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PROFESSOR: Time evolution of
a free particle wave packet.
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So, suppose you
know psi of x and 0.
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Suppose you know psi of x and 0.
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So what do you do
next, if you want
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to calculate psi of x and t?
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Well, the first step, step
one, is calculate phi of k.
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So you have phi of k is equal
1 over square root of 2 pi
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integral dx psi of x,
0 e to the minus ikx.
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So you must do this integral.
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Step two-- step two--
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with this, now rewrite
and say that psi of x, 0
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is 1 over square root
of 2 pi dk e to the--
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no, I'm sorry-- phi
of k, e to the ikx.
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So that has achieved our
rewriting of psi of x and 0,
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which was an arbitrary function
as a superposition of plane
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waves.
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Step three is the
most fun step of all.
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Step three-- you look at
this, and then you say,
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well, I know now what
psi of x and t is.
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Evolving this is as
easy as doing nothing.
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What I must do here is 1
over square root of 2 pi--
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just copy this-- dk,
phi of k, e to the ikx.
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And I put here
minus omega of k, t.
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And I remind you that h bar
omega of k is the energy,
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and it's equal to h
squared k squared over 2m.
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This is our free particle.
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And I claim that,
just by writing this,
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I've solved the
Schrodinger equation
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and I've time-evolved
everything.
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The answer is there--
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I didn't have to solve the
differential equation, or--
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that's it.
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That's the answer.
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Claim this is the answer.
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And the reason is important.
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If you come equipped with a
Schrodinger equation, what
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should you check, that ih bar
d psi dt is equal to h psi--
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which is minus h-- squared over
2m, d second, dx squared psi.
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Well, you can add with
ih d dt on this thing.
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And you remember
all that happens
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is that they all
concentrate on this thing.
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And it solves this,
because it's a plane wave.
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So this thing, this
psi of x and t,
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solves the Schrodinger equation.
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It's a superposition of
plane waves, each of which
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solves the free
Schrodinger equation.
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So, we also mention that since
the Schrodinger equation is
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first ordered in time, if you
know the wave function at one
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time, and you solve it, you get
the wave function at any time.
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So here is a solution that is
a solution of the Schrodinger
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equation.
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But at time equals 0--
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this is 0-- and we reduce
this to psi of x and 0.
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So it has the right condition.
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Not only solve the
Schrodinger equation,
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but it reduces to
the right thing.
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So it is the answer.
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And we could say--
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we could say that there
is a step four, which is--
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step four would be
do the k integral.
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And sometimes it's possible.
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You see, in here, once
you have this phi of k,
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maybe you can just
look at it and say, oh,
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yeah, I can do this k integral
and get psi of x and 0,
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recover what I know.
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I know how to do-- this
integral is a little harder,
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because k appears a
little more complicated.
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But it has the whole
answer to the problem.
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I think one should
definitely focus on this
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and appreciate that, with zero
effort and Fourier's theorem,
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you're managing to solve the
propagation of any initial wave
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function for all times.
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So there will be an exercise in
the homework, which is called
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evolving the free Gaussian--
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Gaussian.
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So you take a psi
a of x and time
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equals 0 to be e to the
minus x squared over 4a
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squared over 2 pi to the 1/4--
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that's for normalization--
square root of a.
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And so what is this?
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This is a psi--
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this is a Gaussian-- and the
uncertainty's roughly a--
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is that right?
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Delta x is about a,
because that controls
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the width of the Gaussian.
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And now, you have a Gaussian
that you have to evolve.
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And what's going
to happen with it?
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This Gaussian, as
written, doesn't
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represent a moving Gaussian.
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To be a moving
Gaussian, you would
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like to see maybe things
of [? the ?] from e
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to the ipx that represent
waves with momentum.
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So I don't see anything like
that in this wave function.
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So this must be a Gaussian
that is just sitting here.
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And what is it
going to do in time?
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Well, it's presumably
going to spread out.
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So the width is going
to change in time.
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There's going to be a time
in which the shape changes.
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Will it be similar to
what you have here?
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Yes.
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The time will be related.
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So time for changes.
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So there will be
some relevant time
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in this problem for which
the width starts to change.
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And it will be related
to ma squared over h bar.
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In fact, you will
find that with a 2,
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the formulas look
very, very neat.
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And that's the relevant time for
the formation of the Gaussian.
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So you will do those four steps.
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They're all doable
for Gaussians.
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And you'll find the
Fourier transform,
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which is another Gaussian.
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Then you will put
the right things
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and then try to do
the integral back.
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The answer is a
bit messy for psi,
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but not messy for psi squared,
which is what we typically
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ask you to find.