When does the wave function spread over the volume of a box?I have heard colloquially...

Ximena Skinner

Ximena Skinner

Answered

2022-07-02

When does the wave function spread over the volume of a box?
I have heard colloquially that for any initial state, a particle enclosed in some volume V will spread itself relatively evenly over that volume after large time, so that | ψ ( x ) | 2 1 / V may be used as the position probability density in the box.
This seems reasonable but of course it is not true, for example, of energy eigenstates. A sine wave with a wavelength fitting into the box will remain a sine wave for all times; this is the case for any energy eigenstate. However one might argue that an exact energy eigenstate is an idealization.
Does anyone know of a condition on the initial wave function ψ 0 ( x ) which is sufficient to guarantee that for large times,
| ψ ( x ) | 2 1 / V   ?
Or else, is there a counterexample of a more realistic state which does not fit the above approximation? My hunch is that any initial state whose fourier transform is supported on all of R 3 might give the approximation above, but I don't know of any such result. Energy eigenstates, for example, have a delta-function fourier transform.

Answer & Explanation

Keely Fernandez

Keely Fernandez

Expert

2022-07-03Added 14 answers

You are correct that for an energy eigenstate there is no time dependence for the probability distribution, hence the name stationary state.
You are also correct that for an initial state that can be written as a superposition of many energy eigenstates, then each of these eigenstates evolves in time with a different frequency, which ultimately leads to a probability distribution that essentially fills the entire box. More precisely, the probability distribution at any given time will not be 1 / V, but if you take an average over a time period then it will tend to 1 / V
However, it is possible to find intermediate situations to these two limits. A simple example is a state made of a superposition of two energy eigenstates only. Say the initial state is:
Ψ ( x , 0 ) = 1 2 [ ψ j ( x ) + ψ k ( x ) ] ,
where ψ n ( x ) are energy eigenstates of energy E n . For simplicity, I will assume that the energy eigenstates are real; if not, you simply need to keep track of some complex conjugates, but the solution does not fundamentally change. The state at a later time t is given by:
Ψ ( x , t ) = 1 2 [ ψ j ( x ) e i E j t / + ψ k ( x ) e i E k t / ] .
It then follows that:
| Ψ ( x , t ) | 2 = 1 2 [ | ψ j ( x ) | 2 + | ψ k ( x ) | 2 + 2 ψ j ( x ) ψ k ( x ) cos ( ( E k E j ) t ) ] .
This is a state that is time dependent (so not a trivial stationary state), but whose time dependence is simply a periodic oscillation of the probability, so it does not fill in the box uniformly at long times.
I recently did a video explaining all this in more detail, and it also comes with some code in a Jupyter notebook for you to play with it.
rjawbreakerca

rjawbreakerca

Expert

2022-07-04Added 5 answers

Here is a partial counterexample based on photons in a laser cavity. The photons fill the cavity, but the emitted beam does not fill space outside the cavity.
Most lasers emit Gaussian beams. This is the output of a cylindrical cavity with spherical mirrors. The emitted beam is nearly columnated, but it does spread out because of diffraction.
A Bessel beam does not spread out. It stays columnated.

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get your answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?