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PROFESSOR: Next is this
phenomenon that when

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you have a wave
packet and it moves

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it can change shape
and get distorted.

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And that is a very
nice phenomenon

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that takes place in
general and causes

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technological complications.

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And it's conceptually
interesting.

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So let's discuss it.

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So it's still wave packets.

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But now we have to go back
and add some time to it.

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So shape changes.

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So we had a psi of x
and t is equal to 1

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over square root of 2
pi phi of k e to the ikx

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e to the minus i omega of kt.

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And what did we do with this
to analyze how it propagates?

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We expanded omega
of k as omega of k0,

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which, again, this quantity is
centered and peaks around k0,

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plus k minus k0 times
d omega dk at k0

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plus 1/2 k minus k0 squared, the
second omega, dk squared at k0.

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And it might seem that
this goes on forever.

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And what did we do before?

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We looked at this thing and we
did the integral with this term

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and ignored the next.

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And with this
term, we discovered

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that the profile moves with this
velocity, the group velocity.

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Now we want to go
back and at least get

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an idea of how this term
could change the result.

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And it would change
the result by deforming

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the shape of the packet.

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So it is of interest to know,
for example, how long you have

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to wait before your packet
gets totally deformed,

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or how do you evolve a packet.

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So we need to recall
these derivatives.

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So the omega vk is the same as
de dp by multiplying by h bar.

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And this you'll
remember, was p over m.

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The edp is p over m and is
equal to h bar k over m.

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So the second omega, dk squared.

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I must differentiate the first
derivative with respect to k.

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So I differentiate the first
derivative with respect k.

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And now I get just h bar
over m, which is quite nice.

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And the third derivative,
the 3 omega, dk cubed, is 0.

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And therefore, I didn't have
to worry about these terms.

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The series terminates.

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The Taylor series
terminates for this stuff.

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Yes?

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AUDIENCE: The
reason this happens

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is because we're [INAUDIBLE].

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PROFESSOR: That's right.

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So of what is it that we get?

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Well, this term is roughly
then 1/2 k minus k0 squared

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times h bar over m.

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And we can go back
to the integral

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that we're trying to do.

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We don't do it again
or not by any means.

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But just observe
what's going on there.

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And we have an e to
the minus i omega of kt

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that we did take into account.

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But the term that
we're dropping now

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is a term that is
minus i omega of k,

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well, whatever we have
here, 1/2 k minus k0

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squared h bar over mt.

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That's the phase that
we ignored before.

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But now we'll just say,
that we expect, therefore,

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that the shape doesn't
change as long as we

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can ignore this phase.

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And this phase would start
changing shape of the object.

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So our statement is going to
be that we have no shapes.

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So let's imagine you
started with a packet

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that sometime t equals 0.

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And then you let time go by.

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Well, there's some numbers
here and time is increasing.

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At some point, this phase is
going to become unignorable.

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And it's going to start
affecting everything.

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But we have no shape change,
or no appreciable shape change,

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as long as this quantity
is much less than 1.

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So as long as say,
k minus k0 squared

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h bar over m absolute value
of t is much less than 1,

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no shape change.

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Now it's convenient to
write it in terms of things

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that are more familiar.

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So we should
estimate this thing.

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Now we're doing estimates in a
very direct and rough way here.

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But look, your
integrals are around k0.

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And as you remember, they
just extend a little bit

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because it has some width.

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So k minus k0, as you
do the integral over k,

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you're basically
saying this thing

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is about the size of
the uncertainty in k.

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So I'll put here
delta k squared.

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Then you'll have h bar t
over m much less than 1.

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Now h bar times
delta k is delta p.

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So this equation is also of
the form delta p squared t

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over h bar m much less than 1.

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There's several forms of
this equation that is nice.

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So this is a
particularly nice form.

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So if you know the uncertainty
and momentum of your packet,

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or wave packet, up to
what time, you can wait

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and there's no big deformation
of this wave packet.

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Another thing you can do is
involve the uncertainty in x.

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Because, well, delta p
delta x is equal to h bar.

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So we can do that.

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And so with delta p times
delta x equal to about h bar,

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you can write t less
than h bar over m

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over a delta p
squared, which would

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be h squared delta x squared.

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I think I'm getting it right.

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Yep, so t much less than m
over h bar delta x squared.

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That's another way you
could write this inequality.

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There is one way to
write the inequality

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that you can
intuitively feel you

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understand what's happening.

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And take this form a from a.

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Write it as delta p t over m is
less than h bar over delta p.

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And h bar over
delta p is delta x.

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So you go delta p over mt
much less than delta x.

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I think this is understandable.

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Why does the packet
change shape?

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The reason it changes shape
is because the group velocity

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is not the same for
all the frequencies.

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The packet mostly moves with k0.

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And we haven't rated the
group velocity in k0.

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But if it would have
a definite velocity,

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we would have a
definite momentum.

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But that's not possible.

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These things have
uncertainty in momentum.

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And they have uncertainty in
k that we use it to write it.

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So different parts
of the wave can

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move with different velocities,
different group velocities.

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The group velocity
you evaluated at k0.

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But some part of the
packet is propagating

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with group velocities that are
near k0 but not exactly there.

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So you have a dispersion
in the velocity, which

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is an uncertainty
in the velocity

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or an uncertainty
in the momentum.

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Think, the momentum divided
by mass is velocity.

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So here it is, an
uncertainty in the velocity.

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And if you multiply
the uncertainty

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in the velocity times this
time that you can wait,

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then the change in
shape is not much

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if this product, which
is the difference of how

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one part moves with respect
to the other, the difference

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of relative term,
is still smaller

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than the uncertainty
that controls

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the shape of the packet.

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So the packet has a delta x.

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And as long as this part, the
left part of the packet, then

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the top of the packet, the
difference of velocities

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times the time, it just still
compared to delta x is small,

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then the thing
doesn't change much.

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So I think this is
one neat way of seeing

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what an equation
that you sometimes

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use in this form, sometimes
use in in this form--

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it's just things that you
can use in different ways.

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So for example, I can do
this a little exercise.

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If you have delta x equals 10
to the minus 10 meters, that's

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atomic size for an electron.

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How long does it
remain localized?

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So you have an electron.

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And you produce a packet.

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You localize it to
the size of an atom.

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How long can you wait
before this electron is just

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all over the room?

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Well, when we say this
t, and we say this time,

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we're basically saying that
it's roughly still there.

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Maybe it grew 20%, 30%.

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But what's the
rough time that you

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can expect that it stays there?

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So in this case, we can
use just this formula.

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And we say the time
could be approximately

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m over h bar delta x squared.

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It's fun to see the numbers.

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You would calculate it
with mc squared over h bar

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c times delta x over
c, this squared.

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The answer is about 10 to the
minus 16 seconds, not much.

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This is a practical issue
in accelerators as well.

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Particle physics
accelerators, they

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concern bunches, a little
bunch of protons in the LHC.

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It's a little cylinder in
which the wave functions

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of the protons are all
collimated very thin, short,

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a couple of centimeters short.

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And after going around many
times around the accelerator,

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they always have to be
compressed and kept back,

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sent back to shape.

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Because just of diffusion,
these things just propagate.

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And so it's a rather
important thing.