# Are all states with a diagonal covariance matrix thermal?

Consider a single-mode bosonic field, with Hamiltonian $$H=\omega a^\dagger a$$. I know that the covariance matrix of the thermal state is diagonal. But is the converse also true?

That is, I have a state $$\rho$$ of the field with the following requirements:

1) The state is Gaussian. That is, the characteristic function is Gaussian.

2) The covariance matrix of the state is diagonal.

I suspect the only state that satisfies these requirements is the thermal state $$\rho_{th}(T)\propto \sum_n e^{-\omega n/(k_B T)}|n\rangle\langle n|$$. Is that true?

I'm also interested in the situation of a multimode field. $$H=\sum_k \omega_k a^\dagger_k a_k$$. Then, if I have the same requirements as above, I suspect the only state that satisfies them is a product state of thermal states, where each mode may have its own temperature, $$\rho=\bigotimes_k \rho_{th}(T_k)$$. Is this true?

A squeezed state, squeezed in $$p$$ or $$q$$ has a diagonal covariance matrix $$\Gamma$$ (se below) and is not thermal. \begin{align}\Gamma&=\begin{bmatrix} S & 0 \\ 0 & \frac{1}{S}\end{bmatrix} & \text{for S>0 and s\neq1} \end{align}.
For a single-mode thermal sate, we have $$\Gamma_{\text{th}}=\gamma\mathbb I=\begin{bmatrix}\gamma&0\\0&\gamma\end{bmatrix}$$, with $$\gamma\ge1$$ depending on the temperature. In the multimode case, with a mode-dependent temperature, the covariance matrix then looks like $$\begin{bmatrix} \gamma_1&0\\ 0&\gamma_1\\ & & \gamma_2&0\\ & & 0&\gamma_2 \\ & & & & \ddots \end{bmatrix},$$ with repeated diagonal terms in the same mode. Hence it is not the most generic diagonal covariance matrix.