# Post-measurement state after homodyne measurements of part of the system

Let us have an $$N$$ mode Gaussian state and B denote the last mode (for simplicity I just consider a two-modes state) It is said in articles (and for example How to find the covariance matrix after a partial homodyne measurement?) that partial measurement of a quadrature (say the $$q$$ one) on the mode $$B$$ will affect the covariance matrix of the mode(s) A.

However I find this contradicting, here is my argument : $$q_A$$ and $$q_B$$ are commutating operators because they operate on different spaces of the tensor product of Hilbert spaces, so according to (https://quantumcomputing.stackexchange.com/q/27912/), a measurement of $$q_A$$ gives the same probabilities as the process of measuring $$q_B$$, discarding the result, then measuring $$q_A$$ therefore the variance of $$q_A$$ should be the same in both cases, which is not the case here, for example in this particular matrix of a two-mode state just for example, the variance switched from a to a-c^2/b:

What am I missing? N.B the example comes from section 7.2.1 of https://arxiv.org/abs/1703.09278

• My guess: This is the CM after the measurement, given that you know the measurement outcome (i.e. don't discard it). -- Same works for qubits: If you measure e.g. half of a maximally entangled state, you know the precise state of the other qubit. Commented Aug 24, 2022 at 13:21
• That would make a lot of sense, without contradicting my linked question, because it's only the conditional variance that changed, not the "whole" variance. Commented Aug 24, 2022 at 14:21
• Exactly, that's what the Schur complement formula (84) shoud describe. Note that this is usually what is meant when people talk about a "post-measurement state". Finally, note that for all post-measurement states, the correlations are the same, but the displacement is different, and depends on the measurement outcome. (Averaging over the displacement gives a correlation matrix with larger fluctuations.) Commented Aug 24, 2022 at 14:52

Thus, if you ignore the measurement outcome (that is, average over it), you obtain another CM, which is the same as you would have obtained by tracing, i.e., $$\Sigma_A$$ (which is indeed an upper bound on the CM in (84)).