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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    $\begingroup$ 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. $\endgroup$ – Luboš Motl Sep 29 '11 at 13:51
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    $\begingroup$ @LubošMotl: $-\text{\Tr}\rho \ln \rho$ is the von Neumann entropy so it is just zero for pure states. $\endgroup$ – Piotr Migdal Sep 29 '11 at 13:56
  • $\begingroup$ 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 ? $\endgroup$ – Frédéric Grosshans Sep 29 '11 at 14:35
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    $\begingroup$ @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). $\endgroup$ – Piotr Migdal Sep 29 '11 at 15:47
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    $\begingroup$ 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. $\endgroup$ – Earl Sep 30 '11 at 16:12

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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    $\begingroup$ Welcome to TP.SE, Earl. Good to see you here. $\endgroup$ – Joe Fitzsimons Oct 1 '11 at 11:54
  • $\begingroup$ @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. $\endgroup$ – Piotr Migdal Oct 3 '11 at 12:23
  • $\begingroup$ 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). $\endgroup$ – John Sidles Oct 4 '11 at 22:59
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    $\begingroup$ @John Sidles I didn't realize that Wigner representations were not unique! Do you have a reference for any examples of this? $\endgroup$ – Earl Oct 5 '11 at 6:54
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    $\begingroup$ 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). $\endgroup$ – John Sidles Oct 6 '11 at 15:58

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.

  • $\begingroup$ 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. $\endgroup$ – 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.

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

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


I found this question by chance yesterday while looking for articles on Werhl entropy. I may have found a possible answer after reviewing properties of the Wigner quasiprobability distribution on http://en.wikipedia.org/wiki/Wigner_quasiprobability_distribution#The_Wigner.E2.80.93Weyl_transformation.

Consider property 7 under the section "Mathematical properties". For the operator $\hat G $, $\langle \hat G \rangle$ is the "phase-space average" of the Wigner transform. Substitute $\hat G = - \ln \hat \rho$. This may reduce to the von Neumann entropy in position representation, but after integrating out momentum and one of the position variables, I get $$ \langle - \ln \hat \rho \rangle = \int dx dy \langle x + \frac{y}{2} | \hat \rho | x - \frac{y}{2} \rangle \langle x - \frac{y}{2} | - \ln \hat \rho | x + \frac{y}{2} \rangle . $$

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    $\begingroup$ If we can make the substitution $u = x + \frac{y}{2}, v = x - \frac{y}{2}$, then we can write $ \langle - \ln \hat \rho \rangle = \int du dv \langle u | \hat \rho | v \rangle \langle v | - \ln \hat \rho | u \rangle , $ which would just be the von Neumann entropy. $\endgroup$ – John May 21 '13 at 21:27

For the second candidate, I assume that the Wigner distribution $ W(q,p) $ is smooth and continuous. This assumption was implicitly assumed with the first candidate entropy.

Given the property $ \lvert W(q,p) \rvert \le 2/h = 1/\pi \hbar$, that $ W(q,p) $ is bounded, and the properties you mentioned of how $ W(q,p) $ relates to $ \langle q \lvert \hat \rho \rvert q \rangle $ and $ \langle p \lvert \hat \rho \rvert p \rangle $, we should have that

$$ W(q,p) \le \int W(q,p) dp = \langle q \lvert \hat \rho \rvert q \rangle , $$ $$ W(q,p) \le \int W(q,p) dq = \langle p \lvert \hat \rho \rvert p \rangle . $$

With the property $ W^*(q,p) = W(q,p) $ and the property I mentioned of how $ W(q,p) $ relates to the expectation value of an operator $ \hat G $, we have the property

$$ 2 \pi \hbar \int \int W(q,p)^2 dq dp = \mathrm {Tr} ( \hat \rho^2 ) \le 1 . $$

Noting that $ \int \langle q \lvert \hat \rho \rvert q \rangle dq = \int \langle p \lvert \hat \rho \rvert p \rangle dp = \mathrm {Tr} ( \hat \rho ) = 1 $, then

$$ \int \int \langle q \lvert \hat \rho \rvert q \rangle \langle p \lvert \hat \rho \rvert p \rangle dq dp = \mathrm {Tr} ( \hat \rho )^2 = 1 . $$

Given that this integrand is in phase space, it stands to reason that we should have a similar inequality to the Cauchy-Schwarz,

$$ 2 \pi \hbar W (q,p)^2 \le \langle q \lvert \hat \rho \rvert q \rangle \langle p \lvert \hat \rho \rvert p \rangle . $$

Referring to the conjecture stated with the first candidate entropy and again using the convex function

$$ f(x) = - x \ln x , $$

we may then have

$$ S_{W^2} \le S_q + S_p , $$

where $ S_q $ and $ S_p $ are defined in the same manner as with the first candidate entropy, and

$$ S_{W^2} = \int \int f ( 2 \pi \hbar W(q,p)^2 ) dq dp = - \ln (2 \pi \hbar) \mathrm {Tr} (\hat \rho^2 ) - 2 \pi \hbar \int \int f ( W(q,p)^2 ) dq dp . $$

If this can be called an entropy, then Cauchy-Schwarz inequality needs to be proven. But as said, this is an attempt at finding and entropy which explicitly uses $ W(q,p) $.

Again reading the question, there may be a third candidate, also explicitly relating to $ W(q,p) $. Given what was stated about separability when $ \langle q \lvert \hat \rho \rvert q \rangle $ is Gaussian, then

$$ W(q,p) \le \lvert W(q,p) \rvert \le \langle q \lvert \hat \rho \rvert q \rangle \langle p \lvert \hat \rho \rvert p \rangle $$

and it may also be the case that

$$ S_{ \lvert W \rvert } \le S_q + S_p , $$


$$ S_{ \lvert W \rvert } = \int \int f ( \lvert W(q,p) \rvert ) dq dp . $$

However, since $ W(q,p) $ isn't necessarily positive definite, we could have that $ \int \lvert W(q,p) \rvert dq dp > 1 $, which means it may be possible to have $ S_{\lvert W \rvert} < 0 $. Perhaps, then, we should also consider $ - \int \int W(q,p) \ln \lvert W(q,p) \rvert dq dp $.

  • $\begingroup$ Thanks. Incidentally, I've tied the approach with $-\int W(q,p) \ln|W(q,p)| dq dp$. Then $S_w\leq S_q + S_p$ looked very promising... until I've found that for the first excited state (i.e Fock state n=1) it is broken by a tiny amount (and in fact it was the only state for which it does not hold). $\endgroup$ – Piotr Migdal Jun 12 '13 at 11:55
  • $\begingroup$ Good to know! Was the result general for any system, or perhaps a Gaussian Wigner distribution in particular? And by how much is a tiny amount? Was the amount in a logarithm? Hope one of these works out for you. If not, please, let us know why! $\endgroup$ – John Jun 12 '13 at 13:13
  • $\begingroup$ For n=1 it is not a Gaussian function. When it comes to "how much" I would need to dig in my notes (I will try to do that this week). $\endgroup$ – Piotr Migdal Jun 12 '13 at 13:18
  • $\begingroup$ When it comes to $2 \pi \hbar W (q,p)^2 \le \langle q \lvert \hat \rho \rvert q \rangle \langle p \lvert \hat \rho \rvert p \rangle$ - unfortunately, it cannot work. For pure stats (and not only being a product in $q$ and $p$, i.e. when the inequality is saturated everywhere) it is 1. $\endgroup$ – Piotr Migdal Jun 12 '13 at 13:29
  • $\begingroup$ Sorry, could you please clarify what you mean by "it is 1"? Do you mean integrating the both sides of the inequality directly? Because then you have $ 2 \pi \hbar \int W(q,p)^2 dq dp = \mathrm{Tr}(\hat \rho^2) \le \int \int \langle q \lvert \hat \rho \rvert q \rangle \langle p \lvert \hat \rho \rvert p \rangle dq dp = \mathrm{Tr}(\hat \rho)^2 = 1^2 = 1 $. I agree that when $ \mathrm{Tr}(\hat \rho^2) = 1 , \hat \rho $ represents a pure state. $\endgroup$ – John Jun 13 '13 at 1:26

Among other things, the entropy discussed by Białynicki-Birula and Mycielski seems to be what Wehrl might refer to as classical entropy. Considering this question further, I have found two possible candidates to your question. I will mention one now and the other later. The first is probably in line with what was mentioned by Grant Teply, and probably corresponds to a joint entropy. I will use the more general density operator formalism, noting that when the density operator $ \hat \rho $ represents a pure state, then $ \langle q \lvert \hat \rho \rvert q \rangle = \lvert \psi (q) \rvert^2 $ and $ \langle p \lvert \hat \rho \rvert p \rangle = \lvert \tilde \psi (p) \rvert^2 $. It was arrived at when considering the Cauchy-Schwarz inequality,

$$ \lvert \langle q \lvert \hat \rho \rvert p \rangle \rvert^2 = \langle q \lvert \hat \rho \rvert p \rangle \langle p \lvert \hat \rho \rvert q \rangle \le \langle q \lvert \hat \rho \rvert q \rangle \langle p \lvert \hat \rho \rvert p \rangle , $$


$$ \langle q \lvert \hat \rho \rvert p \rangle = \frac{1}{\sqrt{2 \pi \hbar}} \int \langle q \lvert \hat \rho \rvert q' \rangle \exp \left ( \frac{i p q'}{\hbar} \right ) dq' = \frac{1}{\sqrt{2 \pi \hbar}} \int \langle p' \lvert \hat \rho \rvert p \rangle \exp \left ( \frac{i p' q}{\hbar} \right ) dp'. $$

I conjecture that with the concave function

$$ f(x) = -x \ln x , $$

we get the inequality

$$ f ( \lvert \langle q \lvert \hat \rho \rvert p \rangle \rvert^2 ) \le f( \langle q \lvert \hat \rho \rvert q \rangle \langle p \lvert \hat \rho \rvert p \rangle ) = \langle p \lvert \hat \rho \rvert p \rangle f ( \langle q \lvert \hat \rho \rvert q \rangle ) + \langle q \lvert \hat \rho \rvert q \rangle f ( \langle p \lvert \hat \rho \rvert p \rangle ) . $$

If this inequality holds, then integrating over the phase space, we then have the subadditivity property for entropy

$$ S_{qp} \le S_q + S_p , $$


$$ S_{qp} = \int \int f ( \lvert \langle q \lvert \hat \rho \rvert p \rangle \rvert^2 ) dq dp = S_{pq} , $$

$$ S_q = \int \int \langle p \lvert \hat \rho \rvert p \rangle f ( \langle q \lvert \hat \rho \rvert q \rangle ) dq dp = \int f ( \langle q \lvert \hat \rho \rvert q \rangle ) dq , $$

$$ S_p = \int \int \langle q \lvert \hat \rho \rvert q \rangle f ( \langle p \lvert \hat \rho \rvert p \rangle ) dq dp = \int f ( \langle p \lvert \hat \rho \rvert p \rangle ) dp . $$

Equality holds when $ \hat \rho $ represents a pure state, or a system that is separable in position $ q $ and momentum $ p $. In other words, if $ S_{qp} $ can indeed be interpreted as a joint entropy, it represents correlation between position and momentum. Strict inequality will come from systems represented by mixed states, and thus when the system cannot be represented by a solution of the Schrödinger equation, which traditionally does not include cross terms between position and momentum.

  • $\begingroup$ Is the Q function (or in Werhl's language, $\langle p,q|\rho|p, q\rangle$ (BTW, a beautiful paper)) the same as $|\langle q | \rho | p \rangle|^2$? $\endgroup$ – Piotr Migdal Jun 11 '13 at 19:26
  • $\begingroup$ As far as I can tell, no. Transforming the density matrix $ \langle q \lvert \hat \rho \rvert q \rangle $ into the Wigner distribution $ W(p,q) $, then using the transform mentioned by Stefan above to arrive at the Husimi Q representation $Q(\alpha)$, seems to be equivalent to taking a [Gabor transform][1] $ \langle q \lvert \hat \rho \rvert q \rangle \to Q(\alpha)$. Whereas $ \langle q \lvert \hat \rho \rvert p \rangle $ is found by simple Fourier transform. It might be equivalent in a coherent state basis. [1] : en.wikipedia.org/wiki/Gabor_transform $\endgroup$ – John Jun 11 '13 at 19:53
  • $\begingroup$ But you asked for an entropy explicitly related to the Wigner distribution. I'll post my second candidate for an entropy which makes an attempt at it later. $\endgroup$ – John Jun 11 '13 at 20:00
  • $\begingroup$ It was a bit open-ended question. My main goal was to find a more general uncertainty principle than the entropic uncertainly relation. Wigner function looks like a good candidate, as it is a "quantum analogue" of the phase-space distribution. $\endgroup$ – Piotr Migdal Jun 18 '13 at 13:51

Regarding the first candidate I proposed, $\lvert \langle q \vert \hat \rho \rvert p \rangle \rvert^2$ has the properties

$$ \int \lvert \langle q \vert \hat \rho \rvert p \rangle \rvert^2 dp = \langle q \lvert \hat \rho^2 \rvert q \rangle , \quad \int \lvert \langle q \vert \hat \rho \rvert p \rangle \rvert^2 dp = \langle p \lvert \hat \rho^2 \rvert p \rangle , \\ \iint \lvert \langle q \vert \hat \rho \rvert p \rangle \rvert^2 dq dp = \mathrm{Tr}(\hat \rho^2) \le 1 , $$

which may invalidate the conjecture that accompanied it.

An allure to finding a classical entropy analogue with the Wigner distribution comes from the properties

$$ \int W(q,p) dp = \langle q \lvert \hat \rho \rvert q \rangle , \quad \int W(q,p) dq = \langle p \lvert \hat \rho \rvert p \rangle , \\ \iint W(q,p) dq dp = \mathrm{Tr}(\hat \rho) = 1 .$$

I think a related question to the one first asked is if there is a joint density distribution $R(q,p)$ with marginal densities $Q(q)$ and $P(p)$ such that

$$ \int R(q,p) dp = \langle q \lvert \hat \rho \rvert q \rangle = Q(q) , \quad \int R(q,p) dq = \langle p \lvert \hat \rho \rvert p \rangle = P(p) , \\ \iint R(q,p) dq dp = \mathrm{Tr}(\hat \rho) = 1 , $$ and as such with $f(x) = - x \ln x$ also satisfies the property of entropy subadditivity,

$$ S_{qp} \le S_q + S_p , $$


$$ S_q = \int f(Q(q)) dq , \quad S_p = \int f(P(p)) dp , \\ S_{qp} = \iint f(R(q,p)) dq dp = S_{pq} . $$

Under these conditions, I propose another candidate

$$ R(q,p) = \lvert \langle q \vert \sqrt{\hat \rho} \rvert p \rangle \rvert^2 = \lvert \langle p \vert \sqrt{\hat \rho} \rvert q \rangle \rvert^2 \ge 0 . $$

As to what is meant by $\sqrt{\hat \rho}$, consider representing $\hat \rho$ diagonally in its discrete eigenbasis $\{ \rvert \lambda_m \rangle \}_{m \in \mathbb{N}}$ (if it has one),

$$ \hat \rho = \sum_m \lambda_m \rvert \lambda_m \rangle \langle \lambda_m \lvert . $$

Given that $\hat \rho^\dagger = \hat \rho, \; \mathrm{Tr}(\hat \rho) = 1,$ and $\langle \lambda_m \lvert \hat \rho \rvert \lambda_m \rangle \ge 0 \; \forall \; \lambda_m$, then $0 \le \lambda_m \le 1 \; \forall \; \lambda_m$. For an orthonormal eigenbasis we have, $\forall \: \lambda_m$,

$$ \hat \rho^2 = \sum_m \lambda_m^2 \rvert \lambda_m \rangle \langle \lambda_m \lvert , \quad 0 \le \lambda_m^2 \le \lambda_m \le 1 , $$

and similarly

$$ \sqrt{\hat \rho} \equiv \sum_m \sqrt{\lambda_m} \rvert \lambda_m \rangle \langle \lambda_m \lvert , \quad 0 \le \lambda_m \le \sqrt{\lambda_m} \le 1 . $$

We choose $\sqrt{\hat \rho}$ to be positive semidefinite because if $\hat \rho = \sqrt{\hat \rho}$, then $\sqrt{\hat \rho}$ can be used to represent a pure state $\left ( \lambda_m = \delta_{mn} , \; n \in \mathbb{N} \right)$. Concluded from this are the properties

$$ \sqrt{\hat \rho}^\dagger = \sqrt{\hat \rho}, \quad \sqrt{\hat \rho} \sqrt{\hat \rho} = \left ( \sqrt{\hat \rho} \right )^2 = \hat \rho , \\ \mathrm{Tr}(\hat\rho^2) \le \mathrm{Tr}(\hat\rho) = 1 \le \mathrm{Tr}(\sqrt{\hat\rho}). $$

In finite basis of dimension $N$, the inequality above for a totally (or completely) mixed state $\left ( \lambda_m = 1/N \; \forall \; \lambda_m \right )$ reads

$$ 1/N \le 1 \le \sqrt N . $$

An example application of the operator $\sqrt{\hat \rho}$ is with fidelity. This clarification was made to compare with other works.

Some comparisons to the first candidate I proposed. From the Cauchy-Schwarz inequality we have

$$ \begin{align*} R(q,p) & \le \langle q \lvert \sqrt{\hat \rho} \rvert q \rangle \langle p \lvert \sqrt{\hat \rho} \rvert p \rangle , \\ {} & \le \langle q \lvert \sqrt{\hat \rho} \sqrt{\hat \rho} \rvert q \rangle \langle p \vert p \rangle = Q(q) \delta(0) = \infty , \\ {} & \le \langle q \vert q \rangle \langle p \lvert \sqrt{\hat \rho} \sqrt{\hat \rho} \rvert p \rangle = \delta(0) P(p) = \infty , \\ {} & \le \langle q \lvert \hat \rho^{1-\varepsilon} \rvert q \rangle \langle p \lvert \hat \rho^\varepsilon \rvert p \rangle , \quad 0 \le \varepsilon \le 1 . \end{align*} $$

The terms $\hat \rho^{1-\varepsilon}$ and $\hat \rho^\varepsilon$ in the last inequality are determined using matrix functions, which also further justifies our choice in $\sqrt{\hat \rho}$.

A way to express $R(q,p)$ should be

$$ \begin{align*} R(q,p) & = \frac{1}{2 \pi \hbar} \iint \langle q \vert \sqrt{\hat \rho} \rvert q_1 \rangle \langle q_2 \vert \sqrt{\hat \rho} \rvert q \rangle \exp \left (\frac{i p (q_1-q_2)}{\hbar} \right ) dq_1 dq_2 & \\ {} & = \frac{1}{2 \pi \hbar} \iint \langle p_1 \vert \sqrt{\hat \rho} \rvert p \rangle \langle p \vert \sqrt{\hat \rho} \rvert p_2 \rangle \exp \left (\frac{i (p_1 - p_2) q}{\hbar} \right ) dp_1 dp_2 . & \end{align*} $$

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    $\begingroup$ John, you seem to have lost contact with your earlier account. See physics.stackexchange.com/help/merging-accounts to fix it and consider using a registered account which will make this kind of occurrence less likely. $\endgroup$ – dmckee Jun 18 '13 at 5:44
  • $\begingroup$ What we seem to be considering is [mutual information][1] $I_{qp}$ for a joint probability density $R(q,p)$ and marginal probability densities $Q(q)$ and $P(p)$. Over the space of random variables $(q,p)$, we have the inequality $I_{qp} = S_q + S_p - S_{qp} \ge 0,$ so $S_q + S_p \ge S_{qp}.$ [1]: en.wikipedia.org/wiki/Mutual_information $\endgroup$ – John Jul 3 '13 at 14:04

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