0
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 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 at 18:21
  • $\begingroup$ Oh yes! Of course! Thanks $\endgroup$ – user85231 Mar 11 at 19:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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