University Physics V · Quantum Spin · 10.1
Intrinsic Angular Momentum & su(2)
Spin is the first observable in this course with no classical quantity standing behind it. This topic teaches the move that makes it respectable: stop asking what is rotating, take the commutation relations as the definition, and read off what the algebra allows — then let experiment say which value each species carries.
Build the model
Connect the measurement to the mechanism.
Orbital angular momentum was constructed: write L = r × p, promote it to operators, derive [Lᵢ, Lⱼ] = iħ εᵢⱼₖ Lₖ. Spin reverses that order. There is no r and no p to build it from, so the commutator is promoted from consequence to definition — an intrinsic angular momentum is any triple of Hermitian operators obeying [Sᵢ, Sⱼ] = iħ εᵢⱼₖ Sₖ on an internal factor of the Hilbert space carrying no dependence on position at all.
The ladder argument runs on that algebra alone: S² commutes with each Sᵢ, S_± = Sₓ ± iSy step m by one, and norm positivity forces the chain to terminate at both ends, giving S²|s, m⟩ = ħ²s(s+1)|s, m⟩ and Sz|s, m⟩ = mħ|s, m⟩ with 2s a non-negative integer. What the orbital case carries extra is not an extra axiom but a realisation: L acts as differential operators on L²(S²), and the l = ½ tower fails there — lowering from the would-be bottom state does not annihilate it, and what it keeps generating leaves the space. Drop the realisation and the half-integers have nothing left to violate, so s = ½ is admitted at last, on a two-dimensional space no function of angle can supply.
The cost is twofold. The algebra fixes the ladder but never says which one a species climbs — ½ for the electron, 0 for the pion, 1 for the W — each an experimental input, as defining as charge or mass. And the picture you would like to keep goes: a sphere of the classical electron radius needs an equator at 171 c to carry ħ/2, and gₛ ≈ 2 is twice what any rotating charge gives.
- Simple definition
- Spin is an intrinsic angular momentum defined by its algebra alone — three Hermitian operators with [Sᵢ, Sⱼ] = iħ εᵢⱼₖ Sₖ acting on an internal space — with no r × p realisation and nothing rotating in space behind it.
- Example
- The electron has s = ½, so S² = ħ²(½)(3/2) = 0.75ħ² while the largest projection is only Sz = 0.5ħ. The magnitude √0.75 ħ = 0.866ħ exceeds it, so the spin sits 54.7° from z even at its most aligned.
The first line is the definition, not a theorem; the second follows from it in three lines, and everything below follows from the pair — and from nothing about what is rotating.
Sᵢ Hermitian on an internal space of dimension 2s+1; εᵢⱼₖ is the Levi-Civita symbol; ħ = 1.055 × 10⁻³⁴ J s carries the units
Buys the whole spectrum — the coefficient dying at m = ±s is what stops the chain at both ends rather than letting it run forever.
S_± are not Hermitian: (S+)† = S_−. The coefficient is real, and vanishes exactly at m = ±s
The half-integers live here: nothing in the algebra excludes them, and the electron takes s = ½ on a two-state space.
m runs s, s−1, …, −s over 2s+1 values, so 2s steps; s is a label on the species, while m is the only part that varies
Names the extra structure the orbital realisation carries and an internal space does not — a differential realisation the ladder has to close on — which is precisely why s = ½ survives here and not there.
not from single-valuedness, which is an assumption about ψ rather than a postulate: the l = ½ tower fails because L_− does not annihilate ψ_(½,−½), and what it returns goes as (sin θ)(−3/2) ∉ L²(S²)
S² varies over nothing inside one species, so the only spin observable that moves is the projection m.
𝟙 is the identity; for s = ½, S² = 0.75ħ² 𝟙 in every state of the species, with no state dependence at all
That is 171 c, and it only worsens as r shrinks toward the experimental bound r < 10⁻¹⁸ m.
from I = (2/5)mₑ r², veq = ωr and L = ħ/2; at r = 2.82 × 10⁻¹⁵ m it demands 5.1 × 10¹⁰ m s⁻¹
Promote the commutator from theorem to definition
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, Sy, Sz satisfying [Sᵢ, Sⱼ] = iħ εᵢⱼₖ Sₖ, and let that be the entire content. Two consequences are worth naming at once. The Casimir S² = Sₓ² + Sy² + Sz² commutes with every Sᵢ, so (S², Sz) 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 = Hspace ⊗ Hₛₚᵢₙ, with every spin operator acting as 𝟙 ⊗ S.
Run the ladder on the algebra alone
Nothing in this argument uses a wavefunction. Write the simultaneous eigenkets as S²|s, m⟩ = ħ²λ|s, m⟩ and Sz|s, m⟩ = ħm|s, m⟩, and define S_± = Sₓ ± iSy. The algebra gives [Sz, S_±] = ±ħS_±, so S_±|s, m⟩ is again an Sz eigenket with m raised or lowered by exactly one. The chain cannot run forever: S_∓S_± = S² − Sz² ∓ ħSz, 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.
Why the orbital realisation loses the half-integers
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 Lz = −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.
The algebra fixes the ladder, not which one a species climbs
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.
No ball is spinning: the numbers that kill the model
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.
Check it in NumPy
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: Sz = 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 Sy = (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 Sy but not Sz, so the commutator comes back a quarter of iħSz, which is the numerical form of the claim that the algebra, not the step size, is what is being posited.
Change one variable at a time
Make the relationship visible.
Set s = 0.25 and watch the ladder stop short of −s: the lowest rung the chain can reach stalls a gap 2s − ⌊2s⌋ above the dashed line, and only values making 2s a whole number — 0, ½, 1, 3/2, 2 — close that gap, which is why half-integer spin is admitted and quarter-integer is not. Then walk k down the rungs: the arrow tilts but never lies along z, because √(s(s+1)) > s.
S² / ħ² = s(s+1)0.75
Sz / ħ = m0.50
LADDER MISS 2s − ⌊2s⌋0.00
TILT FROM z54.7 °
Live interpretationS² / ħ² = s(s+1): 0.75. Sz / ħ = m: 0.50. LADDER MISS 2s − ⌊2s⌋: 0.00. TILT FROM z: 54.7 °
Catch the common trap
Explain before calculating.
Orbital angular momentum never takes a half-integer quantum number, yet the electron has s = ½ — and both obey the same commutation relations. What lets spin take a value orbital motion cannot?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA baryon resonance carries s = 3/2. List its Sz eigenvalues, give the magnitude |S| = ħ√(s(s+1)), find the smallest angle the spin vector can make with the z axis, and evaluate S+|3/2, 1/2⟩.
- The rungs run from +s downward in unit steps and must land on −s: m = 3/2, 1/2, −1/2, −3/2. That is 2s + 1 = 4 states, and it closes exactly because 2s = 3 is a whole number.
- Sz eigenvalues are mħ: +1.5ħ, +0.5ħ, −0.5ħ, −1.5ħ. Nothing in the algebra says which species gets this ladder — that is measured, by counting four beams.
- Magnitude: |S| = ħ√(s(s+1)) = ħ√(1.5 × 2.5) = ħ√3.75 = 1.936ħ, strictly larger than the biggest projection 1.5ħ.
- Smallest tilt: cos θ = mₘₐₓ/√(s(s+1)) = 1.5/1.9365 = 0.7746, so θ = 39.2°. The vector never lies along z, for any s > 0.
- Ladder coefficient at m = 1/2: S+|3/2,1/2⟩ = ħ√(s(s+1) − m(m+1))|3/2,3/2⟩ = ħ√(3.75 − 0.75)|3/2,3/2⟩ = √3 ħ |3/2,3/2⟩ = 1.732ħ |3/2,3/2⟩.
Answerm = ±3/2 and ±1/2, so Sz = ±1.5ħ, ±0.5ħ; |S| = 1.936ħ; the smallest tilt from z is 39.2°; and S+|3/2,1/2⟩ = √3 ħ |3/2,3/2⟩.
MediumUsing only the algebra — no matrices — find ⟨Sₓ⟩, ⟨Sₓ²⟩ and ΔSₓ in the state |s, m⟩ = |1, 1⟩, then test the product ΔSₓ ΔSy against the Robertson bound ½|⟨[Sₓ, Sy]⟩|.
- Write Sₓ = (S+ + S_−)/2. Both terms change m by one, so ⟨1,1|Sₓ|1,1⟩ = 0, and identically ⟨Sy⟩ = 0: in an Sz eigenstate the transverse components average to zero.
- Sₓ² = (S+² + S+S_− + S_−S+ + S_−²)/4. The squared terms shift m by two and cannot return to the same ket, so only the mixed pair survives the expectation value.
- Use S+S_−|s, m⟩ = ħ²(s(s+1) − m(m−1))|s, m⟩ and S_−S+|s, m⟩ = ħ²(s(s+1) − m(m+1))|s, m⟩. At s = 1, m = 1: the first gives ħ²(2 − 0) = 2ħ², the second ħ²(2 − 2) = 0, since m is already at the top rung.
- So ⟨Sₓ²⟩ = (2ħ² + 0)/4 = 0.5ħ² and ΔSₓ = √(0.5ħ² − 0) = ħ/√2 = 0.707ħ; by symmetry ΔSy is the same. Cross-check: ⟨Sₓ²⟩ + ⟨Sy²⟩ + ⟨Sz²⟩ = 0.5 + 0.5 + 1 = 2ħ² = ħ²s(s+1). ✓
- Robertson: [Sₓ, Sy] = iħSz, so the bound is ½|⟨iħSz⟩| = (ħ/2)|⟨Sz⟩| = (ħ/2)(ħ) = 0.5ħ². The product ΔSₓ ΔSy = 0.707ħ × 0.707ħ = 0.5ħ² meets it exactly.
- The stretched state m = s saturates the bound — the closest a spin can come to pointing along an axis. It is still not along it: the transverse variances sum to ⟨Sₓ²⟩ + ⟨Sy²⟩ = sħ² = ħ² here, rather than vanishing, which is the same statement as ħ²s(s+1) − m²ħ² > 0.
Answer⟨Sₓ⟩ = 0, ⟨Sₓ²⟩ = 0.5ħ², ΔSₓ = ΔSy = 0.707ħ, and ΔSₓ ΔSy = 0.5ħ², exactly equal to ½|⟨[Sₓ, Sy]⟩| — the m = s state saturates the Robertson bound.
HardModel the electron as a uniform sphere of radius r carrying angular momentum ħ/2. (a) Derive the equatorial speed veq. (b) Evaluate it at the classical electron radius rₑ = 2.82 × 10⁻¹⁵ m. (c) Find the radius at which veq would equal c. (d) Compare with the experimental bound r < 10⁻¹⁸ m, and add the independent objection from the g factor. Use ħ = 1.055 × 10⁻³⁴ J s, mₑ = 9.109 × 10⁻³¹ kg, c = 2.998 × 10⁸ m s⁻¹.
- For a uniform sphere I = (2/5)mₑ r² and ω = veq/r, so L = Iω = (2/5)mₑ r veq. Setting L = ħ/2 and solving gives veq = 5ħ/(4 mₑ r).
- At r = rₑ: numerator 5ħ = 5.275 × 10⁻³⁴ J s; denominator 4 mₑ r = 4 × 9.109 × 10⁻³¹ × 2.82 × 10⁻¹⁵ = 1.027 × 10⁻⁴⁴ kg m. So veq = 5.13 × 10¹⁰ m s⁻¹, which is 5.13 × 10¹⁰ / 2.998 × 10⁸ = 171 c.
- Set veq = c and solve for r: r = 5ħ/(4 mₑ c) = 5.275 × 10⁻³⁴ / (3.644 × 10⁻³⁰ × 2.998 × 10⁸) = 5.275 × 10⁻³⁴ / 1.092 × 10⁻²¹ = 4.83 × 10⁻¹³ m — that is (5/4) of the reduced Compton wavelength ħ/mₑ c = 3.86 × 10⁻¹³ m, and 171 times bigger than rₑ.
- At the experimental bound r = 10⁻¹⁸ m the demand is veq = 5.275 × 10⁻³⁴ / 3.644 × 10⁻⁴⁸ = 1.45 × 10¹⁴ m s⁻¹, about 4.8 × 10⁵ c. The needed radius exceeds the measured bound by a factor of roughly 5 × 10⁵.
- Independent objection: a rotating body with charge and mass distributed alike has μ = (q/2m)L, i.e. g = 1, and no redistribution of charge inside a rigid rotor gives 2. The electron's spin g factor is 2.0023, which Dirac theory delivers as exactly 2 plus a QED anomaly — not as a shape.
- Conclusion: keep the algebra and drop the ball. S is defined by [Sᵢ, Sⱼ] = iħ εᵢⱼₖ Sₖ on a two-dimensional internal space, which has no radius to shrink and no equator to spin.
Answerveq = 5ħ/(4 mₑ r) = 5.13 × 10¹⁰ m s⁻¹ = 171 c at rₑ; veq = c only at r = 4.83 × 10⁻¹³ m, some 5 × 10⁵ times the measured size bound; and gₛ = 2.0023 against the rotor's g = 1 kills the model independently.