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visit MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: Today we begin
with the harmonic oscillator. And before we get into
the harmonic oscillator, I want to touch on
a few concepts that have been mentioned in class
and just elaborate on them. It is the issue of nodes,
and how solutions look at, and why solutions have
more and more nodes, why the ground state has no nodes. This kind of stuff. So these are just a collection
of remarks and an argument for you to understand a
little more intuitively why these properties hold. So one first thing
I want to mention is, if you have a Schrodinger
equation for an energy eigenstate. Schrodinger equation for
an energy eigenstate. You have an equation of the
from minus h squared over 2m. d second dx squared psi of
x plus v of x psi of x equal e times psi of x. Now the issue of
this equation is that you're trying to solve for
two things at the same time. If you're looking at what
we call bound states, now what is a bound state? A bound state is something that
is not extended all that much. So a bound state will
be a wave function that goes to 0 as the absolute
value of x goes to infinity. So it's a probability function
that certainly doesn't extend all the way to infinity. It just collapses. It's normalizable. So these are bound
states, and you're looking for bound
states of this equation. And your difficulty is
that you don't know psi, and you don't know E either. So you have to solve a problem
in which, if you were thinking oh this is just a plain
differential equation, give me the value of E.
We know the potential, just calculate it. That's not the way it
works in quantum mechanics, because you need to have
normalizable solutions. So at the end of the day,
as will be very clear today, this E gets fixed. You cannot get
arbitrary values of E's. So I want to make a couple of
remarks about this equation. Is that there's this
thing that can't happen. Certainly, if v of x is a smooth
potential, then if you observer that the wave function
vanishes at some point, and the derivative
of the wave function vanishes at that same
point, these two things imply that psi of
x is identically 0. And therefore it
means that you really are not interested in that. That's not a solution of
the Schrodinger equation. Psi equals 0 is obviously solves
this, but it's not interesting. It doesn't represent
the particle. So what I claim here
is that, if it happens to be that you're
solving the Schrodinger problem with some
potential that is smooth, you can take derivatives of it. And then you encounter that
the wave function vanishes at some point, and its slope
vanishes at that same point. Then the wave function
vanishes completely. So you cannot have
a wave function, a psi of x that
does the following. Comes down here. It becomes an inflection
point and goes down. This is not allowed. If the wave function
vanishes at some point, then the wave function
is going to do this. It's going to hit at an angle,
because you cannot have that the wave function is 0 and needs
the derivative 0 at the same point. And the reason is simple. I'm not going to
prove it now here. It is that you have a second
order differential equation, and a second order
differential equation is completely determined
by knowing the function at the point and the the
derivative at the point. And if both are 0s, like
the most trivial kind of initial condition, the only
solution consistent with this is psi equals 0 everywhere. So this can't happen. And it's something good
for you to remember. If you have to do a
plot of a wave function, you should never have this. So this is what we call the
node in a wave function. It's a place where the
wave function vanishes, and the derivative
of the wave function better not vanish at that point. So this is one claim that
it's not hard to prove, but we just don't try to do it. And there's another claim that
I want you to be aware of. That for bound states
in one dimension, in the kind of thing that we're
doing now in one dimension, no degeneracy is possible. What do I mean by that? You will never find
two bound states of a potential that
are different that have the same energy. It's just absolutely impossible. It's a very wonderful
result, but also I'm not going to prove it here. Maybe it will be given as an
exercise later in the course. And it's discussed
in 805 as well. But that's another statement
that is very important. There's no degeneracy. Now you've looked at
this simple potential, the square well infinite one. And how does it look? You have x going from 0 to a. And the potential is 0. From 0 to a is
infinite otherwise. The particle is bound
to stay inside the two walls of this potential. So in here we've plotted the
potential as a function of x. Here is x. And the wave functions are
things that you know already. Yes? AUDIENCE: Is it true if you have
two wells next to each other that there's still no degeneracy
if it's an infinite barrier? PROFESSOR: If there's
two wells and there's an infinite barrier
between them, it's like having two universes. So it's not really a
one dimensional problem. If you have an infinite
barrier like two worlds that can talk to each other. So yes, then you
would have degeneracy. It's like saying you can have
here one atom of hydrogen or something in one energy
level, here another one. They don't talk to each other. They're degenerate states. But in general, we're talking
about normal potentials that are preferably smooth. Most of these things are true
even if they're not smooth. But it's a little more delicate. But certainly two
potentials that are separated by
an infinite barrier is not part of what we
really want to consider. OK so these wave functions
start with m equals 0, 1, 2, 3, and psi ends of x are
square root of 2 over a sin n plus 1 pi x over a. Things that you've seen already. And En, the energies
are ever growing as a function of the integer
n that characterizes them. 2ma squared. And the thing that you
notice is that psi 0 has technically no nodes. That is to say these
wave functions have to varnish at the end, because
the potential becomes infinite, which means a particle
really can go through. The wave function
has to be continuous. There cannot be any wave
function to the left. So it has to vanish here. These things don't
count as nodes. It is like a bound state
has to vanish at infinity. And that's not what
we count the node. A node is somewhere
in the middle of the range of x where
the wave function vanishes. So this is the ground state. This is psi zero has no nodes. Psi one would be something
like that, has one node. And try the next ones. They have more and more nodes. So psi n has n nodes. And the interesting thing
is that this result actually is true for extremely
general potentials. You don't have to just
do the square well to see that the ground
state has no nodes. That first excited state was one
node, and so on and so forth. It's true in general. This is actually a
very nice result, but its difficult to prove. In fact, it's pretty
hard to prove. So there is a nice argument. Not 100% rigorous,
but thoroughly nice and really physical
that I'm going to present to you why
this result is true. So let's try to do that. So here is the general case. So I'm going to take
a smooth v of x. That will be of this kind. The potential, here is
x, and this potential is going to be like that. Smooth all over. And I will actually
have it that it actually goes to infinity as
x goes to infinity. Many of these things are
really not necessary, but it simplifies our life. OK, so here is a result
that it's known to be true. If this thing grows to
infinity, the potential never stops growing, you get
infinite number of bound states at fixed energies. One energy, two
energy, three energy, infinite number of bound states. Number of bound states. That's a fact. I will not try to prove it. We'll do it for the
harmonic oscillator. We'll see those infinite
number of states, but here we can't
prove it easily. Nevertheless, what
I want to argue for you is that these
states, as the first one, will have no nodes. The second state, the
first excited state, will have one node. Next will have two nodes,
three nodes, four nodes. We want to understand what
is this issue of the nodes. OK. You're not trying
to prove everything. But we're trying to
prove or understand something that is
very important. So how do we prove this? Or how do we understand that
the nodes-- so there will be an infinite number
of bound states, psi 0, psi 1, psi 2, up
to psi n, and it goes on. And psi n has n nodes. All right. So what I'm going to do, in
order to understand this, is I'm going to produce what we
will call screened potentials. Screened potentials. I'm going to select the
lowest point of the potential here for convenience. And I'm going to call it x
equals 0 is the lowest point. And the screen potential
will have a parameter a. It's a potential
which is equal to v of x if the absolute
value of x is less than a. And its infinity if the absolute
value of x is greater than a. So I come here,
and I want to see this is this
potential, v of x, what is the screened
potential for sum a? Well, I wanted colored
chalk, but I don't have it. I go mark a here minus a. Here are the points
between x and a. Absolute value of x less than a. Throughout this region
the screened potential is the potential that you have. Nevertheless, for the
rest, its infinite. So the screened
potential is this thing. Is infinite there, and
it's here this thing. So it's just some potential. You take it a screen
and you just see one part of the potential,
and let it go to infinity. So that's a screen potential. So now what I'm going
to do is that I'm going to try to
argue that you could try to find the bound state
of the screen potential. Unless you remove the
screen, you will find, as you let a go to
infinity, you will find the bound states of
the original potential. It's reasonable
that that's true, because as you
remove the screen, you're letting more of
the potential be exposed, and more of the
potential be exposed. And the wave functions
eventually die, so as the time that you're very far away,
you affect the wave functions less and less. So that's the argument. We're going to try
to argue that we're going to look at
the bound states of the screened potentials and
see what happened, whether they tell us about the bound
states the original potential. So for this, I'm going
to begin with a screen potential in which
a goes to 0 and say that a is equal to
epsilon, very small. So what potential so do I have? A very tiny potential here
from epsilon to minus epsilon. Now I chose the original point
down here to be the minimum. So actually, the bottom part of
the potential is really flat. And if you take
epsilon going to 0, well, the potential
might do this, but really at the bottom for
sufficiently small epsilon, this is an infinite square
well with psis to epsilon. I chose the minimum so that you
don't get something like this. If it would be a
point with a slope, this would be an ugly thing. So let's choose the minimum. And we have the screen
potential here, and that's it. Now look what we do. We say all right, here
there is a ground state. Very tiny. Goes like that. Vanishes here. Vanishes there. And has no nodes. Very tiny. You know the two 0s are
very close to each other. And now I'm going
to try to increase the value of the screen a. So suppose we've
increased the screen, and now the potential is here. And now we have a finite screen. Here is the potential. And I look at the wave function. How it looks. Here is psi 0. This ground state psi 0. Well, since this thing in here,
the potential becomes infinite, the wave function
still must vanish here and still must vanish here. Now just for your
imagination, think of this. At this stage, it still more
or less looks like this. Maybe. Now I'm going to ask, as I
increase, can I produce a node? And look what's going to happen. So suppose it might happen
that, as you increase, suddenly you produce a node. So here's what I'm saying here. I'm going to show it here. Suppose up to this
point, there is no node. But then when I double
it, when I increase it to twice the size, when I go
to screen potential like that, suddenly there is a
node in the middle. So if there is a
node in the middle, one thing that could have
happened is that you have this. And now look what must
have happened then. As I stretch this, this slope
must have been going down, and down, and down, until
it flips to the other side to produce a node here. It could have
happened on this side, but it's the same, so
the argument is just done with this side. To produce a node you
could have done somehow the slope here must
have changed sine. But for that to happen
continuously, at some point the this slope must have been 0. But you cannot have
a 0 and 0 slope. So this thing can't
flip, can't do this. Another thing that
could have happened is that when we
are here already, maybe the wave function
looks like that. It doesn't flip at the
edges, but produces something like that. But the only way this
can happen continuously, and this potential is
changing continuously, is for this thing at
some intermediate stage, as you keep stretching the
screen, this sort of starts to produce a depression here. And at some point, to get
here it has to do this. But it can't do this either. It cannot vanish and have
derivative like that. So actually, as you
stretch the screen, there's no way to
produce a node. That property forbids it. So by the time you go and
take the screen to infinity, this wave function has no nodes. So that proves it that the
ground state has no nodes. You could call this
a physicist proof, which means-- not in
the pejorative way. It means that it's
reasonable, it's intuitive, and a mathematician working
hard could make it rigorous. A bad physicist proof is
one that is a little sloppy and no mathematician could
fix it and make it work. So I think this is a good
physics proof in that sense. Probably you can
construct a real proof, or based on this, a
very precise proof. Now look at excited states. Suppose you take now here
this screen very little, and now consider the third
excited state, psi three. I'm sorry, we'll call this psi
2 because it has two nodes. Well, maybe I should do psi 1. Psi 1. One node. Same thing. As you increase
it, there's no way to create another
node continuously. Because again, you have
to flip at the edges, or you have to
depress in the middle. So this one will evolve
to a wave function that will have one node in
the whole big potential. Now stayed does that
state have more energy than the ground state? Well, it certainly begins with
a small screen with more energy, because in the square well
psi 1 has more energy. And that energy should
be clear that it's not going to go below the
energy of the ground state. Why? Because if it went below the
energy of the ground state slowly, at some point for
some value of the screen, it would have the same
energy as the ground state. But no degeneracy is possible
in one dimensional problems. So that can't happen. Cannot have that. So it will always
stay a little higher. And therefore with one node you
will be a little higher energy. With two nodes will
be higher and higher. And that's it. That's the argument. Now, we've argued by this
continuous deformation process that this potential not
only has these bound states, but this is n nodes and En is
greater than En prime for n greater than n prime. So the more nodes,
the more energy. Pretty nice result,
and that's really all I wanted to say
about this problem. Are there any questions? Any? OK. So what we do now is
the harmonic oscillator. That's going to keep us busy
for the rest of today's lecture. It's a very interesting problem. And it's a most famous quantum
mechanics problem in a sense, because it happens to be useful
in many, many applications. If you have any
potential-- so what is the characteristic of
the harmonic oscillator? Harmonic oscillator. Oscillator. Well, the energy operator is p
squared over 2m plus, we write, one half m omega
squared x squared where omega is this
omega that you always think of angular velocity,
or angular frequency. It's more like
angular frequency. Omega has units of 1 over time. It's actually put 2pi over
the period of an oscillation. And this you know from
classical mechanics. If you have a harmonic
oscillator of this form, yeah, it actually oscillates
with this frequency. And E is the energy
operator, and this is the energy of the oscillator. So what defines an oscillator? It's something in which the
potential energy, this term is v of x. v of x is quadratic in x. That is a harmonic oscillator. Then you arrange the
constants to make sense. This has units of
energy, because this has units of length squared. 1 over time squared. Length over time
is velocity squared times mass is kinetic energy. So this term has
the units of energy. And you good with that. And why is this useful? Because actually in any sort
of arbitrary potential, rather general potential
at least, whenever you have a minimum where
the derivative vanishes, then the second derivative
need not vanish. Then it's a good
approximation to think of the potential at the minimum
as a quadratic potential. It fits the potential
nicely over a good region. And therefore when you have
two molecules with a bound or something oscillating,
there is a potential. It has a minimum at the
equilibrium position. And the oscillations
are governed by some harmonic oscillator. When you have photons
in space time traveling, there is a set of
harmonic oscillators that correspond to photons. Many, many applications. Endless amount of applications
for the harmonic oscillator. So we really want to
understand this system quantum mechanically. And what does that mean? Is that we really want
to calculate and solve the Schrodinger equation. This is our first step in
understanding the system. There's going to
be a lot of work to be done even once we have
the solutions of the Schrodinger equation. But the first thing
is to figure out what are the energy
eigenstates or the solutions of the Schrodinger
equation for this problem. So notice that here
in this problem there's an energy quantity. Remember, when you have
a harmontonian like that, and people say so what is
the ground state energy? Well, have to find the
ground state wave function. Have to do things. Give me an hour, I'll find it. And all that. But if you want an
approximate value, dimensional analysis will do it,
roughly what is it going to be. Well, with this constant how
do you produce an energy? Well, you remember
what Einstein did, and you know that h bar
omega has units of energy. So that's an energy associated
with Lagrangian energy like quantity. And we expect that
that energy is going to be the relevant energy. And in fact, we'll find
that the ground state energy is just one half of that. There's another quantity
that may be interesting. How about the length? How do you construct a
length from these quantities? Well, you can start
doing m omega h bar and put powers and struggle. I hate doing that. I always try to find some
way of doing it and avoiding that thing. So I know that
energies go like h over h squared over
m length squared. So I'm going to call
the length a quantity a. So ma squared. That has units of energy. And you should remember
that because energy is b squared over 2m,
and b by De Broglie is h bar over sub lamda. So h bar squared, lambda
squared, and m here, that's units of energy. So that's a length. On the other hand,
we have another way to construct an energy
is with this thing, m omega squared length squared. So that's also m omega
squared a squared. That's another energy. So from this equation I
find that a to the fourth is h squared over m
squared omega squared. And it's a little
complicated, so a squared is h bar over m omega. So that's a length. Length squared. I don't want to take
the square root. We can leave it
for a moment there. But that's important because
of that's a length scale. And if somebody would ask
you in the ground state, how far is this
particle oscillating, you would say probably
about a square root of this. Would be a natural answer
and probably about right. So OK, energy and
units is very important to begin your analysis. So what is the
Schrodinger equation? The Schrodinger
equation for this thing is going to be minus h
squared over 2m, d second psi, dx squared plus the potential,
one half m omega squared x squared psi is equal E psi. And the big problem is I don't
know psi and I don't know E. Now there's so many elegant
ways of solving the harmonic oscillator. You will see those next lecture. Allan Adams will be back here. But we all have to go
through once in your life through the direct, uninspired
method of solving it. Because most of the times
when you have a new problem, you will not come up with
a beautiful, elegant method to avoid solving the
differential equation. You will have to struggle with
the differential equation. So today we struggle with
the differential equation. We're going to just do it. And I'm going to do it slow
enough and in detail enough that I hope you
follow everything. I'll just keep a
couple of things, but it will be one line
computations that I will skip. So this equation is some sort
of fairly difficult thing. And it's complicated and
made fairly unpleasant by the presence of
all these constants. What kind of equation is that
with all these constants? They shouldn't be there,
all this constants, in fact. So this is the first step,
cleaning up the equation. We have to clean it up. Why? Because the nice
functions in life like y double prime is equal to
minus y have no units. The derivatives create no units. y has the same units of that,
and the solution is sine of x, where x must have no units,
because you cannot find the sine of one centimeter. So this thing, we should
have the same thing here. No units anywhere. So how can we do that? This is an absolutely
necessary first step. If you're going to be
carrying all these constants, you'll get nowhere. So we have to clean it up. So what I'm going
to try to see is that look, here is
psi, psi, and psi. So suppose I do the
following thing, that I will clean
up the right hand side by dividing by something
with units of energy. So I'm going to do
the following way. I'm going to divide all
by 1 over h bar omega. And this 2 I'm going
to multiply by 2. So multiply by 2
over h bar omega. So what do I achieve
with that first step? I achieve that
these 2s disappear. Well, that's not too bad. Not that great either, I think. But in the right hand side,
this has units of energy. And the right hand side will
not have units of energy. So what do we get here? So we get minus. The h becomes an h alone over--
the m disappears-- so m omega. The second psi the x squared. The 1/2 disappeared,
so m omega over h bar x squared psi equals
2 E over h bar omega psi. It looks actually
quite better already. Should agree with that. It looks a lot nicer Now
I can use a name for this. I want to call this
the dimensionless value of the energy. So a calligraphic e. It has no units. It's telling me if I find some
energy, that that energy really is this number, this pure
number is how many times bigger is e with respect
to h omega over 2. So I'll write this now as e psi. And look what I have. I have no units here. And I have a psi. And I have a psi. But things have
worked out already. Look, the same factor
here, h over m omega is upside down here. And this factor has
units of length squared. Length squared times d d
length squared has no units. And here's 1 over
length squared. 1 over length squared
times length squared. So things have worked out. And we can now
simply say x is going to be equal to au,
a new variable. This is going to be your new
variable for your differential equation in which is this thing. And then this
differential equation really has cleaned
up perfectly well. So how does it look now? Well, it's all gone
actually, because if you have x equals au, d dx by
chain rule is 1 over a d du. And to derivatives
this with respect to x it's 1 over a squared
times the d second du squared. And this thing is a squared. So actually you
cancel this factor. And when I write x
equals to au, you get an a squared times this. And a squared times this is 1. So your differential
equations has become minus the
second psi du squared, where u is a dimensionless
quantity, because this has units of length, this
has units of length. No units here. You have no units. So minus d second
du squared plus u squared psi is equal to e psi. Much nicer. This is an equation
we can think about without being distracted
by this endless amount of little trivialities. But still we haven't
solved it, and how are we going to solve this equation? So let's again think
of what should happen. Somehow it should happen
that these e's get fixed. And there is some solution
just for some values of e's. It's not obvious at this stage
how that is going to happen. Yes? AUDIENCE: [INAUDIBLE]. PROFESSOR: Here for example,
let me do this term. h bar over m omega is minus,
from that equation, a squared. But dx squared is 1 over
a squared d du squared. So a squared cancels. And here the x is equal
a squared times u, so again cancels. OK so what is the problem here? The problem is that most likely
what is going to go wrong is that this solution for
arbitrary values of e's is going to diverge at
infinity, and you're never going to be able
to normalize it. So let's try to understand
how the solution looks as we go to infinity. So this is the first
thing you should do with an equation like that. How does this solution
look as u goes to infinity? Now we may not be able to solve
it exactly in that case either, but we're going to gain
insight into what's happening. So here it is. When u goes to infinity,
this term, whatever psi is, this term is much
bigger than that, because we're presumably
working with some fixed energy that we still don't
know what it is, but it's a fixed number and,
for you, sufficiently large. This is going to dominate. So the equation
that we're trying to solve as u goes to infinity,
the equation sort of becomes psi double prime-- prime
is for two derivatives-- is equal to u squared psi. OK, so how do we get
an idea what solves this is not all that obvious. It's certainly not a power of u,
because when you differentiate the power of u, you
lower the degree rather than increase the degree. So what function increases
degree as you differentiate? It's not the trivial function. Cannot be a polynomial. If it could be
even a polynomial, if you take two derivatives,
it kind cannot be equal to x squared times a polynomial. It's sort of upside down. So if you think about
it for a little while, you don't have an
exact solution, but you would imagine
that something like this would do it, an e
to the u squared. Because an e to
the u squared, when you differentiate with respect
to us, you produce a u down. When you one derivative. When you take
another derivative, well, it's more complicated,
but one term you will produce another u down. So that probably is quite good. So let's try that. Let's try to see if we
have something like that. So I will try something. I'll try psi equals 2. I'm going to try
the following thing. e to the alpha u squared over
2 where alpha is a number. I don't know how much it is. Alpha is some number. Now could try this
alone, but I actually want to emphasize
to you that if this is the behavior
near infinity, it won't make any difference
if you put here, for example, something
like u to the power k. It will also be
roughly a solution. So let's see that. So for that I have
to differentiate. And let's see what we get. So we're trying to see how the
function behaves far, far away. You might say well
look, probably that alpha should be negative. But let's see what
the equation tells us before we put anything in there. So if I do psi prime,
you would get what? You would get one
term that would be alpha u times this u to
the k into the alpha u squared over 2. I differentiated
the exponential. I differentiated
the exponential. And then you would
get a term where you differentiate the power. So you get ku to the k
minus 1 into the alpha u squared over 2. If you take a second
derivative, well, I can differentiate the
exponential again, so I will get alpha
u now squared, because each derivative
of this exponent produces a factor of alpha u. u to the k into the
alpha u squared over 2. And a couple more terms that
they all have less powers of u, because this term
has u to the k plus-- already has u to the k plus 1. And this has u to the k minus 1. They differ by two powers of u. So for illustration,
please, if you want, do it. Three lines, you should skip
three lines in your notebook if you're taking notes
and get the following. No point in me doing
this algebra here. Alpha u squared over 2. Because actually it's
not all that important. Over alpha 1 over u squared
plus k minus 1 over alpha squared 1 over u to the fourth. That's all you get. Look, this is alpha
squared u squared times psi times these things. 1 plus 2 k plus 1 over
alpha 1 over u squared. So when u goes to
infinity, your solution works, because these
thing's are negligible. So you get a number
times u squared. That is the equation you are
trying to solve up there. And therefore, you get
that the equation if alpha squared is equal to 1. And that means and really that
alpha can be plus minus 1. And roughly this
solution near infinity, probably there's two solutions. This is a second order
differential equation, so even near infinity there
should be two solutions. So we expect as u goes
to infinity psi of u will be some constant A
times u to the k times e to the minus u squared over 2. That's where alpha
equal minus 1. Plus Bu to the k into the
plus u squared over 2. And what is k? Well, we don't know what is k. It seems to work for all k. That may seem a little
confusing now, but don't worry. We'll see other things
happening here very soon. So look at what has happened. We've identified that most
likely your wave function is going to look like
this at infinity. So we're going to want to
this part not to be present. So presumably we're going
to want a solution that just has this, because
this is normalizable. The integral of any power times
a Gaussian is convergence. So this can be normalized. The Gaussian falls so
fast that any power can be integrated
against a Gaussian. Any power however big
doesn't grow big enough to compensate a Gaussian. It's impossible to
compensate a Gaussian. So we hope for this. But we want to
translate what we've learned into some
technical advantage in solving the
differential equation, because, after all, we wanted
be insight how it looks far way, but we wanted to solve
the differential equation. So how can we use
this insight we now have to simplify the solution
of the differential equation? The idea is to change
variables a little bit. So write psi of u to be
equal to h of u times e to the minus u squared over 2. Now you're going to say
wait, what are you doing? Are you making an
approximation now that this is what is
going to look far away? Or what are you putting there? I'm not making
any approximation. I'm just saying whatever
pis is, it can always be written in this way. Why? Because if you have
a psi of u, you can write it as psi of u
over e to the minus u squared over 2 times e minus
u squared over 2. Very trivially this
can always be done. As long as we say
that h is arbitrary, there's nothing,
no constraint here. I have not assume
anything, nothing. I'm just hoping that I have a
differential equation for psi. That because this is a very
clever factor, the differential equation for h will be simpler. Because part of the dependence
has been taken over. So maybe h, for example, could
be now a polynomial solution, because this product
has been taken care. So the hope is that by
writing this equation it will become an
equation for h of u, and that equation
will be simpler. So will it be simpler? Well, here again this
is not difficult. You're supposed to plug
into equation one-- this is the equation
one-- plug into one. I won't do it. It's three lines of
algebra to plug into one and calculate the
equation for h of u. You should do it. It's the kind of thing that
one should do at least once. So please do it. It's three, four lines. It's not long. But I'll just write the answer. So by the time you
substitute, of course, the e to the minus
u squared over 2 is going to cancel
from everywhere. It's very here. You just need to
take two derivatives, so it becomes a second
order differential equation. And indeed, it becomes
a tractable differential equation. The second h, du squared minus
2u dh du plus e minus 1 h equals 0. OK, that is our equation now. So now we face the problem
finally solving this equation. So before we start,
maybe there's some questions of what
we've done so far. Let's see. Any questions? Yes? AUDIENCE: Do you have
right there in the middle would be-- this equation
is linear, so can we just [INAUDIBLE] minus u squared over
2 and you stuck it to that u to the k. PROFESSOR: It's here? This thing? AUDIENCE: Yeah. Could you then just power series
what's going on at 0 with those u to the k terms [INAUDIBLE]? PROFESSOR: No. This is the behavior
as u goes to infinity. So I actually don't know
that the function near 0 is going to behave
like u to the k. We really don't know. It suggest to you that
maybe the solution is going to be near 0 u to
the k times some polynomial or something like that. But it's not that, because
this analysis was just done at infinity. So we really have no information
still what's going on near 0. Other questions? Yes? AUDIENCE: So is k some arbitrary
number or is it an integer? PROFESSOR: At this moment,
actually, it doesn't matter. Is that right? Doesn't matter. The analysis that we did here
suggests it could be anything. That's why I just didn't
put it into h or u. I didn't put it because
would be strange to put here a u to the k. I wouldn't know
what to make of it. So at this moment,
the best thing to say is we don't know what it is,
and maybe we'll understand it. And we will. In a few seconds, we'll
sort of see what's going on. OK, so how does one
solve this equation? Well, it's not a
trivial equation, again. But it can be solved
by polynomials, and we'll see that. But the way we
solve this equation is by a power series expansion. Now you could do
it by hand first, and I did it when
I was preparing the lecture yesterday. I said I'm going to
just write h of u equal a constant a0 plus
a1u plus a2u squared plus a3u cubed. And I plugged it in here. And I just did the
first few terms and start to see what happened. And I found after
a little thinking that a2 is determined
by a0, and a3 is determined by a1
once you substitute. It's not the obvious when you
look at this, but that happens. So when you face a
problem like that, don't go high power
to begin with. Just try a simple series
and see what happens. And you see a little pattern. And then you can do a more
sophisticated analysis. So what would be a more
sophisticated analysis? To write h of u
equal the sum from j equals 0 to infinity
aju to the j. Then if you take a
derivative, because we're going to need the
derivative, dh du would be the sum from
j equals 0 to infinity. j times aju to the j minus 1. You would say that
doesn't look very good because for j equals
0 you have 1 over u. That's crazy. But indeed for j equals 0,
the j here multiplies it and makes it 0. So this is OK. Now the term that we actually
need is minus 2u dh du. So here minus 2u dh du would
be equal to the sum from j equals 0 to infinity, and I
will have minus 2jaju to the j. The u makes this j minus 1 j,
and the constant went there. So here is so far h. Here is this other
term that we're going to need for the
differential equation. And then there's the
last term that we're going to need for the
differential equation, so I'm going to go here. So what do we get
for this last term. We'll have to take
a second derivative. So we'll take-- h
prime was there, so d second h du squared
will be the sum from j equals 0 of j times j minus
1 aju to the j minus 2. Now you have to rewrite this
in order to make it tractable. You want everything
to have u to the j's. You don't want actually to
have u to the j minus 2. So the first thing
that you notice is that this sum actually begins
with 2, because for 0 and 1 it vanishes. So I can write j times j
minus 1 aj u to j minus 2. Like that. And then I can say let j
be equal to j prime plus 2. Look, j begins
with 2 in this sum. So if j is j prime plus 2,
j prime will begin with 0. So we've shifted the sum so it's
j prime equals 0 to infinity. And whenever I have a j I
must put j prime plus 2. So j prime plus 2. j prime plus 1 aj prime
plus 2 u to the j prime. Wherever I had j, I
put j prime plus 2. And finally you say j or
j prime is the same name, so let's call it j. j equals 0. j plus 2. j plus 1. aj plus 2 uj. So we got the series
expansion of everything, so we just plug into the
differential equation. So where is the
differential equation? It's here. So I'll plug it in. Let's see what we get. We'll get some from j
equals 0 to infinity. Let's see the second
derivative is here. j plus 2 times j plus 1 aj
plus 2 uj, so I'll put it here. So that's this second
derivative term. Now this one. It's easy. Minus 2j aj and the uj is there. So minus 2jaj. Last term is just e
minus 1, because it's the function this
times aj as well. That's h. And look, this whole
thing must be 0. So what you learn is that
this coefficient must be 0 for every value of j. Now it's possible to--
here is aj and aj, so it's actually
one single thing. Let me write it here. j plus 2 times j plus
1 aj plus 2 minus 2j plus 1 minus e aj uj. I think I got it right. Yes. And this is the same sum. And now, OK, it's a lot of
work, but we're getting there. This must be 0. So actually that solves for
aj plus 2 in terms of aj. What I had told you that you
can notice in two minutes if you try it a little. That a2 seems to be
determined by a0. And a3 seems to be
determined by a2. So this is saying
that aj plus 2 is given by 2j plus 1 minus e
over j plus 2 j plus 1 aj. A very nice recursive relation. So indeed, if you
put the value of a0, it will determine for you a2,
a4, a6, a8, all the even ones. If you put the value of a1, it
will determine for you a3, a5. So a solution is determined by
you telling me how much is a0, and telling me how much is a1. Two constants, two numbers. That's what you expect from
a second order differential equation. The value of the
function at the point, the derivative at a point. In fact, you are
looking at a0 and a1 as the two constants that
will determine a solution. And this is the value of h at 0. This is the
derivative of h at 0. So we can now write
the following facts about the solution
that we have found. So what do we know? That solutions fixed
by giving a0 and a1. That correspond to the
value of the function at 0 and the derivative
of the function at 0. And this gives one solution. Once you fix a0, you get a2, a4. And this is an even
solution, because it has only even powers. And then from a1, you fixed
a3, a5, all the other ones with an odd solution. OK. Well, we solve the
differential equation, which is really,
in a sense, bad, because we were expecting
that we can only solve it for some values of the energy. Moreover, you have a0,
you get a2, a4, a6, a8. This will go on forever
and not terminate. And then it will be an
infinite polynomial that grows up and doesn't
ever decline, which is sort of
contradictory with the idea that we had before
that near infinity the function was going to be
some power, some fixed power, times this exponential. So this is what we're looking
for, this h function now. It doesn't look
like a fixed power. It looks like it goes forever. So let's see what
happens eventually when the coefficient, the
value of the j index is large. For large j. aj plus 2 is roughly equal
to, for large a, whatever the energy is, sufficiently
large, the most important here is the 2j here, the j and the j. So you get 2 over j aj. So roughly for large j,
it behaves like that. And now you have to ask
yourself the question, if you have a power series
expansion whose coefficients behave like that, how
badly is it at infinity? How about is it? You know it's the
power series expansion because your h was all
these coefficients. And suppose they
behave like that. They grow in that way
or decay in this way, because they're decaying. Is this a solution
that's going to blow up? Or is it not going to blow up? And here comes an
important thing. This is pretty bad
behavior, actually. It's pretty awful behavior. So let's see that. That's pretty bad. How do we see that? Well you could do it
in different ways, depending on whether
you want to derive that this is a bad
behavior or guess it. I'm going to guess something. I'm going to look at how
does e to the u squared behave as a power series. Well, you know as a
power series exponential is 1 over n u squared to the n. Here's n factorial. n equals 0 to infinity. Now these two n's, u to the
2n, these are all even powers. So I'm going to
change letters here, and I'm going to work with j
from 0, 2, 4, over the evens. So I will write u to the j here. And that this correct, because
you produce u to the 0, u to the 2, u to the
fourth, these things. And j is really 2n,
so here you will have one over j
over 2 factorial. Now you might say, j over
2, isn't that a fraction? No, it's not a fraction,
because j is even. So this is a nice factorial. Now this is the
coefficient, cj u to the j. And let's see how this
coefficients vary. So this cj is 1 over
j over 2 factorial. What is cj plus 2 over cj? Which is the analogue
of this thing. Well, this would be 1 over
j plus 2 over 2 factorial. And here is up there,
so j over 2 factorial. Well, this has one more
factor in the denominator than the numerator. So this is roughly
one over j over 2 plus 1, the last value of this. This integer is just
one bigger than that. Now if j is large,
this is roughly 1 over j over 2,
which is 2 over j. Oh, exactly that stuff. So it's pretty bad. If this series
goes on forever, it will diverge like
e to the u squared. And your h will be like e to the
u squared with e to the minus u squared over 2 is going
to be like e to the plus. u squared over 2 is going
to go and behave this one. So it's going to do
exactly the wrong thing. If this series
doesn't terminate, we have not succeeded. But happily, the
series may terminate, because the j's are integers. So maybe for some energies
that are integers, it terminates, and
that's a solution. The only way to get a solution
is if the series terminates. The only way it can
terminate is that the e is some odd number over here. And that will solve the thing. So we actually need to do this. This shows the energy. You found why it's quantized. So let's do it then. We're really done
with this in a sense. This is the most important
point of the lecture, is that the series
must terminate, otherwise it will
blow up horrendously. If it terminates
as a polynomial, then everything is good. So to terminate you can choose
2j plus 1 minus e to be 0. This will make aj
plus 2 equal to 0. And your solution, your
h of u, will begin. aj will be the last
one that is non-zero, so it will be aj
times u to the j, and it will go down like aj
minus 2 u to the j minus 2. It will go down in steps of
2, because this recursion is always by steps of two. So that's it. That's going to be the solution
where these coefficients are going to be fixed by
the recursive relation, and we have this. Now most people
here call j equal n. So let's call it n. And then we have 2n
plus 1 minus e equals 0. And h of u would be an u to
the n plus all these things. That's the h. The full solution is h
times e to the minus u squared over 2 as we will see. But recall what e was. e here is 2n plus 1. But he was the true energy
divided by h omega over two. That was long ago. It's gone. Long gone. So what have you
found therefore? That the energy,
that' we'll call en, the energy of the
nth solution is going to be h omega
over 2 2n plus 1. So it's actually h omega, and
people write it n plus 1/2. Very famous result. The nth level of the harmonic
oscillator has this energy. And moreover, these
objects, people choose these-- you
see the constants are related by steps of two. So just like you could start
with a0, or a1 and go up, you can go down. People call these functions
Hermite functions. And they fix the notation so
that this an is 2 to the n. They like it. It's a nice normalization. So actually h of
n is what we call the Hermite function of u sub n. And it goes like 2 to the
n u to the n plus order u to the n minus 2 plus
n minus 4, and it goes on and on like that. OK, a couple things
and we're done. Just for reference,
the Hermite polynomial, if you're interested
in it, is the one that solves this equation. And the Hermite
sub n corresponds to e sub n, which is 2n plus 1. So the Hermite solution
from that the equation is that the Hermite polynomial
satisfies this minus 2u d Hn du plus 2n. Because en is 2n plus 1. So it's 2n Hn equals 0. That's the equation for
the Hermite polynomial, and interesting thing to know. Actually, if you want to
generate the efficiently the Hermite polynomials, there's
something called the generating function. e to the minus z
squared plus 2zu. If you expand it in
a power series of z, it actually gives you
n equals 0 to infinity. If it's a power series of z,
it will be some z to the n's. You can put a factor here
n, and here is Hn of u. So you can use your
mathematic program and expand this in powers of z. Collect the various powers of
u that appear with z to the n, and that's Hn It's the most
efficient way of generating Hn And moreover, if you want
to play in mathematics, you can show that
such definition of Hn satisfies this equation. So it produces the solution. So what have we found? Our end result is the following. Let me finish with that here. We had this potential, and the
first energy level is called E0 and has energy h omega over 2. The next energy is E1. It has 3/2 h omega. Next one is E2 5/2 h omega. This polynomial is
nth degree polynomial. So it has n zeros,
therefore n nodes. So these wave functions will
have the right number of nodes. E0, the psi 0,
will have no nodes. When you have psi 0, the
Hn becomes a number for n equals zero. And the whole solution
is the exponential of u squared over 2. The whole solution, in
fact, is, as we wrote, psi n Hn of u e to the
minus u squared over 2. In plain English,
if you use an x, it will be Hn u with x over
that constant a we had. And you have minus x
squared over 2a squared. Those are your eigenfunctions. These are the solutions. Discrete spectrum, evenly
spaced, the nicest spectrum possible. All the nodes are there. You will solve this in a
more clever way next time. [APPLAUSE]
