The Angular-Momentum Algebra and Ladder Operators

The spherical-harmonic treatment solved a differential equation and read off the spectrum. That route ties angular momentum to a spatial wavefunction and delivers only integer $\ell$. A purely algebraic method uses nothing but the commutation relations $[J_i,J_j]=i\hslash {\epsilon }_{ijk}{J}_k$, taken as the definition of any angular momentum. It reproduces the orbital spectrum, fixes every matrix element, and additionally admits half-integer quantum numbers. The half-integers are not a mathematical curiosity; they are the values that spin takes, and no differential equation on the sphere can reach them.

Write $\vec{J}$ for a generic angular momentum, meaning any triple of Hermitian operators obeying the algebra. Orbital $\vec{L}$, spin $\vec{S}$, and the total $\vec{J}=\vec{L}+\vec{S}$ are all instances.

The algebra and its Casimir operator

The defining relations and the one combination that commutes with the whole set:

and, as derived for the orbital case and unchanged here,

$J^2$ is the Casimir operator of the algebra: it commutes with every element, so it takes a fixed value on any irreducible multiplet. Since $J^2$ and $J_z$ commute and are both Hermitian, they possess a common orthonormal eigenbasis. Label the joint eigenstates by their eigenvalues, writing them provisionally as

with $a,b$ dimensionless. The task is to determine the allowed pairs $(a,b)$ from the algebra.

The two labels play different roles. $J^2$ is constant across an entire multiplet, so it sorts the state space into blocks; $J_z$ resolves the states within a block. A complete set of commuting observables for a rotational problem therefore reads $\{J^2,J_z\}$ plus whatever radial or internal labels the full Hamiltonian needs. Because $J^2$ commutes with all three components, and in a rotationally invariant problem with $H$ as well, the $2j+1$ states of a multiplet are degenerate in energy: rotating a stationary state about any axis produces another stationary state of the same energy. That degeneracy is the dynamical fingerprint of the symmetry, and lifting it requires an interaction that singles out a direction, such as an external field.

Ladder operators

Define the non-Hermitian combinations

the analogs of the harmonic-oscillator ladder operators. Their commutators with $J_z$ and $J^2$ follow directly from the algebra:

The relation $[J_z,J_{\pm }]=\pm \hslash J_{\pm }$ is the entire mechanism. Apply $J_z$ to the state $J_{\pm }|a,b⟩$ and commute it through:

So $J_{\pm }|a,b⟩$ is again a $J_z$ eigenstate, with eigenvalue raised or lowered by one unit of $\hslash$. Because $[J^2,J_{\pm }]=0$, the value of $a$ is untouched: the ladder moves within a fixed-magnitude multiplet.

Termination and the eigenvalue spectrum

The ladder cannot run forever. The projection is bounded by the magnitude, because $J_x^2+J_y^2=J^2-J_z^2$ is a sum of squares of Hermitian operators and hence has non-negative expectation:

For fixed $a$ the allowed $b$ lie in a bounded interval, so the ladder must terminate at both ends. Let $b_{\max }$ be the largest and $b_{\min }$ the smallest. Termination means the step off each end produces the zero vector:

To turn these into equations for $a$, use the operator identities obtained by multiplying out $J_{\mp }{J}_{\pm }$:

so

Apply the top-rung equation $J_{+}|a,b_{\max }⟩=0$ to $J^2=J_{-}{J}_{+}+J_z^2+\hslash J_z$:

Apply the bottom-rung equation $J_{-}|a,b_{\min }⟩=0$ to $J^2=J_{+}{J}_{-}+J_z^2-\hslash J_z$:

Equating the two expressions for $a$ gives $b_{\max }(b_{\max }+1)=b_{\min }(b_{\min }-1)$, whose only solution with $b_{\min }\le b_{\max }$ is

Finally, $J_{+}$ raises $b$ in integer steps from $b_{\min }$ to $b_{\max }$, so $b_{\max }-b_{\min }=2b_{\max }$ must be a non-negative integer. Writing $j\equiv b_{\max }$:

The relabelling $a=j(j+1)$, $b=m$ matches the orbital result exactly for integer $j$, and each multiplet has dimension $2j+1$.

Matrix elements and representations

The scalars $c_{\pm }$ in the ladder action come from normalization. The squared norm of $J_{+}|j,m⟩$ is

Choosing real, positive phases gives the standard result:

The radical vanishes exactly at $m=+j$ for $J_{+}$ and $m=-j$ for $J_{-}$, confirming that the ladder terminates and no state escapes the multiplet. From $J_x=\frac{1}{2}(J_{+}+J_{-})$ and $J_y=\frac{1}{2i}(J_{+}-J_{-})$, the matrix elements of every component follow:

with $J_y$ carrying the same entries times $\mp i$. In the $|j,m⟩$ basis $J_z$ is diagonal, $J_{\pm }$ are single off-diagonal bands, and $J_x,J_y$ are the symmetric and antisymmetric combinations of those bands. Each fixed $j$ gives a $(2j+1)$-dimensional irreducible representation of the algebra.

The uncertainty relation among components

The non-commutativity has a quantitative face. The generalized uncertainty relation applied to $J_x,J_y$ with $[J_x,J_y]=i\hslash J_z$ reads

Evaluate both sides in a state $|j,m⟩$. The projection has mean $⟨J_z⟩=m\hslash$, while symmetry about the $z$-axis forces $⟨J_x⟩=⟨J_y⟩=0$ and $⟨J_x^2⟩=⟨J_y^2⟩=\frac{1}{2}⟨J^2-J_z^2⟩=\frac{{\hslash }^2}{2}[j(j+1)-m^2]$. The variances are equal, so

the inequality reducing to $j(j+1)\ge m^2+|m|=|m|(|m|+1)$, which holds for all $|m|\le j$ and is saturated at $m=\pm j$. The stretched states $m=\pm j$ are therefore the minimum-uncertainty states of the transverse components, the closest an angular-momentum eigenstate comes to pointing along the axis. This is the algebraic content of the vector-model cones: even the top rung keeps a residual transverse spread ${\sigma }_{J_x}={\sigma }_{J_y}=\hslash \sqrt{j/2}$.

The spin-half and spin-one representations

For $j=\frac{1}{2}$ the basis is $\{|\frac{1}{2},\frac{1}{2}⟩,|\frac{1}{2},-\frac{1}{2}⟩\}$ and the matrices are $\frac{\hslash }{2}$ times the Pauli matrices:

For $j=1$ the basis is $\{|1,1⟩,|1,0⟩,|1,-1⟩\}$. The radical gives $\sqrt{1\cdot 2-1\cdot 0}=\sqrt{2}$ for the $m=0\leftrightarrow \pm 1$ steps, so

Why half-integers survive here but not for orbital motion

Both integer and half-integer $j$ solve the algebra, yet the spherical harmonics carry only integer $\ell$. The algebra is silent about single-valuedness; it knows only the commutators. When $\vec{J}=\vec{L}$ is built from $\vec{r}\times \vec{p}$, the extra requirement that the wavefunction be a single-valued function of the azimuth $\phi$ removes the half-integer rungs, because those states pick up a factor $-1$ under a $2\pi$ rotation. Spin carries no coordinate wavefunction and faces no such constraint, so nature uses the half-integer representations for it. This is the algebraic reason spin cannot be reduced to orbital motion of anything.

The two-index bookkeeping $|j,m⟩$ and the ladder rules developed here are the tools the addition of angular momenta uses to combine two such multiplets into the coupled basis, where the same lowering operator constructs the total-spin states of composite systems.¹²