Higher Spin Systems

Higher-spin systems are spin degrees of freedom with $s>1/2$. They use the same angular-momentum algebra as spin-$1/2$, but the Hilbert space has more than two dimensions, the matrices are larger, and the state geometry contains more information than a single direction on a sphere.

For fixed spin $s$,

$$\mathcal H_s\simeq\mathbb C^{2s+1},$$

with basis states

$$\lvert s,m\rangle, \qquad m=-s,-s+1,\ldots,s.$$

Spin-$1/2$ is the two-state case. Spin $1$ is a triplet. Spin $3/2$ is a quartet. The general formalism is still governed by $SU(2)$.

Canonical Split

This page explains how to think about higher spin as a physical spin degree of freedom. The underlying algebraic derivation of allowed $j,m$ labels belongs to Eigenvalues of J Squared and Jz. The ladder-operator normalization is derived in Ladder Operators. Quick lookup tables for matrices live in Spin Matrices. Spin coherent states, which form a special sphere of classical-looking higher-spin states, have their own page at Spin Coherent States.

Multiplet Dimension

The spin operators satisfy

$$[S_i,S_j] = i\hbar\sum_k\epsilon_{ijk}S_k.$$

A fixed irreducible spin-$s$ multiplet is characterized by

$$S^2\lvert s,m\rangle = \hbar^2s(s+1)\lvert s,m\rangle,$$

and

$$S_z\lvert s,m\rangle = \hbar m\lvert s,m\rangle.$$

The allowed values are

$$m=-s,-s+1,\ldots,s,$$

so the dimension is

$$\dim\mathcal H_s=2s+1.$$

Some common cases are:

Spin $s$	Allowed $m$ values	Dimension	Common name
$0$	$0$	$1$	singlet or scalar
$1/2$	$-1/2,1/2$	$2$	doublet
$1$	$-1,0,1$	$3$	triplet
$3/2$	$-3/2,-1/2,1/2,3/2$	$4$	quartet
$2$	$-2,-1,0,1,2$	$5$	quintet

The words doublet, triplet, quartet, and quintet refer to the number of magnetic sublevels in one irreducible spin multiplet. They do not by themselves say how many particles are present.

Matrix Representations

Choose the standard ordered basis

$$\lvert s,s\rangle,\, \lvert s,s-1\rangle,\, \ldots,\, \lvert s,-s\rangle.$$

Then $S_z$ is diagonal:

$$\langle s,m'|S_z|s,m\rangle = \hbar m\,\delta_{m'm}.$$

The ladder operators are

$$S_\pm=S_x\pm iS_y,$$

with matrix elements

$$\langle s,m'|S_\pm|s,m\rangle = \hbar \sqrt{s(s+1)-m(m\pm1)} \, \delta_{m',m\pm1}.$$

The Cartesian components follow from

$$S_x=\frac{S_++S_-}{2}, \qquad S_y=\frac{S_+-S_-}{2i}.$$

This construction gives a concrete matrix representation for every spin. The matrices are $(2s+1)\times(2s+1)$ and obey the same commutation relations for every $s$.

Spin One

For spin $1$, use the basis

$$\lvert1,1\rangle,\quad \lvert1,0\rangle,\quad \lvert1,-1\rangle.$$

The diagonal component is

$$\frac{S_z}{\hbar} = \begin{pmatrix} 1&0&0\\ 0&0&0\\ 0&0&-1 \end{pmatrix}.$$

The raising operator is

$$\frac{S_+}{\hbar} = \sqrt2 \begin{pmatrix} 0&1&0\\ 0&0&1\\ 0&0&0 \end{pmatrix},$$

and $S_-=S_+^\dagger$. Therefore

$$\frac{S_x}{\hbar} = \frac{1}{\sqrt2} \begin{pmatrix} 0&1&0\\ 1&0&1\\ 0&1&0 \end{pmatrix},$$

and

$$\frac{S_y}{\hbar} = \frac{1}{\sqrt2} \begin{pmatrix} 0&-i&0\\ i&0&-i\\ 0&i&0 \end{pmatrix}.$$

Spin $1$ is the first case where a state can have zero spin expectation value without being spinless. For example, $\lvert1,0\rangle$ has

$$\langle\mathbf S\rangle=\mathbf 0,$$

but it is still in a spin-$1$ multiplet because

$$S^2\lvert1,0\rangle = 2\hbar^2\lvert1,0\rangle.$$

This is one reason the spin-$1/2$ Bloch-vector intuition stops being complete for higher spin.

Spin Three-Halves

For spin $3/2$, the basis is

$$\lvert3/2,3/2\rangle,\quad \lvert3/2,1/2\rangle,\quad \lvert3/2,-1/2\rangle,\quad \lvert3/2,-3/2\rangle.$$

The $z$ component is

$$\frac{S_z}{\hbar} = \begin{pmatrix} 3/2&0&0&0\\ 0&1/2&0&0\\ 0&0&-1/2&0\\ 0&0&0&-3/2 \end{pmatrix}.$$

The raising operator has coefficients $\sqrt3,2,\sqrt3$:

$$\frac{S_+}{\hbar} = \begin{pmatrix} 0&\sqrt3&0&0\\ 0&0&2&0\\ 0&0&0&\sqrt3\\ 0&0&0&0 \end{pmatrix}.$$

Again $S_-=S_+^\dagger$, and $S_x,S_y$ follow from $S_\pm$. The four-state structure is common in nuclear spin, atomic hyperfine manifolds, semiconductor valence-band models, and relativistic field theory. The same abstract representation can appear in very different physical systems.

Integer Versus Half-Integer Spin

A $2\pi$ rotation acts on a spin-$s$ irreducible representation as

$$U_s(2\pi) = (-1)^{2s}I.$$

Thus integer spins return to the same state vector:

$$s=0,1,2,\ldots \quad \Longrightarrow \quad U_s(2\pi)=I,$$

while half-integer spins acquire a minus sign:

$$s=\frac12,\frac32,\frac52,\ldots \quad \Longrightarrow \quad U_s(2\pi)=-I.$$

This is the representation-theoretic form of the distinction between ordinary $SO(3)$ representations and spinorial representations of the double cover $SU(2)$. The spin-$1/2$ sign and its interferometric meaning are discussed in Spinors and 2π Rotations.

Beyond the Bloch Sphere

For spin $1/2$, every pure state is represented by a point on the Bloch sphere. Equivalently, the density matrix can be written as

$$\rho = \frac12 \left( I+\mathbf r\cdot\boldsymbol\sigma \right).$$

For higher spin, the expectation value

$$\langle\mathbf S\rangle$$

is not enough to specify the state. A spin-$1$ density matrix is a $3\times3$ Hermitian matrix with trace one. Besides a vector polarization $\langle S_i\rangle$, it can carry rank-$2$ tensor polarization. One common traceless quadrupole tensor is

$$Q_{ij} = \frac12 \langle S_iS_j+S_jS_i\rangle - \frac13\delta_{ij} \langle S^2\rangle.$$

For spin $1/2$, there is no independent quadrupole structure inside a single irreducible spin multiplet. For spin $1$ and above, quadrupole and higher multipole moments contain real state information. This is why a single arrow can describe a classical-looking spin coherent state, but not an arbitrary higher-spin quantum state.

The systematic language for such tensor structures is introduced in Irreducible Spherical Tensors.

Magnetic and Spectroscopic Context

In a magnetic field, the simplest spin Hamiltonian is

$$H=-\gamma\,\mathbf S\cdot\mathbf B.$$

If $\mathbf B=B\hat z$, then the levels in a fixed spin multiplet split according to

$$E_m=-\gamma\hbar B\,m.$$

For spin $1/2$, this gives two levels. For spin $1$, it gives three equally spaced levels in the simplest Zeeman model. Higher-spin systems may also have quadrupole couplings, crystal-field anisotropies, hyperfine structure, or effective Hamiltonian terms such as

$$H_D=D S_z^2.$$

Those terms are not new angular-momentum algebra; they are additional physical interactions allowed by the system’s symmetry and environment. The basic magnetic-field dynamics are developed in Spin in Magnetic Fields.

Massive and Massless Spin One

A massive spin-$1$ particle has three rest-frame spin projections:

$$m=-1,0,1.$$

A massless spin-$1$ particle, such as the photon, is more subtle. It does not have a rest frame and its physical helicity states are transverse, giving two physical helicities rather than three rest-frame spin projections. This is not a failure of the spin-$1$ representation; it is a consequence of Poincare symmetry and gauge redundancy in relativistic theory. The bridge is discussed in From Angular Momentum to Helicity.

Common Mistakes

• Treating every spin system as a qubit. Only spin-$1/2$ has a two-dimensional spin Hilbert space.
• Thinking $\langle\mathbf S\rangle=\mathbf 0$ implies spin $0$. A spin-$1$ state such as $\lvert1,0\rangle$ has zero vector expectation but nonzero $S^2$.
• Using the Bloch sphere as the full pure-state space for spin $1$ or spin $3/2$.
• Forgetting that spin matrices depend on the basis ordering and phase convention.
• Confusing a spin-$1$ triplet with a three-particle state.
• Assuming “spin one” always means three observed polarizations; massless particles require a separate helicity analysis.
• Applying spin-$1/2$ Pauli-matrix identities directly to higher-spin matrices.

Cross-Links

• Spin as Intrinsic Angular Momentum
• Spin-1/2 Hilbert Space
• Bloch Sphere
• Spin Rotations
• Spinors and 2π Rotations
• Spin Coherent States
• Spin in Magnetic Fields
• Magnetic Moments and g-Factors
• Eigenvalues of J Squared and Jz
• Ladder Operators
• Spin Matrices
• Angular Momentum Algebra Formula Card
• Two Spin-1/2 Particles
• Singlet and Triplet States
• Irreducible Spherical Tensors
• SU(2)
• SU(2) versus SO(3)
• Wigner D-Matrices
• From Angular Momentum to Helicity

References

• A. R. Edmonds, Angular Momentum in Quantum Mechanics, Princeton University Press, 1957.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.
• C. Cohen-Tannoudji, B. Diu, and F. Laloë, Quantum Mechanics, Wiley, 1977.
• D. A. Varshalovich, A. N. Moskalev, and V. K. Khersonskii, Quantum Theory of Angular Momentum, World Scientific, 1988.

Exercises

1. List the allowed $m$ values and dimension for spin $2$.

Solution

For spin $s=2$,

$$m=-2,-1,0,1,2.$$

There are

$$2s+1=5$$

states, so the multiplet is five-dimensional.

2. Build the spin-$1$ raising operator in the ordered basis $\lvert1,1\rangle,\lvert1,0\rangle,\lvert1,-1\rangle$.

Solution

Use

$$S_+\lvert s,m\rangle = \hbar \sqrt{s(s+1)-m(m+1)} \lvert s,m+1\rangle.$$

For $s=1$,

$$S_+\lvert1,-1\rangle = \hbar\sqrt2\,\lvert1,0\rangle,$$

and

$$S_+\lvert1,0\rangle = \hbar\sqrt2\,\lvert1,1\rangle.$$

Also $S_+\lvert1,1\rangle=0$. Therefore

$$\frac{S_+}{\hbar} = \sqrt2 \begin{pmatrix} 0&1&0\\ 0&0&1\\ 0&0&0 \end{pmatrix}.$$

3. What does a $2\pi$ rotation do to spin $1$ and spin $3/2$ state vectors?

Solution

Use

$$U_s(2\pi)=(-1)^{2s}I.$$

For spin $1$, $2s=2$, so

$$U_1(2\pi)=I.$$

For spin $3/2$, $2s=3$, so

$$U_{3/2}(2\pi)=-I.$$

The spin-$1$ vector returns to itself, while the spin-$3/2$ representative changes sign.

4. Show that $\lvert1,0\rangle$ is not a spin-zero state even though $\langle\mathbf S\rangle=\mathbf0$.

Solution

In the state $\lvert1,0\rangle$,

$$S_z\lvert1,0\rangle=0,$$

so $\langle S_z\rangle=0$. By symmetry of the $m=0$ state in the standard spin-$1$ representation, $\langle S_x\rangle=\langle S_y\rangle=0$ as well.

However,

$$S^2\lvert1,0\rangle = \hbar^2\,1(1+1)\lvert1,0\rangle = 2\hbar^2\lvert1,0\rangle.$$

A spin-zero state would have $S^2=0$. Therefore $\lvert1,0\rangle$ has zero vector polarization but belongs to a nonzero spin multiplet.

5. Why is one vector $\langle\mathbf S\rangle$ insufficient to describe a general spin-$1$ state?

Solution

A spin-$1$ density matrix is a $3\times3$ Hermitian trace-one matrix, so it has eight real independent parameters. The vector $\langle\mathbf S\rangle$ has only three components. The remaining information includes tensor polarization, such as quadrupole moments. Thus a spin direction can describe special coherent states, but it cannot describe every spin-$1$ state.