# Raising and lowering operators of orbital angular momentum

For the orbital angular momentum, the raising and lowering operators are given by,

$$L_+ = e^{i\phi} \bigg(\frac{\partial}{\partial\theta} + i\: cot\theta\frac{\partial}{\partial\phi}\bigg)$$ $$L_- = -e^{-i\phi} \bigg(\frac{\partial}{\partial\theta} - i\: cot\theta\frac{\partial}{\partial\phi}\bigg)$$

With this I obtain $$L_+^\dagger = - L_-$$ But with the actual definition in terms of $L_x$ and $L_y$ with $$L_+ = L_x + i L_y$$ $$L_- = L_x - i L_y$$ $$L_+^\dagger = L_-$$

How do I reconcile between these two results ? Or is there any mistake I have committed ?

PS : My professor hinted saying it had something to do with compactness of angular momentum, but I didn't understand !! (My problem is not in obtaining the result of $L_+$ and $L_-$, but in reconciling these two facts).

• Have you considered Euler's formula? – Kyle Kanos Jan 24 '14 at 15:46
• @KyleKanos : I am sorry, I didn't get you !! – user35952 Jan 24 '14 at 15:48
• Did you click the link I provided? – Kyle Kanos Jan 24 '14 at 15:50
• @KyleKanos : Indeed, and I know Euler's formula !! My question is not about obtaining the results for the ladder operators(which I did obtain correctly), but to reconcile between these two facts (that seems contradicting) – user35952 Jan 24 '14 at 15:52
• $(\partial/\partial \theta)^{\dagger} = -\partial/\partial\theta$ and similarly for $\phi$. – higgsss Jan 24 '14 at 16:04

This is the same problem than when one is trying to show that the momentum operator $\hat P$ is hermitian in the position basis $|x\rangle$, where $P=-i\partial_x$. This is because the derivative operators are non-diagonal in the basis used (same thing for the angular momentum operators, that are built from the momentum operator).
Naively, one gets $\hat P^\dagger="i\partial_x$ which seems to be non-hermitian. It's because one is looking at matrix elements, and not the operator itself. The proper way to do that is to look at the matrix element $\langle\psi|\hat P|\phi\rangle$ and show that $\langle\phi|\hat P^\dagger|\psi\rangle=(\langle\psi|\hat P|\phi\rangle)^*$.
By using the same trick to compute $\hat L_-^\dagger$, one can show that it is indeed equal to $\hat L_+$.