Department of Physics

University of Oxford

The Schrodinger Equation for the wavefunction of a particle with Hamiltonian
*H* is

This can be solved by separation of variables. Consider a separable solution:

The separated functions satisfy

The equation for *u* is the time-independent Schrodinger equation, and
is also

the eigenvalue equation for *H*, where *E* is the
eigenvalue.

It has a set of solutions *u _{n}(x)* with
eigenvalues

The equation for

We therefore obtain a set of separated solutions

Because of the linearity of the Schrodinger equation any sum of these with

complex constant coefficients is also a solution.

The general solution
of the Time Dependent Schrodinger Equation

(with a Hamiltonian independent
of time) is thus

The wavefunction contains our knowledge of the system at time *t*, and
using the

standard methods of quantum mechanics we can obtain a probability
distribution for

the result of any measurement at any time. For example

is the probability distribution for the position *x* of the particle at
time *t*.

We can then find the mean and standard deviation of this
probability distribution:

The corresponding results for momentum *p* are:

*Delta x* and *Delta p* are constrained by the uncertainty
principle:

Classical Mechanics predicts a unique *x(t)*, *p(t)*, from precise
initial data *x(0)*, *p(0)*.

In the classical limit of quantum
mechanics we would like to see:

- A compact probability distribution (a
*wavepacket*); - A `small' value of
*Delta x*; - A `small' value of
*Delta p*; - Time evolution such that <
*x(t)*> and <*p(t)*>

follow the classical*x(t)*and*p(t)*.

in fact it always fills the classically available space (i.e. where the potential energy

is less than the energy eigenvalue

are time-independent (that's why it's called a

The classical limit necessarily involves superposition.

In fact quite a modest superposition of 20 states produces quite a compact
wavefunction, and reduces

the *Delta x Delta p* product to close to the
uncertainty principle limit.

The Wavepacket Plotter illustrates the time-evolution of
this wavepacket solution in a 1-dimensional box.

The infinite square well is a simple model of a one-dimensional box of width
*a*:

The potential energy function *V(x)* is given by*V(x)=0* for
*0 < x < a* and *V(x)=*infinity for *x < 0* and *x
> a*.

Any quantum mechanics textbook will show that the eigenvalues and eigenfunctions are given by

The Wavepacket Plotter generates a wavepacket using eigenfunctions with quantum numbers
centred from *n _{min}* to

THE OLD APPLET HASBEEN REPLAXED BY RGE WAVEFUNCTION PLOTTER. THE TEXT HERE NEDS UPDATING

The complex constant coefficients

where *N = 2dn + 2* and *K* ensures that the wavefunction is
properly normalised.

(There is no particular reason for this form, except
that it is a `smooth' function of *n*.)

The phase phi controls where the wavepacket is at *t = 0*.

The
default value is pi/2, entered in degrees as 90., which produces a packet in the
centre of the box

moving in the positive direction at *t = 0*.

1 length unit. The time unit is set such that the classical round trip time in the box for the

particle in quantum state

(hbar) has the value 2/pi

the effect on a fixed particle in a box of changing the size of hbar: the default settings change

hbar from 10

When the Applet is started it will display the position probability
distribution at *t=0*.

Check that it looks roughly normalised (height
times width is about one).

Check that the *<x>* and *Delta
x* values in the text panel correspond with the plot.

Look at the *C _{n}*,

is made up. The

Now return to the *|Psi(x)| ^{2}* plot and note the value of

Click on Increment t button, and check that

(Remember that the mass is one, so

Now click on the Increment t button several times till the wavepacket reaches
the wall of the box.

Notice that *<x(t)>* increases very slightly,
but at the turn-around it suddenly decreases.

Can you work out why? And
what's happened to the probability distribution?

Now go a bit further forward in time and the wavepacket will go into reverse.

What's happened to *<x>* and *<p>* now?

Reset the time to -4.0 and change the plot to *<x>* vs *t*.
(This may take a bit of time).

You will now see the wavepacket position and
width compared to the classical limit.

(The classical prediction is based on
*<p>*=2 exactly.)

Notice that the packet is sharply focussed
around *t=0* and is broader at earlier and later times.

Change back to
*|Psi(x)| ^{2}* and

and

At later times the

and

So the wavepacket represents a fairly classical particle oscillating in the
box, with our

uncertain knowledge of its initial data leading to an
increasing uncertainty of position at later times.

Is the wavepacket with

How does the width of the wavepacket at

The spreading of the wavepacket is obvious in classical terms - can you
explain it in

terms of the wavefunction? (Think about how the phase variable
controls the position of the packet.)

Now for a surprise! Look at the wavepacket at *t=*108. (or whatever is
the current value of *n _{0}*).

Now how did that happen? Look at

In fact the solution is exactly periodic with period

Do your ideas about wavepacket spreading explain this? Is the fact that the revival is exactly

out of phase with the classical motion a clue?

Now for a bigger surprise. Look at the wavepacket around
*t=n _{0}*/2 and

(This may be clearer with a larger

Explore! Explain! Enjoy!

C W P Palmer, University of Oxford.

Wavefunction Plotter © 2017 University of Oxford

Text © 2001 C W P Palmer