For a state $\rho$ acting on single bosonic mode with coherent states $|\alpha\rangle$, one can always define a $P$-function to furnish a diagonal representation of the state in the coherent-state basis $$\rho=\int d^2\alpha |\alpha\rangle\langle \alpha| P(\alpha,\alpha^*).$$ The $P$-function is not unique and does not always exist in terms of well-behaved functions (e.g., Drummond & Gardiner 1980), often containing derivatives of Dirac delta functions. The motivating question is "What set of distributions $P(\alpha,\alpha^*)$ all correspond to the same state?"

The Wikipedia page on this topic, using the notation $f(\alpha,\alpha^*)$ instead of $P(\alpha,\alpha^*)$, says

The function $f$ is not unique. There exists a family of different representations, each connected to a different ordering Ω. The most popular in the general physics literature and historically first of these is the Wigner quasiprobability distribution, which is related to symmetric operator ordering.

This seems to imply that one can, for example, use the Wigner function $W(\alpha,\alpha^*)$ to equivalently write $$\rho=\int d^2\alpha |\alpha\rangle\langle \alpha| W(\alpha,\alpha^*).$$ Is this true? Can one take a quasiprobability distribution like the Wigner or Husimi distribution and use it to directly expand the state in the coherent-state basis? I am almost certain that the answer must be that this is impossible, for many reasons:

  1. The Wigner and Husimi distributions are well-behaved and/or positive for states where the $P$-function is not.
  2. Direct computation with specific states show discrepancies. For example, take the vacuum (i.e., the coherent state with $\alpha=0$). The $P$-function is $P(\alpha,\alpha^*)=\delta(\alpha)$ such that $\rho=|0\rangle\langle 0|$, while the Wigner function is proportional to $\exp(-2|\alpha|^2)$. We can perform the integral \begin{align} \int d^2\alpha |\alpha\rangle\langle \alpha| \exp(-2|\alpha|^2)&=\int_0^\infty r dr\int_0^{2\pi}d\theta |re^{i\theta}\rangle\langle re^{i\theta}| \exp(-2r^2)\\ &=\sum_{m,n\geq 0}\frac{|m\rangle\langle n|}{\sqrt{m!n!}}\int_0^\infty r dr\int_0^{2\pi}d\theta e^{-r^2/2}(r e^{-i\theta})^m(r e^{i\theta})^n e^{-r^2/2} e^{-2r^2}\\ &\propto\sum_{m\geq 0}\frac{|m\rangle\langle m|}{m!}\int_0^\infty r^{2m+1} dr e^{-3r^2}\\ &\propto\sum_{m\geq 0}3^m |m\rangle\langle m|\neq |0\rangle\langle 0|. \end{align}

So the main question is whether one can replace the $P$-function in the diagonal coherent-state representation by the Wigner of Husimi function or another quasiprobability distribution (which Wikipedia seems to imply to be possible but which should not be possible) and the more general question is how to encompass the entire family of distributions that can act as $P$-functions. If the latter is simply answered by "use all equivalent distributions for derivatives of delta functions" then I am happy.

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    $\begingroup$ Linked. $\endgroup$ Commented Oct 18, 2022 at 20:58
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    $\begingroup$ "Is this true?" No, as you demonstrated! Nothing matches the spikiness and simplicity of P. The expression for W is much-much-much messier, and that's why most texts don't bother to write it down. Study up on W Schleich's book and you might come close. $\endgroup$ Commented Oct 18, 2022 at 22:10
  • $\begingroup$ @CosmasZachos good! I needed some verification that the wiki quote per my immediate interpretation couldn't be true. $\endgroup$ Commented Oct 19, 2022 at 0:37
  • $\begingroup$ You are just misreading it. It does not say what you imagine… $\endgroup$ Commented Oct 19, 2022 at 2:14

1 Answer 1


As was previously mentioned in the comments, the decomposition formula $$\rho = \int \mathrm d^2\alpha P_\rho(\alpha) |\alpha\rangle\!\langle\alpha|,$$ uniquely determines the structure of $P_\rho(\alpha)$.

A direct way to see this is to observe we can derive the form of $P_\rho$ without any additional requirements: observe that $$\langle -\alpha|\rho|\alpha\rangle e^{-|\alpha|^2} = \int \mathrm d^2\beta e^{-|\beta|^2}P_\rho(\beta) e^{\alpha\bar\beta-\bar\alpha\beta}.$$ This implies that $$P_\rho(\beta) e^{-|\beta|^2} = \int \frac{\mathrm d^2\alpha}{\pi^2} e^{\beta\bar\alpha-\bar\beta\alpha} e^{-|\alpha|^2} \langle-\alpha|\rho|\alpha\rangle.$$ These relations are a special case of the more general Fourier transform formulas: $$f(\alpha) = \int \mathrm d^2\beta e^{\alpha\bar\beta-\bar\alpha\beta} g(\beta), \qquad g(\beta) = \int \frac{\mathrm d^2\alpha}{\pi^2} e^{\beta\bar\alpha-\bar\beta\alpha} f(\alpha).$$

As shown in [CG1969], the $P$ function can also be written explicitly as $$P_\rho(\nu) = \frac1\pi \lim_{s\to1^-} \operatorname{tr}[T(\nu,s)\rho], \\ T(\nu,s) \equiv \frac{2}{1-s} D(\nu)\left(\frac{s+1}{s-1}\right)^{a^\dagger a} D(-\nu),$$ which also clearly shows that $P$ is uniquely determined by $\rho$. For an example application of the latter approach to thermal states see e.g. my answer here.

  • $\begingroup$ Yes, thanks for this answer. The offending wikipedia statement has been updated since I asked this question, so I am happy to not have misunderstood $\endgroup$ Commented Jun 23 at 19:45

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