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PROFESSOR: Today we want
to discuss wave packets.

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We have studied already
the motivations that

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led to the Schrodinger
equation and we even

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included the potential
that described

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very non-trivial dynamics.

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With that Schrodinger equation,
you could jump and solve

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the hydrogen atom, you could go
and do the harmonic oscillator

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and all kinds of things.

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But we might as
well develop, still,

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our tools and intuition
of what's going on.

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And quite a lot of intuition
comes from the free particle.

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And the free particle
is interesting

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because you can go beyond the de
Broglie wavelengths and the de

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Broglie waves.

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You had a single plane wave
representing a particle

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with momentum and with
energy, but now you

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become more sophisticated and
consider the superposition

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of those waves.

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So we'll begin by discussing the
wave packets and uncertainty.

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So it's our first look into
this Heisenberg uncertainty

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relationships.

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And to begin with, let's
focus it as fixed time, t

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equals zero.

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So we'll work with
packets at t equals zero.

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And I will write a
particular wave function

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that you may have at
t equals 0, and it's

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a superposition of plane waves.

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So it would be e to the ikx.

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You sum over many of them, so
you're going to sum over k,

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but you're going to do it
with a weight, and that's 5k.

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And there's a lot
to learn about this,

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but the physics that is encoded
here is that any wave at time

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equals 0, this psi of
x at time equals 0,

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can be written as a
superposition of states

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with momentum h bar k.

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You remember e to the ikx
represents a particle or a wave

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that carries momentum h bar k.

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So this whole idea here
of a general wave function

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being written in this way
carries physical meaning

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for us.

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It's a quantum
mechanical meaning,

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the fact that this kind
of wave has momentum.

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But this phi of k,
however, suppose

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you know this wave
function at time equals 0.

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Phi of k is then calculable.

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Phi of k can be
determined, and that's

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the foundation of what's called
Fourier's theorem, that gives

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you a formula for phi of k.

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And it's a very similar formula.

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1 over 2 pi, this time
an integral over x.

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So you take this of
psi of x0 that you know

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and then multiply by
e to the minus ikx.

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Integrate over x, and out
comes this function of k.

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So if you know phi of
x0, you know phi of k.

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You can calculate this interval
and you can rewrite phi of x0

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as a superposition
of plane waves.

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So that's how you would do
a Fourier representation.

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So somebody can give you
an initial wave function,

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and maybe it's a sine function
or a Gaussian or something,

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then what you would do if
you wanted to rewrite it

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in this way, is calculate phi
of k, because you know this psi,

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you can calculate this integral,
at least with a computer.

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And once you know
phi of k, you have

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a way of writing psi as a
superposition of plane waves.

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So we've talked
about this before,

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because we were doing
wave packets before

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and we got some intuition about
how you form a wave packet

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and how it moves.

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Now we didn't put the
time dependence here,

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but that can wait.

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What I wish to
explain now is how

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by looking at these
expressions, you

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can understand the uncertainties
that you find on the wave

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function, position, and
momentum uncertainties,

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how they are related.

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So that is our real
goal, understanding

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the role of uncertainties here.

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If phi of k has
some uncertainty,

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how is the uncertainty
in psi determined?

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So that's what
we're looking for.

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So relationship
of uncertainties.

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Now as before, we
will take a phi of k,

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that we've usually be in
writing, that depends on k

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and it's centered
around some value k0.

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It's some sort of nice,
centered function.

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And it has then, we
say, some uncertainty

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in the value of the momentum.

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That is this signal,
this phi of k

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that we're using to
produce this packet.

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It has some uncertainty,
it's not totally sharp,

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it's peaked around k0
but not fully sharp.

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So the uncertainty
is called delta k

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and it's some typical
width over here.

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Delta k is then uncertainty.

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Now it's not the purpose
of today's lecture

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to make a precise definition
of what the uncertainty is.

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This will come later.

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At this moment, you
just want to get

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the picture and the
intuition of what's going on.

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And there is some
uncertainty here,

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perhaps you would say,
look at those points

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where the wave goes from
peak value to half value

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and see what is the width.

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That's a typical uncertainty.

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So all what we're going
to do in these arguments

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is get for you the intuition.

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Therefore, the factors
of 2 are not trustable.

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If you're trying to make
a precise statement,

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you must do precise definitions.

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And that will come
later, probably

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in about one or two lectures.

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So at this moment, that's
the uncertainty, delta k.

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And let's assume that
this phi of k is real.

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And its peaked around
k0 uncertainty delta k.

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Now what happens with psi of x?

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Well, we had our statements
about the stationary phase

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that you already are practicing
with them for this homework.

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If you want to know where
this function peaks,

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you must look where
the phase, this phi--

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we say it's real, so it doesn't
contribute to the phase--

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where the phase, which
is here, is stationary,

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given the condition that
it should happen at k0.

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The only contribution
to the integral

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is basically around k0.

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So in order to
get something, you

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must have a stationary
phase, and the phase

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must be stationary
as a function of k,

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because you're
integrating over k.

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And the phase is
kx, the derivative

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with respect to k of
the face is just x,

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and that must vanish, therefore,
so you expect this to be

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peaked around x equals zero.

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So the x situation, so psi
of x0 peaks at x equals 0.

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And so you have a picture here.

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And if I have a picture, I would
say, well it peaks around the x

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equals 0.

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So OK, it's like that.

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And here we're going to
have some uncertainty.

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Here is psi of x and
0, and here is x.

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And let me mention, I've already
become fairly imprecise here.

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If you were doing
this, you probably

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would run into trouble.

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I've sort of glossed over
a small complication here.

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The complication
is that this, when

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I talk about the peaking
of psi, and you probably

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have seen it already, you have
to worry whether psi is real

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or psi is complex.

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So what is this psi here?

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Should it be real?

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Well actually, it's not real.

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You've done, perhaps,
in the homework already

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these integrals, and you
see that psi is not real.

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So when we say it
peaks at x equals 0,

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how am I supposed to plot psi?

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Am I plotting the real
part, the imaginary part,

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the absolute value?

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So it's reasonable to
plot the absolute value

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and to say that psi absolute
value peaks at x equals 0.

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And there will be some width
again here, delta x, width.

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And that's the
uncertainty in psi of x.

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So the whole point of our
discussion for the next 10

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minutes is to just
try to determine

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the relation between
delta k and delta x

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and understand it intuitively.