Single quantum particle entropy

Consider a wave function of a single particle in free space, whose evolution is described by the (non-dimensional) linear Schrodinger equation $$i\psi _t (t,\underline{x}) + \Delta \psi=V(\underline{x}) \, ,$$

where $V$ is a potential, and $\underline{x}\in \mathbb {R}^d$ with $d=1,2,3$.

My question: Does there exist a notion of entropy for the solution $\psi$? This might be vague, but I'm looking for an integral functional of $\psi$ which increases as time increases. Specifically, I'm interested in the free-space case, i.e., $V\equiv 0$.

What I know: First, for an entropy to be defined we may need some sort of a probabilistic structure which I did not specify, mostly because I don't know how.

Second, I know that in quantum field theory there is a sense of light-entropy, but that's a totally different model equation than that of the linear Schrodinger equation. See e.g., Loudon's Quantum Theory of Light.

(cross posted from physics.se)

• Is ‘non-dimensional’ a typo or you mean something with it? – lcv Jun 5 '18 at 22:45
• @icv not a typo - I mean the Schrodinger equation without physical units. – Amir Sagiv Jun 6 '18 at 5:38

A quantity that might well satisfy the OP is the socalled "entropy of position", $$S(t)=-\int |\psi(r,t)|^2\log|\psi(r,t)|^2 dr.$$ As derived in The entropy of position and the spreading of wave packets (1986), provided that $S(0)<\infty$ this entropy grows faster than $\log t$. One can also define the analogous quantity in momentum space. It has found applications in a variety of quantum mechanical problems, here is a list of references from Google Scholar.

Quantum dynamics of pure states, as described by the Schrodinger equation, is unitary and reversible, so there can be no entropy at that level. In order to formulate entropy you need mixed states. All this is standard material in Quantum Statistical Mechanics.

• However something that they might want to do instead is to pick an observable and give the probability distribution for measuring that quantity. Then calculate the entropy of that distribution. This is reading more information into the question so that it has a nontrivial answer. – AHusain Jun 5 '18 at 19:14
• Thanks @AHusain and Robert. I gather that you have to read more into the original question than what is written there. This is simply because I'm not sure even what is the precise term I'm looking for. – Amir Sagiv Jun 6 '18 at 5:34

I think this question might have been misunderstood. Of course the Von Neumann entropy of a pure state (like the state you described) is 0. But I take it you aren't asking about Von Neumann energy, else there would be nothing to ask.

A notion of entropy that suits your requirements would have to be basis dependent (unlike Von Neumann entropy) otherwise it won't distinguish any state from any other. So the first step might be to choose a specific basis relative to which your entropy will be defined. A reasonable choice would be the position basis. Then, the Born rule gives a probability distribution on this chosen basis, and the entropy of this probability distribution can be calculated. Generically (in an imprecise sense), this entropy should be nondecreasing in time.

Your question is a reasonable one to ask, there is intuition that dispersion causes wave packets to spread, not to contract. This implies that there really is a sort of entropy here that's worth caring about. The question is ultimately a physics one though.

• Thanks. It may be a borderline between physics and math, but the phys.se post didn't receive any attention over a couple of months, and this one received three good answers in less than a day. – Amir Sagiv Jun 6 '18 at 5:31
• As for the content - @RobertIsrael was right. If you have a spreading* packet $\psi$, then $\psi ^{\star}$ is contracting. This is why an "extra structure" needs to be imposed if we want an interesting notion of entropy. – Amir Sagiv Jun 6 '18 at 5:32