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PROFESSOR: So let's
do an example where
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we can calculate
from the beginning
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to the end everything.
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Now, you have to get
accustomed to the idea of even
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though you can
calculate everything,
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your formulas that you get
sometimes are a little big.
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And you look at them
and they may not
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tell you too much unless you
plot them with a computer.
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So we push the calculations
to some degree,
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and then at some
point, we decide
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to plot things using a
computer and get some insight
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on what's happening.
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So here's the example.
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We have a potential up
to distance, a, to 0.
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The wall is always there, and
this number is minus v naught.
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So it's a well,
a potential well.
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And we are producing energy.
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Eigenstates are coming here.
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And the question now is to
really calculate the solution
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so that we can really
calculate the phase shift.
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We know how the
solutions should read,
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but unless you do
a real calculation,
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you cannot get the phase shift.
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So that's what we want to do.
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So for that, we have to solve
the Schrodinger equation.
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Psi of x is equal to what?
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Well, there's a discontinuity.
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So we probably have to write
an answer in which we'll
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have a solution in one piece and
a solution in the other piece.
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But then we say, oh,
we wrote the solution
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in the outside piece already.
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It is known.
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It's always the same.
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It's universal.
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I don't have to think.
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I just write this.
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E to the i delta.
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I don't know what
delta is, but that's
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the answer, E to the i delta
sine kx plus delta should
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be the solution for
x greater than a.
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You know if you were
not using that answer,
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it has all the
relevant information
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for the problem, time
delays, everything,
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you would simply write
some superposition of E
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to the i kx and E to the minus
i kx with two coefficients.
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On the other hand, here, we
will have, again, a wave.
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Now, it could be maybe
an E to the i kx or E
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to the minus i kx.
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Neither one is very good
because the wave function
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must vanish at x equals 0.
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And in fact, the k that
represents the kinetic energy
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here, k is always related to
E by the standard quantity,
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k squared equal to mE
over h squared or E equal
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the famous formula.
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On the other hand, there
is a different k here
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because you have
different kinetic energy.
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There must be a k prime here,
which is 2m E plus v naught.
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That's a total kinetic
energy over h squared.
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And yes, the solutions
could be E to the ik prime
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x equal to minus ik
prime x minus ik prime x,
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but since they must vanish at
0, should be a sine function.
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So the only thing
we can have here
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is a sine of k prime x for x
less than a and a coefficient.
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We didn't put the additional
normalization here.
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We don't want to put
that, but then we
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must put the number here,
so I'll put it here.
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That's the answer, and
that's k and k prime.
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Now we have boundary
conditions that x equals a.
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So psi continues at x equals a.
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What does it give you?
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It gives you a sine of ka
is equal to E to the i delta
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sine of ka plus delta.
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And psi prime continues
at x equals a will give me
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ka cosine ka equal--
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I have primes missing;
I'm sorry, primes--
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equals k E to the i delta
cosine ka plus delta.
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What do we care for?
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Basically we care for delta.
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That's what we want to find
out because delta tells us
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all about the physics
of the scattering.
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It tells us about the scattering
amplitude, sine squared delta.
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It tells us about the time
delay, and let's calculate it.
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Well, one way to
calculate it is to take
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a ratio of these two
equations so that you get rid
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of the a constant.
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So from that side
of the equation,
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you get k cotangent
of ka plus delta
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is equal to k prime
cotangent of k prime a.
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Or cotangent of ka plus
delta is k prime over k.
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We'll erase this.
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And now you can do two things.
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You can display some
trigonometric wizardry,
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or you say, OK, delta is arc
cotangent of this minus ka.
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That is OK, but it's not ideal.
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It's better to do a little bit
of trigonometric identities.
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And the identity
that is relevant
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is the identity for cotangent of
a plus b is cot a cot b minus 1
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over cot a plus cot b.
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So from here, you have
that this expression
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is cot ka cot delta minus 1
over cot ka plus cot delta.
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And now, equating left-hand
side to this right-hand side,
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you can solve for
cotangent of delta.
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So cotangent of delta
can be solved for--
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and here is the answer.
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Cot delta is equal to tan
ka plus k prime over k cot k
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prime a over 1 minus k prime
over k cot k prime a tan ka.
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Now, who would box such
a complicated equation?
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Well, it can't be
simplified any more.
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Sorry.
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That's the best we can do.