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*}


\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?

  • 1
    $\begingroup$ 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 $a=b_\max(b_\max+1)$. $\endgroup$
    – J.G.
    Apr 27, 2023 at 10:36
  • $\begingroup$ @J.G. please see the answer I posted to my question--I believe it's correct, but I would be grateful for your input! $\endgroup$ May 3, 2023 at 5:40
  • $\begingroup$ 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. $\endgroup$ May 3, 2023 at 6:53
  • $\begingroup$ @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? $\endgroup$ May 3, 2023 at 6:54
  • $\begingroup$ 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. $\endgroup$ May 3, 2023 at 6:57

1 Answer 1


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*}


\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*}


\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 $b_{\mathrm{max}}$.


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