= Unitary highest-weight termination
{title2=$\|(J_-)^\ell|j,j\rangle\|^2=\hbar^2\ell(2j-\ell+1)\|(J_-)^{\ell-1}|j,j\rangle\|^2$}
Let the Hermitian <angular momentum operators> obey $[J_+,J_-]=2\hbar J_3$, with $J_+^\dagger=J_-$. A <highest-weight vector> satisfies $J_+|j,j\rangle=0$ and $J_3|j,j\rangle=\hbar j|j,j\rangle$. Commuting $J_+$ through $\ell$ lowering operators gives
$$
J_+(J_-)^\ell|j,j\rangle
=\hbar^2\ell(2j-\ell+1)(J_-)^{\ell-1}|j,j\rangle.
$$
Taking <inner products> therefore yields
$$
\|(J_-)^\ell|j,j\rangle\|^2
=\hbar^2\ell(2j-\ell+1)
\|(J_-)^{\ell-1}|j,j\rangle\|^2.
$$
A positive-definite <inner product> makes squared <norms> nonnegative. If $2j$ were not a nonnegative integer, repeated lowering would encounter a negative factor before a zero factor could terminate the sequence. Thus $2j\in\{0,1,2,\ldots\}$.
The first vanishing <norm> occurs at $\ell=2j+1$, leaving $2j+1$ states in the generated <unitary representation>. With dimensionless generators $J_i/\hbar$, the same recurrence has no explicit $\hbar^2$. Without positivity, an abstract <highest-weight representation> can remain infinite-dimensional even at integral <highest weight>.
Back to article page
