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PROFESSOR: We'll begin by discussing
the wave packets and uncertainty.
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So it's our first look into this Heisenberg
uncertainty relationships.
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And to begin with, let's focus it as fixed
time, t 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
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, but you're going to do it with
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a weight, and that's 5k.
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And there's a lot to learn about this, but
the physics that is encoded here is that any
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wave at time equals 0, this psi of x at time
equals 0, can be written as a superposition
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of states with momentum h bar k.
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You remember e to the ikx represents a particle
or a wave that carries momentum h bar k.
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So this whole idea here of a general wave
function being written in this way carries
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physical meaning for us.
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It's a quantum mechanical meaning, the fact
that this kind of wave has momentum.
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But this phi of k, however, suppose 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 the
foundation of what's called Fourier's theorem,
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that gives 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
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 as a superposition of plane
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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,
and maybe it's a sine function or a Gaussian
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or something, then what you would do if you
wanted to rewrite it in this way, is calculate
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phi of k, because you know this psi, you can
calculate this integral, at least with a computer.
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And once you know phi of k, you have a way
of writing psi as a superposition of plane
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waves.
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So we've talked about this before, because
we were doing wave packets before and we got
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some intuition about how you form a wave packet
and how it moves.
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Now we didn't put the time dependence here,
but that can wait.
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What I wish to explain now is how by looking
at these expressions, you can understand the
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uncertainties that you find on the wave function,
position, and momentum uncertainties, how
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they are related.
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So that is our real goal, understanding the
role of uncertainties here.
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If phi of k has some uncertainty, 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, that
we've usually be in writing, that depends
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on k 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
in the value of the momentum.
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That is this signal, this phi of k that we're
using to produce this packet.
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It has some uncertainty, it's not totally
sharp, it's peaked around k0 but not fully
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sharp.
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So the uncertainty is called delta k 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
to make a precise definition of what the uncertainty
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is.
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This will come later.
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At this moment, you just want to get the picture
and the intuition of what's going on.
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And there is some uncertainty here, perhaps
you would say, look at those points where
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the wave goes from peak value to half value
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
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,
you must do precise definitions.
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And that will come later, probably 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 that you already are practicing with
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them for this homework.
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If you want to know where this function peaks,
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-- where the phase, which is here,
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is stationary, given the condition that it
should happen at k0.
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The only contribution to the integral is basically
around k0.
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So in order to get something, you must have
a stationary phase, and the phase must be
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stationary as a function of k, because you're
integrating over k.
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And the phase is kx, the derivative with respect
to k of the face is just x, and that must
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vanish, therefore, so you expect this to be
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 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 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 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 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
these integrals, and you see that psi is not
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real.
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So when we say it peaks at x equals 0, how
am I supposed to plot psi?
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Am I plotting the real part, the imaginary
part, the absolute value?
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So it's reasonable to plot the absolute value
and to say that psi absolute value peaks at
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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 minutes is to just try to determine
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the relation between delta k and delta x and
understand it intuitively.