# What is the observable corresponding to the $P$ function?

The P-function of a state $$\rho$$ (focusing on the single-mode case) can be written as, using a notation analogous to the one in Gerry&Knight' book, $$P_\rho(\alpha) = \int \frac{d^2\eta}{\pi^2} \chi_N(\eta) e^{\alpha\bar\eta-\bar\alpha\eta}, \quad \chi_N(\eta)\equiv \operatorname{tr}[\rho\exp(\eta a^\dagger) \exp(-\bar\eta a)].$$ This also means that we should be able to write $$P_\rho(\alpha)$$ as the expectation value of some observable on $$\rho$$, i.e. $$P_\rho(\alpha) = \operatorname{tr}(\mathcal O_P(\alpha)\rho)$$, where I defined $$\mathcal O_P(\alpha) \equiv \int \frac{d^2\eta}{\pi^2} \exp(\eta A_\alpha^\dagger)\exp(-\bar\eta A_\alpha), \quad A_\alpha\equiv a-\alpha.$$ For reference, the same procedure is especially straightforward with the $$Q$$ function instead, and leads to $$\mathcal O_Q(\alpha) \equiv \int \frac{d^2\eta}{\pi^2} e^{-\bar\eta A_\alpha}e^{\eta A_\alpha^\dagger} = \int\frac{d^2\beta}{\pi}\int\frac{d^2\eta}{\pi^2} \mathbb{P}_\beta e^{-\bar\eta(\beta-\alpha)+\eta(\beta-\alpha)^*} = \frac1\pi \mathbb{P}_\alpha,$$ where I integrated over $$\eta$$ in the last step, and used the shorthand notation $$\mathbb{P}_\beta\equiv|\beta\rangle\!\langle\beta|$$. Of course, this recovers the well-known fact that $$Q_\rho(\alpha)=\frac{\langle\alpha|\rho|\alpha\rangle}{\pi}$$.

What I'm trying to figure out is whether there's any kind of similar calculation/simplification that can be done for $$\mathcal O_P$$. We can't simply add a $$\mathbb{P}_\beta$$ in between the operator exponentials, as they're in the wrong order. We can add two of them on left and right, but that would lead to $$\mathcal O_P(\alpha) = \int\frac{d^2\eta}{\pi^2} \int\frac{d^2\beta d^2\gamma}{\pi^2} \mathbb{P}_\beta\mathbb{P}_\gamma e^{\eta(\beta-\alpha)^*-\bar\eta(\gamma-\alpha)}.$$ The problem is that now the exponential over $$\eta$$ is problematic, and generally leads to exponential divergences unless $$\beta,\gamma$$ are suitably chosen. An alternative approach would be to exploit the usual commutation rules to write $$\mathcal O_P(\alpha) = \int \frac{d^2\eta}{\pi^2} e^{-\bar\eta A_\alpha}e^{\eta A_\alpha^\dagger} e^{|\eta|^2} = \int\frac{d^2\beta}{\pi}\mathbb{P}_\beta \int\frac{d^2\eta}{\pi^2} e^{\eta(\beta-\alpha)^*-\bar\eta(\beta-\alpha)+|\eta|^2}.$$ This looks slightly more "elegant", but throws the divergence problem in my face even stronger with the $$\exp(|\eta|^2)$$ term. Decomposed in real and imaginary parts, the argument in the exponential reads $$2i[\eta_2(\beta_1-\alpha_1)-\eta_1(\beta_2-\alpha_2)] + \eta_1^2+\eta_2^2,$$ which seems like it could never converge. Though I'm guessing convergence is still possible thanks to the exponential terms hidden $$\mathbb{P}_\beta$$ which might kill the $$\eta$$ terms. After all, we know that, for example, $$\operatorname{tr}(\mathbb{P}_\gamma \mathcal O_Q(\alpha))=\delta^2(\alpha-\gamma)$$.

Now, I'm aware that the $$P$$ function is highly irregular (can be more singular than a $$\delta$$ etc), so it is absolutely to be expected that $$\mathcal O_P$$ would be precisely as irregular as $$P$$. Still, singularities can often be dealt with and written concisely, at least at a formal level. Is there any way to simplify these expressions, or maybe put them in a form that makes it more evident that the singularities disappear in at least some cases?

• doesn't answer the question, but I realised a similar way to define Wigner and other quasiprobability distributions in terms of expectation value wrt an observable is found in Ferrie's review, arxiv.org/abs/1010.2701, see chapter IIIA there.
– glS
Commented Feb 25 at 18:24

I think I found the answer in one of the original 1969 papers by Kahill and Glauber, via some expressions I also discussed in Can we get quasiprobability distributions other than $P,Q,W$ from generalised characteristic functions?. In (Kahill and Glauber 1969, PhysRev.177.1857) the authors define (Eq. 6.6) $$T(\alpha,s) = \int\frac{d^2\xi}{\pi} \exp\left[ (\alpha-a)\bar\xi- (\bar\alpha-a^\dagger)\xi+\frac{s}{2}|\xi|^2\right],\tag{6.6}$$ and prove that this can be given the concise formal expression (Eq. 6.24) $$T(\alpha,s) = \frac{2}{1-s} \left(\frac{s+1}{s-1}\right)^{(a^\dagger-\bar\alpha)(a-\alpha)}.\tag{6.24}$$ Note that this $$T$$ function is essentially (a generalisation of) the observables $$\mathcal O_P,\mathcal O_Q$$ I defined above, and they are tied to the quasiprobability distributions via $$W_s(\alpha) \equiv\int\frac{d^2\eta}{\pi^2} \chi_s(\eta) e^{\alpha\bar\eta-\bar\alpha\eta} =\frac1\pi\operatorname{tr}[\rho T(\alpha,s)], \\ \chi_s(\eta) \equiv \operatorname{tr}[\rho\exp(\eta a^\dagger-\bar\eta a + s|\eta|^2/2)].$$ In particular, $$W_0(\alpha)=W(\alpha)$$, $$W_1(\alpha)=P(\alpha)$$, and $$W_{-1}(\alpha)=Q(\alpha)$$, and thus $$\mathcal O_P(\alpha)=\frac1\pi T(\alpha,1)$$ and $$\mathcal O_Q(\alpha)=\frac1\pi T(\alpha,-1)$$. Which is compatible with them showing (Eq. 6.29) that $$T(\alpha,-1)=|\alpha\rangle\!\langle\alpha|$$.
It's clear from (6.24) that the $$s=1$$ case is particularly "troubling". This is mentioned by the authors around Eq. (7.22), and then later in their following paper (PhysRev.177.1882), where they define in (3.13, 3.15) the $$P$$ as $$P(\alpha) = \frac1\pi \operatorname{tr}[\rho T(\alpha,1)] \equiv \frac1\pi \lim_{s\to1^-} \operatorname{tr}[\rho T(\alpha,s)],$$ including also a discussion of its convergence.