University Physics IV · Angular Momentum and Spin · 9.2
Ladder Operators & the Angular-Momentum Spectrum
This is the derivation you can run before knowing anything else about the system — no potential, no wavefunction, not even a sphere to separate variables on. Anything that satisfies the angular-momentum commutators inherits this spectrum, which is why one page of algebra covers orbital motion, spin, and everything you will later couple.
Build the model
Connect the measurement to the mechanism.
Take nothing but the commutation relations [Jx, Jy] = iħJz and its cyclic partners — forget where they came from — and the entire spectrum of angular momentum follows. Build J± = Jx ± iJy and the algebra makes them rung-climbers: acting on an eigenstate of Jz, they shift its eigenvalue by exactly ħ while leaving J² untouched. But every state must have a non-negative norm, and the norm of a stepped state is ħ²[λ − m(m±1)], which turns negative if the ladder runs too far; the only escape is a top rung that J₊ annihilates and a bottom rung that J₋ annihilates.
Those two conditions force the bottom to sit at −j when the top sits at +j, and since rungs are one unit apart, 2j must be a whole number: j = 0, ½, 1, 3⁄2, …, with J² = j(j+1)ħ² and 2j+1 projections mħ. The price of this generality is threefold. The algebra cannot say which j values nature uses — restricting orbital l to integers takes the spherical harmonics of the next topic, and spin ½ enters on Stern–Gerlach evidence.
It delivers eigenvalues and matrix elements but no wavefunctions. And its eigenstates fix only the length and one projection: the other two components keep an irreducible spread, so the familiar cone diagram records what is known about the state, and nothing on it precesses.
- Simple definition
- The ladder operators J± = Jx ± iJy raise or lower the Jz eigenvalue by one unit of ħ without changing J², and demanding that every stepped state keep a non-negative norm forces the eigenvalues J² = j(j+1)ħ² and Jz = mħ with m running from −j to +j.
- Example
- For j = 2 the magnitude is |J| = √6 ħ ≈ 2.45ħ, while Jz takes only the five values −2ħ, −ħ, 0, +ħ, +2ħ — the largest projection, 2ħ, falls 0.45ħ short of the full length, so J can never point along the axis.
One application of J₊ raises m by 1, one of J₋ lowers it. Neither touches J², so a whole ladder shares one length.
Jx, Jy are the Cartesian components; J± are not Hermitian — each is the other's adjoint; units of J s, like ħ
They make Jz(J±ψ) = (m ± 1)ħ(J±ψ): the stepped state is again an eigenstate, exactly one rung away.
both follow from [Jx, Jy] = iħJz and cyclic partners; ħ = 1.055 × 10⁻³⁴ J s
This is the quantisation: the ladder ends not at a wall but where the next rung would have negative norm.
must be ≥ 0 for every state; it hits zero exactly at m = +j (raising) and m = −j (lowering)
Half-integers are legitimate outputs here; only the orbital representation later strikes them out.
j = 0, ½, 1, 3⁄2, …; for each j, m = −j, −j+1, …, +j — 2j+1 values in all
Every sequential Stern–Gerlach amplitude and every Clebsch–Gordan table is assembled from these numbers.
square root of the norm condition; the phase choice is the Condon–Shortley convention
The cone's rim is this spread, not a hidden direction — which is why the largest m still leaves J tilted.
in the state |j, m⟩; the spread vanishes only when j = 0
Only the commutators go in
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.
J± steps m by exactly one rung
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.
Positive norms terminate the ladder at both ends
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 each raising step increases m(m+1) and 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 mbot with λ = mbot(mbot − 1). Equating the two gives mbot = −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 − mbot = 2mₜₒₚ must be a non-negative integer. Write j for mₜₒₚ: j ∈ (0, ½, 1, 3⁄2, …).
Read off the spectrum — and the strange length
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.
What the algebra permits, the representation decides
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.
The cone records ignorance, not motion
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.
Change one variable at a time
Make the relationship visible.
Push k until m sits on the top rung and see the arrow stop short of the pole; then grow j from ½ to 5 and watch that top-rung cone angle shrink from 54.7° to 24.1° — the shortfall √(j(j+1)) − j stays below ½, so it stops mattering as the ladder grows tall.
MAGNITUDE |J|2.449 ħ
RUNG m1.0
CONE ANGLE65.9 °
TRANSVERSE RMS ΔJx1.58 ħ
Live interpretationMAGNITUDE |J|: 2.449 ħ. RUNG m: 1.0. CONE ANGLE: 65.9 °. TRANSVERSE RMS ΔJx: 1.58 ħ
Catch the common trap
Explain before calculating.
An electron occupies the orbital state |l = 2, m = 1⟩. When the raising operator L₊ acts on it, what is the magnitude of the coefficient multiplying the resulting state |2, 2⟩?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA p electron has l = 1. Find the magnitude of its orbital angular momentum in units of ħ and in SI units, list the allowed values of Lz, and find the smallest angle L can make with the z-axis.
- The magnitude comes from the L² eigenvalue: |L| = √(l(l+1)) ħ = √(1 × 2) ħ = √2 ħ ≈ 1.414ħ — not 1ħ, which would be the classical reading of "l = 1".
- In SI units: |L| = 1.414 × 1.055 × 10⁻³⁴ J s = 1.49 × 10⁻³⁴ J s.
- The ladder holds 2l + 1 = 3 rungs: m = −1, 0, +1, so Lz can only be measured as −ħ, 0, or +ħ.
- The tightest alignment is the top rung: cos θmin = m/√(l(l+1)) = 1/√2 = 0.7071, so θmin = 45.0° — even at maximum m the vector sits a full 45° off the axis.
Answer|L| = √2 ħ ≈ 1.49 × 10⁻³⁴ J s; Lz ∈ (−ħ, 0, +ħ); θmin = 45°.
MediumStarting from the top state |2, 2⟩, apply the lowering operator L₋ twice. Give the coefficient produced at each step and the resulting normalised state, then verify that lowering the bottom rung |2, −2⟩ gives zero.
- The lowering coefficient is ħ√(j(j+1) − m(m−1)). First step, from m = 2: ħ√(6 − 2 × 1) = ħ√4 = 2ħ, so L₋|2,2⟩ = 2ħ |2,1⟩.
- Second step, from m = 1: ħ√(6 − 1 × 0) = √6 ħ ≈ 2.449ħ, so L₋|2,1⟩ = √6 ħ |2,0⟩.
- Chain them: (L₋)²|2,2⟩ = 2√6 ħ² |2,0⟩ ≈ 4.90 ħ² |2,0⟩. The normalised state after two steps is |2,0⟩ — the operator supplies the number, and the state stays a unit vector only after that number is divided back out.
- Bottom-rung check, m = −2: ħ√(6 − (−2)(−3)) = ħ√(6 − 6) = 0. The ladder cannot step off the end into an unphysical m = −3; the coefficient itself kills the state.
Answer(L₋)²|2,2⟩ = 2√6 ħ² |2,0⟩ ≈ 4.90 ħ² |2,0⟩; the normalised result is |2,0⟩; L₋|2,−2⟩ = 0.
HardFor the stretched state |j, m⟩ = |2, 2⟩, find ⟨Jx⟩ and ⟨Jx²⟩, form the product ΔJx ΔJy, compare it with the uncertainty bound (ħ/2)|⟨Jz⟩|, and give the cone half-angle of the state.
- Write Jx = (J₊ + J₋)/2. Both J₊ and J₋ move m by one, so ⟨2,2|J±|2,2⟩ = 0, giving ⟨Jx⟩ = 0 — and ⟨Jy⟩ = 0 the same way.
- For the square, use J₊J₋ + J₋J₊ = 2(J² − Jz²). The J₊² and J₋² terms move m by two and average to zero, so ⟨Jx²⟩ = (⟨J²⟩ − ⟨Jz²⟩)/2 = (6 − 4)ħ²/2 = ħ². By symmetry ⟨Jy²⟩ = ħ² as well.
- Check the books: ⟨Jx²⟩ + ⟨Jy²⟩ + ⟨Jz²⟩ = ħ² + ħ² + 4ħ² = 6ħ² = ⟨J²⟩ — the part of the squared length missing from Jz² lives entirely in the transverse spread.
- ΔJx = ΔJy = ħ, so ΔJx ΔJy = ħ². The generalised uncertainty bound is (ħ/2)|⟨Jz⟩| = (ħ/2)(2ħ) = ħ². The product equals the bound: the top rung is a minimum-uncertainty state — as classical as this j allows.
- Cone half-angle: cos θ = m/√(j(j+1)) = 2/√6 = 0.8165, so θ = 35.3°. Pointing as far up as it can, J still keeps a full ħ of rms spread in each transverse component.
Answer⟨Jx⟩ = 0; ⟨Jx²⟩ = ħ²; ΔJx ΔJy = ħ², exactly saturating the bound (ħ/2)|⟨Jz⟩| = ħ²; θ = 35.3°.