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 $|\psi(\vec{x})|^2\approx 1/V$ 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 $\psi_0(\vec{x})$ which is sufficient to guarantee that for large times,

$$|\psi(\vec{x})|^2\approx 1/V~?$$

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 $\mathbb{R}^3$ might give the approximation above, but I don't know of any such result. Energy eigenstates, for example, have a delta-function fourier transform.

  • $\begingroup$ perhaps starting with a Gaussian distribution wave packet. There one can see that the width of the package increases with time and would eventually fill the box. As you said, it is not a pure wave but rather a Fourier superposition. $\endgroup$ May 6, 2020 at 20:18

2 Answers 2


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:

$$ \Psi(x,0)=\frac{1}{\sqrt{2}}\left[\psi_j(x)+\psi_k(x)\right], $$

where $\psi_n(x)$ 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:

$$ \Psi(x,t)=\frac{1}{\sqrt{2}}\left[\psi_j(x)e^{-iE_jt/\hbar}+\psi_k(x)e^{-iE_kt/\hbar}\right]. $$

It then follows that:

$$ |\Psi(x,t)|^2=\frac{1}{2}\left[|\psi_j(x)|^2+|\psi_k(x)|^2+2\psi_j(x)\psi_k(x)\cos\left(\frac{(E_k-E_j)t}{\hbar}\right)\right]. $$

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. You can find it here.

  • $\begingroup$ I took a short look at the video but it doesn't show the 1/V. But yes it's true that a superposition of 2 eigenstates (or really probably any small number of eigenstates) is not going to have that behavior, and in general it may be necessary to average over some time scale. Do you have supporting evidence for the first 2 paragraphs? $\endgroup$ Jun 11, 2020 at 23:23
  • $\begingroup$ For the first paragraph, this is explained in the first slide of the video: time dependence in a single energy eigenstate is equivalent to a global phase multiplication, so no time dependence in the absolute value squared. $\endgroup$
    – ProfM
    Jun 12, 2020 at 6:31
  • $\begingroup$ For the second paragraph, I imagine that you would need the full limit of infinitely many eigenstates to ensure a true 1/V. A simple example is to look at high energy eigenstates of an infinite square well potential, which do tend to 1/V as the energy increases. For the time dependence, I do show that in the video (towards the end) for the time evolution of an initially localised wave function. In that simulation I only use 100 states, so it may not quite yet be the full limit 1/V, but there is a link to the code that you can modify to see it. $\endgroup$
    – ProfM
    Jun 12, 2020 at 6:34
  • $\begingroup$ I looked at the video more in detail now, I'm sorry I missed the last bit. This was helpful and interesting (for anyone reading the comments it is a ~gaussian wave packet expressed in basis of energy eigenstates truncated at a finite term which goes to ~1/V). I hesitate because I really was the most interested in a proof for a condition on the initial wave function which guarantees 1/V behavior, rather than a working example, but this was helpful nonetheless. For that reason I am still waiting for a final answer but I wish I had 3 upvotes to give you because you deserve more than just one.. $\endgroup$ Jun 12, 2020 at 11:43

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.


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