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BARTON ZWIEBACH:
Scattering in dimension.

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And we will consider
a world that

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is just one dimensional, x.

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And, in fact, there's an
infinite barrier at x equals 0.

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Infinite barrier, nothing
goes beyond there.

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On the other hand, in here,
up to some distance R,

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there could be a
potential V of x.

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So we will have a
potential V of x,

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it will have the following
properties-- it will

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be identically 0 for x
greater than R, which

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is the range of the potential;
it will be some function

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V of x for x in
between R and 0; and it

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will be infinity for x
less than or equal than 0.

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This is your potential, it's
a potential of range R--

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range of the potential.

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And the experiment
that we think about

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is somebody at x
equal plus infinity

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throwing waves into
this potential.

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And this observer can only
get back a reflected wave,

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and from that reflected
wave, the observer

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wants to deduce the type of the
potential that you have there.

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And that's absolutely
the way physics

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goes in particle physics.

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In LHC, you throw protons
or electrons together

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and you just catch
what flies out

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of the collision with
all the detectors

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and read that they
then deduce what

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happened to the collision,
what potential was there,

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what forces were there,
was there a new particle?

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It's all found by looking at
what comes out and flies away.

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So there's enormous
amount of information

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on the potential from
the data that comes out

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of you throwing
in some particles

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in and waiting to see
what comes back to you.

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So we will always have
this infinite wall.

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And this infinite
wall at x equals 0

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means that x less than
0 is never relevant.

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And this is analogous
to R, the variable R

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in radial coordinates for which
the radial distance is never

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negative.

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So in fact, what we'll do here
has immediate applications when

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we will consider--
not in this course--

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scattering in three dimensions.

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So to begin this, we'll solve
the simplest case where you

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have no potential whatsoever.

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Now no potential means
still the barrier

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at x equals 0, the
infinite barrier,

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but in between 0 and R,
nothing is happening.

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So you have just a
case of no potential.

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Is the case where you
have the barrier here

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and x is over there and up
to R, nothing's happening,

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it's just the wall.

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That's all there
is, just one wall.

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So this is V--

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no potential is V is
equal to V of x is equal 0

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for x greater than 0.

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And it's infinity for x
less or equals than 0.

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So in this case, let's assume
we have an incident wave.

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An incident wave must be
propagating in this way,

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so an incident wave is
an e to the minus ikx.

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And if you have
an outgoing wave,

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it would be some
sort of e to the ikx.

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These are the only
things that can be there.

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They correspond to
energy eigenstates,

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this is the de
Broglie wave function

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of a particle with
momentum, in one direction

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or in the other direction.

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But let's combine them in a way
to produce a simple solution.

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So this solution, phi of
x, will be the solution.

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Will be a combination
that's similar--

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e to the ikx and e
to the minus ikx,

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and I should make the wave
function vanish at x equals 0.

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At x equals 0, both
exponentials are equal to 1,

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so if I want them to cancel,
I should put a minus.

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So in order to simplify
this the best possible way,

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we can put a 1 over
to 2i's over there

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so that we have a sine
function, and the sine function

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is particularly nice.

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So we'll have e to the--

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output like this--
e to the minus ikx

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plus e to the ikx over 2i, and
this is just the sine function

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side of kx, which
you would admit,

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it's a V solution over
here, a sine of kx.

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On the other hand, I can
think of this solution

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as having an incoming
wave, which is minus

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e to the minus ikx over 2i,
and an outgoing wave of e

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to the ikx over 2i.

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So this is the representation
of the solution when nothing

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is happening, and the good
thing about this solution

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is that it tells us
what we should write--

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gives us an idea of what
we should write when

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something really is happening.

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So now let's
consider how we would

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write the general
experiment in which you

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send in a wave but this time,
there is really a potential.

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So let's consider now, if no
potential was there and now

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yes, potential--

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so no potential here, so what
does it mean yes potential?

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Well, it means you have this and
you have some potential there

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up to some distance R,
and then it flattens out,

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and something happens.

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So in order to
compare, we'll take

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an incoming wave, the
same as the one where

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there was no potential.

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So let's take an incoming
wave, which is of the form e

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to the minus ikx
times minus over 2i.

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But I must say here, I
must write something more--

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I must say that x
is greater than R,

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otherwise this is
not the solution.

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You see, in the region
where the potential really

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exists, where the--

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goes up and down, you
don't know the solution.

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It would take solving
the Schrodinger equation.

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You know the solution
where the potential is 0,

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so yes, this incoming wave is
the solution of the Schrodinger

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equation in this potential
when x is greater than R.

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And how about the outgoing wave?

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Well, we would like
to write it like that.

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So we'll say 1 over 2i e to the
ikx is also an outgoing wave,

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and we have no hope
of solving it here,

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finding what's
happening here unless we

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solve a complicated equation,
but then let's look outside--

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we're still looking outside.

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But that cannot be
the outgoing wave.

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This is the same as the other
one and there is a potential,

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so something must be different.

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On the other hand, if
you think about it,

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very little can be
different because you

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must have a solution
with 0 potential and--

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you know these plane
waves going out

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are the only things that exist.

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And now you decide,
oh, if that's the case,

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I cannot put another function
of x in there because that's not

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a solution.

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The best I can do is
multiply by a number,

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because maybe there's
very little outgoing wave

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or there is not, but then
I think of another thing--

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remember if you had e
to the A, e to the minus

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ikx plus B e to the ikx,
well, the probability current

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was proportional to A
squared minus B squared.

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And this time,
however, you have--

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you're sending in a wave and
you're getting back a wave

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and this is a
stationary state-- we're

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trying to get energy eigenstate,
solutions of some energy

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just like this energy.

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And the only way it can happen
is if they carry the same

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amount of probability--
probability cannot be

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accumulating here, nor it can
be depleted there as well,

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so the currents associated to
the two waves must be the same.

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And the currents are
proportional to those numbers

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that multiply these things
squared, so in fact,

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A squared must be
equal to B squared,

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and therefore we cannot
have like a 1/3 here,

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it would just ruin everything.

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So the only thing I
can have is a phase.

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It's only thing--
cannot depend on x,

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because that was an
unsolved equation.

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Cannot be a number that is
less than 1 or bigger than 1.

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The only thing we can
put here is a phase.

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So we'll put an e
to the 2i delta.

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And this delta will depend on
k or will depend on the energy,

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and it will depend on
what your potential is,

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but all the information
of this thing

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is in this delta
that depends on k.

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And you say, well, that's
very little, you just

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have one phase, one number that
you could calculate and see,

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but remember, if you
have a delta of k,

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you could measure it
for all values of k

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by sending particles
of different energies

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and get now a whole function.

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And with a whole
function delta of k,

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you have some probability of
getting important information

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about the potential.

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So we'll have a phase there,
e to the i delta of k.

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And let's summarize here, it's
due to current conservation--

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the current of this wave and the
current of the outgoing waves

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should be the same.

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And also note that no extra
x dependents is allowed.

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So this will produce
the J incident

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will be equal to J reflected.

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Now you could say, OK, very
good, so there's delta,

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there's a phase--

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should I define
it from 0 to 2 pi?

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From minus pi to pi?

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It's kind of natural to
define it from minus pi to pi,

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and you could look
at what it is,

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but as we will see from another
theorem, Levinson's theorem,

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it will be convenient
to just simply say,

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OK, you fixed the phase
delta at k equals 0.

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At 0 energy scattering, you
read what is your delta--

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unless you increase the
energy, the phase will change.

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So if you have a phase,
for example, on a circle,

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and the phase starts to grow and
to grow and to grow and to grow

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and to go here, well, should you
call this pi and this minus pi?

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No.

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You probably should just--

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if it keeps growing
with energy--

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and it might happen,
that the phase

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keeps growing with energy--
well, pi, 2 pi, 3 pi, 4 pi,

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just keep the phase continuous.

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So keeping the phase continuous
is probably the best way

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to think about the phase.

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You start at some
value of the phase

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and then track it continuously.

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There is always a problem
with phases and angles,

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they can be pi or minus
pi's the same angle,

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but try for continuity in
defining the phase when

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we'll face that problem.

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So let's write the solution.

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We have this, so
the total solution.

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We call the solution
with no potential phi,

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this one we'll
call psi of x will

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be 1 over 2i, the first term--
e to the ikx plus 2i delta minus

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00:16:34,760 --> 00:16:36,330
e to the minus ikx.

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00:16:39,410 --> 00:16:44,700
It's convenient to
pull out of i delta

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to make the two terms
have opposite arguments,

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so e to the ikx plus
i delta, and-- or e

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to the ikx plus
delta parenthesis

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minus e to the minus
ikx plus delta.

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So this is e to the i delta
times sine of kx plus delta.

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So that's this full scattered
wave, not the full reflected--

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well, that word again.

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This is the full
wave that you have

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for x greater than R. So let's
write it here-- psi of x.

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It's not the reflected wave
nor that it covers everything.

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We include-- it's
for x greater than R,

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but we include the incoming
and outgoing things,

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because both are defined
for x greater than R,

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so the total wave is this one.

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And you notice that if
the phase shift is 0,

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you are nicely back to
the wave function phi

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that we found before.