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The Hamiltonian of a system is $H=\frac{L^2_z}{2I}+gBL_z $. The initial state is $\Psi(0)=A\sin^2\phi $. I want to find the expectation value of $H$ for $t=0.$

I think that I should express $\Psi(0)$ in terms of the $Y_l^m$ but I cant since there is no $Y_l^m$ depending only on $\phi$. Is it that $\theta = \frac{\pi}{2} $ or should I act with $L_z=-i\hbar \frac{\partial }{\partial \phi }$ ?

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closed as off-topic by ACuriousMind, Martin, yuggib, LDC3, Kyle Kanos Jun 10 '15 at 14:33

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You need to calculate $<\psi|\hat H|\psi>$. First, I would calculate $\hat H|\psi>$:

$$\hat H\psi = \left(\frac{L^2_z}{2I}+gBL_z\right)\psi = -\frac{\hbar^2}{2I}\frac{\partial^2 \psi}{\partial \phi^2}-i\hbar gB \frac{\partial \psi}{\partial \phi}$$

$$=-\frac{2A\hbar^2}{2I}\cos(2\phi)-2Ai\hbar gB \sin\phi\cos\phi$$.

Then you'll need to multiply this by the conjugate of $\psi$ and integrate over $\phi$ from $0$ to $2\pi$... can you continue?

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You could decompose your state into spherical harmonics, but I think that would be unnecessarily difficult. I would do what you mentioned, and that is to write $L_z$ in the coordinate representation (cylindrical coordinates), $$L_z=-i\hbar\frac{\partial}{\partial \phi}$$

Then, you should calculate the expectation value: $$\langle H\rangle=\int_0^{2\pi}d\phi A^*\sin^2\phi(-i\hbar)\frac{\partial}{\partial \phi}A\sin^2 \phi$$

You might also want to normalize your wave function to determine $A$.

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