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I have some eigenstate $|\psi\rangle$, a smooth function $f$ and an observable $\hat{A}$. I want to compute the expected value of $f(\hat{A})$: $$ \langle \psi|f(\hat{A})|\psi\rangle. $$

I would do that by expanding $f(\hat{A})$ in a Taylor series. However my situation is quite simple and there might be a somewhat easier way of doing it. I have eigenstates of the form $|\ell, m\rangle$ such that $\langle \theta,\phi|\ell,m\rangle=Y_\ell^m(\theta,\phi)$. The observable in my case is $\hat{A} = \hat{y}/\hat{r} = \sin(\hat{\theta})\sin(\hat{\phi})$. I'm having trouble here. For instance, if $\hat{A}=\hat{\theta}$ then I would do:

$$ \langle \psi|\hat{\theta}|\psi\rangle = \langle \psi|\int_0^\pi \theta|\theta\rangle\langle\theta||\psi\rangle. $$

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    $\begingroup$ If $\psi$ is an eigenstate of $\hat A$ then $\langle\psi|f(\hat A)|\psi\rangle = f(\langle\psi|\hat A|\psi\rangle)$. $\endgroup$ – Phoenix87 Nov 13 '16 at 21:40
  • $\begingroup$ @Phoenix87 thanks, in my case how would I "separate" the function sine (applied to the 2 angular operators) from the angles? $\endgroup$ – Vladimir Vargas Nov 13 '16 at 21:43
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    $\begingroup$ I'm not sure I understand your question. Are you asking how to compute the average of $\sin(\theta)\sin(\phi)$ over the state $\psi$ in order to compute $\langle\psi|\hat A|\psi\rangle$? $\endgroup$ – Phoenix87 Nov 13 '16 at 21:46
  • $\begingroup$ @Phoenix87 exactly. specifically if $|\psi\rangle = |\ell, m\rangle$ $\endgroup$ – Vladimir Vargas Nov 13 '16 at 21:50
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    $\begingroup$ @VladimirVargas $\int d\Omega \, Y^{-m}_\ell (\sin \theta \, \sin \phi) Y^m_\ell = 0.$ $\endgroup$ – JamalS Nov 13 '16 at 23:11
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If the state $|\psi \rangle = |\ell, m\rangle$, where $\langle \theta,\phi | \ell, m\rangle = Y^m_\ell(\theta,\phi)$ are the normalised spherical harmonics, then the expectation value of an operator $\mathcal O$ is given by,

$$\langle \mathcal O \rangle = \int d\Omega \, Y^{-m}_\ell (\theta,\phi) \, \mathcal{O} \, Y^m_\ell(\theta,\phi)$$

over the unit sphere, $d\Omega = \sin \theta \, d\theta d\phi$. In your case, applying the definition of the harmonics,

$$= \frac{(2\ell+1)}{4\pi}\int \, d\Omega \, P^{-m}_\ell(\cos \theta) \, (\sin \theta \sin \phi) \, P^m_\ell(\cos \theta) = 0$$ since luckily, the easier integration over $\phi$ is simply, $$\int_0^{2\pi} d\phi \, \sin \phi = 0,$$

and there is no need to do the daunting integration over $\theta$. (Though, if you did need to do the integration over $\theta$, since Mathematica was unable to produce a closed-form answer, off the top of my head I would either replace the Legendre polynomials by a series representation, or maybe write them in terms of hypergeometric functions and make use of some of their identities.)

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