Unitary highest-weight termination ($\Vert (J_{-}{)}^{\ell }|j,j⟩{\Vert }^2={\hslash }^2\ell (2j-\ell +1)\Vert (J_{-}{)}^{\ell -1}|j,j⟩{\Vert }^2$)

Let the Hermitian angular momentum operators obey $[J_{+},J_{-}]=2\hslash J_3$, with $J_{+}^{†}=J_{-}$. A highest-weight vector satisfies $J_{+}|j,j⟩=0$ and $J_3|j,j⟩=\hslash j|j,j⟩$. Commuting $J_{+}$ through $\ell$ lowering operators givesTaking inner products therefore yieldsA 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 $2j\in \{0,1,2,\dots \}$.

The first vanishing norm occurs at $\ell =2j+1$, leaving $2j+1$ states in the generated unitary representation. With dimensionless generators $J_i/\hslash$, the same recurrence has no explicit ${\hslash }^2$. Without positivity, an abstract highest-weight representation can remain infinite-dimensional even at integral highest weight.