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PROFESSOR: Let's
look at the magnitude

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squared of those waves that
we've already defined here.

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We have two solutions,
one for no potential

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and one for a real potential.

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Both are for some
finite range potential.

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We have phi of x squared is
equal to sine squared of kx.

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And psi of x squared is equal to
sine squared of kx plus delta.

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Even from this information
you can get something.

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Think of x plotted here.

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Here's x equals 0.

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And there's this sine squared.

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This wave for phi of x squared.

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Suppose you're looking
at some feature--

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a maximum, a minimum--

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of this function.

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Suppose the feature happens
when the argument, kx, is

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equal to some number, a0.

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Whatever feature-- this
number a0 could be 0,

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in which case you're
looking at a minimum,

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it could be pi over 2, in
which case you're looking

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at a maximum--

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some feature of sine squared.

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Well the same feature
will appear in this case

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when the whole argument
is equal to a0.

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So while this one happens
at x equals a0 over k,

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here it happens at x still
equals a0 over k minus delta

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over k.

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If this is the probability
density associated

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to the solution for
no potential and it

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has a maximum here, the
maximum of the true solution--

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say, here-- would appear at a
distance equal to delta over k.

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Earlier-- so this is like the x,
and this is like the x tilde--

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that feature would appear,
delta over k in that direction.

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So this is psi.

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This is psi squared.

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So we conclude, for example,
that when delta is greater

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than 0, the wave is pulled.

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Delta equals 0, the two shapes
are on top of each other.

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For delta different from 0,
the wave function is pulled in.

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So delta greater than
0, psi is pulled in.

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What could we think of this?

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The potential is attractive.

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It's pulling in
the wave function.

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Attractive.

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Delta less than 0, the
wave is pushed out.

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It would be in the
other direction,

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and the psi is pushed out.

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Potential is repulsive.

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So a little bit of information
even from the signs

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of this thing.

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We want to define one last
thing, and then we'll stop.

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It's the concept of
the scattered wave.

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What should we call
the scattered wave?

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We will define
the scattered wave

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psi s as the extra
piece in the solution--

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the psi solution-- that would
vanish without potential.

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So we say, you have a psi, but
if you didn't have a potential,

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what part of this
psi would survive?

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Think of writing the
psi of x as the solution

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without the potential
plus the extra part,

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the scattered wave, psi x.

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So this is the definition.

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The full scattering
solution, the full solution

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when you have a potential,
can be written in a solution

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without the potential
and this scattered thing.

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Now, you may remember--

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we just did it a second ago--

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that this original solution
and the psi solution

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have the same incoming wave.

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The incoming wave up there is
the same for the psi solution

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as for this one.

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So the incoming
waves are the same.

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So only the outgoing
waves are different.

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And this represents how much
more of an outgoing wave

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you get than from what you
would have gotten with psi.

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So this must be
an outgoing wave.

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We'll just plug in
the formula here.

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psi s is equal to psi minus phi.

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And it's equal to 1 over 2i
e to the ikx plus 2i delta

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minus e to the minus
ikx minus 1 over 2i--

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the phi-- into the ikx
minus e to the minus ikx.

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So the incoming
waves wee the same.

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Indeed, they cancelled.

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But the outgoing waves are not.

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You can factor an e
to the ikx, and you

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get e to the 2i delta minus 1.

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Which is equal to e
to the ikx e to the i

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delta times sine delta.

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There we go.

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We have the answer for
the scattered wave.

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It's proportional to sine delta,
which, again, makes sense.

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If delta is equal to 0,
there is no scattering.

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It's an outgoing
wave and all is good.

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So I'll write it like this. psi
s is equal to As e to the ikx,

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with As equal to e to
the i delta sine delta.