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