Show completeness of general Gaussian POVM

Question

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.
$$int d^2alpha ;  Pi (alpha) = I$$
I am stuck. I know that the trace of $Pi^0$ is normalized to unity, i.e.
$$  frac{1}{pi}int d^2alpha ; langle alpha|Pi^0|alpharangle = 1$$
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 $T=int d^2alpha Pi(alpha) = lambda I$. To fix the constant, it is suggested to do a spectral decomposition of $Pi^0=sum_j p_j |psi_jrangle 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_irangle &= lambdasum_i langle psi_i| I|psi_irangle 
sum_{ij} p_j int frac{d^2alpha}{pi}   langle psi_i| D(alpha) |psi_jrangle langle psi_j | D^{dagger} (alpha) |psi_irangle&= lambdasum_i langle psi_i|psi_irangle 
sum_{ij} p_j int frac{d^2alpha}{pi}   |langle psi_i| D(alpha) |psi_jrangle|^2 &= lambdasum_i 1 
end{align}
So, in order to complete the proof, I would need to show that
$$int frac{d^2alpha}{pi}   |langle psi_i| D(alpha) |psi_jrangle|^2=1$$
which is the transition probability from $|psi_jrangle$ via $D(alpha)$ to $|psi_irangle$ integrated over $alpha$. I do not know how to show this.

flippiefanus · Answer

Perhaps you can use the property that the coherent states can resolve the identity $$ frac{1}{pi} int |alpharanglelanglealpha| d^2alpha = I . $$
The $1/pi$ comes from the fact that the coherent states form an over-complete basis.

Emilio Pisanty · Answer

This essentially follows from Schur's lemma:

Let $G$ be a group and let $R_1:Gto mathrm{end}(V_1)$ and $R_2:Gto mathrm{end}(V_2)$ be two irreducible representations of $G$ on vector spaces $V_1$ and $V_2$. Moreover, let $T:V_1to V_2$ be a linear transformation such that $$Tcirc R_1(g) = R_2(g) circ T qquad forall gin G.$$ 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 $R:Gto mathrm{end}(V)$ be an irreducible representation of $G$ on a vector space $V$. Moreover, let $T:Vto V$ be a linear operator such that $$R(g)^{-1}circ Tcirc R(g) = T qquad forall gin G.$$ 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
$T= int mathrm d^2alpha : Pi(alpha)$.

To apply the lemma, you only need to show that
$$
D(beta)^{dagger}: T: D(beta) = T
$$
for arbitrary $betainmathbb C$, 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 $Pi^0=sum_i p_i |psi_iranglelangle psi_i|$, calculate the expectation values $langle psi_i|T|psi_irangle$ over all of the eigenstates, and add.
For more details, see

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

Show completeness of general Gaussian POVM

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