PROFESSOR: Time evolution of
a free particle wave packet.
So, suppose you
know psi of x and 0.
Suppose you know psi of x and 0.
So what do you do
next, if you want
to calculate psi of x and t?
Well, the first step, step
one, is calculate phi of k.
So you have phi of k is equal
1 over square root of 2 pi
integral dx psi of x,
0 e to the minus ikx.
So you must do this integral.
Step two-- step two--
with this, now rewrite
and say that psi of x, 0
is 1 over square root
of 2 pi dk e to the--
no, I'm sorry-- phi
of k, e to the ikx.
So that has achieved our
rewriting of psi of x and 0,
which was an arbitrary function
as a superposition of plane
waves.
Step three is the
most fun step of all.
Step three-- you look at
this, and then you say,
well, I know now what
psi of x and t is.
Evolving this is as
easy as doing nothing.
What I must do here is 1
over square root of 2 pi--
just copy this-- dk,
phi of k, e to the ikx.
And I put here
minus omega of k, t.
And I remind you that h bar
omega of k is the energy,
and it's equal to h
squared k squared over 2m.
This is our free particle.
And I claim that,
just by writing this,
I've solved the
Schrodinger equation
and I've time-evolved
everything.
The answer is there--
I didn't have to solve the
differential equation, or--
that's it.
That's the answer.
Claim this is the answer.
And the reason is important.
If you come equipped with a
Schrodinger equation, what
should you check, that ih bar
d psi dt is equal to h psi--
which is minus h-- squared over
2m, d second, dx squared psi.
Well, you can add with
ih d dt on this thing.
And you remember
all that happens
is that they all
concentrate on this thing.
And it solves this,
because it's a plane wave.
So this thing, this
psi of x and t,
solves the Schrodinger equation.
It's a superposition of
plane waves, each of which
solves the free
Schrodinger equation.
So, we also mention that since
the Schrodinger equation is
first ordered in time, if you
know the wave function at one
time, and you solve it, you get
the wave function at any time.
So here is a solution that is
a solution of the Schrodinger
equation.
But at time equals 0--
this is 0-- and we reduce
this to psi of x and 0.
So it has the right condition.
Not only solve the
Schrodinger equation,
but it reduces to
the right thing.
So it is the answer.
And we could say--
we could say that there
is a step four, which is--
step four would be
do the k integral.
And sometimes it's possible.
You see, in here, once
you have this phi of k,
maybe you can just
look at it and say, oh,
yeah, I can do this k integral
and get psi of x and 0,
recover what I know.
I know how to do-- this
integral is a little harder,
because k appears a
little more complicated.
But it has the whole
answer to the problem.
I think one should
definitely focus on this
and appreciate that, with zero
effort and Fourier's theorem,
you're managing to solve the
propagation of any initial wave
function for all times.
So there will be an exercise in
the homework, which is called
evolving the free Gaussian--
Gaussian.
So you take a psi
a of x and time
equals 0 to be e to the
minus x squared over 4a
squared over 2 pi to the 1/4--
that's for normalization--
square root of a.
And so what is this?
This is a psi--
this is a Gaussian-- and the
uncertainty's roughly a--
is that right?
Delta x is about a,
because that controls
the width of the Gaussian.
And now, you have a Gaussian
that you have to evolve.
And what's going
to happen with it?
This Gaussian, as
written, doesn't
represent a moving Gaussian.
To be a moving
Gaussian, you would
like to see maybe things
of [? the ?] from e
to the ipx that represent
waves with momentum.
So I don't see anything like
that in this wave function.
So this must be a Gaussian
that is just sitting here.
And what is it
going to do in time?
Well, it's presumably
going to spread out.
So the width is going
to change in time.
There's going to be a time
in which the shape changes.
Will it be similar to
what you have here?
Yes.
The time will be related.
So time for changes.
So there will be
some relevant time
in this problem for which
the width starts to change.
And it will be related
to ma squared over h bar.
In fact, you will
find that with a 2,
the formulas look
very, very neat.
And that's the relevant time for
the formation of the Gaussian.
So you will do those four steps.
They're all doable
for Gaussians.
And you'll find the
Fourier transform,
which is another Gaussian.
Then you will put
the right things
and then try to do
the integral back.
The answer is a
bit messy for psi,
but not messy for psi squared,
which is what we typically
ask you to find.
