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I'm provided with the covariance matrix of a three-mode cluster state (a state which is obtained by entangling three momentum squeezed states). I want to see what happens to this state when I measure the third node, either in the position quadrature, either in the momentum quadrature. How does the covariance matrix change? And how does the displacement vector, the vector of the first statistical momenta, change?

I found two references:

[1] https://arxiv.org/pdf/quant-ph/0204085.pdf

[2] http://cds.cern.ch/record/546624/files/0204052.pdf

and a related question: How to find the covariance matrix after a partial homodyne measurement?

According to these references, considering an $n+m$ mode gaussian state with covariance matrix \begin{equation} \gamma=\begin{pmatrix} A&C\\C^T&B \end{pmatrix} \end{equation} and projecting (thus performing a POVM measurement) on the gaussian state $\gamma_p$, the new statistical moments are \begin{equation} \gamma'=A-C\frac{1}{B+\gamma_p}C^T\\ d'=\frac{1}{2}C\frac{1}{B+\gamma_p}d \end{equation} These are precisely Eqs. (15a) and (15b) of Ref. [1]. For what I understood $d$ should be the displacement vector of the measure. Then an homodyne detection correspond to the limiting case where $\gamma_p=\mathrm{diag}(1/a, a)$ with $a\to 0$. In this case we may substitute \begin{equation} \frac{1}{B+\gamma_p}\to (\pi B \pi)^{MP} \end{equation} where $\pi=\mathrm{diag}(1,0)$ and $MP$ denotes the Moore-Penrose inverse.

My problem

When I try to compute the displacement vector of the state after the measurement, I always get zero. This is not reasonable to me, as, being the nodes entangled, I should find a displacement in the entangled node and for this reason I perform a feed-forward operation in a Measurement-Based-Quantum-Computing framework, to remove this displacement and return to a state with zero first statistical momenta.

As an example, I implemented the thee-mode state: \begin{equation} \gamma=\left( \begin{array}{cccccc} 1.675 & 0. & 0. & 0.825 & -0.825 & 0. \\ 0. & 0.85 & 0.825 & 0. & 0. & 0.825 \\ 0. & 0.825 & 0.85 & 0. & 0. & 0.825 \\ 0.825 & 0. & 0. & 1.675 & 0.825 & 0. \\ -0.825 & 0. & 0. & 0.825 & 1.675 & 0. \\ 0. & 0.825 & 0.825 & 0. & 0. & 0.85 \\ \end{array} \right) \end{equation} where the canonical operators are in the order $(x_1, p_1, x_2, p_2, \dots)$. If I measure the momentum of the third node, I get as pseudo-inverse $(\pi B \pi)^{MP}$ the matrix \begin{equation} \left( \begin{array}{cc} 0.597015 & 0. \\ 0. & 0. \\ \end{array} \right) \end{equation} My vector $d$ of the POVM should be $d=(0, \bar{p})$, for a fixed value of $\bar{p}$. Clearly, I will always obtain $d'=0$ as displacement, as $(\pi B \pi)^{MP}d=0$ no matter the value of $\bar{p}$. This is counterintuitive to me.

Can someone see a mistake or does someone have an explanation?

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  • $\begingroup$ Quick shot: IIRC one has to multiply with the symplectic form in the Schur complement (which doesn't appear in the above). This should swap x and p. But I'll have to look a bit more closely. $\endgroup$ – Norbert Schuch Aug 1 '18 at 16:45
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I think the issue is with your definition of $\pi$. $\pi$ is defined as diag(1, 0) if you are performing a measurement of the q-quadrature. If you are measuring the p-quadrature it is defined as diag(0, 1). Similarly for $d$, we have $d=(0, \bar{p})$ for the p-quadrature and $d=(\bar{q}, 0)$ for the q-quadrature. See this review (Weedbrook et al "Gaussian Quantum Information) for more details.

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