# What is the state after a projective measurement?

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 $$|v_m\rangle$$?

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.
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.
• 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. Mar 23 at 13:04
• @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. Mar 23 at 13:54