if i move things up a
little bit and get myself some space u of x and t what we originally wanted
to find is now going to be the product of these two functions so i have a normalization constant in front and i
have e times root a t and e times root a over c x now if this doesn't look like a wave
and that surprises you because i told you this was the wave equation it's because we have in principle some
freedom for what we want to choose for our normalization constant and for what we
want to choose for our separation constant this constant a and the value of that constant will in
principle be determined by the boundary conditions a and a
are determined by boundary conditions the consideration of boundary conditions and initial conditions in partial
differential equations is subtle and i don't have a lot of time to fully explain it here but
if what you're concerned with is why this doesn't look like a wave equation what actually happens when you plug in
to your initial conditions and your boundary conditions to find your normalization constants and your actual
value for the separation constant you'll find that a is complex and when you do
when you substitute in the complex value for a into these expressions you'll end up with
e to the i omega t sort of behavior which is going to give you effectively cosine of omega t
up to some phase shifts as determined by your normalization constant and your initial conditions
so this is how we actually solve a partial differential equation the wave equation in particular
separates easily into these two ordinary differential equations which have solutions that you can go and
look up pretty much anywhere you want finding the actual value of the
constants that match this general solution to the specific circumstances you're concerned with can be a little
tricky but in the case of the wave equation if what you want is say a traveling wave solution you can find it
there are appropriate constants that produce traveling waves in this expression
so to check your understanding what i'd like you to do is go through that exercise again performing separation of
variables to convert this this equation into again two ordinary differential equations
this equation is called the heat equation and it's related to the diffusion of heat throughout a material
if you have say a hot spot and you want to know how that hot spot will spread out with time
since this is a quantum mechanics course let's move on to the time dependent schrodinger equation
this is the full schroedinger equation in all of its glory except i've just written it in terms of
the hamiltonian operator now h hat is the hamiltonian
the hamiltonian is related to the total energy i evidently can't spell
total energy of the system
meaning it's you know kinetic energy plus potential
and we have a kinetic energy operator and we have well we will soon have a potential energy operator
what h hat actually looks like is it's the kinetic energy operator which if you recall correctly is minus h
bar squared over 2m times the second derivative with respect to position and the potential energy operator is
just going it looks a lot like the position operator it's just multiplying by
some potential function which here i'll consider to be a function of x now this is an operator which means it
acts on something so i need to substitute in a wave function here
and when you do that in the context of the schrodinger equation you end up with the form that we've seen before i h bar
d psi dt equals minus h bar squared
over 2m d squared psi dx squared
plus v of x psi so that's our short energy equation
how can we apply separation of variables to this well
we make the same sort of guess as we made before namely
psi is going to be x t where x is a big x is a function of
position and big t is a function of time if i substitute psi equals x t into this equation
you get pretty much what you would expect i h bar now when i substitute x t in here
big x big t big x is a function only of position
so i don't need to worry about the time derivative acting on big x so i can pull big x out
and what i'm left with then is a time derivative of big t this is then going to be equal to
minus h bar squared over 2m times the same thing when i substitute x t in here
the second derivative with respect to position is not going to act on the time part
so i can pull the time part out t
second derivative of big x with respect to position and substituting in x t here doesn't
really do anything there's no derivatives here so this is not a real it's not a particularly interesting term
so we've got we're getting v x t on the right now the next step in separation of
variables is to divide through by your solution x t assuming it's not zero that's okay
and you end up with i h bar one over big x sorry one over big t
canceling out the x and you're just left with big t one over t partial of t dt
and then on the right hand side we have minus h bar over two m sorry h bar squared over two m one over big x
second partial of x with respect to position plus
v x and t are fully cancelled out in this term
now as before this is a function of time only and this is a function of space
only which means both of these functions have to be
constant and in this case the constant we're going to use
is e and you'll see why once we get into talking about the
energy in the context of the wave function so we have
our two equations one i h bar over t first partial derivative of big t
with respect to time is equal to e and on the right hand side from the
right hand side we get minus h bar squared over 2m one over big x second partial of big x
with respect to position plus v is equal to the energy
so these are our two equations now i've written these with partial derivatives but
since as i said before these functions big t and big x are only functions of a single variable there's effectively no
reason to use partial derivative symbols i could use d's instead of partials essentially there's no difference if you
only have a function of a single variable whether you take the partial different partial derivative or the
total derivative so let's take these equations one by one
the first one the time part this we can simplify by multiplying through by big t as
before and you end up with i h bar d big t d t equals
e times t taking the derivative of something and getting it back multiplied by a constant
again should suggest two exponentials let me move this i h bar to the other side
so we would have divided by i h bar and 1 divided by i is minus i so i'm going to erase this from here and say
minus i in the numerator so first derivative with respect to time of our function
gives us our function back with this out front immediately this suggests exponentials
and indeed our general solution to this equation is some normalization constant times e to the minus i
e over h bar times time
so if we know what the separation constant capital e is we know the time part of the evolution
of our wave function this is good what this tells us is that our time
evolution is actually quite simple it's in principle a complex number t is in principle a complex number
it has constant magnitude time evolving this doesn't change the absolute value of capital t
and essentially it's just rotating about the origin in the complex plane so if this is my complex plane real axis
imaginary axis wherever capital t starts as time evolves
it just rotates around and around and around and around in the complex plane
so the time evolution that we'll be working with for the most part in quantum mechanics is quite simple
the space part of this equation is a little more complicated all i'm going to be able to do now is
rearrange it a little bit by multiplying through by capital x
just to get things on top and change the order of terms a little bit to make it a little more recognizable
minus h bar squared over 2m second derivative
of capital x with respect to position plus v times capital x
is equal to e times capital x and this is all the better we can do we can't solve this equation because we
don't know what v is yet v is where the physics enters this equation and where the wave function
from one scenario differs from the wave function for another scenario essentially the potential is where you
encode the environment into the schrodinger equation
now if you remember back a ways when we were talking about the schrodinger equation on the very first slide of this
lecture what we had was the hamiltonian operator acting on the wave function and this is that same hamiltonian this
is h hat not acting on psi now just acting on x so you can also express the schrodinger
equation as h times x equals e times x the hamiltonian operator acting on your spatial part
is the energy of sorry is the separation constant e which is related to the energy
times the spatial part so this is another expression of the schrodinger equation
this equation itself is called the time-independent schrodinger equation or t-i-s-e if i ever use that abbreviation
and this is really the hard part of any quantum mechanics problem to summarize what we've said so far
starting with the schrodinger equation which is this time derivatives with complex parts in
terms of hamiltonians and wave functions gives you this substituting in the actual definition of the hamiltonian
including a potential v and applying separation of variables gets us this pair of ordinary
differential equations the time part here gave us numbers that just basically spun
around in the complex plane not the imaginary part
this is traditionally the real part and this is the imaginary part so the time evolution is basically
rotation in the complex plane and the spatial part well we have to solve this this equation being the time
independent schrodinger equation we have to solve this for a given potential
the last comment i want to make in this lecture is a comment about notation my notation is admittedly sloppy and if
you read through the chapter griffiths calls my notation sloppy um in griffiths since it has the luxury
of being a book and not the handicap of having my messy handwriting they use capital psi to denote the function of x
and time and when they do separation of variables they re-express this as
lowercase psi as a function of position and lowercase phi as a function of time so for this i used capital x sorry i
should put things in the same order i use capital t of t and capital x
of x because i have a better time distinguishing my capital letters from
my lowercase letters than trying to well you saw how long it took me to
write that symbol i'm not very good at writing capital size
there is a lot of sloppiness in the notation in quantum mechanics namely because
oops geez i have two functions of time this is griffith's function of position
sorry about that this here and this here these are really the interesting parts the functions of
position the solutions to the time-independent schrodinger equation what that gives us
well what that means is that a lot of people are sloppy with what they call the wave
function this is the wave function
this is the spatial part or the solution to the time independent schrodinger equation this is not the wave function
but i mean i've already made this sloppy mistake a couple of times in problems that i've given to you guys in class
namely i'll ignore the time domain part and just focus on the spatial part since that's the only interesting part
so perhaps that's my mistake perhaps i need to relearn my handwriting but at any rate be aware that
sometimes i or perhaps even griffis or whoever you are talking to will use the term the
wave function when they don't actually intend to include the time dependence the time dependence is in some sense
easy to add on because it's just this rotation in complex number space but hopefully things will be clear from
the context what is actually meant by the wave function so we're still moving toward solutions
to the schrodinger equation and the topic of this lecture is what you get from separation of variables and
the sorts of properties it has to recap what we talked about last time the schrodinger equation i h bar
partial derivative of psi with respect to time is equal to minus h bar squared over 2m second partial derivative of psi
with respect to position plus v times psi where this is the essentially the
kinetic energy and this is the potential energy as part of the hamiltonian operator
we were able to make some progress towards solving this equation by writing psi
which is in principle a function of position and time as some function of position multiplied
by some function of time why did we do this well it makes things easier we can make
some sort of progress but haven't we restricted our solution a lot by writing it this way
well really we have but
it does make things easier and it turns out that these solutions that are written as products that result from
solving the ordinary differential equations you get from separation of variables with the schrodinger equation
can actually be used to construct everything that you could possibly want to know
so let's take a look at the properties of these separated solutions
first of all these solutions are called stationary states
what we've got is psi as a function of position and time
is equal to some function of position multiplied by some function of time and i wrote that
as capital t on the last slide but if you remember from the previous lecture the time eq evolution equation was a
solvable and what it gave us was a simple exponential e to the there we go minus
e sorry i times e times t divided by h bar so this is our time evolution part and
this is our spatial part what does it mean for these states to be stationary
well consider for instance the probability density for the outcome of position measurements
hopefully you remember this is equal to the squared absolute magnitude of psi which is equal to the complex conjugate
of psi times psi now if i plug this
in for psi and its complex conjugate i end up with the complex conjugate of
big x as a function of position times the complex conjugate of this the only part that's complex about this is the i
here and the exponent so we need to flip the sign on that and we'll have e to the
i positive i now e t over h bar
that's for the complex conjugate of psi and for the science south well x of x e to the minus i
e t over h bar now multiplying these things together
there's nothing special about the multiplication here and this and this are complex conjugates of each
other so they multiply together to give the magnitude of the squared magnitude of each of these
numbers together which since these are just complex exponentials
is magnitude 1. so what we end up with here is
x star x essentially the squared magnitude of
just the spatial part of the wave function there's now no time dependence here
which means the probability density here does not change as time evolves
so that's one interpretation of the or one meaning of these things being called stationary states
the fact that i can write a wave function as a product like this and
the only time dependence here comes in a simple complex exponential means that that time dependence drops out when i
find the probability distribution another interpretation of these things as stationary states comes from
considering expectation suppose i want to calculate the
expectation value of some generic operator capital q the expression for the expectation of an
operator is an integral of the wave function times the operator acting on the wave
function so complex conjugate wave function operator wave function now i'm going to go straight to the wave
function as expressed in terms of x and t parts so complex conjugate of the spatial part
times the complex conjugate of the time part which from the last slide is e to the plus i e t over h bar
our operator gets sandwiched in between the complex conjugate of the wave function and the wave function itself
so this is again no no stars anymore come on brett just x and then e to the minus i e t
over h bar this is all integrated dx so this is psi star
and this is psi and this is our operator sandwiched between them as in the expression for
the expectation now provided this operator does not act
on time it doesn't have anything to do with the time coordinate and that will be
true for basically all of the operators we will encounter in this course now we talked about how the schrodinger
equation can be split by separation of variables into a time independent schrodinger equation in a relatively
simple time dependent part what that gave us is provided we have
solutions to that time independent schrodinger equation we have something called a stationary
state and it's called a stationary state because nothing ever changes the probability densities are constant the
expectation values are constant in the state effectively since it has a precise exact no uncertainty energy has to live
for an infinite amount of time that doesn't sound particularly useful from the perspective of physics we're
often interested in how things interact and how things change with time so how do we get things that actually change
with time in a non-trivial way well it turns out that these stationary states while their time dependence is
trivial the interaction of their time dependence when added together in a superposition is not trivial and that's
where the interesting time dynamics of quantum mechanics comes from superpositions of stationary states
now we can make superpositions of stationary states because of one fundamental fact and that fact is the
linearity of the schrodinger equation so the schrodinger equation as you hopefully remember it by now is i h bar
partial derivative of psi with respect to time is equal to minus h bar squared over 2m
second derivative of psi with respect to x and that's a really ugly sign must fix
second derivative of psi with respect to position plus
v times psi so this is our hamiltonian operator applied to the wave function and this is
our time dependence part now in order for an equation to be linear
what that means is that if psi solves the equation psi plus some other psi that also solves
the equation we'll solve the equation so if say let's call it a solves the schrodinger equation
and b solves the schrodinger equation and uh let me write this out in a little
more detail first of all i'm talking about a as a is a function of position and time
as is b if a and b both solve the schrodinger equation
then a plus b must also solve the schrodinger equation and we can see that pretty easily
let's substitute psi equals a plus b into this equation the first step
i h bar partial derivative respect to time of a plus b is equal to minus
h bar squared over 2m second partial derivative with respect to space of a plus b
plus the potential v times a plus b now the partial derivative of the sum is
the sum of the partial derivatives that goes for the second partial derivative as well
and well this is just just the uh product of the potential with the sum is the sum of the product of the potential
with whatever you're multiplying out i'm going to squeeze things a little bit more here
so i can write that out i h bar d by dt of a plus i h bar
db dt equals
minus h bar squared over 2m second derivative of a with respect to space
forgot my squared on the second derivative minus
h bar squared over 2m second derivative of b with respect to position
plus v times a plus v times b that's just following those fundamental rules
now you can probably see where this is going this
this and this this
these three terms together make up the schrodinger equation the
time dependent schroedinger equation for a
fo a for a
and this this and this altogether that's the time
dependent schroedinger equation for b so if a satisfies the time-dependent
schrodinger equation which is what we supposed when we got started here then this term this term
and this term will cancel out they will obey the equality likewise for the parts with b in them
so essentially if a solves the schrodinger equation b solves the schrodinger equation a plus b also
solves the schrodinger equation the reason for that is the partial derivatives here partial
derivative of the sum is the sum of the partials and the product with the sum is the sum with the product
these are linear operations so we have a linear partial differential equation
and the linearity of the partial differential equation means well essentially that if a solves and b
solves then a plus b will also solve it that allows us to construct solutions that are surprisingly complicated and
actually the general solution to the schrodinger equation is
psi of position and time is equal to
the sum and i'm going to be vague about the sum here you're summing over some index j
x sub j as a function of position these are solutions now to the time independent
schrodinger equation the spatial part of the schrodinger equation times
your time part and we know the time part from the well from us back from when we discussed
separation of variables is minus i e now this is going to be e sub j
t over h bar so this is a general expression that
says we're we're summing up a whole bunch of stationary state solutions to the time independent schrodinger
equation and we're getting psi now oh i've left something out and i've
left and what i've left out is quite important here we need some
constant c sub j that tells us how much of each of these
stationary states to add in so this is actually well it's going to be a solution to the
schrodinger equation since it's constructed from solutions to destroying your equation and this
is completely general that's a little surprising what that
means is that this can be used to express not just a subset of solutions to the schrodinger
equation but all possible solutions to the showing of your schrodinger equation all the solutions to the short injury
equation can be written like this that's a remarkable fact
and it's certainly not guaranteed you can't just write down any old partial differential equation
apply separation of variables and expect the solutions that you get to be completely general and super posable to
make any solution you could possibly want the reason this works
for the schrodinger equation is because the schroeder equation is well just to drop some mathematical terms if
you're interested in looking up information later on the schroedinger equation is an instance of what's called
a sturm liuval problem stormley oval problems are a class of linear operator equations for instance
partial differential equations or ordinary differential equations that have a lot of really nice
properties and this is one of them so the fact that the schrodinger equation is a sternly oval equation but the fact
that the time independent schrodinger equation is a sternly oval equation means that this will work
so if you go on to study you know advanced mathematical analysis methods in physics
you'll learn about this but for now you just need to sort of take it on faith the general solutions to the schrodinger
equation look like this superpositions of stationary states so if we can superpose stationary states
what does that actually give one example i would like to do here is and this is just an example of the sorts
of analysis you can do given superpositions of stationary states is to consider the energy
suppose i have two solutions to the time independent schroedinger equation which i'm just going to write as h hat x1
equals e1 x1 and hat x2 equals
e2 x2 so x1 and x2 are solutions to the time independent
schrodinger equation and their distinct solutions e1
not equal to e2 i'm going to use these to construct a
wave function let's say psi of x and at time t equals 0
let's say it looks like this c1 times x1 as a function of position plus
quantum mechanics is really all about solving the schrodinger equation that's a bit of an oversimplification
though because if there was only one schrodinger equation we could just solve it and be done with it and that would be
it for quantum mechanics the reason this is difficult is that the schrodinger equation
isn't just the schrodinger equation there are many schrodinger equations each physical scenario for which you
want to apply quantum mechanics has its own schrodinger equation they're all slightly different and they all require
slightly different solution techniques the reason there are many different schrodinger equations is that the
situation over under which you want to solve the schrodinger equation enters the schrodinger equation as a
potential function so let's talk about potential functions and how they
influence well the physics of quantum mechanics first of all where does potential appear
in the schrodinger equation this is the time dependent schrodinger equation and the right hand side here you know is
given is giving the hamiltonian operator acting on the wave function now the hamiltonian is related to the
total energy of the system and you can see that by looking at the parts this is the kinetic energy
which you can think of as the momentum operator squared over 2m
sort of a quantum mechanical and now analog of p squared over 2m in classical mechanics
and the second piece here is in some sense the potential energy this v of x
is the potential energy as a function of position as if this were a purely classical system for instance if the
particle was found at a particular position what would be its potential energy that's what this function v of x
encodes now we know in quantum mechanics we don't have
classical particles that can be found at particular positions everything is probabilistic and uncertain but you can
see how this is related this is the time dependent schrodinger equation which is a little bit
unnecessarily complicated most of the time we work with the time-independent schrodinger equation
which looks very similar again we have a left-hand side given by the hamiltonian we have a kinetic energy here
and we have a potential energy here if we're going to solve this time-independent equation note now that
the wave functions here are expressed only as functions of position not as functions of time
this operator gives you the wave function itself back multiplied by e which is just a number
this came from the separation of variables it's just a constant and we know from considering the expectation
value of the hamiltonian operator which is related to the energy for solutions to this time independent
schrodinger equation that we know this is essentially the energy of the state now what does it mean here in this
context or in this potential context well you have a potential function of
position and you have psi the wave function so this v of x
psi of x if that varies as a function of position
and it will if the wave function has a large value a large
magnitude in a certain region and the potential has a large value in a certain region
that means that there is some significant probability the particle will be found in a region with high
potential energy that will tend to make the potential energy of the state higher
now if psi is zero in some region where the potential energy is high that means the particle will never be found in a
region where the potential energy is high that means
the state likely has a lower potential energy
this is all very sort of heuristic qualitative argument and we can only really do better once we know what these
solutions are and what these actual potential functions look like um
what i'd like to do here before we move on is to rearrange this a little bit to show you what effect the potential
energy related to the energy and how it's related to the energy of the state what
effect that has on the wave function and in order to do that i'm going to multiply through by this h bar squared
over 2m and rearrange terms a little bit what you get when you do that
is the second derivative of psi with respect to x there's my eraser
with respect to x being equal to 2m
over h bar squared times v of x minus e psi
so this quantity here relates the second derivative of psi to psi itself
for instance if the potential is larger than the energy of the state you'll get one overall sign relating the
second derivative in psi whereas if energy is larger than potential then you'll end up with a
negative quantity here relating the second derivatives of psi with itself
so keep this in the back of your mind and let's talk about some example potential functions
this is what we're going to be doing or this is what the textbook does in all of chapter two write different potential
functions and solve the schrodinger equation the first example potential we do and
this is section 2.2 is what i like to call the particle in a box the textbook calls it an infinite square
well the particle in a hard box for instance you can think of as a potential function
that looks like this get myself some coordinate systems here you have a potential function v of x
oops turn off my ruler that looks something like this
this is v of x as a function of x the potential goes to infinity for x larger than some size let's call
this you know minus a to a if you're inside minus a to a you have zero potential energy if you're
outside of a you have infinite potential energy it's a very simple potential function it's a
little bit non-physical though because while infinite potential energy what does that really mean it means it would
require infinite energy to force the particle beyond a if you had some infinitely dense material that just
would not tolerate the electron ever being found inside that material and you made a box out of that material this is
the sort of potential function you would get much more realistically
we have the harmonic oscillator potential the harmonic oscillator potential
is the same as what you would get in classical physics it's a parabola this is something you
know proportional to x squared uh v of x being proportional to x squared is what i mean
this is what you would get if you had a particle attached to a spring connected to the origin if you move the particle
to the right you stretch the spring put quantum mechanically if you happen to find the particle at a large
displacement from the origin the spring would be stretched quite a large amount and would have a large
amount of potential energy associated with it from a more physical down to earth sort
of perspective this is what happens when you have any sort of equilibrium position for a particle to be in the
particle is sitting here near the origin where there is a flat potential but any displacement
from the origin makes the potential tend to increase in either direction this is a like a
an electron in a particle trap or an atom in a particle trap harmonic oscillator potentials show up
all over the place and we'll spend a good amount of time talking about them
the next potential that we consider is the delta function potential and what that looks like
now i'm starting going to start at zero and draw it going negative but it's effectively an infinitely sharp
infinitely deep version of this particle in a box potential
instead of going to infinity outside of your realm it's at zero and instead of being a zero inside your
realm it goes to minus infinity there this now continues downwards it doesn't bottom out here
the overall behavior will be different now because the particle is no longer disallowed from being outside of the
domain there is no longer an infinite potential energy here and we'll talk about that as well these
are all sort of weird non-physical potentials the particle in a soft box potential is
a little bit more physical if i have my coordinate system here the particle in a soft box potential
looks something like this to keep things simple it still changes instantaneously at say minus a and a but
the potential energy is no longer infinity this is for instance a box made out of a
material that has some pores in it the electron or whatever particle you're considering to be in the box doesn't
like being in those pores so there's some energy you have to add in order to push the particle in once
it's in it doesn't really matter where it is you've sort of made that energy investment to push the particle into the
box and we'll talk about the quantum mechanical states that are allowed by
this potential as well finally we will consider what happens when
there's no potential at all essentially your potential function is constant that actually has some interesting
implications for the form of the solutions of the schrodinger equation and we'll well we'll talk about that in
more detail to map this onto textbook sections this is section 2.2 the harmonic
oscillator section 2.3 the delta function potential is section 2.5 the particle in a box is section 2.6
particle in a soft box is 2.6 and particle with no potential or an overall constant potential everywhere in space
is section 2.4 so these are some example potentials that we'll be talking about in this
chapter what do these potentials actually mean though how do they influence the
schrodinger equation and its solutions well the way i wrote the schrodinger equation
a few slides ago second derivative of psi with respect to x
is equal to two m over h bar squared just a constant times v of x minus
e psi this is now the time independent schrodinger equation so we're just
talking about functions of position here and e keep in mind is really is the energy of the state
if we're going to have a solution to the time independent schrodinger equation this e exists and it's just a number
so what does that actually mean let's think about it this way we have a left-hand side determined by the
right-hand side of this equation the left-hand side is just the second derivative with respect to position of
the wave function this is related to the curvature of the wave function i could actually write this as a total
derivative since this is just psi is only a function of position now so there's no magic going on with this
partial derivatives it's going to behave same as the ordinary derivative that you're used to from calculus class
the second derivative is related to the concavity of a function whether something's concave up or concave down
so let's think about what this means if you have a potential v of x that's
greater than your energy if v of x is greater than e
what does that mean that means v of x minus e is a positive quantity that means the right hand side here will
have whatever sign psi has and i'm being a little sloppy since psi here is in general complex function
but if we consider it to just be say positive which isn't as meaningful for a complex
number as it is for a real number you would have psi of x
if psi of x is positive and this number is positive then the second derivative is positive
which means that if we're say if psi is say here psi is positive when it's multiplied by is positive then
the second derivative is positive it curves like this whereas if psi is down here
psi is negative this is positive second derivative of psi is negative it curves like this
what this means is that psi
curves away from our axis
away from this psi equals 0 line on the other hand if v of x is less than the energy
this quantity will be negative and we get the opposite behavior if psi is up here positive it's
multiplied by a negative number and the second derivative is negative you
get something that curves downwards if psi is on the other side of the axis it curves upwards
psi curves toward
the axis so this helps us understand a little bit about the shape
of the wave function for instance let me do an example here in a little
bit more detail suppose i have i'll do it over here
coordinate system if i have a potential function let's do the sort of soft particle in a
box i can do better than that soft particle in a box so v of x is constant
outside your central region and constant inside your central region and has a step change at the boundaries of your
region let's think about what our wave function might look like
under these circumstances so we have our boundaries of our region here
the other thing that we need to know to figure out what the wave function might look like is a hypothetical energy and
i'm just going to set an energy here i'm going to do the interesting case let's say this is the energy
i'm plotting energy on the same axis as the potential which is fine this is the energy of the state this is the
potential energy as a function of position so they have the same units what this energy hypothetically means is
that outside here the potential energy is greater than the energy of the state and inside here the potential energy is
less than the energy of the state so we'll get different signed sort of behaviors different curvatures of the
wave function so do my wave function in blue here
if i say start my wave function this is all hypothetical now this may not work if i
start my wave function here at some point on the positive side of the axis
at the origin we know the energy of the state is larger than the energy of or than the potential energy
so this quantity is negative and psi curves towards the axis so since psi is positive here i'm looking at this
sort of curvature so i could draw my wave function out sort of like this
maybe that's reasonable maybe that's not this is obviously not a quantitative calculation this is just sort of the
sort of curvature that you would expect now i only continue these curving lines out to the boundaries since at the
boundaries things change outside our central region here the potential energy is larger than the
energy of the state and you get curvature away from the axis what might that look like
well something curving away from the axis it's going to look sort of like that
but where do i start it do i start it going like that do i start it going like that what does this actually look like
well if you think about this we can say a little bit more about what
happens to our wave function when it passes a boundary like this and the key fact is that if v of x
is finite then while we might have
the second derivative of psi with respect to x being discontinuous maybe
might not be in this case the second derivative of psi is just set by this difference so
when we have a discontinuous discontinuity in the potential we have a discontinuity in the second derivative
the first derivative of psi will be continuous think about integrating a function that
looks like this i integrate it once i get something maybe with large
positive slope going to slightly smaller positive slope there will be no discontinuity in the
first derivative what this means for psi is that it's effectively smooth
and that i just by that i just sort of mean no corners the first derivative psi
won't ever show a corner like this it will be something
like that for example no sharp corners to it what that means in the context of a
boundary like this is that if i have psi going downwards at some angle here i have to keep that angle as i cross the
boundary now once i'm on the other side of the boundary here
i have to curve and i have to curve according to the rules that we had here so depending on
what i actually chose for my initial point here and what the actual value of the energy was and what the actual value
of the potential is outside in this region i may get differing degrees of curvature i may get something that
happens like this curves up very rapidly or i may get something that doesn't curve very rapidly at all
perhaps it's curving upwards very slowly but it crosses the axis now as it crosses the axis the sine on
psi here changes the curvature is also determined by psi as psi gets smaller and smaller the
curvature gets smaller and smaller the curvature becoming zero as psi crosses the axis
then when psi becomes negative the sine of the curvature changes so this would start curving the other
direction curving downwards it turns out that there is actually a state
right in the middle sort of a happy medium state where psi
curves curves curves curves curves and just kisses the axis
comes towards the axis and when it comes towards the axis and reaches the axis with zero slope and zero curvature it's
stuck it will never leave the axis again and these are the sorts of states that you might actually associate with
probability distributions you know if psi is blowing up like this going to positive infinity or to negative
infinity that your your wavefunction will not be normalizable but the wavefunction here denoted by
these green curves has finite area therefore is sort of normalizable
so these are the sorts of things that the potential function tells you about the
wave function in general what direction it curves how much it curves and how quickly
of course doing this quantitatively requires a good deal of mathematics but i wanted to introduce the mat or
before i introduced the math i wanted to give you some conceptual framework with which to understand what exactly this
potential means if the potential is larger than the energy
you expect things that curve upwards and when you get things that curve upwards you'll have a curve away from the axis
you tend to have things blow up unless they just sort of go down and kiss the axis like this so there will be a lot of
things approaching the axis and never leaving so that we have normalizable wave
functions on the other hand if the potential energy is less than the energy of the
state you get things that curve towards and well if you have something that curves
towards it tends to do this always curving towards always curving towards always
curving towards the axis you get these sort of wave-like states so that's a
very hand-waving discussion of the sorts of behavior you get from in this case
uh step discontinuous potential and we'll see the sort of behavior throughout this chapter
to check your understanding take this discontinuous potential and tell me which of these hypothetical wave
functions is consistent with the schrodinger equation now i did
not actually go through and solve the schrodinger equation here to make sure these things are quantitatively
accurate they're probably all not quantitatively accurate what i'm asking you asking you
to do here is identify the sort of qualitative behavior of these systems is the curvature right
and let's see yeah is the are the boundary conditions
right uh in particular does the wave function behave as you would expect as it passes
from the sort of interior region to the exterior region we've been talking about solving the
schrodinger equation and how the potential function encodes the scenario under which we're solving
the schrodinger equation the first real example of a solution to the schrodinger equation and a realistic
wave function that we will get comes from this example the infinite square well which i like to
think of as a particle in a box the infinite square well is called that because its potential is
infinite and well square what the potential ends up looking like is
if i plot this going from zero to a
the potential is infinity if you're outside the ray the region between 0 and a and
at 0 if you're in between the region if you're in between 0 and a so what does this look like
when it comes to the schrodinger equation well what we'll be working with now is
the time independent schrodinger equation the t i s e which reads
minus h bar squared over 2m times the second derivative of sorry i'm getting ahead of myself the second derivative of
psi with respect to x plus potential as a function of x times
psi is equal to the energy of the stationary state that results from the solution of
this equation times psi now this equation doesn't quite look
right if we're outside the region bad things happen
you end up with an infinity here for v of x if x is not between 0 and a the only reason this the only way this
equation can still make sense under those circumstances is if psi
of x is equal to zero if
x is less than zero or x is greater than a so outside this region we already know
what our wavefunction is going to be it's going to be zero and that's just a requirement on the basis of
infinite potential energy can't really exist in the real world now what if we're inside
then v of x is zero and we can cancel this entire term out of our equation
what we're left with then is minus h bar squared over two m second partial derivative of psi with
respect to x is equal to e times psi just dropping that term entirely
so this is the time independent schrodinger equation that we want to solve
so how do we solve it well we had minus h bar squared over 2m times
the second derivative of psi with respect to x being equal to e times psi
we can simplify that just by rearranging some constants what we get
minus second derivative of psi with respect to x equal to minus k squared
psi and this is the sort of little trick that people solving differential
equations employ all the time knowing what the solution is you can define a constant that makes a little more sense
in this case using a square for k instead of just some constant k but in this circumstance k
is equal to root would go root 2 m times e
over h bar so this is our constant which you just
get from rearranging this equation this equation you should recognize
this is the equation for a simple harmonic oscillator a mass on a spring for instance
now as i said before the partial derivatives here don't really matter we're only only talking about one
dimension and we're talking about the time independent schrodinger equation so the wave function here psi is just a
function of x not a function of x and time so this is the ordinary the ordinary
differential equation that you're familiar with for things like masses on springs and
what you get is oscillation psi as a function of x is going to be a
sine kx plus
b cosine kx
and that's a general solution a and b here are constants to be determined by the actual scenario
under which you're trying to solve this equation this equation now not the original
schrodinger equation so these are our solutions sines and cosines
sines and cosines that's all well and good but that
doesn't actually tell us what the wave function is because well we don't know what a is
we don't know what b is and we don't know what k is either we know k in terms of
the mass of the particle that we're concerned with plunk's constant and the e separation constant we got from
deriving the time independent schrodinger equation while that might be related to the
energy we don't know anything about these things these are free parameters still
but we haven't used everything we know about the situation yet in particular we haven't used the boundary conditions and
one thing the boundary conditions here will determine is the form of our solution
now what do i mean by boundary conditions well the boundary conditions are what you get
from considering the actual domain of your solution and what you know about it in particular at the
edges now we have a wave function
that can only be non-zero between zero and a outside that it has to be zero so we
know right away our wave function is zero here and zero
here so whatever we get for those unknown constants a b and k it has to somehow
obey this we know a couple of things about the general form of the wave function
in particular just from consideration of things like the hamiltonian operator or the momentum operator we know that the
wave function itself psi must be continuous we can't have wave functions that look
like this the reason for that is this discontinuity here would do very strange
things to any sort of physical operator that you could think of for example the momentum operator
is defined as minus i h bar partial derivative with respect to x the derivative with respect to x here would
blow up and we would get a very strange value for the momentum that can cause problems
by sort of contradiction then the wave function itself must be continuous we'll come back to talking about the
boundary conditions on the wave function later on in this chapter but for now all we need to know is that the wave
function is continuous what that means is that since we're zero here
we must go through zero there and we must go through 0 there since we're 0 here
so what that means wrong color
means psi of 0 is equal to 0. and psi of a is equal to zero
what does that mean for our hypothetical solution psi of x equals a sine kx
plus b cosine kx well first of all consider
this one the wave function at 0 equals 0. when i plug 0 into this the sine of 0 k
times 0 is going to be 0. the sine of 0 is 0. but the cosine of 0 is 1. so what i'll
get if i plug in 0 for psi is 1 times b so i'll get b now if i'm going to get 0 here
that means b must be equal to 0.
so we have no cosine solutions no cosine part to our solutions so everything here is going to start
like sines it's going to start going up like that that's not the
whole story though because we also have to go through zero when we go through a
so if i plug a into this what i'm left with
is psi of a is equal to
capital a times the sine of k a if this is going to be equal to zero
then i know something about ka in particular the sine function goes through 0 for particular values of k
particular values of its argument sine of x is 0 for x equals integer multiples of pi
what that actually looks like on our plot here is things like this
our wave functions are going to end up looking like this
so let me spell that out in a little more detail our
psi of a wave function is a times the sine
of k times a and if this is going to be equal to zero ka
has to be either 0 plus or minus pi
plus or minus 2 pi plus or minus 3 pi etc
this is just coming from all of the places where the sine of something crosses zero crosses the axis
now it turns out this this is not interesting this means psi
is 0 everywhere since the sine of 0 is well sine k times a if ka is going to be 0 then everything
if ka is 0 k is 0. so the sine of k times x is going to be 0 everywhere
so that's not interesting this is not a wavefunction that we can work with another fact here is that these plus or
minuses the sine of minus x is equal to minus the sine of x sine is an odd function
since what we're looking at here has a normalization constant out front we don't necessarily care whether there's a
plus or a minus sign coming from the sine itself we can absorb that into the
normalization constant so essentially what we're working with then
is that ka equals pi
2 pi 3 pi et cetera which i'll just write as n
times pi now if k times a is going to equal n times pi
we can figure out what um well let's substitute in for k
which we had a few slides ago was root 2 m capital e over h bar
so that's k k times a is equal to n pi
this is interesting we now have integers coming from n here as part of our solution so we're no
longer completely free we in fact have a discrete set of values now a that's a property of the system
we're not going to solve for that m that's a property of the system h bar that's a physical constant the only
thing we can really solve for here is e so let's figure out what that tells us about e
and if you solve this for e you end up with n squared pi squared h bar squared over 2m
a squared this is a discrete set
of allowed energies i keep talking about solutions to the time independent schrodinger equation
and how they have nice mathematical properties what that actually means
is well what i'm referring to are the orthogonality and completeness of solutions to the time-independent
schrodinger equation what that actually means is the topic of this lecture
to recap first of all these are what our stationary states look like for the infinite square well
potential this is the potential such that v of x is infinity
if x is less than 0 or x is greater than a and 0
for x in between 0 and a
so if this is our potential you express the time independent schrodinger equation you solve it you
get sine functions for your solutions you properly apply the boundary
conditions mainly that psi has to go to zero at the ends of the interval because the potential goes to infinity there
and you get n pi over a times x as your argument to the sine functions and you normalize them properly you get
a square root of 2 over a out front the energies associated with these wave functions and this energy now is the
separation constant in from in the conversion from the time dependent schrodinger equation to time independent
schrodinger equation are proportional to n that index the wave functions themselves look like sine
functions and they have an integer number of half wavelengths or half cycles in between 0 and a
so this orange curve this is n equals 1. blue curve is n equals two
the purple curve is n equals three and the green curve is n equals four
if you calculate the squared magnitude of the wavefunctions they look like this one hump for n equals one
two humps for the blue curve n equals two three humps for the purple curve n
equals three and four humps for the green curve n equals four so you can see just by looking at these
wave functions that there's a lot of symmetry one thing we talked about in class is
that these wave functions are either even or odd about the middle of the box and this is a consequence of the
potential being an even function about the middle of the box if i draw a coordinate system here
going between 0 and a either the wave functions have a maximum or they have
a 0. at the middle of the box so for n equals one we have a maximum
for n equals two we have a zero and this pattern continues the number of nodes
is another property that we can think about and this is the number of points where the wave function goes to zero for
instance the blue curve here for n equals two has one node this trend continues
as well if i have a wave function that for instance let me draw it in some absurd
color has one two three four five six seven nodes you know this would be for n equals
eight this would be sort of like the wave function for n equals eight these symmetry properties are nice they
help you understand what the wave function looks like but they don't really help you calculate
what helps you calculate are the orthogonality and completeness of these wave functions
so what does it mean for two functions to be orthogonal let's reason to at this from a
perspective which are more familiar the orthogonality of vectors we say two vectors are orthogonal if
they're at 90 degrees to each other for instance so if i had a two-dimensional coordinate system
and one vector pointing in this direction let's call that a and another vector pointing in this direction let's
call that b i would say those two vectors are orthogonal if they have a 90 degree
angle separating them now that's all well and good in two dimensions it gets a little harder to
visualize in three dimensions and well what does it mean for two vectors to be separated by 90 degrees if you're
talking about a 17 dimensional space in higher dimensions like that it's more convenient to define orthogonality in
terms of the dot product and we say two vectors are orthogonal in that case if the dot product of those two vectors is
zero now in two dimensions you know the dot product is given by the x components of
both vectors ax times bx plus the y component of so those two vectors multiplied together
a y times b y if this is zero we say these two vectors are orthogonal
in three dimensions we can say plus a z times b z and if this is equal to zero we say the
vectors are orthogonal and you can continue this multiplying together like components or
same dimension of the components of vectors in each dimension multiplying them together a1 b1 a2 b2 a3 b3 a4 b4
all added up together and if this number is zero we say the vectors are orthogonal
we can extend this notion to functions but what does it mean to multiply two functions like this
in the case of vectors we were multiplying like components both x components both y components both z
components in the case of functions we can multiply both functions values at
particular x coordinates and add all those up and what that ends up looking like is an integral say the
integral of f x g of x dx
so i'm scanning over all values of x instead of scanning over all dimensions and
i'm multiplying the function values at each individual point at each individual x together
and adding them all up instead of multiplying the components of each vector together at each individual
dimension and adding them all up the overall concept is the same
and you can think about this as in some sense a dot product of two functions now in quantum mechanics since we're
working with complex functions it turns out that we need to put a complex conjugate here
on f in order for things to make sense this should start to look familiar now you've seen expressions like the
integral of psi star of x times psi of x dx is equal to one our normalization condition this is essentially the dot
product of psi with itself psi of course is not orthogonal to itself
but it is possible to make a fun pair of functions that are orthogonal and we say
functions are orthogonal if orthogonal or forgone
so we've been working with solutions to the time independent schrodinger equation for the infinite
square well potential the particle in a box case how do these things actually work though
in order to give you guys a better feel for what the solutions actually look like and how they behave i'd like to do
some examples and use a simulation tool to show you what the time evolution of the schrodinger equation in this
potential actually looks like so the general procedure
that we've followed or will be following in this lecture is once we've solved the time independent chosen schrodinger
equation we get the form of the stationary states knowing the boundary conditions
we get the actual stationary states the stationary state wave functions and their energies
these can then be normalized to get true stationary state wave functions that we can actually use
these stationary state wave functions will for the most part form an orthonormal set
psi sub n of x we can add the time part knowing the time dependent schrodinger equation or the time part
that we got when we separated variables in the time dependent schrodinger equation
we can then express our initial conditions as a sum of these stationary state wave functions
and use this sum then to determine the behavior of the system so what does that actually look like in
the real world not like not like very much unfortunately because the infinite
square will potential is not very realistic but a lot of the features that we'll see in
this sort of potential will appear in more realistic potentials as well so this is our example these are our
stationary state wave functions this is what we got from the solution to the time independent schrodinger equation
this was the form of the stationary states these were the energies and then this was the normalized solution with
the time dependence added back on since the time dependence is basically trivial the initial conditions that i'd like to
consider in this lecture are the wave function evaluated at zero
is either zero if you're outside the sorry this should be a
if you're outside the domain you're zero if you're inside the domain you have
this properly normalized wave function we have an absolute value in this which means this is a little difficult
to work with but what the plot actually looks like if i draw a coordinate system here
going from zero to a is this it's just a tent
a properly built tent with straight walls going up to a nice peak in the middle
our general procedure suggests that we express this initial condition in terms of these stationary states with their
time dependence and that will tell us everything we need to know one thing that will make this a little
easier to work with is getting rid of the absolute values we have here so let's express psi
of x time t equals 0 as a three part function first we have root three over a one
minus now what we should substitute in here is what we get if say zero is less than x
is less than a over two sort of the first half integral interval going out to a over 2 here
in this case we have something sloping upwards which is going to end up in this context
being 1 minus a over 2 minus x over a over 2. so
to say another word or two about that if x is less than a over 2.
this quantity here will be negative so i can get rid of the absolute value if i know that this quantity in the
numerator is positive so i multiply the quantity in the numerator by a minus sign
which i can express more easily just by writing it as a over 2 minus x a over 2 minus x
that will then ensure that this term here this term here is positive for x is in this range
1 minus that is then this term in our wave function for the other half of the range root 3
over a 1 minus something and this is now from a over 2
is less than x is less than a the second half of the interval for the second half of the interval x is larger than a over
2. so x minus a over 2 is positive so i can take care of this absolute value just by leaving it as x minus a over 2.
i don't need to worry about the absolute value in this range so this is x minus a over two all over a over two
and of course if we're outside that we get zero this technique of splitting up absolute
values into separate ranges makes the integrals a little easier to express and a little easier to think about
so that is our initial conditions
how can we express these initial conditions as a sum of stationary state wave functions
evaluated at time t equals 0. this is where fourier's trick comes in if i want to express my initial
conditions as a sum of stationary state wave functions
i know i can use this sort of an expression this is now my initial conditions
and my stationary state wave functions are being left multiplied complex conjugated integrated over the domain
and that gives us our constants c sub n that go in this
expression for the initial conditions in terms of the stationary state wave functions
the notation here is that if psi appears without a subscript that's our initial condition that's our actual wave
function and if psi appears with a subscript it's a stationary state wave function
so what does this actually look like well we know what these functions are
first of all we know that this function which has an absolute value in it is best expressed if we split it up in two
so we're going to split this integral up into a one going from zero to a over two and one going from zero to a
so let's do that we have c sub n equals the integral from 0 to a over 2 of
our normalized stationary state wave function which is root 2 over a
times the sine of n pi x over a that's this psi sub n star
evaluated at time t equals zero i'm ignoring time for now so
even if i had my time parts in there i would be evaluating e to the zero where time is zero
so i would get one from those parts then you have psi our initial conditions and our initial conditions for the first
half of our interval plus root 3 over a 1 minus
a over 2 minus x over a over 2. and i'm integrating that dx
the second half of my integral integral from a over 2 to a looks much the same root 2 over a
sine n pi x over a that part doesn't change the only part that changes is the fact that we're
dealing with the second half of the interval so the absolute value gives me a minus sign up here more or less
root 3 over a 1 minus x minus a over 2 over a over 2 dx
so substitute in for n and do the integrals this
as you can imagine is kind of a pain in the butt so what i'd like to do at this point is
give you a demonstration of one way that you can do these integrals without really having to think all that hard
and that's doing them on the computer you can of course use all from alpha to do these you can of course use
mathematica but the tool that i would like to demonstrate is called sage sage is different than
wolfram alpha and mathematica and that sage is entirely open source and it's entirely freely available you can
download a copy install it on your computer and work with it whenever you want
it's a very powerful piece of software unfortunately it's not as good as the commercial alternatives of course but it
can potentially save you a couple hundred dollars the interface to the software that i'm
using is their notebook web page so you can use your google account to log into this notebook page
and then you have access to this sort of an interface so if i scroll down a little bit here
i'm going to start defining the problem a here that's our
domain our domain goes from 0 to a h bar i'm defining equal to 1 since that number is a whole lot more convenient
than 10 to the minus 31st n x and t those are just variables and i'm defining them as variables given by
these strings and x and t now we get into the physics the energy
that's a function of what index you have what your which particular stationary state you're
talking about this would be psi sub n this would be e sub n e sub n is equal to n squared pi squared h bar squared
over 2 m a squared that's an equation that we've derived psi
of x and t psi sub n of x and t in particular is given by this it's square root of 2
over a times the sine function times this complex exponential which now uses the energy which i just defined here
psi star is the complex conjugate of psi which i've just done by hand by removing the minus sign here
more or less just to copy paste g of x is what i've defined
the initial conditions to be which is square root of 3 over a times this 1 minus absolute value expression
and c sub n here that's the integral of g of x times psi from 0 to a over 2 plus g of x times psi
going from a over 2 to a that's all well and good now i've left off the psi stars but
since i'm evaluating at time t equals 0 it doesn't matter psi is equal to psi star at t equals 0.
i did have to split up the integral from 0 to a over 2 and a over 2 to a because otherwise sage got a little too
complicated in terms of what it thought the integral should be but given all this i can plot for
instance g and if i click evaluate here momentarily a plot appears this is the plot of g of x
as a function of x now i define a to be equal to one so we're just going from zero to one this is that tent function i
mentioned if i scroll down a little bit we can evaluate c of n
this is what you would get if you plugged into that integral that i just wrote
on the last slide you can make a list evaluating c of n for x going from one to ten
and this is what you get you get these sorts of expressions four times square root of six over pi squared
or minus four root six over pi squared divided by nine four root six over pi squared over 25
4 root 6 over pi squared over 49 you can see the sort of pattern that we're working with
some number divided by an odd number raised to the nth power squared we can approximate these things
just to get a feel for what the numbers are actually like and we have
0.99 minus 0.11 plus 0.039 etc moving on down so that's the sort of thing that we can
do relatively easily with sage get these types of integral expressions and their values
um you can see i've done more with this sage notebook and we'll come back to it in a moment but for now
these are the sorts of expressions that you get for c sub n
so our demo with sage tells us c sub n equals some messy expression
and it can evaluate that messy expression and tell us what we need to know
now the actual form of the evaluated c sub n was not actually all that complicated and if we truncate our sum
instead of summing from now this is expressing psi of x t our wave function as an infinite sum n
equals 1 to infinity of c sub n psi sub n of x and t if i truncate this sum at say n equals 3
i'll just have a term from psi 1 and psi 3. recall back from the sage results that
psi 2 the coefficient of psi 2 c sub 2 was equal to 0. so let's find the expectation of x
squared knowing the form of these functions and now knowing the values of these c sub n
from sage you can write out what x squared should be
this is the expected value of x squared and it's going to be an integral of these numbers 4 root 6
over pi squared times psi 1 which was root 2 over a sine
n so they're just dealing with psi 1. now we have pi x over a
we have to include the time dependence now since i'm looking for the expected value of x squared as a function of time
now then we have e to the minus
i times pi squared h bar squared t
over 2 m a squared all divided by h bar or i could just cancel out one of the h bars here
that's our first term in our first term of our expression the next term we have
4 root 6 over 9 pi squared from this coefficient now psi 3 is root 2 over a
sine of 3 pi x over a times again complex exponential e to the
minus i pi squared h bar squared t over 2
m sorry 9 pi squared h bar squared t over 2 m a squared all divided by h bar now what is this
this whole thing needs to be complex conjugated because this is psi
star what's next well i need to multiply this by x squared
and i need to multiply that by the same sort of thing
e to the plus this
minus same sort of thing e to the plus this so these
this is the term in orange brackets here is psi star
this is our x the term in blue brackets here is our psi so we're just using the same sort of
expression only you can certainly see just how messy it is this is the integral
of psi star x squared
psi this is psi star this is x squared
and this stuff is psi we have to integrate all of this dx from 0 to a
this is pretty messy as well messy but doable now since i was working with sage anyway
i thought let's see how the time dependence in this expression plays out in sage
so going back to sage we know these c sub n's
these these are the c sub n's that i chose for c sub one and c sub three and
c sub n of x gives me some digits or um sorry c sub n evaluated
gave me these numbers in uh just in decimal form now i can use these c sub n's to express
that test function where i truncated my sum at psi sub 3. so this is our test function in fact if
you evaluate it it's a lot more simple when you plug in the numbers sine 3 pi x and sine pi x
when h bar is one and a is one these these expressions are a lot easier to work with which gives you a feeling for
why quantum mechanics quantum mechanics often we assign h bar equal to one
the expected value of x squared here is then the integral of the conjugate of
my test function times x squared times times my test function
integrated from 0 to a and sage can do that integral
it just gives you this sage can also plot what you get as a result
now you notice sage has left complex exponentials in here if you take this expression and manually
simplify it you can turn this into something with just a cosine there is no complex part to this expression but sage
isn't smart enough to do that numerically so if i ha so i have to take the absolute value of this expression to
make the complex parts the tiny tiny complex parts go away and if i plot it over some reasonable
range this is what it looks like it's a sinusoid or a cosine you saw it actually
and what we're looking at here on the y axis is the expected value of x
squared this is related to the variance in x so it's a measure of more or less the
uncertainty in position so our uncertainty in position is oscillating with time
what does this actually look like in the context of the wave function well
the wave function itself is going to be a sum you know c sub 1 times psi 1 c sub 3 times psi c 3 c sub 5 times 5 c