What operator lowers the total angular momentum?

Question

Assume states $|j,m\rangle$, say $j\in\{3,2,1,0\}$, initially at $3$.

Is there any "lowering" operator I could apply such that $L_-|j,m\rangle = |j-1,m\rangle $?

How to express it in the $J_z$ basis?

Answer 1

First you need to define the operator whose eigenvalues are j. As your avatar invites me to, I skip the fine formal fussing and just calculate instead. Such operators are routine, e.g., Curtright and Zachos (1990) PhysLettB243. For the Casimir $\vec J \cdot \vec J$, with eigenvalues j(j+1), define $$ L_0\equiv \frac{\sqrt{1+4\vec J \cdot \vec J}-1}{2}, $$ with eigenvalues j, $$ L_0| j,m\rangle = j| j,m\rangle . $$

Your target operator $L_-$ should distinctly fail to commute with the Casimir, and hence $L_0$, so as to take you from a fixed j to a lower one, just as the LRL vectors do in the Hydrogen atom! That is $$ [L_-,L_0]= L_- \\ [L_-,J_z]=0, $$ so that $$ L_0(L_- |j,m\rangle)= (j-1)(L_- |j,m\rangle),\\ J_z (L_- |j,m\rangle) =m(L_- |j,m\rangle), $$ as per your posit.

You might construct such operators, with pain, out of pieces of LRL invariants, such as $A_z$, but without context this readily slips into the slough of mootness.

Answer 2

In principle, any vector operator, i.e. any tensor operator with $\ell=1$ will shift you $\Delta \ell=0,\pm 1$. Moreover, if your operator is parity-odd, then by Laporte’s rule or by simple properties of the CG coefficient $\Delta l=0$ is excluded. Thus, for instance, the operator $\hat z$ has the property \begin{align} \langle \alpha’\ell’m’\vert z\vert \alpha \ell m\rangle = C^{\ell’m’}_{\ell m; 10}\frac{\langle \alpha’\ell’\Vert r\Vert \alpha \ell\rangle}{\sqrt{2\ell’+1}}\delta_{m’m}\delta_{\ell’,\ell\pm 1} \end{align}
by the Wigner-Eckart theorem, with $C^{\ell’m’}_{\ell m;10}$ a Clebsch-Gordan coefficient.

This is not quite what you want since it will shift up or shift down the value of $\ell$. However, with \begin{align} \hat V_\pm=\mp \frac{1}{\sqrt{2}}(\hat x\pm i\hat y)\, ,\qquad \hat V_0=\hat z \end{align} then the sum \begin{align} \sum_{km} C^{\ell-1,m’}_{\ell m;1k} \hat V_k\vert\alpha \ell m\rangle \tag{1} \end{align} will have non-zero matrix element with only $\vert\alpha’ \ell-1,m’\rangle$. This can be seen by invoking again the Wigner-Eckart theorem: \begin{align} \langle \alpha’\ell’m’\vert\sum_{km} C^{\ell-1,m’}_{\ell m;1k} \hat V_k\vert\alpha \ell m\rangle&= \sum_{km}. \sum_{km} C^{\ell-1,m’}_{\ell m;1k} \langle\alpha’ \ell’m’\vert \hat V_k\vert\alpha \ell m\rangle\, ,\\ &=\sum_{km} C^{\ell-1,m’}_{\ell m;1k} C^{\ell’ m’}_{\ell m;1k} \frac{\langle\alpha’ \ell’ \Vert \hat V_k\Vert\alpha \ell \rangle}{\sqrt{2\ell’+1}}\, ,\\ &=\frac{\langle\alpha’ \ell’\Vert \hat V_k\Vert\alpha \ell \rangle}{\sqrt{2\ell’+1}}\delta_{\ell’,\ell-1} \end{align} by orthogonality of the CGs. (1) is not a single operator but it does contain a projector to the correct $\ell’$ subspace, and you can freely choose the $m’$ value of your target state.