The following content is
provided under a Creative Commons license. Your support will help
MIT Open Courseware continue to offer high quality
educational resources for free. To make a donation or to
view additional materials from hundreds of MIT courses,
visit MIT Open Courseware at ocw.mit.edu. ROBERT FIELD: That's the outline
of what we're going to cover. But before we get
started on that, I want to talk about
a couple of things. First of all, last time, we
talked about the two slit experiment. And it's mostly classical. There is only a little
bit of quantum in it where we talk about
momentum as being determined by h over lambda. OK. But what were the
two surprising things about the two slit experiment? There are two of
them, two surprises. Yes? AUDIENCE: [INAUDIBLE]
That a single particle can interfere with itself. ROBERT FIELD: Yes. That's the most
surprising thing. And when you go to
really low intensity-- so there's only one
photon, which is quantum, in the apparatus,
somehow, it knows enough to interfere with itself. And this is the most
mysterious aspect. But then there's
one other aspect, which is how it communicates
that interference with itself. What is that? You're hot. You want to do another one? AUDIENCE: What do you mean
by how it communicates? ROBERT FIELD: Well, here
we have the screen on which the information is deposited. And you have possibly some sort
of probability distribution, which is a kind of
a continuous thing. But you don't observe that. What do you observe? So you could say the
state of a particle has amplitude everywhere
on this screen. But what do you see
in the experiment? Yes? AUDIENCE: So you just see
one [INAUDIBLE] point. [INAUDIBLE] thousands
and thousands. And eventually, you see
it mimic that probability [? sequence. ?] ROBERT FIELD: That's
exactly right. So this state of the
system which is distributed collapses to a single point. We say that what
we are observing-- and this is mysterious. And it should bother you now. We're seeing an eigenvalue
of the measurement operator. So we have this sort of thing. It goes into the
measurement operator. And the measurement
operator says, this is one of the answers
I'm permitted to give you. And another aspect that's
really disturbing or puzzling is that you do many
identical experiments. And the answer is
always different. There is no determinant. It's all probabilistic. So this really
should bother you. And eventually, it won't. OK. Now, there's another
question I have. When I described the two slit
experiment, I intentionally put something up on the
diagram that should bother you, that should have said,
this is ridiculous. I used a candle in the lecture. And I used a light
bulb in the notes. And why is that ridiculous? Yes? AUDIENCE: You said
many frequencies. ROBERT FIELD: That's right. So the sources of light that I
misled you with intentionally have a continuous
frequency distribution. And the interference depends
on the same frequency. So the only way you would
see any kind of diffraction pattern, any kind of pattern
on the two slit experiment, is if you had
monochromatic light. It would all wash out. You'd still get dots. But the dots would
never give you anything except perhaps a
superposition of two or three or an infinite
number of patterns. OK. This is, again, something
that's really bothersome. Now, I also use the crude
illustration of an uncertainty principle, that the
uncertainty in position z-axis and the uncertainty in the
momentum along the z-axis was greater than h. And I froze. And I didn't realize that I
had delta s with the slit. But it really is the
same thing as the z. Because the slit
is how you define the position in the z-axis. And so this is the first taste
of the uncertainty principle. And I said I didn't like it. OK. In quantum mechanics,
you're not allowed to look inside small stuff. You're not allowed to see
the microscopic structure. You're only able
to do experiments, usually thought experiments,
an infinite number of identical experiments. And they reveal the structure in
some complicated, encoded way. And this is really not what
the textbooks are about. Textbooks don't tell
you how you actually think about a problem
with quantum mechanics. They tell you, here are some
exactly solved problems. Memorize them. And I don't want you-- I don't want to do that. OK. Last part of the
introduction here-- suppose we have a circular
drum, a square drum, and a rectangular drum. Have you ever seen a square
drum or a rectangular drum? Do you have an idea how a
square drum would sound? Yes. You do have an idea. It would sound terrible. Because the frequencies, you
get are not integer multiples. It would just sound
amazingly terrible. But if you had a square
drum or a rectangular drum, you could do an
experiment with some kind of acoustic
instrument to find out what the frequency distribution
is of the noise you make. And you would be able to tell. It's not round. It might be square. Or it might have a certain
ratio of dimensions. This is what we're talking about
as far as internal structure is concerned. And it's very much like what
you would do as a musician. I mean, certainly, when
a musical instrument is arranged correctly, it's
not like a square drum. It sounds good. And that's because
you get harmonics or you get integer multiples
of some standard frequency. OK. So now, we're going to talk
about the classical wave equation, which is not quantum. But it's a very similar sort
of equation to the Schrodinger equation. And so the methods for solving
this differential equation are on display. And so the trick is--
well, first of all, where does this equation come from? And it's always force
is equal to mass times acceleration in disguise. OK. And then you have tricks
for how you solve this. And one of the most frequently
used and powerful tricks is separation of variables. You need to know how that works. Then once you solve the problem,
you have the general solution. And you then say, well,
OK, for the specific case we have, like a string
tied down at both ends, we have boundary conditions. And we impose those
boundary conditions. And then we have
basically what we would call the normal
modes of the problem. And then we would
ask, OK, well, suppose we're doing a
specific experiment or doing a specific
preparation of the system. And we can call that
the pluck of the system. And you might pluck
several normal modes. You get a superposition state. And that superposition state
behaves in a dynamic way. And you want to be able to
understand that dynamics. And the most important
thing that I want you to do is, instead of trying to draw
the solutions to a differential equation, which is a
mathematical equation, I want you to draw
cartoons, cartoons that embody your
understanding of the problem. And I'm going to be
trying to do that today. OK. In this course, for the
first half of the course, most of what we're
going to be doing is solving for exactly
soluble problems-- the particle in a box,
the harmonic oscillator, the rigid rotor, and
the hydrogen atom. With these four problems,
most of the things that we will encounter
in quantum mechanics are somehow related to these. And in the textbooks, they
treat these things as sacred. And they say, OK, well, now
that you've solved them, you understand
quantum mechanics. But these are really
tools for understanding more complicated situations. I mean, you might have
a particle in a box. Instead of with a square bottom,
it might have a tilted bottom. Or it might have
a double minimum. But if you understand
that, you then can begin to build
an understanding of, what are the things in the
experiment that tell you about these distortions
of the standard problem? And the same thing for
a harmonic oscillator. Almost everything
that's vibrating is harmonic approximately. But there's a little
bit of distortion as you stretch it more. And again, you
can understand how to measure the distortions from
harmonicity by understanding the harmonic oscillator. We did rotor, H atom. It's all the same. So I would like to tell you
that these standard problems are really important. But nothing is like that. And what's important is how
it's different from that. And this is my
unique perspective. And you won't get that from
McQuarrie or any textbook. But this is MIT. So there are templates for
understanding real quantum mechanical system. And the big thing, the
most important technique for doing that is
perturbation theory. And so perturbation
theory is just a way of building beyond
the oversimplification. And it's mathematically
really ugly. But it's tremendously powerful. And it's where you get insight. OK. Now, many people have complained
that they found 5.61 hard. Because it's so mathematical. And maybe this is going to be
the most mathematical lecture in the course. But I don't want it to be hard. Now, chemists usually derive
insights from pictorial rather than mathematical
views of a problem. So what are the pictures that
describe these differential equations? How do we convert
what seems to be just straight mathematics to
pictures that mean something to us? And that's my goal, to
get you to be drawing freehand pictures that
embody the important features of the solutions
to the problems. OK. So we're going to be looking
at a differential equation. And one of the first
questions you ask, well, where did that equation come from? And you're not going to derive
a differential equation ever in this course. But you're going to
want to think, well, I pretty much understand what's
in this differential equation. And then we'll use
standard methods for solving that equation. And one such differential
equation is this-- second derivative
of some function with respect to a variable
is equal to a constant times that function. Now, that you know. You know sines and cosines
are solutions to that. And you know that exponentials
are solutions to that. Now, that pretty much takes
you through a lot of problems in quantum mechanics. But now, one of the
important things is this is a second-order
differential equation. And that means that
there are going to be two linearly
independent solutions. And you need to
know both of them. I'll talk about this
some more later. Now, sometimes, the
differential equations look much more
complicated than this. And so the goal is
usually to rewrite it in a form which corresponds to
a differential equation that is well known and
solved by mathematicians whose business is doing that. But we won't be doing that. OK. But usually, when you have
a differential equation, the function is of
more than one variable. And frequently, it's
position and time. And so the first thing you do is
you try to separate variables. And so that's what
we're going to do. So we have a
differential equation. And the first thing
is a general solution. And one of the things that this
solution will have is nodes. And the distance between nodes-- here's a node. Here's a node. That's half the wavelength. And we know that in
quantum mechanics, if you know the wavelength,
you know the momentum. So nodes are really important. Because it's telling you
how fast things are moving. We can also look
at the envelope. And this would be some
kind of classical, as opposed to a quantum
mechanical, probability distribution. And so it might look like this. But the important thing
about the envelope is that it's always positive. Because it's
probability, as opposed to a probability
amplitude, which can be positive and negative. Interference is really
important in quantum mechanics. But sometimes, the envelope
tells you all you need to know. And the other thing is the
velocity of a stationary phase. So you have a wave. And it's moving. And you sit at a
point on that wave. And you ask, how fast
does that point move? And I did that last time. And OK. So I've already talked a little
bit about what we do next. But the important thing
is always, at the end, you draw a cartoon. And you endow that cartoon
with your insights. And that enables you to
remember and to understand and to organize questions
about the problem. OK. So let's get to work
on a real problem. So we have a string that's
tied down to two points. And so let's look at the
distortion of that string. And so we chop this region
of space up into regions. So this might be the
region at x minus 1. And this might be
the region at x0. And this might be
the region of x1. And we're interested in-- OK, suppose we have the value
of the displacement of the wave here at x minus 1
and here and here. OK, so these would be
the amount that the wave is displaced from equilibrium. And we call those u of x. And so the first segment
here, the minus 1 segment, this segment is pulling down
on this segment of the string by this amount. And this one is pulling up on
the segment by that amount. So we want to know, what is the
force acting on each segment? And so we have the
force constant times the displacement at x0 minus
the displacement at x minus 1. So this is the difference
between the displacements. And this is the force constant. We're talking about Hooke's law. Hooke's law is the
force is equal to minus k times the displacement. And so we collect the forces
felt by each particle. And the forces felt by each
particle are, again, the force constant times the difference
in u at 0 and minus 1 minus the difference-- plus 1 minus the difference
in u at 0 and plus 1. And this is a second derivative. This is the second derivative
of u with respect to x. So we've derived
a wave equation. And we know it's going to
involve a second derivative. So force is equal to
mass times acceleration. Well, we already
know the force is going to be related to
the second derivative of u with respect to x. And now, this is something. And we know what this is. This is going to be
the time derivative. OK. And this is just something
that gets the units right. And it has physical
significance. But in the case of this
particle on a string-- this wave on a
string, it's related to the mass of the string and
the tension of the string. And it's also related to the
velocity that things move. OK. So we have a
differential equation that is related to forces equal
to mass times acceleration. And the differential equation
has the form second derivative of u with respect
to x is equal to 1 over v squared times the
second derivative of u with respect to t. That's the wave equation. So it is really
f is equal to ma. But OK. And now, the units of this-- this is x. And this is t. In order to be
dimensionally consistent, this has to be something
that is x over t, OK? And so this may be a velocity. But it has units of velocity. That's the differential
equation we want to solve. OK. Well, the original differential
equation that I wrote-- but I'm getting ahead of myself. OK. So this U of x and t-- we'd like it to be
X of x times T of t. We think we could separate
the variables in this way. So we try it. If we fail, it says
you can't do that. Failure is usually
going to be a result that the solution to the
differential equation in this form is nothing. It's a straight line. Nothing's happening. So failure is acceptable. But if we're
successful, we're going to get two separate
differential equations. OK. So what we do then is take
this differential equation, substitute this in,
and divide on the left. So we have 1 over X of
x times T of t times the second derivative
with respect to x of xt is equal to 1 over xt times the
second derivative with respect to t of xt. OK. Well, on this side
of the equation, the only thing that
involves time is here. This doesn't operate on time. And so we can cancel the time
dependence from this side. And over on this
side, this derivative operates on t but not x. And so we can cancel the x part. And so what we have
now is an equation X of x second derivative
with respect to x squared of x is equal to this constant,
1 over v squared, times 1 over t times the derivative of
T with respect to little t, OK? So this is interesting. We have a function
of x on this side and a function of
t on this side. They are independent variables. This can only be
true if both sides are equal to the same constant. So now, we have to
differential equations. We have 1 over x second
derivative with respect to x of x is equal to a concept. And we have 1 over v squared
1 over t second derivative with respect to t
of T is equal to K. So now, we're on firmer ground. We know about solutions
to this kind of equation. And so there's two cases. One is this K is greater than 0. And the other is K less than 0. So let's look at this equation. If K is greater than 0, then if
we plug in a sine or a cosine, we get something
that's less than 0. Because the derivative of sine
with respect to its variable is negative cosine. And then we do it again,
we get back to sine. And so sines and cosines are
no good for this equation if K is greater than 0. But exponentials-- so we can
have e to some constant x or e to the minus
some constant x. Or here, for the
negative value of K, we could have sine
some constant x and cosine of some constant x. We know that. So we have two cases. So to make life
simple, we say K is going to be equal to
lowercase k squared. Because we want to use this
lowercase k in our solutions. All right. So I may have confused matters. But the solution for the time
equation and the position equation are clear. And so depending on whether
K is positive or negative, we're dealing with sines
and cosines or exponentials. OK. So I don't want to belabor
this, but the next stage is boundary conditions. We don't know whether K positive
or K negative is possible. But we do know
boundary conditions. And so if we have a string
which is tied down at the end-- so this is x equals 0. And this is x equals
L. Then we input impose the boundary conditions. Well, the boundary conditions
are u of 0t is equal to 0. And u of Lt is equal to 0. So if we take the K greater than
0 case, u of 0t is equal to-- well, let's just do this again. 0t. We have x of 0 times T of t. OK, we don't really
care about this. But x of 0 has to be 0-- I'm sorry. OK. So we have two solutions. If K is greater than 0, we
have the exponential terms. So we have Ae to the 0
plus B to the minus 0. And this has to be equal to 0. That's x of 0. Well, e to the 0 is 1. E to the minus 0 is 1. And so this is good. A has to be equal to
minus B. And then we have the other boundary
condition, X of L. We have A e to the L plus B e
to the minus L. Well, I'm sorry. Let's just put what
we already know. Minus A. So we can write
this as A e to the L minus e to the minus L. And
that has to be equal to 0. Can't do it. This can never be 0. So that means the K greater
than 0 solutions are illegal. Well, that's kind of bad news. Because it sounds like
separation of variables is failing. But it doesn't. Because the K less
than 0 solution works. So for K less than 0, X of 0 is
equal to C sine 0 plus D cosine 0. And so that means
D is equal to 0. Things are dying. And X of L, boundary condition,
is C sine KL is equal to 0. And this we can solve. So sine is equal to 0 when KL
is equal to 0, 0 pi, 2 pi, et cetera. And so we have KL
is equal to n pi. So we can write this
as Kn is equal to n pi over L. We get quantization. This isn't quantum mechanics. But there are certain allowed
values of this K constant. And we have a
bunch of solutions. And so what do they look like? So n equals 0, n
equals 1, n equals 2. So what does the 0
solution look like? Yes? AUDIENCE: No node. ROBERT FIELD: Nothing. So we don't even
think about this. We say, n equals 0
is not a solution. Because the wave isn't there. No nodes, one node. And if we look at
this carefully, the node is always
at a cemetery point. It's in the middle. If we have the next one,
we'll have two nodes. And they'll be at
the 1/3 2/3 point. And so we know
where the nodes are. We also know that the amplitude
of each loop of the wave function is the same. But it alternates in sine. So you can draw
cartoons now at will. Because this spatial part of the
solution to this wave is clear. Any value of the quantum
number, or the n, gives you a picture that
you can draw in seconds. And there are a lot of quantum
mechanical problems like that. But sometimes, you
have to keep in mind that the node separation,
in other words-- well, let's just say node
separation is lambda over 2. And lambda over 2 is
equal to 1/2 h over p. So if we know what
the momentum is, or we know what the kinetic energy is,
we know what the momentum is. We know how momentum is
encoded in node separations. So everything we want to know
about a one-dimensional problem is expressed in the
spacing of nodes and the amplitude between nodes. And the amplitude between
nodes have something to do with the momentum, too. Because if you're
going from here to here at some high velocity,
there's not much amplitude. And at a lower velocity,
you get more amplitude. And so the amplitude
in each of these node to node separate
sections is related to the average momentum
of the classical particle in that section. So the classical
mechanics is going to be extremely important
in drawing cartoons for quantum mechanical systems. Not in the textbooks. We supposedly know classical
mechanics pretty well, and especially here at MIT. So you might as well
use that in order to get an idea of how all of
the quantum mechanical problems you're facing will be behaving. OK. So the next thing we want
to do is finish the job. And so I can simply write down
the time-dependent solutions. They are E sine vkn t
plus F cosine vkn t. And we can say that-- rather than carrying
around all this stuff, we can say omega n is vkn. Isn't that nice? So we have a frequency for
the time-dependent part, which is an integer multiple of
this constant V times this [? vector. ?] Or this you
can think of as just k sub n. So we can rewrite this in
a frequency and phase form. We have now the full solution. We have A n sine n pi L over x. And then this e N sine n pi-- I'm so used to the
pictures I don't even want to look at the
equations anymore-- n omega t plus F n
cosine n omega t. OK. So we can also take
this and rewrite it in a simpler form, E n prime
cosine n omega t plus phi n. So we can combine these two
terms as a single cosine with a phase vector. OK. So now, we're ready to actually
go to the specific thing that you do in a real experiment
or a real musical instrument. We say, OK, here is the
actual initial condition, the pluck of the system. And the pluck usually
occurs at t equals 0. But I'll just specify it here. And what we have is now a
sum over as many normal modes as you want. We have A n E n prime times sine
n pi over L x times cosine N omega t plus phi N. So we have a bunch of terms
like this, a spacial factor, and a temporal factor. And you can draw
pictures of both. Now, but there is
another simplification. From trigonometry, sine A cosine
B-- we have sine and cosine-- can be written as 1/2
times sine n pi L x plus n omega t plus phi
n plus sine n pi L x minus n omega t minus phi n. So these are the two
possible solutions. And we can write them now
in terms of position factor at a time factor in the same
sine or cosine function. So these are the things. Now, we're ready
to make a picture. OK. So these are the
actual things that you make by exciting the system
not in an eigenfunction. But it's a superposition
of eigenfunctions. And again, there are
certain things you learn. If you have a pure
eigenfunction, you have standing waves. There's no left-right motion. There's no breathing motion. There's only up-down motion of
each loop of the wave function. If you have a superposition of
two or more functions, which are all of even n, then what
happens is you have no motions, you just have breathing. In other words,
you have a function that might look sort of
like this at one time and like that at another time. So amplitude is moving. So it can be moving. And in between, it's
sort of like this. Now, if you have
a function which involves both even and odd n,
you have left-right motion. This is true in
quantum mechanics, too. So you only get motion if you're
making superposition of eigen-- Yes? AUDIENCE: What is the
difference between breathing and the standing
wave with no nodes? ROBERT FIELD: Well, for this
picture, this is over-simple. So I mean, you could have-- basically, what's happening is
amplitude is moving from middle to the edges and back. And so yes. But you want to develop
your own language, your own set of drawings so that
you understand these things. And the important thing is
the understanding, the ability to draw these
pictures which contain the critical information about
node spacings, amplitudes, shapes, and to
anticipate when you're going to have left-right
motion or when you're just going to have
complicated up-down motions. Because there could be nodes. But there is no motion of the
center of this wavepacket. Now, this is fantastic. Because I just said wavepacket. Quantum mechanics--
eigenfunctions don't move. Superpositions of
eigenfunctions do move. If we make a superposition
of many eigenfunctions, it is a particle-like state. What that particle-like
state will do is exactly what you
expect from 8.01. The particle-like
states-- the center of the wavepacket for a
particle-like state moves according to Newton's equations. So I'm saying I'm
taking away your ability to look at microscopic stuff. And I'm going to
give it back to you. By the end of the
course, we're going to have the time-dependent
Schrodinger equation. We're going to be able
to see things move. And we're going to
see why they move and how they encode that motion. Not in the textbooks,
but I think it's something you really
want to be able to do. If you're going to
understand physical systems and use it to guide
your understanding, you have to be able
to draw these pictures and build a step at a time. And so this way the equation,
the classical wave equation, gives you almost all of
the tools for artistry as well as insight. OK. Now, in the notes, there is
a time-lapse movie that shows what a two-state wave function-- what a two-state solution
to the wave equation looks like if you have even and odd
or only even and even terms. OK. Now, I'm going to make
some assertions at the end. We're coming back
to the drum problem. And suppose we have
a rectangular drum. Well, solving the
differential equation for this rectangular
drum gives you a bunch of normal
mode frequencies that depend on two indices. And so this is the geometric
structure of the drum. And these are the
quantum numbers. And these are the frequencies. And it's going to make a whole
bunch of frequencies that are not integer
multiples of each other or of any simpler thing. And that's why it
sounds horrible. It's perhaps a little bit like
playing a violin with a saw. It will sound terrible. You would never do it. But you would also never build
a square or rectangular drum. But the noise that you
make tells immediately not just what the shape
of the instrument is but how it was played. For example, suppose you
had an elliptical drum. That'll sound terrible, too. But here are the two foci. I'm not so sure
it'll sound terrible if you hit it here or here. And certainly, if you
are a circular drum, if you hit it in the middle
as opposed to on the edges, it'll sound different. The spectral content
will be the same. But the amplitudes of each
component will be different. And so in quantum mechanics, you
use the same sort of instinct as you develop as
a musician in order to figure out how this
system is going to respond to what you do to it. And that's pretty powerful. So many of you are musicians. And you know
instinctively what's wrong when you do something
that's not quite right, or your instrument
is out of tune. But in quantum mechanics,
all of those insights will come to bear. Not in the textbooks. Because the textbooks tell you
about exactly solved problems. And then they tell
you how to do spectra that are too perfect for
anybody else to observe. And you won't see those spectra. And they don't tell you what
the spectra you will observe tell you about the
system in question. OK. So I should stop now. And I'm going to be generating
Problem Set 2, which will be posted on Friday. Problem Set 1 is
due this Friday. And-- Good. Thank you.
