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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:

enter image description here

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

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    $\begingroup$ 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. $\endgroup$
    – Norbert Schuch
    Commented Aug 24, 2022 at 13:21
  • $\begingroup$ 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. $\endgroup$
    – eternalstudent
    Commented Aug 24, 2022 at 14:21
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    $\begingroup$ 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.) $\endgroup$
    – Norbert Schuch
    Commented Aug 24, 2022 at 14:52

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The formula (84) describes the covariance matrix (CM) of the state after a homodyne measurement of part of the system, given you know (and use) the measurement outcome.

More precisely, after measurement, the state of the unmeasured part of the system has always the same CM (the one of (84)), but displaced by a displacement which is determined by the measurement outcome.

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)).

On the other hand, if you use the measurement outcome, you can correct for the displacement, and obtain a smaller uncertainty.

Note that this is not surprising: If you measure a maximally (or, for bosons, highly) entangled state, you will know the state of A to arbitrary precision: It will be equal/opposite to the state of the measured system.

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