# Proof of inability to obtain global phase

I'm curious if there's a quick proof of the inability to obtain global phase from a quantum state, since they're supposedly indistinguisable. I suppose to measure this, you would need a Hermitian operator whose eigenvalues are some function of $\theta$.

So if you have $e^{i\theta}|\psi\rangle$, then you need some $\hat{A}$ such that $\hat{A}(e^{i\theta}|\psi\rangle) = f(\theta)(e^{i\theta}|\psi\rangle)$. Right? And the goal is to prove that $\hat{A}$ can't exist. And... I don't even know how to begin with this. (This isn't HW; I'm just curious.)

• By linearity, if $|\psi\rangle$ is an eigenvector of some operator, so is $e^{i\theta}|\psi\rangle$, with the same eigenvalue. – Robin Ekman Oct 24 '14 at 22:14
• Oh... I'm always missing these really obvious things – Nick Oct 24 '14 at 22:16

Let $$|\psi'\rangle = e^{i\theta}|\psi\rangle.$$ The probability to be in the $n$:th eigenstate of the observable $\hat O$ is given by the expectation value of $P_n$ where $P_n$ is the projector onto the $n$:th eigenstate. Since the expectation value in the state $|\psi'\rangle$ is $$\langle \psi'|P_n|\psi'\rangle = \langle \psi|e^{-i\theta} P_n e^{i\theta} |\psi\rangle = \langle\psi|P_n|\psi\rangle$$ it is not possible to distinguish between $|\psi\rangle$ and $e^{i\theta}|\psi\rangle$ with observable quantities.