Show completeness of general Gaussian POVM In Continuous-Variable quantum information, a general Gaussian measurement is described by the POVM elements (say on a single mode for simplicity)
$$ \Pi(\alpha) = \frac{1}{\pi} D(\alpha) \Pi^0 D^{\dagger}(\alpha)$$
where $D(\alpha)$ is the displacement operator and $\Pi^0$ is any (generally mixed) density matrix of a zero-mean Gaussian state.
See for example https://arxiv.org/abs/0706.2799, Eq. 1 and 2.
or in this review https://arxiv.org/abs/1401.4679, section 3.4
Show that this POVM is complete, i.e.
$$\int d^2\alpha \;  \Pi (\alpha) = I$$
I am stuck. I know that the trace of $\Pi^0$ is normalized to unity, i.e.
$$  \frac{1}{\pi}\int d^2\alpha \; \langle \alpha|\Pi^0|\alpha\rangle = 1$$
but I just cannot see how to make use of this.
Edit: As explained in the answer below, Schur's Lemma can be used to conclude that $T=\int d^2\alpha \Pi(\alpha) = \lambda I$. To fix the constant, it is suggested to do a spectral decomposition of $\Pi^0=\sum_j p_j |\psi_j\rangle \langle \psi_j| $ then basically taking the trace on both side of $T=\lambda I$
I get the following
\begin{align}
\mathrm{Tr}(T) &= \lambda \mathrm{Tr}(I)\\
\sum_i \langle \psi_i| T |\psi_i\rangle &= \lambda\sum_i \langle \psi_i| I|\psi_i\rangle \\
\sum_{ij} p_j \int \frac{d^2\alpha}{\pi}   \langle \psi_i| D(\alpha) |\psi_j\rangle \langle \psi_j | D^{\dagger} (\alpha) |\psi_i\rangle&= \lambda\sum_i \langle \psi_i|\psi_i\rangle \\
\sum_{ij} p_j \int \frac{d^2\alpha}{\pi}   |\langle \psi_i| D(\alpha) |\psi_j\rangle|^2 &= \lambda\sum_i 1 \\
\end{align}
So, in order to complete the proof, I would need to show that
$$\int \frac{d^2\alpha}{\pi}   |\langle \psi_i| D(\alpha) |\psi_j\rangle|^2=1$$
which is the transition probability from $|\psi_j\rangle$ via $D(\alpha)$ to $|\psi_i\rangle$ integrated over $\alpha$. I do not know how to show this.
 A: This essentially follows from Schur's lemma:

Let $G$ be a group and let $R_1:G\to \mathrm{end}(V_1)$ and $R_2:G\to \mathrm{end}(V_2)$ be two irreducible representations of $G$ on vector spaces $V_1$ and $V_2$. Moreover, let $T:V_1\to V_2$ be a linear transformation such that $$T\circ R_1(g) = R_2(g) \circ T \qquad \forall g\in G.$$ Then either $T=0$ or $T$ is an isomorphism.

To apply it to this problem, you choose $R_1=R_2$ and invert the second representation, so it reads as follows:

Let $G$ be a group and let $R:G\to \mathrm{end}(V)$ be an irreducible representation of $G$ on a vector space $V$. Moreover, let $T:V\to V$ be a linear operator such that $$R(g)^{-1}\circ T\circ R(g) = T \qquad \forall g\in G.$$ Then $T = \lambda \mathbb I$ is a multiple of the identiy.

In this specific application, you choose:

*

*$G=(\mathbb C,{+})$ is the additive group of the complex numbers

*$V=\mathcal H$, the Hilbert space, and $R=D$ is the displacement operator

*$T= \int \mathrm d^2\alpha \: \Pi(\alpha)$.

To apply the lemma, you only need to show that
$$
D(\beta)^{\dagger}\: T\: D(\beta) = T
$$
for arbitrary $\beta\in\mathbb C$, and this follows directly from its definition. (Basically, join the $D(\beta)$ with the $D(\alpha)^\dagger$, and vice versa, and then do an origin shift in the integration.) This then tells you that $T=\lambda \mathbb I$, and you just need to fix $\lambda$; to do that, do a spectral decomposition of $\Pi^0=\sum_i p_i |\psi_i\rangle\langle \psi_i|$, calculate the expectation values $\langle \psi_i|T|\psi_i\rangle$ over all of the eigenstates, and add.
For more details, see

*

*A non group theoretic proof of completeness of arbitrary coherent states $D(α)∣f\rangle$. G.S.Agarwal and S. Chaturvedi. Mod. Phys. Lett. A 11, 2083 (1996), arXiv:quant-ph/9608033.

A: I have neglected the $\pi$ factors in the expressions, but they will cancel out even if you include them.
You need to show that:
$$\int d^2\alpha \; D(\alpha) \Pi_0 D(-\alpha) = I$$
Inserting the resolution of identity on both sides of $\Pi_0$ gives:
$$\int d^2\alpha \; d^2\beta \; d^2\gamma \; D(\alpha) \; | \beta \rangle \langle \beta | \; \Pi_0 | \; \gamma \rangle \langle \gamma | \; D(-\alpha) $$
Now, you can carry out the $\alpha$ integration using the following identity:
$$ I \langle \gamma | \beta \rangle = \int d^2\alpha \; D(\alpha) \; | \beta \rangle \langle \gamma | D(-\alpha) $$
Which gives:
$$ \int \; d^2\beta \; d^2\gamma \; \langle \gamma | \beta \rangle  \langle \beta | \; \Pi_0 | \; \gamma \rangle $$
Carry out the $\beta$ integration, which is just the identity, and carrying out the $\gamma$ integral will just give 1, because $\Pi_0$ is normalized, as you yourself stated.
I got the identity from the following paper, equation (3.20):
https://journals.aps.org/pr/abstract/10.1103/PhysRev.177.1857
