# Variance of a hermitian operator

Take an hermitian operator $$O$$ such that $$O|\psi\rangle = x|\psi\rangle$$. The variance of an operator $$O$$ is defined as $$(\Delta O)^2 = \langle{O^2}\rangle - \langle{O}\rangle^2.$$ Let's consider the first term, I would write it as

$$\langle{O^2}\rangle = \langle\psi|O O|\psi\rangle = x\langle\psi|O|\psi\rangle = x^2\langle\psi|\psi\rangle = x^2.$$ But then for the second term I get the same result $$\langle{O}\rangle^2 = \langle\psi| O|\psi\rangle^2 = (x\langle\psi|\psi\rangle)^2 = x^2.$$ Therefore I end up with $$(\Delta O)^2 = 0$$ which is obviously wrong. What's the problem here?

The only idea I have is that $$O^2|\psi\rangle \neq O(O|\psi\rangle)$$, but i cannot understand why... What am I doing wrong?

As you have written things, the variance is indeed $$0$$ because $$\vert\psi\rangle$$ is an eigenstate of $$O$$: thankfully this is so as it means the outcome with eigenvalue $$x$$ is not uncertain and we can use the eigenvalue $$x$$ to label the state.
• @user85231 The coherent state is an eigenstate of $\hat a$, which is not hermitian. The variance is not $0$ in this case. See physics.stackexchange.com/questions/158849/… – ZeroTheHero Mar 11 '19 at 18:21