2
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

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}

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – commutatertot Feb 29 '16 at 19:42
  • 1
    $\begingroup$ @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 $(a a^\dagger)_N = a^\dagger a + 1$. $\endgroup$ – ACuriousMind Feb 29 '16 at 20:21
  • $\begingroup$ Ah I see; in my question I was using the "ignore commutation relations" kind of normal ordering. Thank you! $\endgroup$ – commutatertot Feb 29 '16 at 22:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.