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If I want to know the expectation value of an operator O in the quantum trajectory formalism, I average over $N$ trajectories, where I call one such trajectory $\Psi_n$:

\begin{equation} \langle O \rangle = \frac{1}{N}\sum_{n=1}^N \langle \Psi_n | O |\Psi_n \rangle, \end{equation} correct?

If so, my question is: Is that still correct if the different $\Psi_n$ are not orthogonal/parallel to each other?

For example: If I am interested in the average probabiltiy density of the position distribution $\rho$, do I get it by calculating:

\begin{equation} \rho(x) = \frac{1}{N}\sum_{n=1}^N \Psi^*_n(x) \Psi_n(x), \end{equation}

even if $\langle \Psi_n(x)|\Psi_m(x) \rangle \neq \delta_{n,m} $?

If not, what would be the correct formula?

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  • $\begingroup$ yes, i believe what i wrote should be correct, but i want to be sure $\endgroup$ – Luke Oct 11 '20 at 18:55
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Yes, it is still correct. It is even the common case.

Provided you have sampled a sufficient amount of trajectories to have reached statistical significance, of course. This is the thing, you are sampling the density matrix randomly, en each of these samples (the trajectories) can contain contributions to all possible eigenvectors simultaneously, in a complete basis of choice.

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