PROFESSOR: Square well. So what is this problem? This is the problem of having a
particle that can actually just move on a segment, like it
can move on this eraser, just from the left to the right. It cannot escape here. So the way we represent
it is the interval 0 to a on the x-axis. And there's going to be two
walls, one wall to the left and one wall to the right,
and no potential in between. That is, I write the potential
V of x as 0, for x in between a and 0, and infinity for x
less than or equal to 0, and x greater than
or equal to a. So basically the particle
can move from 0 to a, and nowhere else. The potential is infinity. Now, this problem, meaning
that the wave function-- the particle cannot be
outside the interval, means that the wave
function must vanish outside the interval. And you could say,
how do you know? Well, if the potential is close
to infinite amount of energy to be there, so the particle
cannot really be there if it's really infinite
energy that you need. You will see in the finite
square well that the particle has probability to be in regions
where it classically cannot be. But that probability will go to
0 if the potential is infinite. So we can think of it as a limit
and we will reconfirm that. But in fact, if the
potential is infinity, we will take it to mean that
psi of x is equal to 0 for x less than 0 and for
x greater than a. I am putting this equals or-- there are many
ways of doing this. If this function,
as this continues, you have a wall at a,
is the potential 0 at a or is it infinity? Well it doesn't quite matter. The issue is that the
wave function is 0 here, is 0 there, and we've
said that the wave function must be continuous. So it should be 0 by that
time you're at 0 or at a. So therefore we will
take psi of 0 to be 0, and psi of a to be
0 by continuity. So we discuss why the wave
function has to be continuous. If the wave function
is not continuous, the second derivative
of the wave function is terribly singular. It's like a derivative of
a delta function, which is an impossible situation. So the wave function,
we will take it to vanish at
these two places, and this is what is
called a hard wall. So what is the
Schrodinger equation? The Schrodinger equation is,
again, a free Schrodinger equation. Nothing, no potential here,
so it's the same Schrodinger equation we had there, psi
double prime equals minus 2mE over h squared, psi of x. Or, again, minus k
squared psi of x. Let's solve this. So how do we do it? Well it's, again, a
very simple equation, but this time it's
conveniences-- we don't have a
circle or periodicity to use sines and cosines. So I'll take psi of x
to be c1 cosine of kx plus c2 sine of kx. But the wave function
must vanish at 0. And at 0, the cosine
is 1, so you get c1. And the sine is 0, so this
must be 0, so c1 is gone. There's no c1 contribution
to the solution. So psi of x is c2 sine of kx. But we're not done. We need this function to
vanish at the other side. So psi of x equals a must
be 0, and that c2 sine of ka must be 0. And therefore we realize
that ka must be equal to a multiple of pi because sine
vanishes for 0, pi, 2 pi, 3 pi, minus pi, minus 2 pi, minus 3
pi, all the multiples of pi. And therefore we will
write kn equals 2 pi n-- not 2 pi n. Pi n over a. OK, well, let me ask you,
what should we take for n? All integers? Should we skip some? We took all integers
for the circle, but should we take
all integers here? So what happens
here, n equals 0. What's the problem
with n equals 0? n equals 0, k equals 0,
the wave function vanishes. Well, wave function
vanishing is really bad because there is
no particle then. There is nowhere
in the probability to find the particle. So n equals 0 is not
allowed, for sure. n equals 0, no. So why did we allow it, n
equals 0, in the circle? In the circle for n equals 0,
exponential doesn't vanish. It's a constant
and that constant is a fine wave function. 0 is not fine, but
the constant is good. But n equals 0 is not. So how about positive
ends or negative ends. And here comes the problem,
see we're getting to it. For n equals minus
2 or for n equals 2. So in one case, k is a number. And in the other case, k is
the opposite sign number. And sine of a number, or minus
a number, that number goes out. So if you have a sine of minus
kx, that's minus sine of kx. And two wave functions
that differ by a sign are the same wave
function, physically. There's nothing different. They could differ by
an i and other things. So when you pick
negative n minus 1, or pick n equals plus 1, you
get the same wave function, but just different by a sign. So it's not new. So in this case,
it's very interesting that we must restrict ourselves. We can correct all this
and just say n equals 1, 2, 3, all the way to infinity. The wave function,
then, is psi n of x, is proportional to
sine of n pi x over a. And you look at it and you
say, yes, that looks nice. For x equals 0, it vanishes. For x equals a, it vanishes. n and minus n would
give me the same wave function up to a sine. So this is good. I just have to normalize it. And normalizing it would be
done by putting an n here. And then the integral psi n
squared dx from 0 to a only would be n squared integral
from 0 to a dx of sine squared n pi x over a. Now, you can do this
integral by calculation. And our sine squared
is written in terms of a double angle cosine
of double angle plus a 1/2. The intuition with
these things are that if you're integrating
over the right interval that contains an integer number of
cycles of the sine squared, then the sine squared
has average 1/2. Because sine squared plus
cosine squared is equal to 1. So you don't have to do
the interval in general. This is n squared times
1/2 times the length of the interval, which is a. And therefore n squared,
this is equal to 1, and therefore n is equal
to square root of 2/a and we can write
now our solutions. Our solutions are m psi n
of x equals the square root of 2/a sine n pi x over a. And n equals 1,
2, up to infinity. And En is equal to h
bar squared, k squared, so pi squared, n
squared, a squared, to m. That's it for the solutions of-- are there degeneracies? No. Every energy state is
different because there's any single 1, 2, 3,
infinity, each one has more energy than the next. No, I'm sorry, the energy
increases as you increase n. The energy levels actually
become more and more spaced out. And the last thing I
want to do with this box is to look at the states and
see how they look and gather some important
properties that are going to be very relevant soon.
