1
$\begingroup$

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).

$\endgroup$
  • $\begingroup$ Have you considered Euler's formula? $\endgroup$ – Kyle Kanos Jan 24 '14 at 15:46
  • $\begingroup$ @KyleKanos : I am sorry, I didn't get you !! $\endgroup$ – user35952 Jan 24 '14 at 15:48
  • $\begingroup$ Did you click the link I provided? $\endgroup$ – Kyle Kanos Jan 24 '14 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 Jan 24 '14 at 15:52
  • 2
    $\begingroup$ $(\partial/\partial \theta)^{\dagger} = -\partial/\partial\theta$ and similarly for $\phi$. $\endgroup$ – higgsss Jan 24 '14 at 16:04
2
$\begingroup$

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 $\hat L_+$.

$\endgroup$
  • $\begingroup$ Thanks, it is the matrix-representation that is Hermitian, it makes sense !! $\endgroup$ – user35952 Jan 25 '14 at 3:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.