3
$\begingroup$

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 $|v_m\rangle$?

$\endgroup$

2 Answers 2

4
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
2
  • 1
    $\begingroup$ 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. $\endgroup$ Commented Mar 23, 2022 at 13:04
  • 2
    $\begingroup$ @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
    Commented Mar 23, 2022 at 13:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.