PROFESSOR: We'll begin by discussing
the wave packets and uncertainty.
So it's our first look into this Heisenberg
uncertainty relationships.
And to begin with, let's focus it as fixed
time, t equals zero.
So we'll work with packets at t equals zero.
And I will write a particular wave function
that you may have at t equals 0, and it's
a superposition of plane waves.
So it would be e to the ikx.
You sum over many of them, so you're going
to sum over k, but you're going to do it with
a weight, and that's 5k.
And there's a lot to learn about this, but
the physics that is encoded here is that any
wave at time equals 0, this psi of x at time
equals 0, can be written as a superposition
of states with momentum h bar k.
You remember e to the ikx represents a particle
or a wave that carries momentum h bar k.
So this whole idea here of a general wave
function being written in this way carries
physical meaning for us.
It's a quantum mechanical meaning, the fact
that this kind of wave has momentum.
But this phi of k, however, suppose you know
this wave function at time equals 0.
Phi of k is then calculable.
Phi of k can be determined, and that's the
foundation of what's called Fourier's theorem,
that gives you a formula for phi of k.
And it's a very similar formula.
1 over 2 pi, this time an integral over x.
So you take this of psi of x0 that you know
and then multiply by e to the minus ikx.
Integrate over x, and out comes this function
of k.
So if you know phi of x0, you know phi of
k.
You can calculate this interval and you can
rewrite phi of x0 as a superposition of plane
waves.
So that's how you would do a Fourier representation.
So somebody can give you an initial wave function,
and maybe it's a sine function or a Gaussian
or something, then what you would do if you
wanted to rewrite it in this way, is calculate
phi of k, because you know this psi, you can
calculate this integral, at least with a computer.
And once you know phi of k, you have a way
of writing psi as a superposition of plane
waves.
So we've talked about this before, because
we were doing wave packets before and we got
some intuition about how you form a wave packet
and how it moves.
Now we didn't put the time dependence here,
but that can wait.
What I wish to explain now is how by looking
at these expressions, you can understand the
uncertainties that you find on the wave function,
position, and momentum uncertainties, how
they are related.
So that is our real goal, understanding the
role of uncertainties here.
If phi of k has some uncertainty, how is the
uncertainty in psi determined?
So that's what we're looking for.
So relationship of uncertainties.
Now as before, we will take a phi of k, that
we've usually be in writing, that depends
on k and it's centered around some value k0.
It's some sort of nice, centered function.
And it has then, we say, some uncertainty
in the value of the momentum.
That is this signal, this phi of k that we're
using to produce this packet.
It has some uncertainty, it's not totally
sharp, it's peaked around k0 but not fully
sharp.
So the uncertainty is called delta k and it's
some typical width over here.
Delta k is then uncertainty.
Now it's not the purpose of today's lecture
to make a precise definition of what the uncertainty
is.
This will come later.
At this moment, you just want to get the picture
and the intuition of what's going on.
And there is some uncertainty here, perhaps
you would say, look at those points where
the wave goes from peak value to half value
and see what is the width.
That's a typical uncertainty.
So all what we're going to do in these arguments
is get for you the intuition.
Therefore, the factors of 2 are not trustable.
If you're trying to make a precise statement,
you must do precise definitions.
And that will come later, probably in about
one or two lectures.
So at this moment, that's the uncertainty,
delta k.
And let's assume that this phi of k is real.
And its peaked around k0 uncertainty delta
k.
Now what happens with psi of x?
Well, we had our statements about the stationary
phase that you already are practicing with
them for this homework.
If you want to know where this function peaks,
you must look where the phase, this phi--
we say it's real, so it doesn't contribute
to the phase-- where the phase, which is here,
is stationary, given the condition that it
should happen at k0.
The only contribution to the integral is basically
around k0.
So in order to get something, you must have
a stationary phase, and the phase must be
stationary as a function of k, because you're
integrating over k.
And the phase is kx, the derivative with respect
to k of the face is just x, and that must
vanish, therefore, so you expect this to be
peaked around x equals zero.
So the x situation, so psi of x0 peaks at
x equals 0.
And so you have a picture here.
And if I have a picture, I would say, well
it peaks around the x equals 0.
So OK, it's like that.
And here we're going to have some uncertainty.
Here is psi of x and 0, and here is x.
And let me mention, I've already become fairly
imprecise here.
If you were doing this, you probably would
run into trouble.
I've sort of glossed over a small complication
here.
The complication is that this, when I talk
about the peaking of psi, and you probably
have seen it already, you have to worry whether
psi is real or psi is complex.
So what is this psi here?
Should it be real?
Well actually, it's not real.
You've done, perhaps, in the homework already
these integrals, and you see that psi is not
real.
So when we say it peaks at x equals 0, how
am I supposed to plot psi?
Am I plotting the real part, the imaginary
part, the absolute value?
So it's reasonable to plot the absolute value
and to say that psi absolute value peaks at
x equals 0.
And there will be some width again here, delta
x, width.
And that's the uncertainty in psi of x.
So the whole point of our discussion for the
next 10 minutes is to just try to determine
the relation between delta k and delta x and
understand it intuitively.
