Spin-j Systems as Finite Operator Models

A fixed spin-$j$ multiplet is a finite-dimensional, irreducible model of angular momentum with no unbounded-domain ambiguity. Its three spin components generate the full matrix algebra, although their three expectation values generally do not determine a state. The model also provides a precise example of why small commutators in a large-dimension limit do not make every state classical.

Required background. Angular Momentum Algebra develops the ladder construction and allowed spin labels. Higher Spin Systems supplies the standard matrices and physical examples.

A fixed multiplet has bounded generators

Fix $j\in\{0,\frac12,1,\frac32,\ldots\}$ and let $\mathcal H_j$ have orthonormal basis $|m\rangle$, $m=-j,-j+1,\ldots,j$. Use the standard representation,

$$J_z|m\rangle=\hbar m|m\rangle,$$ $$J_\pm|m\rangle =\hbar\sqrt{(j\mp m)(j\pm m+1)}\,|m\pm1\rangle,$$

with the endpoint action zero. Set $J_x=(J_++J_-)/2$ and $J_y=(J_+-J_-)/(2i)$. These are everywhere-defined Hermitian matrices, so their symmetry is self-adjointness.

The representation identities are

$$[J_a,J_b]=i\hbar\epsilon_{abc}J_c,\qquad J_x^2+J_y^2+J_z^2=\hbar^2j(j+1)I.$$

Rotations make each component unitarily equivalent to $J_z$, hence $\|J_a\|=\hbar j$. Any polynomial Hamiltonian in the components is bounded on this fixed multiplet. Its unitary evolution is therefore defined on every vector without an extra domain choice.

This conclusion applies to fixed $j$. A direct sum over arbitrarily large $j$ can produce unbounded generators and requires a new domain analysis.

Spectral projectors and ladders generate matrix units

Since the eigenvalues of $J_z$ are distinct, each one-dimensional spectral projector is a polynomial:

$$P_m=\prod_{\substack{r=-j\\r\ne m}}^j \frac{J_z-\hbar rI}{\hbar(m-r)}.$$

For $m<j$, multiplying by $J_+$ isolates a matrix unit,

$$|m+1\rangle\langle m| =\frac{J_+P_m} {\hbar\sqrt{(j-m)(j+m+1)}}.$$

Products of neighboring matrix units, together with their adjoints, give every $|m\rangle\langle r|$. Therefore the unital star algebra generated by $J_x,J_y,J_z$ is all of $M_{2j+1}(\mathbb C)$.

This proves irreducibility constructively. Any invariant subspace for the star algebra is invariant under all matrix units, so a nonzero vector generates the entire multiplet. It also identifies the observable algebra without importing a general irreducibility theorem.

The result does not mean measuring three mean values is informationally complete. Algebra generation uses products, while a state is linear rather than multiplicative. Higher moments encode additional information when $j>1/2$.

Angular momentum commutators evade the finite-trace obstruction

Every component has trace zero because its eigenvalues sum to zero. Consequently taking traces of $[J_x,J_y]=i\hbar J_z$ gives $0=0$.

This differs from the impossible finite-dimensional relation $[Q,P]=i\hbar I$, whose right side has nonzero trace. A spin multiplet is an exact representation of the angular momentum algebra, not an exact finite-dimensional canonical position–momentum pair.

Near a highly polarized large spin, suitably rescaled transverse components can approximate canonical variables on a restricted family of states. Replacing $J_z$ by a scalar requires control of fluctuations and of the region of the multiplet being used. It is an approximation with a specified state regime, not an operator identity on the whole multiplet.

Small rescaled commutators do not localize every state

For $j>0$, define dimensionless $S_a=J_a/(\hbar j)$. Then

$$[S_x,S_y]=\frac{i}{j}S_z,\qquad \|[S_x,S_y]\|=\frac1j.$$

The commutators tend to zero in norm as $j\to\infty$. Nevertheless, consider the maximally mixed state $\rho_j=I/(2j+1)$. Rotational invariance and the Casimir identity give

$$\operatorname{Tr}(\rho_j S_a)=0,\qquad \operatorname{Tr}(\rho_j S_a^2)=\frac{j+1}{3j}.$$

The component variance approaches $1/3$, not zero. This family remains broadly distributed; small commutators do not force concentration around one classical spin direction.

A classical limit therefore needs both a scaling of observables and a choice of states or probability measures. Spin Coherent States supplies a different family with directional concentration. The two limits serve different physical questions.

Exercise: construct a spin-one event without eigenvectors

For $j=1$, express the projector onto the outcome $J_z=0$ as a polynomial in $J_z$. What is its probability in the maximally mixed state?

Solution

The polynomial must equal one at zero and zero at $\pm\hbar$, so $P_0=I-J_z^2/\hbar^2$. Its trace is one, hence the maximally mixed state’s probability is $1/3$. This uses a spectral polynomial inside the generated matrix algebra and illustrates that a measurement event depends on more than the first moment of $J_z$.

References

• Hall, Brian C., Quantum Theory for Mathematicians, Graduate Texts in Mathematics 267, Springer, 2013, doi:10.1007/978-1-4614-7116-5 — angular momentum and finite-dimensional representations.
• Teschl, Gerald, Mathematical Methods in Quantum Mechanics, 2nd ed., Graduate Studies in Mathematics 157, American Mathematical Society, 2014, author’s text, Section 8.2 — angular momentum operators and algebraic methods.