WEBVTT
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PROFESSOR: Let me
demonstrate now
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with plain doing the
integral that, really,
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the shape of this wave is
moving with that velocity.
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So in order to do
that, I basically
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have to do the integral.
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And of course, if it's a general
integral, I cannot do it.
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So I have to figure out
enough about the integral.
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So here it is.
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We have psi of x and t.
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It's integral dk phi of k e
to the ikx minus omega of kt.
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OK.
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It's useful for us to look
at this wave at time equals 0
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so that we later compare it
with the result of the integral.
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So phi psi at time equals 0 is
just dk phi of k e to the ikx.
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Only thing you know is that
phi has peaked around k0.
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You don't know more than that.
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But that's psi of x
and time equals 0.
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Let's look at it later.
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So we have this thing here.
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And I cannot do the
integral unless I do some
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approximations.
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And I will approximate omega.
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Omega of k, since we're anyway
going to integrate around k0,
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let's do a Taylor series.
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It's omega of k0 plus k
minus k0, the derivative
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of omega with respect to k at k0
plus order k minus k0 squared.
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So let's--
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do this here.
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So if I've expanded
omega as a function of k,
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which is the only
reasonable thing to do.
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k's near k0 are the only
ones that contribute.
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So omega of k may be
an arbitrary function,
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but it has a Taylor expansion.
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And certainly, you've
noted that you get back
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derivative that somehow
is part of the answer,
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so that's certainly a bonus.
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So now we have to plug
this into the integral.
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And this requires a
little bit of vision
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because it suddenly seems
it's going to get very messy.
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But if you look at
it for a few seconds,
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you can see what's going on.
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So psi of x and t, so far,
dk phi of k e to the ikx.
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So far so good.
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I'll split the exponential so
as to have this thing separate.
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Let's do this. e to the minus i.
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I should put omega of k times t.
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So I'll begin.
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Omega of k0 times t.
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That's the first factor.
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e to the minus i,
the second factor.
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k--
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k d omega dk.
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k0 times t.
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And the third factor is
this one with the k0.
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e to the minus-- it
should be e to the plus.
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i k not d omega dk.
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k0 t.
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Plus order--
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higher up.
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So e to the negligible--
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negligible until you
need to figure out
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distortion of wave patterns.
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We're going to see
the wave pattern move.
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If you want to see
the distortion,
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you have to keep
that [INAUDIBLE].
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We'll do that in
a week from now.
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This is the integral.
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And then, you probably
need to think a second.
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And you say, look.
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There's lots of things making it
look like a difficult integral,
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but it's not as
difficult as it looks.
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First, I would
say, this factor--
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doesn't depend on k.
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It's omega evaluated at k0.
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So this factor is
just confusing.
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It's not-- doesn't
belong in the integral.
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This factor, too.
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k0 is not a function of k.
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d omega dk evaluated at
k0 is not a function of k.
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So this is not really
in the integral.
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This is negligible.
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This is in the integral
because it has a k.
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And this is in the integral.
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So let me put here, e to
the minus i omega of k0 t
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e to the minus--
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to the plus i k0 d omega dk--
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at k0 t.
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Looks messy.
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Not bad.
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dk.
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And now I can put phi of k.
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e to the i k x minus
these two exponentials,
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d omega dk at k0 times t.
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And I ignore this.
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So far so good.
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For this kind of wave, we
already get a very nice result
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because look at this thing.
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This quantity can be written
in terms of the wave function
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at time equals 0.
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It's of the same
form at 5k integrated
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with ik and some
number that you call x,
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which has been changed to this.
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So to bring in this and to
make it a little clearer--
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and many times it's useful.
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If you have a
complex number, it's
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a little hard to see the bump.
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Because maybe the bump
is in the real part
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and not in the imaginary
part, or in the imaginary part
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and not in the real part.
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So take the absolute value,
psi of x and t, absolute value.
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And now you say, ah, that's why.
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This is a pure phase.
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The absolute value of
a pure phase is that.
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So it's just the absolute value
of this one quantity, which
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is the absolute value of psi at
x minus d omega dk k0 t comma
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0.
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So look what you've proven.
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The wave function-- the
norm of the wave function--
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or the wave.
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The new norm of the
wave at any time t
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looks like the
wave looked at time
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equals 0 but just
displaced a distance.
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If there was a peak at x
equals 0, at time equals 0.
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If at time equals 0, psi had a
peak when x is equal to zero,
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it will have a peak--
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This function, which is the
wave function at time equals 0,
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will have a peak when this
thing is 0, the argument.
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And that corresponds to x
equals to d omega dk times t,
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showing again that the wave has
moved to the right by d omega
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dk times t.
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So I've given two
presentations, basically,
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of this very important
result about wave packets
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that we need to understand.