# What operator lowers the total angular momentum?

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?

• The lowering operator is just the adjoint operator (hermitian conjugate for a matrix operator) of the raising operator. – Paul Childs Dec 11 '19 at 2:45
• You ask how to define an operator and express it in the $J_z$ basis... but you literally already did that in the question. Writing $L_- |j, m \rangle = |j-1, m \rangle$ is a definition of an operator that lowers $j$ in the basis you wanted (though it isn't the usual definition of "$L_-$"). What more do you want? – knzhou Dec 11 '19 at 2:54
• The Rungeâ€“Lenz vector has components failing to commute with the Casimir, so taking you out of a fixed j multiplet. You must first calculate the freak operator with eigenvalues j in terms of the Casimir whose eigenvalues are j(j+1), and then construct a lowering operator for it. Possibly out of LRL pieces. Context? – Cosmas Zachos Dec 11 '19 at 11:59
• @CosmasZachos so, we need a bigger degenerate state space? but we should be able to do that with physical meanings consider all the states were there already. – ShoutOutAndCalculate Dec 11 '19 at 14:35
• Consider Burkhardt & Leventhal 2004. – Cosmas Zachos Dec 12 '19 at 20:35

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.
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 Laporteâ€™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 \alphaâ€™\ellâ€™mâ€™\vert z\vert \alpha \ell m\rangle = C^{\ellâ€™mâ€™}_{\ell m; 10}\frac{\langle \alphaâ€™\ellâ€™\Vert r\Vert \alpha \ell\rangle}{\sqrt{2\ellâ€™+1}}\delta_{mâ€™m}\delta_{\ellâ€™,\ell\pm 1} \end{align}
by the Wigner-Eckart theorem, with $$C^{\ellâ€™mâ€™}_{\ell m;10}$$ 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,mâ€™}_{\ell m;1k} \hat V_k\vert\alpha \ell m\rangle \tag{1} \end{align} will have non-zero matrix element with only $$\vert\alphaâ€™ \ell-1,mâ€™\rangle$$. This can be seen by invoking again the Wigner-Eckart theorem: \begin{align} \langle \alphaâ€™\ellâ€™mâ€™\vert\sum_{km} C^{\ell-1,mâ€™}_{\ell m;1k} \hat V_k\vert\alpha \ell m\rangle&= \sum_{km}. \sum_{km} C^{\ell-1,mâ€™}_{\ell m;1k} \langle\alphaâ€™ \ellâ€™mâ€™\vert \hat V_k\vert\alpha \ell m\rangle\, ,\\ &=\sum_{km} C^{\ell-1,mâ€™}_{\ell m;1k} C^{\ellâ€™ mâ€™}_{\ell m;1k} \frac{\langle\alphaâ€™ \ellâ€™ \Vert \hat V_k\Vert\alpha \ell \rangle}{\sqrt{2\ellâ€™+1}}\, ,\\ &=\frac{\langle\alphaâ€™ \ellâ€™\Vert \hat V_k\Vert\alpha \ell \rangle}{\sqrt{2\ellâ€™+1}}\delta_{\ellâ€™,\ell-1} \end{align} by orthogonality of the CGs. (1) is not a single operator but it does contain a projector to the correct $$\ellâ€™$$ subspace, and you can freely choose the $$mâ€™$$ value of your target state.