Prove that integrating a displacement operator with a Gaussian gives $\int d^2\gamma e^{-|\gamma|^2/2}D(\gamma)=\pi|0⟩\!⟨0|$

Question

I'm looking for "nice" ways to prove the following identity for displacement operators: $$\int d^2\gamma e^{-|\gamma|^2/2}D(\gamma)=\pi|0⟩\!⟨0|,$$ with $|0\rangle$ the vacuum state and $D(\gamma)\equiv \exp(\gamma a^\dagger-\bar\gamma a)$.

This is used at the end of Wilde's notes (Link to the pdf; they actually say they're going to prove the statement in the following lecture but I can't find said lecture) as a step to prove the formula to decompose density matrices in terms of displacement operators: $$\rho= \frac1\pi\int d^2\alpha\operatorname{tr}(D(\alpha)\rho) D(-\alpha).$$

I'm looking for "nice" approaches to prove this.

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As an example of a "not nice" approach, one can do the following: use the antinormally ordered expression for $D(\gamma)$ to transform the LHS in $\int d^2\gamma e^{-\bar\gamma a} e^{\gamma a^\dagger}$. Then compute the matrix elements with respect to Fock states, obtaining $$ I_{n,m}\equiv \langle n|\int d^2\gamma e^{-\bar\gamma a} e^{\gamma a^\dagger} |m\rangle = \sum_{s,t=0}^\infty \int d^2\gamma \frac{\gamma^t (-\bar\gamma)^s}{t!s!} \langle n|a^s a^{\dagger t}|m\rangle. $$ Switching to polar coordinates the integral over the phase of $\gamma$ becomes $\int d\phi e^{i\phi(t-s)}=2\pi \delta(t-s)$, and thus $$I_{n,m} = 2\pi\sum_{s=0}^\infty \frac{(-1)^s}{s!^2}\int_0^\infty r^{2s+1}\langle n|a^s a^{\dagger s}|m\rangle. $$ This shows that $I_{n,m}=0$ if $n\neq m$, and then $$I_{n,n} = 2\pi\sum_{s=0}^\infty \frac{(-1)^s (s+n)!}{n! s!^2} \int_0^\infty dr\, r^{2s+1}.$$ While this would seem to obviously diverge due to the integral, it turns out that doing the series first dampens the function enough to get a meaningful result. Mathematica tells me that $$\sum_{s=0}^\infty \frac{(-1)^s (s+n)!}{n! s!^2} r^{2s+1} = r\,\, {}_1 \!F_1(1+n, 1; -r^2),$$ whose integral over $r$ is apparently $0$ whenever $n>0$, and is $1/2$ for $n=0$.

Obviously, it's not the nicest solution, hence why I'm curious about alternative approaches.

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For completeness let me also include what you get following a similar reasoning but using the normally ordered expansion for $D(\alpha)$. This is somewhat better, in that at least there's less stuff diverging, and we don't need to involve hypergeometric stuff.

The integral to solve is in this case $$I_{n,m} = \int d^2\gamma \langle n|e^{\gamma a^\dagger} e^{-\bar\gamma a}|m\rangle e^{-\frac12|\gamma|^2} \\= 2\pi \delta_{n,m} \sum_{s=0}^n \frac{n!(-1)^s}{s!^2(n-s)!} \int_0^\infty dr \, r^{2s+1} e^{-r^2} \\= \pi \delta_{n,m} \sum_{s=0}^n (-)^s \binom{n}{s}.$$ There isn't much new here compared to the previous calculation: I switched to polar coordinates, used the integral on the phase to get the $2\pi\delta_{n,m}$ term, ending up with only an integral over the radial variable $r$. The differences are that (1) now there's a Gaussian factor ensuring convergence of all integrals, and (2) now the series only gets up to $n$, because $a|0\rangle=0$.

We thus see that our conclusion amounts to the combinatorial identity $$\sum_{s=0}^n (-)^s \binom{n}{s} = \delta_{n,0},$$ which is quite interesting of its own right, but again I suspect this route isn't the ideal one to prove the original statement.

Answer 1

OP's operator is $$\begin{align}P_0~:=~& \int_{\mathbb{C}} \!\frac{\mathrm{d}\bar{z} \wedge \mathrm{d}z}{2 \pi i} e^{-\bar{z}z/2}D(z), \qquad D(z)~:=~e^{a^{\dagger}z-\bar{z}a},\cr ~=~& \int_{\mathbb{C}} \!\frac{\mathrm{d}\bar{z} \wedge \mathrm{d}z}{2 \pi i} \underbrace{AN(e^{a^{\dagger}z-\bar{z}a})}_{\text{antinormal order}}\cr ~=~& \int_{\mathbb{C}} \!\frac{\mathrm{d}\bar{z} \wedge \mathrm{d}z}{2 \pi i} e^{-\bar{z}z}\underbrace{N(e^{a^{\dagger}z-\bar{z}a})}_{\text{normal order}}\cr ~=~& \int_{\mathbb{C}} \!\frac{\mathrm{d}\bar{z} \wedge \mathrm{d}z}{2 \pi i}N( e^{-(\bar{z}-a^{\dagger})(z+a)-a^{\dagger}a})\cr ~\stackrel{(2/3)}{=}& N( e^{-a^{\dagger}a})\cr ~=~& |0\rangle\langle 0|, \tag{1} \end{align} $$ where we used the fact that $$\forall \mu,\nu~\in~\mathbb{C}:~~\int_{\mathbb{C}} \!\frac{\mathrm{d}\bar{z} \wedge \mathrm{d}z}{2 \pi i}~e^{-(\bar{z}-\mu)(z-\nu)} ~=~1, \tag{2}$$ or equivalently $$\int_{\mathbb{C}} \!\frac{\mathrm{d}\bar{z} \wedge \mathrm{d}z}{2 \pi i}~e^{-\bar{z}z}\bar{z}^nz^m ~=~n!\delta_{n,m}. \tag{3}$$ For the last equality in eq. (1), see e.g. my Phys.SE answer here.

Answer 2

Another way to understand it is to remember that the Wigner function of a vacuum state is a Gaussian at the origin. The characteristic function is the Fourier transform of the Wigner function, which is again a Gaussian at the origin. But the characteristic function can be directly obtained from the density operator of the vacuum state by tracing it with the adjoint of the displacement operator. Since the density operators provide a complete basis for operators, one can then reproduce the operator as an expansion of density operators with the characteristic function as coefficient function. This is exactly the first expression of the OP.

Answer 3

A slight rewording on the great approach given in this other answer (just because I like having things written explicitly): observe that $$\int \frac{\mathrm d^2\alpha}{\pi} e^{-|\alpha|^2/2} e^{\alpha a^\dagger-\bar\alpha a} = \int \frac{\mathrm d^2\alpha}{\pi} e^{-|\alpha|^2} e^{\alpha a^\dagger} e^{-\bar\alpha a} = \sum_{n,m} \int \frac{\mathrm d^2\alpha}{\pi} \frac{\alpha^n (-\bar\alpha)^m}{n!m!} e^{-|\alpha|^2} a^{\dagger n}a^m. $$ Switching to polar coordinates and integrating over the phase we thus get a $2\pi \delta_{n,m}$ factor, and the leftover Gaussian integral then gives $$= \sum_{n=0}^\infty \frac{(-1)^n}{n!} a^{\dagger n}a^m = N(e^{-a^\dagger a}). $$ Up to this point everything works at the formal level where we do not need to apply the CCRs. The last step is then simply to apply the CCRs and use the identity $N(e^{-a^\dagger a})= |0\rangle\langle0|$ discussed here.

Generalisation of the result

Let's now the tackle the more general case of $$I_t \equiv \int \frac{\mathrm d^2\gamma}{\pi} e^{s|\gamma|^2/2} D(\gamma),$$ for $t>0$. Using the same expansion and integral as above we get, using the shorthand $t\equiv \frac{1-s}{2}$ and assuming $s\le -1$, $$ I_t = \sum_{n=0}^\infty \frac{(-1)^n}{n!} t^{-1-n} a^{\dagger n}a^n = \frac1 t N(e^{-a^\dagger a/t}). $$ Remembering now the identity $\lambda^{a^\dagger a} =N(e^{(\lambda-1)a^\dagger a})$, we then conclude $$I_t = \frac{2}{1-s}\left(\frac{s+1}{s-1}\right)^{a^\dagger a}.$$