Gaussian density matricies are nice because they are fully characterised by its 2-point correlation function. Consider a free fermionic theory with creation/annihilation ops $c_i,c^{\dagger}_i$, the 2-point correlator is given by $$ C_{ij} =\langle c_i^{\dagger} c_j\rangle \equiv \text{Tr} (c_i^{\dagger}c_j \rho). $$ It admits the following representation (quadratic in fermionic ops) $$ \rho = \mathcal{K}\exp(-\sum_{ij}A_{ij}c_i^{\dagger}c_j), $$ where $\mathcal{K},A_{ij}$ are some constants.

Is the linear combination of gaussian matrices also gaussian? i.e. $$ \rho_1 + \rho_2 = \mathcal{K}\exp(-\sum_{ij}A_{ij}c_i^{\dagger}c_j) + \mathcal{Q}\exp(-\sum_{ij}B_{ij}c_i^{\dagger}c_j) =? \mathcal{W}\exp(-\sum_{ij}C_{ij}c_i^{\dagger}c_j), $$ for some suitable choice of $\mathcal{W},C_{ij}$. It might be useful to note that one can diagonalise the exponent, i.e. $$ \rho_1 = \mathcal{K}\exp(-\sum_{ij}A_{ij}c_i^{\dagger}c_j) = \mathcal{K}\exp(-\sum_{k}E_{k}d_k^{\dagger}d_k) $$

  • 1
    $\begingroup$ I'm pretty sure the corresponding statement isn't true for gaussian distributions in probability theory so would be very surprised if it were true here. $\endgroup$
    – jacob1729
    Apr 20, 2021 at 17:06

1 Answer 1



Just take two states with opposite 2-point correlators (like the all-zero and the all-one state). ("Opposite" in a suitable normalization.)

Then, their sum has correlators zero (again, zero in a suitable normalization - but in any normalization, the correlation matrix of the maximally mixed state). Yet, it is not the maximally mixed state (unless you consider a single mode).

  • $\begingroup$ I guess this is regardless of the base no? $\endgroup$ Apr 20, 2021 at 19:57
  • $\begingroup$ @FriendlyLagrangian What do you mean? $\endgroup$ Apr 20, 2021 at 21:20
  • $\begingroup$ Regardless of the basis one chooses for representing $\rho$ I meant. $\endgroup$ Apr 20, 2021 at 21:29
  • 1
    $\begingroup$ @FriendlyLagrangian I don't get it. You mean if you choose different bases for rho1 and rho2? Also, when two things are not equal in one case, it usually means they are never equal except in very special (fine-tuned) cases, as long as you consider properties which have some kind of continuity property (so basically anything in physics). So the fact that I give some counterexample essentially implies that this is almost never true. $\endgroup$ Apr 20, 2021 at 21:40
  • $\begingroup$ Thank you as always @Norbert Schuch! $\endgroup$ Apr 21, 2021 at 8:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.