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An entropy of the Wigner function

+ 17 like - 0 dislike

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: $$1+\ln\pi\leq\text{some_entropy}\left[ W\right]\leq -\int |\psi(q)|^2 \ln\left[|\psi(q)|^2\right]dq-\int |\tilde{\psi}(p)|^2 \ln\left[|\tilde{\psi}(p)|^2\right]dp.$$

This post has been migrated from (A51.SE)
asked Sep 29, 2011 in Theoretical Physics by Piotr Migdal (1,255 points) [ no revision ]
retagged Mar 7, 2014 by dimension10
Most voted comments show all comments
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.

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@LubošMotl: $-\text{\Tr}\rho \ln \rho$ is the von Neumann entropy so it is just zero for pure states.

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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 ?

This post has been migrated from (A51.SE)
@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).

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Would a complex valued entropy be bad? I forgot the exact requirements on Shannon Entropy, but if you don't require maximum entropy but just a stationary, you could _maybe_ still obtain a useful equation of motion, and the inequality you mention is perhaps still valid if considering the absolute value. Also, what about $\int - (\int W(q,p) dp) \ln\left[\int W(q,p) dp\right] dq$ itself?

This post has been migrated from (A51.SE)
Most recent comments show all comments
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.

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@Piotr, I see. So let me guess that you won't find any entropy depending on the full distribution in a 2-dimensional way that satisfies your last inequality. Why do you expect such a thing to exist? The vanishing entropy of a pure state is a "real" thing. Maybe if you just add $1+\ln\pi$ to von Neumann entropy, it would work?

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2 Answers

+ 9 like - 0 dislike

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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answered Oct 1, 2011 by Earl (405 points) [ no revision ]
Welcome to TP.SE, Earl. Good to see you here.

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@Earl: Thanks. Similar treatment (i.e. $\int W^2$) is in http://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.

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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).

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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?

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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 Generalized Coherent States and Their Applications (1986) discusses this topic (as I recall).

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Helpful indeed, thanks. A first no-nonsense and really accessible explanation I encounter of the relation between the Shannon entropy and the uncertainty relation, thanks. (This is just a subjective comment based on my limited knowledge of the field).

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@John Sidles: the Wigner representation is indeed unique (it is a bijective mapping from states to functions on phase space). I should add that the Piotr asked another question [here](http://theoreticalphysics.stackexchange.com/q/245/493) which addresses the non-uniqueness of people's _preference_ for quasi-probability functions.

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+ 4 like - 0 dislike

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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answered Oct 6, 2011 by Stefan (40 points) [ no revision ]
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.

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