PROFESSOR: Last time, we talked
about the Broglie wavelength. And our conclusion was,
at the end of the day, that we could write the
plane wave that corresponded to a matter particle, with some
momentum, p, and some energy, E. So that was our
main result last time, the final form for the wave. So we had psi of
x and t that was e to the i k x minus i omega t. And that was the matter
wave with the relations that p is equal to h bar k. So this represents a
particle with momentum, p, where p is h bar times
this number that appears here, the wave number,
and with energy, E, equal to h bar
omega, where omega is that number that appears
in the [? term ?] exponential. Nevertheless, we
were talking, or we could talk, about
non-relativistic particles. And this is our
focus of attention. And in this case, E is
equal to p squared over 2m. That formula that expresses
the kinetic energy in terms of the momentum, mv. So this is the wave function
for a free particle. And the task that
we have today is to try to use this insight,
this wave function, to figure out what
is the equation that governs general wave functions. So, you see, we've been
led to this wave function by postulates of the Broglie
and experiments of Davisson, and Germer, and
others, that prove that particles like electrons
have wave properties. But to put this
on a solid footing you need to obtain this
from some equation, that will say, OK, if you have
a free particle, what are the solutions. And you should
find this solution. Perhaps you will
find more solutions. And you will understand
the problem better. And finally, if you understand
the problem of free particle, there is a good chance you
can generalize this and write the equation for a
particle that moves under the influence
of potentials. So basically, what
I'm going to do by trying to figure out how this
wave emerges from an equation, is motivate and
eventually give you, by the middle of this lecture,
the Schrodinger equation. So that's what we're
going to try to do. And the first thing is
to try to understand what kind of equation this
wave function satisfies. So you want to think of
differential equations like wave equations. Maybe it's some kind
of wave equation. We'll see it's kind
of a variant of that. But one thing we
could say, is that you have this wave function here. And you wish to know, for
example, what is the momentum. Well you should look at k, the
number that multiplies the x here, and multiply by h bar. And that would give
you the momentum. But another way of doing it
would be to do the following. To say, well, h bar over
i d dx of psi of x and t, calculate this thing. Now, if I differentiate
with respect to x, I get here, i
times k going down. The i cancels this
i, and I get h bar k. So, I get h bar k
times the exponential. And that is equal to the value
of the momentum times the wave. So here is this wave actually
satisfies a funny equation, not quite the differential
equation we're looking for yet, but you can act with a
differential operator. A derivative is something
of a differential operator. It operates in functions,
and takes the derivative. And when it acts on
this wave function, it gives you the momentum
times the wave function. And this momentum
here is a number. Here you have an operator. An operator just means something
that acts on functions, and gives you functions. So taking a derivative of a
function is still a function. So that's an operator. So we are left here to
think of this operator as the operator that
reveals for you the momentum of the free particle, because
acting on the wave function, it gives you the momentum
times the wave function. Now it couldn't be that
acting on the wave function just gives you the momentum,
because the exponential doesn't disappear after the
differential operator acts. So it's actually
the operator acting on the wave function gives you a
number times the wave function. And that number is the momentum. So we will call
this operator, given that it gives us the momentum,
the momentum operator, so momentum operator. And to distinguish it
from p, we'll put a hat, is defined to be
h bar over i d dx. And therefore, for
our free particle, you can write what we've
just derived in a brief way, writing p hat
acting on psi, where this means the
operator acting on psi, gives you the momentum of this
state times psi of x and t. And that's a number. So this is an operator
state, number state. So we say a few things,
this language that we're going to be using all the time. We call this wave function,
this psi, if this is true, this holds, then we
say the psi of x and t is an eigenstate of
the momentum operator. And that language comes from
matrix algebra, linear algebra, in which you have a
matrix and a vector. And when the matrix
on a vector gives you a number times the
same vector, we say that that vector is an
eigenvector of the matrix. Here, we call it an eigenstate. Probably, nobody
would complain if you called it an eigenvector,
but eigenstate would be more appropriate. So it's an eigenstate of p. So, in general, if you have an
operator, A, under a function, phi, such that A acting
on phi is alpha phi, we say that phi is an
eigenstate of the operator, and in fact eigenvalue alpha. So, here is an eigenstate
of p with eigenvalue of p, the number p, because
acting on the wave function gives you the number p
times that wave function. Not every wave function
will be an eigenstate. Just like, when you have a
matrix acting on most vectors, a matrix will rotate
the vector and move it into something else. But sometimes, a matrix
acting in a vector will give you the same
vector up to a constant, and then you've
got an eigenvector. And here, we have an eigenstate. So another way of
expressing this, is we say that psi of x
and t, this psi of x and t, is a state of definite momentum. It's important terminology,
definite momentum means that if you measured it, you
would find the momentum p. And the momentum-- there
would be no uncertainty on this measurement. You measure, and
you always get p. And that's what,
intuitively, we have, because we decided
that this was the wave function for a free
particle with momentum, p. So as long as we
just have that, we have that psi is a state
of definite momentum. This is an interesting statement
that will apply for many things as we go in the course. But now let's consider another
aspect of this equation. So we succeeded with that. And we can ask if there
is a similar thing that we can do to figure out
the energy of the particle. And indeed we can
do the following. We can do i h bar d dt of psi. And if we have that,
we'll take the derivative. Now, this time,
we'll have i h bar. And when we differentiate that
wave function with respect to time, we get minus i omega
times the wave function. So i times minus i is 1. And you get h bar omega psi. Success, that was the energy
of the particle times psi. And this looks quite
interesting already. This is a number, again. And this is a time derivative
of the wave function. But we can put more physics into
this, because in a sense, well, this differential
equation tells you how a wave function
with energy, E, what the time dependence
of that wave function is. But that wave function
already, in our case, is a wave function
of definite momentum. So somehow, the information
that is missing there, is that the energy
is p squared over 2m. So we have that the energy
is p squared over 2m. So let's try to think of
the energy as an operator. And look, you could
say the energy, well, this is the energy
operator acting on the function gives you the energy. That this true, but it's too
general, not interesting enough at this point. What is really interesting is
that the energy has a formula. And that's the physics
of the particle, the formula for the energy
depends on the momentum. So we want to capture that. So let's look what
we're going to do. We're going to do a
relatively simple thing, which we are going to walk back this. So I'm going to
start with E psi. And I'm going to invent
an operator acting on psi that gives you this energy. So I'm going to invent an O. So how do we do that? Well, E is equal to p
squared over 2m times psi. It's a number times psi. But then you say, oh,
p, but I remember p. I could write it as an operator. So if I have p times
psi, I could write it as p over 2m h bar
over i d dx of psi. Now please, listen
with lots of attention. I'm going to do a simple
thing, but it's very easy to get confused
with the notation. If I make a little typo
in what I'm writing it can confuse you
for a long time. So, so far these are numbers. Number, this is a
number times psi. But this p times
psi is p hat psi which is that operator, there. So I wrote it this way. I want to make one more-- yes? AUDIENCE: Should that say E psi? PROFESSOR: Oh yes,
thank you very much. Thank you. Now, the question is, can I
move this p close to the psi. Opinions? Yes? AUDIENCE: Are you asking
if it's just a constant? PROFESSOR: Correct,
p is a constant. p hat is not a constant. Derivatives are not. But p at this
moment is a number. So it doesn't care
about the derivatives. And it goes in. So I'll write it as
1 over 2m h/i d dx, and here, output p psi,
where is that number. But now, p psi, I can
write it as whatever it is, which is h/i d dx, and p
psi is again, h/i d dx psi. So here we go. We have obtained, and
let me write the equation in slightly reversed form. Minus, because of the two
i's, 1 over 2m, two partials derivatives is a second order
partial derivative on psi, h bar squared over
2m d second dx psi. That's the whole right-hand
side, is equal to E psi. So the number E
times psi is this. So we could call this
thing the energy operator. And this is the energy operator. And it has the property that
the energy operator acting on this wave function
is, in fact, equal to the energy times
the wave function. So this state again is
an energy eigenstate. Energy operator on the
state is the energy times the same state. So psi is an energy eigenstate,
or a state of definite energy, or an energy
eigenstate with energy, E. I can make it clear for
you that, in fact, this energy operator, as you've
noticed, the only thing that it is is minus h squared
over 2m d second dx squared. But where it came
from, it's clear that it's nothing else but
1 over 2m p hat squared, because p hat is
indeed h/i d dx. So if you do this computation. How much is this? This is A p hat times p
hat, that's p hat squared. And that's h/i d dx h/i d dx. X And that gives you the answer. So the energy operator
is p hat squared over 2m. All right, so actually,
at this moment, we do have a Schrodinger
equation, for the first time. If we combine the
top line over there. I h bar d dt of psi
is equal to E psi, but E psi I will write it
as minus h squared over 2m d second dx squared psi.
