According to e.g. Serafini (Quantum Continuous Variables), the Hilbert-Schmidt product ('overlap') of two multimode Gaussian states $\rho_1,\rho_2$ is


where $r_{1,2}$ are the displacements in phase space and $\sigma_1,\sigma_2$ the covariance matrices.

My question: is there a similar formula for $\langle\psi_1|\psi_2\rangle$ itself, retaining information on their relative phase?


There cannot be such an expression because the covariance matrix and displacement don't contain the relative phase information. This is easy to see, since they are computed from the reduced density matrix (which does not depend on the phase of the state).

A way around this can be to include a third reference state and $\vert\chi\rangle$ and consider the overlap $$ \langle \psi_1|\psi_2\rangle\langle\psi_2\vert\chi\rangle\langle\chi\vert\psi_1\rangle $$ which only depends on the density matrices, and is thus a function of the covariance matrices & displacement.

Note, however, that also this quantity is (necessarily!) invariant under changing phases, so whether it gives you the desired information depends on the scenario you consider.

  • $\begingroup$ Thanks, but I'm a bit confused because the density matrix should contain all relevant information and also for the total one $\rho=\frac{1}{N}\sum_i |\psi_i\rangle\langle\psi_i|$? What is confusing me more is now the difference between states $|-\alpha>$ and $exp(i\pi)*|\alpha>$, which should both be $|\alpha>$ rotated over $\pi$, but for the first one the density matrix changes as well, and for the second phasefactor it doesn't. $\endgroup$ – Wouter Sep 10 '19 at 7:58
  • $\begingroup$ What I'm after in the end is calculating the total von Neumann entropy, as you have given me already the very useful answer in physics.stackexchange.com/q/498790/25292 . Now, if the DMs of the individual Gaussians, and hence the covariance matrices and displacement, loose the phase information; I would expect that if I'm working with a stochastic unraveling (quantum trajectories), there should be phase symmetry in the ensemble steady state and hence I can in retrospect assign a phase to each Gaussian from a uniform $[0,2\pi]$ distribution? $\endgroup$ – Wouter Sep 10 '19 at 8:04
  • $\begingroup$ Well, in fact the Hamiltonian explicitly breaks the $U(1)$ symmetry in each of the modes, so I know this affects the phase that is reflected in DM and displacement (my $|-\alpha>$ example), but I've assumed now this doesn't affect these additional phasefactors? $\endgroup$ – Wouter Sep 10 '19 at 9:48
  • $\begingroup$ Ok, I think now that this phasefactor should be irrelevant for all practical purposes, see e.g. the situation on the link above $\endgroup$ – Wouter Sep 10 '19 at 13:40
  • $\begingroup$ You can compute the von Neumann entropy directly from the covariance matrix, no need for all this. Why don't you post a question how to compute the vN entropy from the CM? $\endgroup$ – Norbert Schuch Sep 10 '19 at 14:44

If the Gaussians are not too far from coherent states, a useful approximation is obtained by calculating the phasefactors of a coherent state, for which analytical formulas do exist (this comes down to taking into accound the displacement but not the squeezing/covariance matrix). According to e.g. the book Quantum Noise (Gardiner-Zoller), the scalar product of two single-mode coherent states is


from which one immediately gets

$$\frac{\langle\alpha|\beta\rangle}{|\langle\alpha|\beta\rangle|}=\exp(\frac{\alpha^*\beta-\beta^*\alpha}{2})$$ for the phasefactor.

As a coherent state factorizes over all modes, the overall phasefactor under this approximation is the product of the phasefactors over these individual modes.


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