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There are many ways to consider the Husimi ($Q$) quasi-probability distribution function, e.g. as the expectation of the density operator in a coherent state or as the Weirstrass transform of the Wigner function. I am going to use the "measurement" of a wavefunction by a coherent state (I will drop all normalization constants for simplicity):

$$ Q(x_0, k_0; \sigma) = |\left\langle \psi | \mathcal{C}(x_0, k_0; \sigma) \right\rangle |^2 $$

where the coherent state is parameterized by $\sigma$ (squeezed state), where $$ \left\langle x | \mathcal{C}(x_0, k_0; \sigma) \right\rangle = e^{-\frac{(x - x_0)^2}{4\sigma^2}}e^{ik_0x} $$ in position representation.

In my studies about quasi-probability distributions of wavefunctions, I have encountered the statement:

The $Q$ function does not allow calculating the marginals i.e. integrating over one of the phasespace variables to get the full distribution function of the other one, unlike the Wigner function or some similar variant. For example: the book "Quantum Optics in Phasespace" by Wolfgang Schleich, chapter 12.

When calcuating Husimis of a given wavefunction, I can always see this statement and make sense out of it. For example, the Husimi function of a plane wave $P = \exp(ikx)$ can be straightforwardly calculated, $$ Q_P(x_0, k_0; \sigma) \propto \sigma^2 e^{-(k-k_0)^2\sigma^2} $$

It is immediately clear that the marginal over $x_0$ makes no sense here.

Question: Is there a general proof, or some intuitive way to understand why the marginals do not make sense for any wavefunction?

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The stuff is all in Husimi's original 1940 paper, but you have to work at it... So I'll just illustrate the point based on our concise summary of the material in Wolfgang's outstanding treatise, eqns (1) and (122) here. I'll ignore normalizations, and worse yet, I'll set $\hbar=1$ and $\sigma=1/4$ to drive the qualitative point home.

So the Husimi Q is a Gaussian smoothing, Gaussian low-pass filtering, or Weierstrass transform of the Wigner function f, $$ f(x,p)\propto \int\! dy~\psi^* \left (x- y \right )~e^{-i2yp} ~ \psi \left (x+ y \right ) ,\\ Q(x,p)=T(f)= \exp \left({ 1\over 4}(\partial_x^2 +\partial_p^2 ) \right) f \\ \propto \int dx' dp' \exp \left( -{(x'-x)^2-(p'-p)^2 } \right) ~ f(x',p') . $$

Consequently, the p marginal of Q is a quadruple integral, $$ \int dp ~ Q(x,p)\propto \int dp dy dp'dx' ~ e^{ -{(x'-x)^2-(p'-p)^2 } } ~e^{-i2yp'} ~ \psi^* \left (x'- y \right )\psi \left (x'+ y \right ) . $$ One can now leave p' alone, and shift p by p', and do the unlinked Gaussian integral in p. Then do the integral in p' to obtain a δ(y), which then collapses the y integral, to yield a mere Weierstrass transform of the space probability density, $$ \int dp ~ Q(x,p)\propto\int dx' ~ e^{ - (x'-x)^2 } \psi^* \left (x' \right )\psi \left (x' \right )= \int dx' ~ e^{ - (x'-x)^2 } \rho(x'), $$
so, clearly not a probability density, unless ρ is an eigenfunction of T, so a Gaussian. This is in sharp contrast to $\int\! dp ~f(x,p)=\rho(x)$.

The evident intuitive visualization of this is that (even though it is positive semidefinite) the Husimi Q is not a plain phase-space probability measure for functions, but must be star-multiplied with them to yield expectation values . And, unlike the Wigner-Moyal product $\star$, the Husimi star product, $\circledast$ (124) in our summary, cannot be entirely integrated by parts out of the integral$\langle {{\mathfrak G}}\rangle = \int\! dx dp~ g(x,p) \exp\!\left(-{\hbar\over 4}(\partial_x^2+\partial_p^2 ) \right) Q(x,p)$: it will emphasize and de-emphasize parts of Q and the observable which only exceptional visualizers can intuit. This is the best illustration of the elusive picture the Husimi provides for intuiting where the action is probability-wise.

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  • $\begingroup$ I think the last equation of your answer is more than strong enough to convince me. I also appreciate the extra last paragraph and the reference to your concise summary! $\endgroup$ – George Datseris Dec 18 '17 at 23:39
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    $\begingroup$ OK, thanks. If it is not a digression, I added a couple of formulas to make it more explicit. $\endgroup$ – Cosmas Zachos Dec 19 '17 at 13:44
  • $\begingroup$ Just one final question regarding the Weirstrass transform of $f$. In the case where one wants to use squeezed states (the generalized transform), then I would guess that the exponentials $\exp(-(x'-x)^2 -(p'-p)^2)$ would not have equal factors; the second would have $\sigma^2$ and the first $1/\sigma^2$? $\endgroup$ – George Datseris Dec 19 '17 at 21:05
  • $\begingroup$ To answer my own comment: yes. The answer is also in the paper "Wigner and Husimi Functions in Quantum Mechanics" by Kin'ya Takahashi. $\endgroup$ – George Datseris Dec 19 '17 at 21:21

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