Covariance matrix after projection on gaussian state

I cannot understand the proof in Eisert article (https://arxiv.org/abs/quant-ph/0204052) about finding the covariance matrix of a state after projecting some of its modes onto a gaussian state.

We have the characteristic function of the remaining (unprojected) modes being $$$$\chi(\xi_A)=\frac{1}{\pi^2}\int d\xi_5\dots d\xi_8 e^{-\xi^T\Gamma'\xi/2}e^{-\xi_B^T \gamma\xi_B}$$$$ where $$\xi=(\xi_A, \xi_B)$$, the labels $$A$$ and $$B$$ referring respectively to the subsystem that ''remains'' and to the subsystem that is projected to a gaussian state with covariance matrix $$\gamma=\mathrm{diag}(1/d,d,1/d,d )$$ and where $$$$\Gamma'=\begin{pmatrix}C_1 & C_3\\C_3^T &C_2 \end{pmatrix}$$$$ He then states that the resulting covariance matrix, after carrying out the integration, is $$$$M_d=C_1-C_3(C_2+\gamma^2)^{-1}C_3^T$$$$ but I have no idea of how he manages to carry out the integration, since the term $$\xi\Gamma'\xi$$ has mixed terms in it reading $$\xi_B^T C^T \xi_A$$ and $$\xi_A^T C \xi_B$$ and I don't know how to treat them.

Any hint?

• Can't one just complete the squares for the A and B variables respectively? Commented Oct 3, 2018 at 4:07
• I am looking at the same article and I am confused with the notation. In $\Gamma '$, to me it looks like $C_1$ corresponds to the modes $A_1$ and $B_1$ mentioned in the article. Is that correct? I.e. in your notation introduced here, $C_1$ is the associated with remaining modes (which are actually across the two parties mentioned in the article)? This seems weird, since before the basis for all the expressions seem to have been fixed to $(A_1,A_2,B_1,B_2)$ while $\Gamma '$ seems to be written in the basis $(A_1,B_1,A_2,B_2)$. Commented Jan 26, 2020 at 21:46

1 Answer

As suggested by @flippiefanus, it’s a case of completing the square. (Or, alternatively, it is a well known formula for Gaussian integration, but this sounds a bit like cheating)

If I’m not mistaken, you made a typo in your first equation, which is equation (8) in the Eisert, Scheel, Plenio article. You should replace $$γ$$ by $$\frac{γ^2}2$$. We have then \begin{align} χ(ξ_A)=\frac1{π^2}∫dξ_B\exp(-\tfrac12ξ_A^TC_1ξ_A -\tfrac12ξ_B^TC_3^Tξ_A-\tfrac12ξ_A^TC_3ξ_B-\tfrac12ξ_B^T(C_2+γ^2)ξ_B)\\ =\frac{\exp(-\tfrac12ξ_A^TC_1ξ_A )}{π^2}∫dξ_B\exp( -\tfrac12ξ_B^TC_3^Tξ_A-\tfrac12ξ_A^TC_3ξ_B-\tfrac12ξ_B^T(C_2+γ^2)ξ_B). \end{align} The idea is to express the argument of the integrand as $$-\tfrac12(ξ_B+Δ)^TM(ξ_B+Δ)+\tfrac12Δ^TMΔ$$, the “$$+Δ$$” translation of $$\xi_B$$ being irrelevant in the integration. This expression becomes $$-\tfrac12ξ_B^T M Δ - \tfrac12 Δ^T M ξ_B - \tfrac12 ξ_B^T M ξ_B$$ which, when identified with the argument of the exponential, leads to \begin{align} M& =C_2+γ^2 & M\Delta&=C_3^Tξ_A & Δ^T&=ξ_A^TC_3M^{-1}\\ && && &=ξ_A^TC_3(C_2+γ^2)^{-1} \end{align}

We have then \begin{align} χ(ξ_A)&=\frac{\exp(-\tfrac12ξ_A^TC_1ξ_A )}{π^2} ∫dξ_B\exp( -\tfrac12(ξ_B+Δ)^TM(ξ_B+Δ)+\tfrac12Δ^TMΔ)\\ &=\frac{\exp(-\tfrac12ξ_A^TC_1ξ_A +\tfrac12Δ^TMΔ)}{π^2} ∫dξ_B\exp( -\tfrac12(ξ_B+Δ)^TM(ξ_B+Δ))\\ &=\frac{\exp(-\tfrac12ξ_A^TC_1ξ_A +\tfrac12Δ^TMΔ)}{π^2} \underbrace{∫dξ_B\exp( -\tfrac12ξ_B^TMξ_B)}_{\text{scalar independent of ξ_A}}\\ &∝\exp(-\tfrac12ξ_A^TC_1ξ_A +\tfrac12ξ_A^TC_3(C_2+γ^2)^{-1}C_3^Tξ_A)\\ &=\exp(-\tfrac12ξ_A^T \underbrace{\left(C_1+C_3(C_2+γ^2)^{-1}C_3^T\right)}_{M_d} ξ_A) \end{align}

• Hey Frédéric, could you explain where the square on the $\gamma$ comes from. In the article, it says that the state projected on has covariance matrix $\gamma$ but then in the Gaussian integration suddenly $\gamma^2$ turns up in the exponent. This does not make sense to me. Commented Jan 26, 2020 at 21:57