# Ladder Operators for Quantum Angular Momentum

Consider the operators

$$\begin{equation*} \mathbf{J}^{2} = J_{x}J_{x}+J_{y}J_{y}+J_{z}J_{z} \end{equation*}$$

where the $$J_{i}$$ are the generators of infinitesimal rotations. Choosing $$\mathbf{J}^{2}$$ to be simultaneously diagonalizable with $$J_{z}$$, we have

$$\begin{equation*} \mathbf{J}^{2}|a,b\rangle = a |a,b\rangle \end{equation*}$$

and

$$\begin{equation*} J_{z}|a,b\rangle = b |a,b\rangle. \end{equation*}$$

Now define as usual the ladder operators

$$\begin{equation*} J_{\pm} = J_{x} \pm iJ_{y}. \end{equation*}$$

Noting that

$$\begin{equation*} \mathbf{J}^{2}-J_{z}^{2} = \frac{1}{2}\left(J_{+}J_{+}^{\dagger}+J_{+}^{\dagger}J_{+}\right), \end{equation*}$$

we can see that

$$\begin{equation*} \langle a,b|(\mathbf{J}^{2}-J_{z}^{2})|a,b\rangle \geq 0, \end{equation*}$$

and so

$$\begin{equation*} a\geq b^{2}. \end{equation*}$$

In the text I'm using (the 2$$^{\mathrm{nd}}$$ edition of Modern Quantum Mechanics by Sakurai and Napolitano), the authors claim that this implies there exists $$b_{\mathrm{max}}$$ such that

$$\begin{equation*} J_{+}|a,b_{\mathrm{max}}\rangle = 0. \end{equation*}$$

It's clear to me that there must exist a $$b_{\mathrm{max}}$$ such $$b$$ can't exceed $$b_{\mathrm{max}}$$, but why is it the case that the $$J_{+}$$ eigenket corresponding to this $$b_{\mathrm{max}}$$ is the null ket?

• Hint: if $J_+|v\rangle=\lambda|v\rangle$, find with proof a value of $\lambda'$ such that $J_+J_z|v\rangle=\lambda'J_z|v\rangle$, making $J_z|v\rangle$ an eigenket of $J_+$ or else the zero ket. By the way, the minimum of $a-b^2$ is non-zero because $a=b_\max(b_\max+1)$.
– J.G.
Apr 27, 2023 at 10:36
• @J.G. please see the answer I posted to my question--I believe it's correct, but I would be grateful for your input! May 3, 2023 at 5:40
• It is actually not obvious. The raising past the top state could well have resulted not in a zero, but rather a non-normalisable entity, is also possible. It just so happens that we get a sensible theory if it is zero, and we do not yet know how to get a sensible theory if, say, it is infinity. May 3, 2023 at 6:53
• @naturallyInconsistent if we arbitrarily impose the requirement that we eventually obtain a null ket rather than some other non-normalizable ket, is my reasoning correct? May 3, 2023 at 6:54
• Well, the standard reasoning is that we insist upon getting the null and thereby build a sensible theory, so that part is likely to be correct. May 3, 2023 at 6:57

Suppose $$|a,b\rangle$$ is such that

$$\begin{equation*} \mathbf{J}^{2}|a,b\rangle = a|a,b\rangle \hspace{1pc}\mbox{ and }\hspace{1pc} J_{z}|a,b\rangle = b|a,b\rangle. \end{equation*}$$

Because $$\mathbf{J}^{2}-J_{z}^{2}$$ is positive semi-definite, we must have $$a\geq b^{2}$$. Application of $$J_{+}$$ to $$|a,b\rangle$$ yields another eigenket, $$|a,b+\hbar\rangle$$ of $$\mathbf{J}^{2}$$ and $$J_{z}$$ such that

$$\begin{equation*} \mathbf{J}^{2}|a,b+\hbar\rangle = \mathbf{J}^{2}J_{+}|a,b\rangle = a|a,b+\hbar\rangle \end{equation*}$$

and

$$\begin{equation*} J_{z}|a,b+\hbar\rangle = J_{z}J_{+}|a,b\rangle = (b+\hbar)|a,b+\hbar\rangle \end{equation*}$$

In this way, we obtain that

$$\begin{equation*} \mathbf{J}^{2}|a,b+n\hbar\rangle = \mathbf{J}^{2}J_{+}^{n}|a,b\rangle = a|a,b+n\hbar\rangle \end{equation*}$$

and

$$\begin{equation*} J_{z}|a,b+n\hbar\rangle = J_{z}J_{+}^{n}|a,b\rangle = (b+n\hbar)|a,b+n\hbar\rangle \end{equation*}$$

for $$n\in \mathbb{N}$$. But for some $$N-1$$, we then have $$(b+(N-1)\hbar)^{2}\leq a$$ while $$(b+N\hbar)^{2}>a$$, violating the positive semi-definite nature of $$\mathbf{J}^{2}-J_{z}^{2}$$. A way of circumventing this contradiction that leads to physically reasonable conclusions is to require $$J_{+}|a,b+(N-1)\hbar\rangle = 0$$ and label $$b+(N-1)\hbar$$ as $$b_{\mathrm{max}}$$.