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Given an observable $M = \sum_m \lambda_m P_m$ and assuming that $P_m = |v_m\rangle \langle v_m|$, the state after measurement after getting result $\lambda_m$ is given as $$ \frac{P_m |\psi\rangle}{||P_m|\psi\rangle||} = \frac{|v_m\rangle\langle v_m|\psi\rangle}{|| |v_m\rangle\langle v_m|\psi\rangle||} = \frac{\langle v_m|\psi\rangle}{|\langle v_m|\psi\rangle|} |v_m\rangle$$

Can we safely assume that $\langle v_m|\psi\rangle = |\langle v_m|\psi\rangle|$, such that the state after measurement is simply $|v_m\rangle$?

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The way I'd put it is that the state after the measurement is just $|v_m\rangle$, regardless of what's the phase of $\langle v_m|\psi\rangle$. This is because states are defined up to a global phase, so the vector $\frac{\langle v_m|\psi\rangle}{|\langle v_m|\psi\rangle|} |v_m\rangle$ and the vector $|v_m\rangle$ describe the same identical physical state.

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Technically, there could be a phase difference between $\langle v_m|\psi\rangle$ and $|\langle v_m|\psi\rangle|$. However, this is just a global phase that you can neglect. So, the net effect is the same as what you want to achieve, although the reasoning is a little different.

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    $\begingroup$ Global phases again! Maybe one can argue that the global phase of $|v_m\rangle$ is not specified by the projector $P_m$ and thus one can always set of the global phase to make $\langle v_m|\psi\rangle$ real and positive. $\endgroup$ Mar 23 at 13:04
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    $\begingroup$ @QuantumMechanic Yes, I guess that depends on whether you're given $P_m$ or $|v_m\rangle$. But this is a valid argument either way. $\endgroup$
    – DaftWullie
    Mar 23 at 13:54

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