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.

  • $\begingroup$ Thanks, make sense, but then I have a problem somewhere else: I'm trying to calculate the variance of the field quadratures on a coherent state, with your reasoning the coherent state is an eigenstate of the quadrature operators and therefore the variance is 0, which is not what I expect.... $\endgroup$ – user85231 Mar 11 '19 at 17:43
  • $\begingroup$ @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/… $\endgroup$ – ZeroTheHero Mar 11 '19 at 18:21
  • $\begingroup$ Oh yes! Of course! Thanks $\endgroup$ – user85231 Mar 11 '19 at 19:04

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