Intrinsic angular momentum and su(2)

Every angular-momentum statement so far descended from L = r × p: the operators came from the classical quantity, and [Lᵢ, Lⱼ] = iħ εᵢⱼₖ Lₖ was derived out of [xᵢ, pⱼ] = iħ δᵢⱼ. The Stern–Gerlach result leaves nothing to descend from — a silver atom in its ground state has l = 0, yet its beam splits in two — so the derivation is run backwards. Define an angular momentum to be any triple of Hermitian operators Sₓ, S_y, S_z satisfying [Sᵢ, Sⱼ] = iħ εᵢⱼₖ Sₖ, and let that be the entire content. Two consequences are worth naming at once. The Casimir S² = Sₓ² + S_y² + S_z² commutes with every Sᵢ, so (S², S_z) is a complete set of commuting observables on the internal space and its simultaneous eigenkets |s, m⟩ are the working basis. And U(θ, n̂) = exp(−iθ n̂⋅S/ħ) is unitary, so the operators that label states also rotate them: the algebra says S generates rotations, whatever it is that turns. The full state space factorises as H = H_space ⊗ Hₛₚᵢₙ, with every spin operator acting as 𝟙 ⊗ S.

Nothing in this argument uses a wavefunction. Write the simultaneous eigenkets as S²|s, m⟩ = ħ²λ|s, m⟩ and S_z|s, m⟩ = ħm|s, m⟩, and define S_± = Sₓ ± iS_y. The algebra gives [S_z, S_±] = ±ħS_±, so S_±|s, m⟩ is again an S_z eigenket with m raised or lowered by exactly one. The chain cannot run forever: S_∓S_± = S² − S_z² ∓ ħS_z, so ‖S_±|s, m⟩‖² = ħ²(λ − m² ∓ m) ≥ 0 bounds m from both sides, and termination demands S₊|s, mₘₐₓ⟩ = 0 and S₋|s, mₘᵢₙ⟩ = 0. Those two conditions give λ = mₘₐₓ(mₘₐₓ + 1) = mₘᵢₙ(mₘᵢₙ − 1), hence mₘᵢₙ = −mₘₐₓ ≡ −s and λ = s(s+1). Now the step that matters here: the rungs are spaced by exactly one, so mₘₐₓ − mₘᵢₙ = 2s must be a non-negative integer, and s ∈ (0, ½, 1, 3/2, …). Every coefficient follows from the same norm, S_±|s, m⟩ = ħ√(s(s+1) − m(m±1))|s, m±1⟩ — for s = ½ that reads S₊|½,−½⟩ = ħ|½,½⟩ and S₊|½,½⟩ = 0.

The identical algebra applied to L returned l(l+1) and unit steps, so why is l never ½? Not because of the commutators — they are the same three lines — but because L arrives with a realisation the abstract case does not have: it acts on L²(S²) as L_z = −iħ ∂/∂φ and L_± = ħe^(±iφ)(±∂_θ + i cot θ ∂_φ). Most texts stop the question with single-valuedness — e^(imₗφ) must survive φ → φ + 2π, so mₗ ∈ ℤ — and it is worth knowing, but treat it as a mnemonic rather than a derivation: nothing in the postulates requires ψ itself to be single-valued, only |ψ|² and the ray, and spin-½ is the standing counterexample to demanding more. The sound version is a closure argument, and it is short. Assume l = ½ and solve L₊ψ = 0: the top state is ψ_(½,½) ∝ e^(iφ/2)√(sin θ), which is perfectly square-integrable on the sphere, so L² membership rejects nothing yet. Lower it once, getting ψ_(½,−½) ∝ e^(−iφ/2) cos θ/√(sin θ) — also square-integrable. Now lower again: sitting at m = −l, the coefficient ħ√(l(l+1) − m(m−1)) is zero, so L₋ must annihilate this state. It does not. It returns ħe^(−3iφ/2)[√(sin θ) + cos²θ (sin θ)^(−3/2)], and the ∫|ψ|² sin θ dθ of that diverges. The tower never closes and what it generates leaves the space, so no l = ½ representation lives on functions of angle. Spin faces none of it: its operators act on C^(2s+1), a finite-dimensional internal space with no θ, no φ and no differential realisation to be consistent with, so the algebra's own answer stands unamended. The group-theoretic statement of the same fact is worth carrying forward: exp(−2πi n̂⋅S/ħ) = (−1)^(2s) 𝟙, so a half-integer state comes back with a minus sign after 2π and needs 4π to come home. Functions on the sphere carry a genuine representation of SO(3), which has no such tower; spin-½ carries a representation of SU(2), its double cover.

Ask the algebra which s an electron has and it will not answer. Every s with 2s a non-negative integer labels an irreducible representation of su(2), all equally consistent, and which one a species carries is an experimental fact, as basic as its charge or mass. Measurement is a count: a beam split into 2s+1 components fixes s, so two silver beams give s = ½, three components give s = 1, and no splitting at all gives s = 0 — with massless species the exception, since a photon has helicity ±1 and only two states, not three. Nature's list is short — electrons, protons, neutrons and neutrinos at ½; pions and the Higgs at 0; photons, W and Z at 1; the Δ and Ω⁻ baryons at 3/2. Two structural points sit beneath that list. Unlike l, which changes from state to state inside one atom, s is fixed for a species forever, because no operator in this theory connects different irreducible spin spaces. And S² = ħ²s(s+1) 𝟙 there is a multiple of the identity by Schur's lemma, so S² carries no information inside a species at all: the only spin observable that varies is the projection m.

The name invites a picture — a small charged sphere turning — so test it. A uniform sphere of mass mₑ and radius r has I = (2/5)mₑ r², and with equatorial speed v = ωr its angular momentum is L = (2/5)mₑ r v, so carrying ħ/2 needs v = 5ħ/(4 mₑ r). At the classical electron radius rₑ = 2.82 × 10⁻¹⁵ m that is 5.1 × 10¹⁰ m s⁻¹, or 171 c; the equator would reach c only at r = 4.8 × 10⁻¹³ m, over a hundred times larger than rₑ, while scattering experiments bound the electron below 10⁻¹⁸ m, where the demand climbs past 10⁵ c. A second, independent failure: a rotating body whose charge and mass are distributed alike gives μ = (q/2m)L, that is g = 1, while the electron's spin g factor measures 2.0023. The deepest objection is structural rather than numerical. Rotation through space is generated by r × p, and that realisation supports only integer towers — the l = ½ ladder does not close on functions of angle — so whatever spin-½ is, it admits no realisation as a rotation of anything through space.

The algebra is finite-dimensional, so every claim above is testable in a dozen lines. Build the (2s+1)-square matrices in the |s, m⟩ basis with m ordered s, s−1, …, −s: S_z = ħ⋅diag(m), and S₊ with the single superdiagonal ⟨s, m+1|S₊|s, m⟩ = ħ√(s(s+1) − m(m+1)), then S₋ = S₊†, Sₓ = (S₊ + S₋)/2 and S_y = (S₊ − S₋)/(2i). For s = 1 both superdiagonal entries are ħ√2, and np.allclose(Sx@Sy − Sy@Sx, 1j*ℏ*Sz) returns True to machine precision, while Sx@Sx + Sy@Sy + Sz@Sz comes back as 2ħ² times the identity, matching ħ²s(s+1) at s = 1. Two diagnostics are worth running while you are there. Diagonalising n̂⋅S for any unit n̂ returns the same eigenvalues ħm, because the algebra is isotropic and no axis is special. And halving the ladder coefficients — the intuitive 'spin steps by ½' error — rescales Sₓ and S_y but not S_z, so the commutator comes back a quarter of iħS_z, which is the numerical form of the claim that the algebra, not the step size, is what is being posited.