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).
Best Answer
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 $\hat L_+$.