Ladder operators and the angular-momentum spectrum

Promote L = r × p to operators and a short calculation gives [Lx, Ly] = iħLz, with cyclic partners. Now make the decisive move: keep the commutators and throw the definition away. Call any triple of Hermitian operators Jx, Jy, Jz satisfying [Jx, Jy] = iħJz (and cyclically) an angular momentum. The square J² = Jx² + Jy² + Jz² commutes with each component, so J² and one chosen component — conventionally Jz — can share eigenstates, while the commutators forbid sharpening Jx and Jy alongside them. The target is therefore the pair of labels in J²|ψ⟩ = λħ²|ψ⟩ and Jz|ψ⟩ = mħ|ψ⟩. Everything that follows uses the algebra alone: no wavefunction, no sphere, no potential. That is the point — spin has no position representation at all, yet it obeys the same commutators, so it will inherit the same spectrum without a single new calculation.

Form J± = Jx ± iJy — not Hermitian, and adjoints of one another. Two commutators do all the work: [J², J±] = 0 and [Jz, J±] = ±ħJ±. Apply the second to an eigenstate: Jz(J₊|λ, m⟩) = J₊Jz|λ, m⟩ + ħJ₊|λ, m⟩ = (m+1)ħ (J₊|λ, m⟩). The raised state is again an eigenstate of Jz, one rung higher, and the first commutator says its λ has not changed. So within one value of λ the eigenstates form a ladder with rungs one ħ apart, climbed by J₊ and descended by J₋. The harmonic oscillator's a† did the same job on n, but with one structural difference that decides everything here: that ladder was bounded below only, while a component of a vector of fixed length must be bounded at both ends — and the algebra is about to enforce exactly that.

Nothing yet stops the ladder running forever — the norm does. Using J∓J± = J² − Jz² ∓ ħJz, the squared length of a stepped state is ‖J±|λ, m⟩‖² = ħ²[λ − m(m±1)]. A squared length cannot be negative, yet once m is positive each raising step increases m(m+1), and once m is negative each lowering step increases m(m−1); run far enough either way and the bracket goes negative. The only escape is a rung where the coefficient is exactly zero, so that the step produces the zero vector and the ladder ends. Raising must therefore halt at some mₜₒₚ with λ = mₜₒₚ(mₜₒₚ + 1), and lowering at some m_bot with λ = m_bot(m_bot − 1). Equating the two gives m_bot = −mₜₒₚ (the other root, mₜₒₚ + 1, would put the bottom rung above the top). And since the top is reached from the bottom in unit steps, mₜₒₚ − m_bot = 2mₜₒₚ must be a non-negative integer. Write j for mₜₒₚ: j ∈ (0, ½, 1, 3⁄2, …).

The termination fixed λ = j(j+1), so the eigenvalues are J² = j(j+1)ħ² and Jz = mħ, with m taking the 2j+1 values −j, −j+1, …, +j. Note what the algebra insists on: the squared magnitude is j(j+1)ħ², not j²ħ². For j = 2 the length is √6 ħ ≈ 2.449ħ while the largest projection is only 2ħ. This is not a bookkeeping convention — it is measurable: rotational energies go as ħ²l(l+1)/2I, and molecular rotational spectra fit l(l+1), never l². The step coefficients ħ√(j(j+1) − m(m±1)) carry the same physics. They are largest in the middle of the ladder — √6 ħ for the step up from m = 0 when j = 2 — and fall to exactly zero at the ends, which is the termination condition showing up as a number inside every matrix element you will ever compute with these states.

The ladder argument delivered j ∈ (0, ½, 1, 3⁄2, …) — and the half-integers are not a defect. The commutators genuinely allow them, and nature genuinely uses them: the electron's spin realises j = ½, its 2j+1 = 2 projections appearing as the two beams of a Stern–Gerlach magnet. What the algebra cannot do is tell you which j values a particular degree of freedom realises. Orbital motion, where Lz becomes the differential operator −iħ ∂/∂φ acting on functions over a sphere, supports only integer l — but that exclusion is the next topic's argument about the half-integer tower failing on the sphere, not anything derived here, and not the single-valuedness slogan either. Keep the division of labour straight: the commutators fix the menu of possible spectra; the representation, or the experiment, picks from the menu. The algebra also fixes no energies — energy needs a Hamiltonian, which is precisely the ingredient this derivation never used.

Draw a vector of length √(j(j+1))ħ with vertical component mħ and you get the standard cone diagram. Use it as bookkeeping only. In the state |j, m⟩ the transverse components have ⟨Jx⟩ = ⟨Jy⟩ = 0 with equal spreads ⟨Jx²⟩ = ⟨Jy²⟩ = ħ²[j(j+1) − m²]/2 — the cone's rim is a variance, not a set of positions, and a stationary state has nothing precessing around it. The same numbers say why the vector never aligns with the axis: at m = j the tilt is cos θ = j/√(j(j+1)), which is 54.7° for j = ½, 35.3° for j = 2, and 1.8° for j = 1000. Since √(j(j+1)) ≈ j + ½ at large j, the length exceeds the top projection by a fixed ≈ ħ/2 while both grow without bound — so the misalignment matters enormously for one electron and not at all for a flywheel. That is the classical limit arriving as a ratio, with the ħ/2 still sitting underneath.