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Is there an entropy that one can use for the Wigner quasi-probability distribution? (In the sense of a phase-space probability distribution, not - just von Neumann entropy.)

One cannot simply use $\int - W(q,p) \ln\left[ W(q,p) \right] dq dp$, as the Wigner function is not positively defined.

The motivation behind the question is the following:

A paper I. Białynicki-Birula, J. Mycielski, Uncertainty relations for information entropy in wave mechanics (Comm. Math. Phys. 1975) (or here) contains a derivation of an uncertainty principle based on an information entropy: $$-\int |\psi(q)|^2 \ln\left[|\psi(q)|^2\right]dq-\int |\tilde{\psi}(p)|^2 \ln\left[|\tilde{\psi}(p)|^2\right]dp\geq1+\ln\pi.$$ One of the consequences of the above relation is the Heisenberg's uncertainty principle. However, the entropic version works also in more general settings (e.g. a ring and the relation of position - angular momentum uncertainty).

As $|\psi(q)|^2=\int W(q,p)dp$ and $|\tilde{\psi}(p)|^2=\int W(q,p)dq$ and in the separable case (i.e. a gaussian wave function) the Winger function is just a product of the probabilities in position an in momentum, it is tempting to search for an entropy-like functional fulfilling the following relation: $$1+\ln\pi\leq\text{some_entropy}\left[ W\right]\leq -\int |\psi(q)|^2 \ln\left[|\psi(q)|^2\right]dq-\int |\tilde{\psi}(p)|^2 \ln\left[|\tilde{\psi}(p)|^2\right]dp.$$

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I think that the right formula is obtained from the naive $-\int W\ln W$ by replacing the product (between the two things, and inside the logarithm, when e.g. Taylor-expanded) by the star-product relevant for quantum mechanics. In other words, you first calculate the density operator $\rho$ to the Wigner distribution and then calculate $-{\rm Tr} \rho \ln \rho$ out of it. – Luboš Motl Sep 29 '11 at 13:51
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@LubošMotl: $-\text{\Tr}\rho \ln \rho$ is the von Neumann entropy so it is just zero for pure states. – Piotr Migdal Sep 29 '11 at 13:56
In the Gaussian case, the entropic uncertainty relation is saturated (at least when the state is pure and "squeezed" in the canonical directions). In that case, $\text{some_entropy}[w]=1+\ln \pi$. Of course, there is still room for optimization in the direction, but it is not very interesting. What application do you have in mind ? – Frédéric Grosshans Sep 29 '11 at 14:35
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@FrédéricGrosshans: For (squeezed) Gaussian case it is simple. When it comes to applications - well, for now I don't have any particular in my mind, besides of making a generalization. Maybe the exact way written above is a blind path and one needs to try playing with $W^2$ (which is just proportional to $\text{Tr}[\rho^2]$) or work with deconvolving Husimi Q distribution (which is positively defined and precisely bounded from the von Neumann entropy plus a constant). – Piotr Migdal Sep 29 '11 at 15:47
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The entropy like functional you describe is interesting because it tells us the uncertainty if we make measurements of position and momentum. If you want to make find a functional with a smaller value, you want to consider a wider class of measurements. In the phase space picture it is natural to consider measurements of conjugate pairs of quadratures that are simply rotation of p and q. As such I would consider the minimization of the entropy functional over all Gaussian operations. This could give a smaller value but clearly not zero. – Earl Sep 30 '11 at 16:12
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3 Answers

The Wigner Function is simply a particular representation of a quantum state and so it only has an entropy in so far as the state does. One can ask are there any entropic quantities that have an elegant representation in terms of the Wigner function, and there may well be such quantities. Indeed, the linear entropy $1-\mathrm{tr}(\rho^{2})$ where $\mathrm{tr}(\rho^{2}) \propto \int W (p,q)^{2}$ has a neat form. However, like the von Neumann entropy this will give zero for a pure state! You already stated that you would like an entropy that does not give zero always so that you can make interesting statements like the above uncertainty relation. However, uncertainty relations crop up when you sum two or more entropy quantities.

I will expand on my remarks a bit more formally. The classical entropies, like the Shannon entropy, are defined on bit strings. We can define a quantum mechanical entropy by defining a measurement that gives us a bit string. For an observable $M$ with eigenvalues $\lambda_{j}$ and projectors $P_{j}$ onto the corresponding subspace, we can define a bit string $X_{M}(\rho)=\{ x_{1}, x_{2},... x_{j}... \}$ where $x_{j}=\mathrm{tr} ( \rho P_{j} )$. Now we can covert this into an entropy by classical means such as taking the Shannon entropy $S( X_{M}(\rho))$. If the state is an eigenstate of the measurement basis then the entropy will be zero.

How does the von Neumann entropy fit into this picture. Well an equivalent definition to the usual one is the following: $S_{vonN}(\rho)=\min \{ S( X_{M}(\rho)) |M=M^{\dagger} \}$ which is simply the minimum possible measurement entropy.

Where do uncertainly relations come in? Well to have an uncertainty relation we must have 2 measurement observables that do not commute. If they have no common eigenstates an uncertainty relations follows by simply adding the 2 entropies. The inequality you have cited is simply $S_{P}(\rho)+S_{X}(\rho)$ for position plus momentum uncertainty. As noted in the comments this inequality can be saturated and so there is no hope of improving on it.

My opinion is that it is meaningless to ask for an entropy outside of a measurement context, and so this is what you need to decide on first. If it really is just position and momentum your interested in then I think the cited inequality says all there is!

This is my first attempt at an answer on stack exchange so I hope it is useful!

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Welcome to TP.SE, Earl. Good to see you here. – Joe Fitzsimons Oct 1 '11 at 11:54
@Earl: Thanks. Similar treatment (i.e. $\int W^2$) is in arxiv.org/abs/quant-ph/0203102. When it comes to 'meaninglessness' - well, I don't require this 'entropy' to be physical - just to give a lower bound for the sum entropies of position and momentum densities. – Piotr Migdal Oct 3 '11 at 12:23
To extend this answer, we remark that the Wigner representation of a quantum state is not unique. Thus if we associate an entropy measure to Wigner representations, and are given a state, and wish to assign an entropy to that state in a physically and mathematically natural way, then (it seems to me) we have maximize (minimize?) that entropy measure over all possible Wigner representations of the given state (a task that perhaps is not computationally easy). – John Sidles Oct 4 '11 at 22:59
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@John Sidles I didn't realize that Wigner representations were not unique! Do you have a reference for any examples of this? – Earl Oct 5 '11 at 6:54
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Earl, as I recall (perhaps imperfectly) non-unique Wigner representations can be regarded as the large-$j$ limit (for spin-$j$ Hilbert spaces) of non-unique $P$-representations. Perelemov's text <i>Generalized Coherent States and Their Applications</i> (1986) discusses this topic (as I recall). – John Sidles Oct 6 '11 at 15:58
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The entropy your are looking for may be what is known as the Wehrl entropy. You obtain it by replacing the Wigner function $W_\psi(q,p)$ in $-\int W \log W$ by the Husimi function, which is the convolution of the Wigner function with a Gaussian, $$ H_\psi(q,p) = \frac{1}{\pi} \int W_\psi(q',p') \exp\left( -(q-q')^2 -(p-p')^2 \right) \mathrm{d}q' \mathrm{d} p' . $$ (I've set $\hbar=1$.) Since the Husimi function is non-negative $$ S_\psi := - \int H_\psi(q,p) \log H_\psi(q,p) \mathrm{d}q \, \mathrm{d} p $$ is well-defined. A good staring point may be Gnutzmann & Zyczkowski: Rényi-Wehrl entropies as measures of localization in phase space, J. Phys. A 34 10123.

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Thanks Stefan. Actually I know the approach but wanted to go a step further. From Wherl entropy one can obtain entropic uncertainty for smeared densities in position in momentum (i.e. $\frac{1}{\sqrt{\pi}}\int |\psi(q')|^2 \exp(-(q-q'))dq'$ and $\frac{1}{\sqrt{\pi}}\int |\tilde{\psi}(p')|^2 \exp(-(p-p'))dp'$, respectively). Unfortunately, such smeared uncertainty principle is weaker that the usual one and I don't know if there is a way to derive the not-smeared uncertainty principle from the Husimi Q function. – Piotr Migdal Oct 6 '11 at 17:04

There is a way to handle this for a different phase space function, $f(x,k)=\langle x|\hat{\rho}|k\rangle$. I am told this is sometimes known as the Mehta function, mostly used when operators are in standard order. It turns out that this function is its own 2D Fourier transform (up to phase), which lets you write an entropic uncertainty principle in terms of just one entropy similar to what you had in mind. It works for mixed states, too. The only reference I can find comes from Bialynicki-Birula and Rudnicki's chapter in Statistical Complexity, "Entropic Uncertainty Relations in Quantum Physics," section 1.5.3. Rudnicki tells me this was original research noticed while they were writing the chapter. In their notation, the inequality is just like their relation 1.24,

$H^{(x,p)}>1-\ln2-\ln\left(\frac{\delta x \delta p}{h}\right)$.

It's not stronger than the original entropic uncertainty principle, but unlike the Wehrl entropy version it's not weaker.

Of course if you really want to go quantum, you should probably be more concerned with von Neumann entropy in the end. A couple references that compare the two entropy inequalities for the Wigner function are Braunss and Zachos. They take the approach of defining the classical entropy as a quantum entropy with offset, similar to what Luboš suggested, then with the limit $\hbar \rightarrow 0$. This ends up giving relation 7 or 9 of Zachos (copied below), either of which may address your question.

$0 \le S_q \le S_{cl}-\ln h$

$0 \le S_q \le \ln \left(\frac{e \sigma^2}{2 \hbar}\right)$

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