# 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?

# 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}

• Thanks! I'm still a little confused: if $\rho_N$ is normal ordered, I see that $\langle\alpha\vert\rho_N(a,a^\dagger)\vert\alpha\rangle = \rho_N(\alpha,\alpha^*)$, and for arbitrarily ordered $\rho$ we have the definition $Q_\rho(\alpha)=\langle \alpha\vert\rho\vert\alpha\rangle$, but I don't see how these are related. In particular, $\langle\alpha\vert\rho\vert\alpha\rangle$ will not necessarily equal $\langle\alpha\vert :\rho: \vert\alpha\rangle$, right? Apologies if I am being dense. – commutatertot Feb 29 '16 at 19:42
• @commutatertot: As operators, $\rho(a,a^\dagger) = \rho_N(a,a^\dagger)$. I'm not using $:\rho :$, which is normal ordering the operators in it and just discarding the terms you pick up from the commutation relations. I.e. $: a a^\dagger : = a^\dagger a$, but $(a a^\dagger)_N = a^\dagger a + 1$. – ACuriousMind Feb 29 '16 at 20:21
• Ah I see; in my question I was using the "ignore commutation relations" kind of normal ordering. Thank you! – commutatertot Feb 29 '16 at 22:17