# Why is a state Gaussian if and only if its covariance matrix satisfies $\boldsymbol\sigma+i\boldsymbol\Omega\ge0$?

Let $$\rho$$ be a Gaussian state, described by the $$2N\times 2N$$ covariance matrix $$\newcommand{\bs}[1]{{\boldsymbol{#1}}}\bs\sigma$$. Denote with $$\bs\Omega$$ the $$N$$-mode symplectic form associated with the space: $$\bs\Omega\equiv \bigoplus_{k=1}^N\bs\omega, \qquad \bs\omega\equiv \begin{pmatrix}0&1\\-1&0\end{pmatrix}.$$ It is then the case that $$\bs\sigma+i\bs\Omega\ge0$$. Moreover, a matrix $$\bs\sigma$$ is the covariance matrix of some state if and only if this condition is satisfied.

This is mentioned often in this context, for example in (Weedbrook et al. 2011), eq. 17 in page 6, or in (Adesso et al. 2014), eq. (29) in page 11. In both cases, they cite Simon and collaborators (1987 and 1994) for this result. These papers seem however quite dense in content, and use a formalism I'm not familiar with.

Is there a simple way to see where this result comes from? A relevant modern review or similar resource proving this result would also be welcome.

From the definition of the covariance matrix, $${\sigma}_{ij}=\left\langle \frac{x_i x_j+ x_j x_i}{2}\right\rangle -\langle x_i\rangle\langle x_j\rangle,$$ where we define $$\boldsymbol{x}=(q_1,p_1,\cdots, q_N, p_N)^\top.$$ Given that $$2i\Omega_{jk}=[x_j,x_k],$$ we learn that $${\sigma}_{ij}=\left\langle x_i x_j\right\rangle -\langle x_i\rangle\langle x_j\rangle-i \Omega_{ij}\quad\Rightarrow \quad \sigma_{ij}+i\Omega_{ij}=\left\langle x_i x_j\right\rangle -\langle x_i\rangle\langle x_j\rangle=\left\langle\left(x_i-\left\langle x_i\right\rangle\right)\left(x_j-\left\langle x_j\right\rangle\right)\right\rangle.$$

Next, the right hand side must be greater than or equal to zero due to standard properties of covariance matrices. We have that $$\boldsymbol{\sigma}+i\boldsymbol{\Omega}=\langle (\boldsymbol{x}-\langle\boldsymbol{x}\rangle)(\boldsymbol{x}-\langle\boldsymbol{x}\rangle)^\top\rangle\equiv\langle\boldsymbol{X}\boldsymbol{X}^\top\rangle.$$ We can follow a simple proof using an arbitrary real-valued vector $$\boldsymbol{w}$$: $$\boldsymbol{w}^\top \langle\boldsymbol{X}\boldsymbol{X}^\top\rangle\boldsymbol{w}= \langle \boldsymbol{w}^\top\boldsymbol{X}\boldsymbol{X}^\top\boldsymbol{w}\rangle= \langle \left(\boldsymbol{w}^\top\boldsymbol{X}\right)^2\rangle= \langle \left|\boldsymbol{w}^\top\boldsymbol{X}\right|^2\rangle\geq 0.$$ In the final inequality we used that each of the $$x_i$$ is Hermitian. Any matrix $$\boldsymbol{M}$$ that satisfies $$\boldsymbol{w}^\top\boldsymbol{M}\boldsymbol{w}\geq 0$$ for all real vectors $$\boldsymbol{w}$$ is by definition positive semidefinite. Thus $$\boldsymbol{\sigma}+i\boldsymbol{\Omega}\geq \boldsymbol{0}\boldsymbol{0}^\top.$$

This relation essentially translates between the symmetrized version of the covariance $$\boldsymbol{\sigma}$$ to the standard non-symmetrized version of the covariance matrix $$\langle\boldsymbol{X}\boldsymbol{X}^\top\rangle$$.

To answer the question in the title, all states have their covariance matrices satisfy $$\boldsymbol{\sigma}+i\boldsymbol{\Omega}\geq 0$$. This is not only a property of Gaussian states! The property of Gaussian states is that they are completely characterized by their covariance matrices (and their mean value vectors $$\langle \boldsymbol{x}\rangle$$).

So maybe you want to ask why one only needs to specify the covariance matrix if one wants to completely describe a Gaussian state? The answer is given by Isserlis's theorem (sometimes more familiar in terms of Wick's theorem). To completely describe any state, one could list all of its moments $$\langle x_i x_j\cdots\rangle$$. Given knowledge of the mean values $$\langle\boldsymbol{x}\rangle$$, we can instead describe a state by all of its zero-mean moments $$\langle X_i X_j\cdots\rangle$$. The Isserlis theorem says that, for Gaussian probability distributions, all of the higher-order moments can be found from sums of products of all of the different pairings of the the second-order moments (ie, elements of the covariance matrix), such as $$$$\langle X_i X_j X_k X_l\rangle = \langle X_i X_j\rangle \langle X_k X_l\rangle+\langle X_i X_k\rangle \langle X_l X_l\rangle + \langle X_i X_l\rangle \langle X_k X_j\rangle .$$$$ Gaussian state $$\Rightarrow$$ Gaussian probability distribution $$\Rightarrow$$ all moments characterized by second-order moments $$\Rightarrow$$ state completely characterized by covariance matrix. In general, the Isserlis theorem applies to distributions with $$\langle X_i X_j\rangle=\langle X_j X_i\rangle$$, so the symmetrized version of the covariance matrix $$\boldsymbol{\sigma}$$ is necessary here.

• thanks, this makes it quite clear
– glS
Jun 14 at 19:03
• You're welcome - I initially misunderstood the thrust of the question Jun 14 at 19:04
• eh, I think the answer was on point also before the addition, I just didn't find the time to read it properly. Indeed, you are right that I misunderstood the statement, as this relation defines the covariance matrix of any state. Still, the addendum is interesting information
– glS
Jun 14 at 19:12