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PROFESSOR: I've put on the
blackboard here the things
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we were doing last time.
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We began our study
of stationary states
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that are not normalizable.
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These are scattering states.
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Momentum eigenstates
were not normalizable,
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but now we have more
interesting states that
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represent the solutions of
the Schrodinger equation, that
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are stationary states
with some energy e.
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Because they are
not normalizable,
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but we cannot directly
interpret any of these solutions
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as the behavior of a particle.
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I kind of tell you a story, OK.
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So this is-- a particle
is coming, colliding,
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doing something.
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These are not
normalizable states.
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So part of what we're going
to be trying to do today
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is connect to the
picture of wave packets
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and see how this is
used to really calculate
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what would happen if you send
in a particle of potential.
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Nevertheless, we
wrote a solution
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that has roughly that
interpretation, at least
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morally speaking.
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We think of a wave that is
coming from the left, that's
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ae to the ikx.
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Now, in order to
have a wave, you
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would have to have
time dependents,
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and this is a
stationary solution.
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And there is some
time dependents.
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There is exponential of e to
the minus iet over h bar, where
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e is the energy.
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So this could be
added here to produce
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the full stationary
state psi of x and t.
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But we'll leave it understood--
it's a common phase
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factor for both the
solutions of x less than 0,
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and x greater than 0, because
the whole solution represents
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a single solution.
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It's not like a
solution on the left
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and a solution on the right.
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It's a single solution
over all of x 4 a psi that
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has some definite energy.
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So we looked at the
conditions of continuity
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of the wave function
and continuity
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of the derivative of the
wave function, at x equals 0.
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Those were two conditions.
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And they gave you
this expression
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for the ratios of c
over a and b over a.
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We could even imagine, since
we can't normalize this,
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setting a equal to 1, And.
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Then calculating b and
c from those numbers.
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We have two case a k in a k bar.
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The k is relevant to the wave
function for x less than 0.
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The k bar is relevant
for x greater than 0.
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And they have to be different
because this represents
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a DeBroglie wavelength, and
the DeBroglie wavelength
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encodes the momentum
of the particle,
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and the momentum of the
particle that we imagine here
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classically is
different here where
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this has this much kinetic
energy, and in the region
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on the right, where
the particle only has
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a much smaller kinetic energy.
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So that's represented by k bar.
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And k bar, being the
energy proportional
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to the energy minus v 0, while
k squared has just the energy,
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it's smaller than k.
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So this is what we did.
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We essentially
solved the problem,
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and this qualifies
as a solution,
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but we still haven't learned
anything very interesting
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from it.
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We have to understand
more what's going on.
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And one thing we can do is
think of a particular limit.
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The limit, or the case,
when e is equal to v 0,
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exactly equal to
b 0, what happens?
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Well, k bar would be equal to 0.
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And if k bar is
equal to 0, you're
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going to have just the
constant c in here.
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But if k bar is
equal to 0, first b
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is equal to a, because then it's
k over k, so b is equal to a.
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And c is equal to 2a.
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And the solution would
become psi of x equals--
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well, a is equal to b.
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So this is twice a cosine
of kx, when a equal to b
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is a common factor, call it a.
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And this thing is the
sum of two exponentials
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with opposite signs.
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That gives you the cosine.
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And for c, you have 2a.
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And since k bar is equal
to 0, well, it's just 2a.
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It's a number.
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This is unnormalizable, but
even the original solution
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is unnormalizable,
so we wouldn't
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worry too much about it.
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So how does that
look for x over here?
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You have a cosine
of kx, so it's going
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to be doing this to the left.
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That's for x less than 0.
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And at this point,
it goes like that.
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Just flat side
effects, and here's 2a.
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so there's nothing wrong with
the solution in this case.
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It's kind of a little strange
that it becomes a constant,
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but perfectly OK.
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What really helps
you here is to find
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some conditions that express
the conservation of probability.
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You see, you have a
stationary state solution.
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Now, stationary states
are funny states.
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They're not static states,
completely static states.
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For example, if you have a
loop, and you have a current
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that never changes in time.
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This is a stationary condition,
even in electromagnetism.
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So what we imagine
here is that we're
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going to have some current,
probability current,
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that is coming from the
left, and some of it
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maybe bounces back, and
some of it goes forward.
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But essentially, if you think
of the barrier, whatever--
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you look a little to
the left of the barrier
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and a little to the
right of the barrier.
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Whatever is coming in, say,
must be going out there,
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because probability cannot
increase in this region.
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It would be like saying that
the particle suddenly requires
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larger and larger
probability to be
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in this portion of the graph.
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And that can't happen.
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So probabilistic
current gives you
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a way to quantify some of the
things that are happening.
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So probability current
plus j effects,
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which was h bar over m,
imaginary part of psi star
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d psi dx.
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So let's compute the
probability current.
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Let's compute for x less than 0.
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What is the probability
current j of x?
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We then call it the probability
current on the left side.
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I would have to substitute
the value of the wave function
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for x less than 0, which
is the top line there,
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into this formula.
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And see what is the current.
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In fact, I believe you've
done that in an exercise
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some time ago.
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And as you can
imagine, the current
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is proportional to the
modulus of a squared,
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the length of a squared,
the length of b squared also
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enters.
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And the funny thing is,
between those two waves,
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one that is going to the right
and one is going to the left,
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that that's very
visible in the current.
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This is h bar k over m.
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A squared minus b squared.
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It's a short computation, and it
might be an OK and a good idea
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to do it again.
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It was done in the homework.
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And x greater than 0.
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J right of x would be
equal to h bar k bar.
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Now that, you can almost
do it in your head.
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This c into the ik bar x.
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Look what's happening.
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From psi star you get a c star.
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From psi you get the c, so
that's going to be a c squared.
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The face is going in a
cancel between the one here
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and the one on psi,
but the derivative
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will bring down an ik bar,
and the imaginary part of that
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is just k bar.
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So the answer is this.
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And these are the two currents.
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Now, if we are doing
things correctly,
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the two currents
should be the same.
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Whatever current
exists to the left,
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say a positive current that
is coming in, to the right
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must be the same.
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Another pleasant thing
is that this current
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doesn't depend on the value
of x, as x is less than 0.
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Nor of the value of x
when x is greater than 0.
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And that's good
for conservation.
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It would be pretty
bad as well, if you
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look at two places
for x less than 0,
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and you find that the
current is not the same.
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So where is it accumulating?
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What's going on?
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So the independence
of these things--
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from x, this is
constant, and a constant
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is very important, because this
constant should be the same.
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Now, whether or not about
current conservation,
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it's encoded in
Schrodinger's equation.
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And we solved
Schrodinger's equation.
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That's how we got these
relations between b, a and c
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So it better be that these
two things are the same.
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So Jay elsewhere--
jl, for example.
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I won't put off x,
because it's just clear.
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It doesn't depend on x.
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1 minus b over a
squared a squared.
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I'm starting to
manipulate the left one,
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and see if indeed the
currents are the same.
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And now, I get h bar k over m.
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1 minus b over a squared.
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I put the modulus squared,
but so far everything
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is real in this--
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k, k bar, b, a, c--
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all real.
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Is that right?
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No complex numbers there.
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So I don't have to
be that careful.
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B over a squared is
k minus k bar over k
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plus k bar squared a squared.
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And I gfit that here, maybe.
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H bar k over m.
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Now what?
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This is squared,
so the si, squared
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passes here to the numerator
minus the difference squared,
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that's going to be
a 4k k bar over k
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plus k bar squared a squared.
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And yes, this seems to
be working quite well.
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Flip these k's-- these
two k's, flip them around.
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So the answer for
jl so far is h bar,
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now k bar, because
I flipped that, m,
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and now I would have 4 k
squared over this thing, which
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is 2k over k plus k bar
squared of a squared.
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And this quantity,
if I remember right,
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is just c squared, as
you can see from there.
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So indeed, this is j
right, and it worked out.
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This is j right.
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So j left is equal to j right
by current conservation.
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So that's nice.
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That's another way
of getting insight
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into these coefficients.
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And that kind of
thing that makes you
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feel that there's no
chance you got this wrong.
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It all works well.