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).

  • $\begingroup$ Have you considered Euler's formula? $\endgroup$
    – Kyle Kanos
    Commented Jan 24, 2014 at 15:46
  • $\begingroup$ @KyleKanos : I am sorry, I didn't get you !! $\endgroup$
    – user35952
    Commented Jan 24, 2014 at 15:48
  • $\begingroup$ Did you click the link I provided? $\endgroup$
    – Kyle Kanos
    Commented Jan 24, 2014 at 15:50
  • $\begingroup$ @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) $\endgroup$
    – user35952
    Commented Jan 24, 2014 at 15:52
  • 2
    $\begingroup$ $(\partial/\partial \theta)^{\dagger} = -\partial/\partial\theta$ and similarly for $\phi$. $\endgroup$
    – higgsss
    Commented Jan 24, 2014 at 16:04

1 Answer 1


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 $\hat L_+$.

  • $\begingroup$ Thanks, it is the matrix-representation that is Hermitian, it makes sense !! $\endgroup$
    – user35952
    Commented Jan 25, 2014 at 3:58

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