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