Normal Ordering and Smearing I read on Wikipedia two different descriptions of the "Husimi-Q representation."
One is that it is the Wigner function convolved with a Gaussian, which in particular results in a positive definite function.
The other is that it is "essentially" (their words) the density matrix put into normal order. I had some trouble understanding why these are the same.
For instance, if we let $H=\omega a^\dagger a$, then the thermal state at inverse temperature $\beta$ is $$\rho=N\exp(-\beta H),$$ which (if I did everything right) normal-orders to $$:\rho:=N\exp(-\beta'H),$$ where $$\beta'=-\log(1-\beta\omega)/\omega.$$ This seems fairly pathological to me (ignoring the issue that this new density matrix doesn't seem to be normalized): we have $\beta'>\beta$, so the system is at a colder temperature, so I would expect the distribution in phase space to be "less blurry" rather than "more blurry," (certainly classically this is true), and at $\beta\omega=1$ we have singular behavior: for any temperature colder than $\omega$, it seems like the object we'll get out will assign negative probabilities to certain states.
Did I do the computation wrong? Does normal ordering here mean something different than pushing $a$'s to the right of $a^\dagger$'s? Are there other contexts in which we can think of normal ordering as smearing out distribution functions?
 A: Q-function and P-function
The Husimi Q-function  of a density matrix $\rho$ is defined by
$$ Q_\rho(\alpha) = \frac{1}{\pi}\langle \alpha \vert \rho \vert \alpha \rangle$$
where 
$$\rho = \frac{1}{\pi}\int\rho_\text{coh}(\alpha,\alpha^\ast)\lvert \alpha \rangle\langle\alpha\lvert\mathrm{d}\alpha\mathrm{d}\alpha^\ast$$
in the coherent states. $P_\rho(\alpha,\alpha^\ast) := \frac{1}{\pi}\rho_\text{coh}(\alpha,\alpha^\ast)$ is the Glauber–Sudarshan P-function. It follows with $\langle \beta\vert \alpha\rangle = \mathrm{e}^{-\beta^\ast\beta/2-\alpha^\ast\alpha/2 + \beta^\ast\alpha}$ that
\begin{align}
Q_\rho(\beta) & = \int P_\rho(\alpha,\alpha^\ast)\langle\beta\vert  \lvert \alpha \rangle\langle\alpha\lvert\vert \beta\rangle\mathrm{d}\alpha\mathrm{d}\alpha^\ast\\
& = \int P_\rho(\alpha,\alpha^\ast)\mathrm{e}^{-\beta^\ast\beta-\alpha^\ast\alpha + \beta^\ast\alpha + \beta\alpha^\ast}\mathrm{d}\alpha\mathrm{d}
\alpha^\ast \\
& = \int P_\rho(\alpha,\alpha^\ast)\mathrm{e}^{-\lvert \beta-\alpha\rvert^2}\mathrm{d}\alpha\mathrm{d}
\alpha^\ast
\end{align}
Normal ordering and anti-normal ordering
The Q-function naturally normal orders $\rho$. Since $a\lvert \alpha \rangle = \alpha \lvert\alpha\rangle$, we have that $f(a)\lvert \alpha\rangle = f(\alpha)\lvert \alpha\rangle$ and $\langle \alpha \rvert f(a^\dagger) = \langle\alpha\rvert \alpha^\ast$, so for a normal-ordered symbol $f_N(a,a^\dagger)$ with all annihilators to the right and all creators to the left, we have $\langle \alpha\rvert f_N(a,a^\dagger)\lvert \alpha\rangle = f_N(\alpha,\alpha^\dagger)$, and so
$$ Q_\rho(\alpha) = \frac{1}{\pi}\rho_N(\alpha,\alpha^\ast)$$
The P-function naturally anti-normal orders $\rho$. Expand
$$ \rho_A(a,a^\dagger) = \sum_{i,j}\rho_{i,j}a^i (a^\dagger)^j$$
and insert the completeness relation $\mathbf{1} = \frac{1}{\pi}\int\lvert\alpha\rangle\langle\alpha\rvert\mathrm{d}\alpha\mathrm{d}\alpha^\ast$ to get
$$ P_\rho(\alpha,\alpha^\ast) = \frac{1}{\pi}\rho_A(\alpha,\alpha^\ast)$$
What is a bit confusing is that this ordering prescription is exactly opposite to what it does on observables. One finds that anti-normal ordered expectation values are computed with the Q-function and normal ordered expectation values are computed with the P-function, i.e.
\begin{align}
\langle \mathcal{O}_A(a,a^\dagger) \rangle & = \int Q(\alpha,\alpha^\ast) \mathcal{O}_A(\alpha,\alpha^\ast)\mathrm{d}\alpha\mathrm{d}\alpha^\ast \\
\langle \mathcal{O}_N(a,a^\dagger) \rangle & = \int P(\alpha,\alpha^\ast) \mathcal{O}_N(\alpha,\alpha^\ast)\mathrm{d}\alpha\mathrm{d}\alpha^\ast 
\end{align}
