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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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e to the minus ikx.
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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.