# Ladder operators applied to magnetic quantum number [duplicate]

I cannot solve this simple equation, since I cannot find nowhere how ladder operators works on the magnetic quantum number:

$$\langle m|\hat a^\dagger \hat a+\hat a\hat a^\dagger)|m\rangle$$

• What are $a$ and $a^\dagger$? If you mean creation and destruction, they don't do anything to $m$ but to energy quanta. Sep 17, 2018 at 11:20
• Possible duplicate of How To Use Ladder Operators? Sep 18, 2018 at 10:17

## 1 Answer

Hey and welcome to Physics SE! Which exactly are these ladder operators $\hat a$ and $\hat a^{\dagger}$? Do you mean the usual ladder operators $\hat a=\hat J_+=\hat J_1+i\hat J_2$ and $\hat a^{\dagger}=\hat J_-=\hat J_1-i\hat J_2$, that act on the joint $\hat J^2$- $\hat J_3$ eigenstates? If yes, then it is a standard procedure to show that: $$\hat J_+|j,m\rangle=\hbar\sqrt{j(j+1)-m(m+1)}|j,m+1\rangle$$ $$\hat J_-|j,m\rangle=\hbar\sqrt{j(j+1)-m(m-1)}|j,m-1\rangle$$

If, in the contrary, you refer to the ladder operators of the q.h.o. acting on Hamiltonian eigenstates, i.e. $\hat a=\sqrt{\frac{m\omega}{2\hbar}}\left(\hat x+\frac{i}{m\omega}\hat p\right)$ and $\hat a^{\dagger}=\sqrt{\frac{m\omega}{2\hbar}}\left(\hat x-\frac{i}{m\omega}\hat p\right)$, then it is again a trivial exercise to show that: $$\hat a|m\rangle=\sqrt{m}|m-1\rangle$$ $$\hat a^{\dagger}|m\rangle=\sqrt{m+1}|m+1\rangle$$

• Hello GK! Thanks a lot for your answer :) The exercise was written by our tutor and we do not have clear details about it. We suppose that the a and a* operators are the creation and annhilation of the harmonic oscillator. Does it make sense? Sep 17, 2018 at 14:20
• That doesn't make sense to me.. The q.h.o. ladder operators act on energy eigenstates and have nothing to do with angular momentum. However, given your notation with the "sandwich" between $\langle m|$ and $|m\rangle$, I suppose that your tutor suggested the ang. momentum ladder operators that I wrote you at my answer.
– Ozz
Sep 17, 2018 at 14:37
• @ValentinaTi If your tutor didn't give any details, then why do you think that $|m\rangle$ denotes an angular momentum state? It's simpler to assume that it's just an eigenstate of the harmonic oscillator Hamiltonian with eigenvalue $(m+1/2)\hbar\omega$. Sep 17, 2018 at 15:06
• @MarkMitchison Well, I didn't think about that probability. Thanks! I edited my answer a little bit.
– Ozz
Sep 17, 2018 at 15:13