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A general (pure) Gaussian state has the form $\newcommand{\on}[1]{\operatorname{#1}}\newcommand{\ket}[1]{\lvert #1\rangle}\ket{\alpha,\xi}\equiv D(\alpha)S(\xi)\ket{\on{vac}}$, with $\ket{\on{vac}}$ the vacuum state, $D(\alpha)\equiv\exp(\alpha a^\dagger-\alpha^* a)$ the displacement operator, and $S(\xi)\equiv\exp(\frac12 \xi a^{\dagger 2}-\frac12\xi^* a^2)$ the squeezing operator.

I'm interested in the Fock state probabilities of such a state as a function of $\alpha,\xi\in\mathbb C$, that is, the values $$P_k\equiv \langle k|\alpha,\xi \rangle \equiv \frac{1}{\sqrt{k!}}\langle \on{vac}| a^k|\alpha,\xi\rangle, \qquad k\in\mathbb N$$ What are good ways to compute these?

Two "natural" approaches are (1) the direct calculation decomposing the operators as a sum of creations and annihilation operators, and (2) passing through the Wigner representation of Gaussian states. Other "better" routes are likely also viable.

Following the former path, I get, applying BCH and other standard techniques to rearrange the exponential expressions, to the expression: $$ \langle n|\alpha,\xi\rangle = \frac{e^{|\alpha|^2/2}}{\sqrt{n!\cosh(|\xi|)}} \alpha^n \sum_{\ell,k\ge0} (-|\alpha|^2)^\ell (C/\alpha^2)^k \frac{(n+\ell)!}{\ell! k! (\ell+n-2k)!} $$ with $C\equiv \frac12 e^{i\theta_\xi} \tanh(|\xi|)$ and $\xi=|\xi|e^{i\theta_\xi}$. I cannot get this into a useful closed formula. Assuming this is possible at all, knowing the solution obtained from another route (e.g. via the Wigner representation or similar) might shed light into how to handle this.

glS
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  • Schleich does not do it in his book? – Cosmas Zachos Sep 07 '20 at 20:56
  • Suggestion for a possible "better route": if you can swap $D(\alpha)$ with $a^k$ by knowing $[D(\alpha), a^k]$, then you can express your state as: $(D(\alpha)|0)^a^kS(E)|0\rangle$. Since $D(\alpha)|0\rangle$ and $S(E)|0\rangle$ both have explicit fock state expansions, you just need to simplify the product at the end. – Steven Sagona Sep 07 '20 at 22:44
  • Actually...looking at your final expression it looks pretty similar to what would come out of my suggested "better route" - so I guess it ends up being the same in the end? – Steven Sagona Sep 07 '20 at 22:58
  • Maybe try the math stackexchange to see if there are any binomial coefficient identities that can get that double sum to converge? – Steven Sagona Sep 07 '20 at 23:22
  • @CosmasZachos I can't find any mention of Gaussian states in the version I could find. Can you point to where this is discussed? – glS Sep 08 '20 at 05:38
  • @StevenSagona the way I did it was writing $D(\alpha)=e^{-\alpha^* a}e^{\alpha a^\dagger} e^{|\alpha|^2/2}$ and $S(\xi)=e^{C a^{\dagger 2}}e^{-\ln\cosh(|\xi|)(N+1/2)}e^{-C^* a^2}$. Then the expval automatically contains operators in antinormal order and you get something like $\frac{e^{|\alpha|^2/2}}{\sqrt{n!\cosh(|\xi|)}}\langle a^n e^{-\alpha^* a}e^{\alpha a^\dagger}e^{C a^{\dagger 2}}\rangle$. Expanding the exponentials and doing some index juggling I get to the expression in the post. I'd guess methods that just take care of the ordering of the ops should lead to the same expression – glS Sep 08 '20 at 05:43
  • It was a question, not a hint. There is so much stuff in there, including in the problems... – Cosmas Zachos Sep 08 '20 at 18:13
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    Gerry and Knight, Chapter 7. https://i.imgur.com/fpdOC1l.png – Kall Sep 08 '20 at 20:35
  • @Kamil thanks, I'll have a look at the derivation there – glS Sep 08 '20 at 21:18

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