University Physics V · Angular Momentum in Quantum Mechanics · 9.3

Ladder Operators & the Quantisation of j and m

No differential equation gets solved here. Take three Hermitian operators obeying [Jᵢ, Jⱼ] = iħ εᵢⱼₖ Jₖ, build J± from two of them, and watch one inequality — a squared length cannot be negative — force every allowed value of J² and J_z. Learn the argument well enough to run it on spin, where no wavefunction exists to fall back on.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Angular momentum in quantum mechanics is defined by an algebra, not by r × p. Any three Hermitian operators with [Jᵢ, Jⱼ] = iħ εᵢⱼₖ Jₖ qualify, and because J² commutes with each component, a state can be labelled by simultaneous eigenvalues of J² and one component, conventionally J_z. The non-Hermitian combinations J± = Jₓ ± iJ_y satisfy [J_z, J±] = ±ħJ±, so they move a J_z eigenvalue up or down by exactly ħ while leaving J² untouched: they walk along a ladder of rungs.

What quantises the ladder is one inequality. The squared length of J±|λ, m⟩ works out to ħ²[λ − m(m ± 1)], and a squared length cannot be negative, so m is trapped between two bounds and the only way the ladder can obey both is to end — a top rung annihilated by J₊ and a bottom rung annihilated by J₋. The two end conditions give λ = j(j+1), a bottom rung at −j, and 2j an integer, hence j = 0, ½, 1, 3⁄2, … with 2j+1 rungs each.

The cost of that generality is that the algebra names every possible j and selects none: which j a system realises is a fact about the system. Orbital L = r × p on functions of angle admits only integer l, giving L² = ħ²l(l+1) and L_z = mₗħ, while spin-½ realises j = ½ with no wavefunction at all. A second cost is baked into the numbers: |J| = ħ√(j(j+1)) exceeds the largest J_z = jħ for every j > 0, so no state ever has its angular momentum fully along z.

Simple definition: The ladder operators J± = Jₓ ± iJ_y raise or lower the J_z eigenvalue by ħ without changing J², and the requirement that their output have a non-negative norm forces J² = ħ²j(j+1) with j = 0, ½, 1, … and J_z = mħ with m running from −j to j in unit steps.
Example: For j = 2 and m = 1, ‖J₊|2,1⟩‖² = ħ²[2⋅3 − 1⋅2] = 4ħ², so J₊|2,1⟩ = 2ħ|2,2⟩; one more step gives ‖J₊|2,2⟩‖² = ħ²[6 − 6] = 0, and the ladder stops on its fifth rung.

Ladder operatorsJ± = Jₓ ± iJ_y · (J₊)† = J₋

They are the algebraic tools that step between eigenstates of J_z; nothing is ever measured with them.

Jₓ, J_y Hermitian, carrying units of ħ (J s); J± are neither Hermitian nor observables

Ladder commutators[J_z, J±] = ±ħJ± · [J₊, J₋] = 2ħJ_z · [J², J±] = 0

The first says J± shift m by exactly ±1; the last says they never leave the J² eigenspace.

Three lines each from [Jᵢ, Jⱼ] = iħ εᵢⱼₖ Jₖ; εᵢⱼₖ the Levi-Civita symbol

Product identities and the normJ∓J± = J² − J_z² ∓ ħJ_z → ‖J±|j, m⟩‖² = ħ²[j(j+1) − m(m±1)]

This inequality is the whole quantisation argument: it traps m between bounds and forces the ladder to end.

Left side an operator identity; right side a number that is ≥ 0 because it is a squared length

The spectrumJ²|j, m⟩ = ħ²j(j+1)|j, m⟩ · J_z|j, m⟩ = mħ|j, m⟩

|J| = ħ√(j(j+1)) exceeds jħ for every j > 0, so J_z never reaches the full length of J.

j = 0, ½, 1, 3⁄2, …; m = −j, −j+1, …, j; 2j+1 states for each j

Normalised rungs (Condon–Shortley)J±|j, m⟩ = ħ√((j∓m)(j±m+1)) |j, m±1⟩

The single input from which every (2j+1)-square matrix Jₓ, J_y, J_z is assembled.

Coefficients real and non-negative; identical to ħ√(j(j+1) − m(m±1))

Orbital specialisationL² = ħ²l(l+1) · L_z = mₗħ · l = 0, 1, 2, …

Integer l is a property of the operator's domain, not of the commutators; spin-½ takes the half-integer entry instead.

L = r × p acting on functions of θ and φ; the same algebra, a restricted list of j

Start from the algebra and pick the basis it allows

The input is three Hermitian operators with [Jᵢ, Jⱼ] = iħ εᵢⱼₖ Jₖ and nothing else — no r, no p, no wavefunction. Form J² = Jₓ² + J_y² + J_z² and check [J², Jᵢ] = 0 for each i: the commutator of Jₓ² with J_z, for instance, is Jₓ[Jₓ, J_z] + [Jₓ, J_z]Jₓ = −iħ(JₓJ_y + J_yJₓ), which the J_y² term cancels exactly. So J² is compatible with any one component, but the components are not compatible with each other: [Jₓ, J_z] ≠ 0, so Jₓ and J_z share no eigenbasis. The largest commuting set is therefore J² and one component, and convention picks J_z. Write the simultaneous eigenvectors as |λ, m⟩ with J²|λ, m⟩ = ħ²λ|λ, m⟩ and J_z|λ, m⟩ = mħ|λ, m⟩, where λ and m are for now just real numbers — real because both operators are Hermitian. Everything that follows is about which pairs (λ, m) the algebra permits.

J± move a state one rung, and only one rung

Define J± = Jₓ ± iJ_y. They are adjoints of each other and not Hermitian, so they are not observables; they are tools. Two commutators do the work. From [J_z, Jₓ] = iħJ_y and [J_z, J_y] = −iħJₓ, [J_z, J₊] = iħJ_y + ħJₓ = ħJ₊, and likewise [J_z, J₋] = −ħJ₋. From [J², Jᵢ] = 0 follows [J², J±] = 0. Now act on a rung: J_z(J₊|λ, m⟩) = (J₊J_z + ħJ₊)|λ, m⟩ = (m + 1)ħ (J₊|λ, m⟩), and J²(J₊|λ, m⟩) = ħ²λ (J₊|λ, m⟩). So J₊|λ, m⟩ is either the zero vector or an eigenvector with the same λ and with m raised by exactly one; J₋ lowers by exactly one. Applying J₊ repeatedly generates rungs m, m+1, m+2, … all inside one J² eigenspace. Nothing yet stops the climb — that needs the norm.

Norm positivity closes the ladder at both ends

Compute the squared length of the raised vector. Since (J₊)† = J₋, ‖J₊|λ, m⟩‖² = ⟨λ, m|J₋J₊|λ, m⟩, and multiplying out, J₋J₊ = (Jₓ − iJ_y)(Jₓ + iJ_y) = Jₓ² + J_y² + i[Jₓ, J_y] = J² − J_z² − ħJ_z. So ‖J₊|λ, m⟩‖² = ħ²[λ − m(m+1)], and by the same route ‖J₋|λ, m⟩‖² = ħ²[λ − m(m−1)]. A squared length is never negative, so λ ≥ m(m+1) and λ ≥ m(m−1): for fixed λ, m is bounded above and below. But J₊ raises m by one each time at the same λ, so an unbounded climb would eventually break the bound. The only escape is a rung mₘₐₓ with J₊|λ, mₘₐₓ⟩ = 0, that is λ = mₘₐₓ(mₘₐₓ + 1); call mₘₐₓ = j, so λ = j(j+1). Descending, there is a rung mₘᵢₙ with λ = mₘᵢₙ(mₘᵢₙ − 1), and the two conditions give mₘᵢₙ = −j (the other root, j + 1, lies above the top). The rungs run from −j to j in unit steps, so 2j is a non-negative integer: j = 0, ½, 1, 3⁄2, …, with 2j + 1 rungs. Take λ = 3.75 = (3⁄2)(5⁄2): four rungs at m = ±½ and ±3⁄2, and nothing else.

Normalise the rungs and read off the matrix elements

The norm calculation also fixes the coefficients. With the Condon–Shortley convention of real, non-negative phases, J±|j, m⟩ = ħ√(j(j+1) − m(m±1)) |j, m±1⟩ = ħ√((j∓m)(j±m+1)) |j, m±1⟩. Run it for j = 1: J₊|1, −1⟩ = ħ√((1+1)(1−1+1)) |1, 0⟩ = ħ√2 |1, 0⟩, J₊|1, 0⟩ = ħ√((1−0)(1+0+1)) |1, 1⟩ = ħ√2 |1, 1⟩, and J₊|1, 1⟩ = ħ√(0⋅3) |1, 2⟩ = 0. The factor is not ħ and it is not constant along the ladder: J± are not unitary shifts, and a raised vector must be renormalised before it is a state. Because Jₓ = (J₊ + J₋)/2 and J_y = (J₊ − J₋)/2i, these numbers are the only input to the matrices of topic 9.4: in the basis (|1,1⟩, |1,0⟩, |1,−1⟩) the j = 1 matrix Jₓ is (ħ/√2) times [[0,1,0],[1,0,1],[0,1,0]], whose eigenvalues are 0 and ±ħ — the same set as J_z, because no direction is special.

Check the algebra numerically before trusting it

The derivation is short enough to test in NumPy in a dozen lines, and the test catches the ordering and phase slips that hand calculation hides. Order the basis m = j, j−1, …, −j, so m = np.arange(j, −j-1, −1). Then Jp = ℏ*np.diag(np.√(j*(j+1) - m[1:]*(m[1:]+1)), k=1) places ⟨m+1|J₊|m⟩ on the superdiagonal, Jm = Jp.conj().T, Jz = ℏ*np.diag(m), Jx = (Jp + Jm)/2 and Jy = (Jp - Jm)/(2*1j), where 1j is Python's imaginary unit and not the angular momentum j. Three checks close the loop: np.allclose(Jx@Jy - Jy@Jx, 1j*ℏ*Jz) should return True with residuals near 10⁻¹⁶; Jx@Jx + Jy@Jy + Jz@Jz should equal ħ²j(j+1) times the identity; and np.linalg.eigvalsh of n⋅J for any unit vector n should return the same m values as J_z. A residual of order one instead of 10⁻¹⁶ almost always means the superdiagonal was filled from the wrong end of m — the same slip as raising with the lowering coefficient.

What the algebra decides, and what it leaves to the system

The result is a catalogue of representations, one for each j = 0, ½, 1, …, each of dimension 2j + 1. The commutators cannot say which entries a given system uses; that depends on what the operators J actually are. For orbital motion L = r × p acts on functions of θ and φ, L_z = −iħ ∂/∂φ has eigenfunctions e^(imφ), and only the integer ladders close on the sphere — the half-integer tower fails because lowering below its would-be bottom rung does not give zero, so it never terminates (topic 9.6 finishes that argument). Hence L² = ħ²l(l+1) and L_z = mₗħ with l = 0, 1, 2, …. Spin is the opposite case: S on C² is the j = ½ entry, with no coordinate representation at all. Two consequences carry forward. For every j > 0, m ≤ j < √(j(j+1)), so ⟨Jₓ²⟩ + ⟨J_y²⟩ = ħ²[j(j+1) − m²] ≥ jħ² > 0 on every rung — the transverse components never vanish, which is the origin of the cone picture in topic 9.4. And a physical state need not be a single rung: it can be a superposition over several j. Which j appear is a property of the state, not of the algebra; a Hamiltonian that commutes with J² only guarantees that the weight on each j stays fixed as the state evolves.

Change one variable at a time

Make the relationship visible.

Interactive model

Angular momentum j1.5

Rungs climbed k (m = k − j, capped at j)1

Set j = 2.5 and push k up from 0: both norms rise to 3ħ near the middle of the ladder, then the J₊ circle and bar shrink to exactly nothing when m reaches j while J₋ is still √5 ħ; drop j to 1 and the same ladder closes after three rungs. The dashed |J| line never comes down to meet the top rung.

J² EIGENVALUE j(j+1)3.75 ħ²

RUNG m-0.5

‖J₊|j, m⟩‖2.000 ħ

‖J₋|j, m⟩‖1.732 ħ

Live interpretationJ² EIGENVALUE j(j+1): 3.75 ħ². RUNG m: −0.5. ‖J₊|j, m⟩‖: 2.000 ħ. ‖J₋|j, m⟩‖: 1.732 ħ

Catch the common trap

Explain before calculating.

Misconception check

The half-integer solutions are not a mathematical leftover, and the integer ones are not a proof that all angular momentum is orbital. Students read j = 0, ½, 1, … in one of two wrong ways: either every angular momentum can take every value on the list, so an electron's orbital l could be ½; or, since L = r × p produces only integer l, the half-integers should be struck out. Both misread what was derived. The only input was the commutation relations, so the output is the list of representations that algebra permits — a catalogue, not a selection. Which entries a system realises is fixed by what its operators are: L on functions of angle closes only the integer ladders (the l = ½ tower fails because it never terminates, not because of single-valuedness), while S on C² realises j = ½ with no wavefunction to be single-valued. The algebra rules out j = 0.3; it cannot rule in j = ½.

An eigenstate of J² and J_z has J² eigenvalue 6ħ² and J_z eigenvalue ħ. What is the norm of J₊ acting on it, ‖J₊|j, m⟩‖?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyFor j = 3⁄2, list the allowed m values and the dimension of the space, then find ‖J₋|3⁄2, ½⟩‖ and the magnitude |J| = ħ√(j(j+1)). Show that J₊ annihilates the top rung.

m runs from −j to j in unit steps: m = −3⁄2, −½, ½, 3⁄2, so there are 2j + 1 = 4 rungs and the space is four-dimensional.
J² eigenvalue: ħ²j(j+1) = ħ²(3⁄2)(5⁄2) = 3.75ħ², so |J| = ħ√3.75 = 1.94ħ — larger than the biggest J_z, which is 1.5ħ.
Lowering: ‖J₋|3⁄2, ½⟩‖² = ħ²[j(j+1) − m(m−1)] = ħ²[3.75 − (½)(−½)] = ħ²[3.75 + 0.25] = 4ħ², so the norm is 2ħ. Factored check: (j+m)(j−m+1) = (2)(2) = 4 ✓.
Top rung: ‖J₊|3⁄2, 3⁄2⟩‖² = ħ²[3.75 − (3⁄2)(5⁄2)] = ħ²[3.75 − 3.75] = 0, so J₊|3⁄2, 3⁄2⟩ = 0 and the ladder has nowhere to go.

Answerm = −3⁄2, −½, ½, 3⁄2 (dimension 4); ‖J₋|3⁄2, ½⟩‖ = 2ħ; |J| = 1.94ħ; J₊|3⁄2, 3⁄2⟩ = 0.

MediumUsing the Condon–Shortley matrix elements, build J₊ and Jₓ for j = 1 in the basis (|1,1⟩, |1,0⟩, |1,−1⟩), find the eigenvalues of Jₓ, and check Jₓ² + J_y² + J_z² against ħ²j(j+1).

Non-zero elements of J₊: ⟨1,1|J₊|1,0⟩ = ħ√((1−0)(1+0+1)) = ħ√2 and ⟨1,0|J₊|1,−1⟩ = ħ√((1+1)(1−1+1)) = ħ√2. So J₊ = ħ√2 × [[0,1,0],[0,0,1],[0,0,0]], and J₋ = J₊† is its transpose.
Jₓ = (J₊ + J₋)/2 = (ħ/√2) × [[0,1,0],[1,0,1],[0,1,0]]; J_y = (J₊ − J₋)/2i = (ħ/√2) × [[0,−i,0],[i,0,−i],[0, i,0]]; J_z = ħ diag(1, 0, −1).
Characteristic equation of [[0,1,0],[1,0,1],[0,1,0]]: −μ³ + 2μ = 0, so μ = 0, ±√2. Multiplying by ħ/√2 gives Jₓ eigenvalues 0 and ±ħ — identical to those of J_z, as rotational symmetry demands.
Squares: Jₓ² = (ħ²/2)[[1,0,1],[0,2,0],[1,0,1]], J_y² = (ħ²/2)[[1,0,−1],[0,2,0],[−1,0,1]], J_z² = ħ² diag(1,0,1). Sum: ħ² diag(1,2,1) + ħ² diag(1,0,1) = 2ħ² I = ħ² × 1 × (1+1) I ✓.

AnswerJₓ = (ħ/√2)[[0,1,0],[1,0,1],[0,1,0]] with eigenvalues −ħ, 0, +ħ; Jₓ² + J_y² + J_z² = 2ħ² I, the j = 1 value of ħ²j(j+1).

HardAn eigenstate |ψ⟩ = |j, m⟩ of J² and J_z satisfies ‖J₊|ψ⟩‖² = 8ħ² and ‖J₋|ψ⟩‖² = 9ħ². Determine m and j, the dimension of the representation, the raising coefficients up to the top rung, and ⟨ψ|Jₓ² + J_y²|ψ⟩.

Write both norms with λ = j(j+1): λ − m(m+1) = 8 and λ − m(m−1) = 9. Subtracting, m(m+1) − m(m−1) = 2m = 1, so m = ½.
Then λ = 8 + (½)(3⁄2) = 8.75 = 35⁄4. Solve j(j+1) = 35⁄4: 4j² + 4j − 35 = 0, j = (−4 + √(16 + 560))/8 = (−4 + 24)/8 = 5⁄2. The other root is negative, so j = 5⁄2 and the space has 2j + 1 = 6 rungs.
Steps to the top: j − m = 2. Coefficients from ħ√(j(j+1) − m(m+1)): from m = ½, ħ√(8.75 − 0.75) = ħ√8 = 2.83ħ; from m = 3⁄2, ħ√(8.75 − 3.75) = ħ√5 = 2.24ħ; from m = 5⁄2, ħ√(8.75 − 8.75) = 0 — the top rung, annihilated.
Transverse part: Jₓ² + J_y² = J² − J_z², so ⟨Jₓ² + J_y²⟩ = ħ²[j(j+1) − m²] = ħ²[8.75 − 0.25] = 8.5ħ². Cross-check: Jₓ² + J_y² = (J₊J₋ + J₋J₊)/2, giving (9 + 8)ħ²/2 = 8.5ħ² ✓.
Sanity check on the data: had the norms been 8ħ² and 12ħ², then m = 2 and λ = 14. An integer m needs an integer j, but 3⋅4 = 12 and 4⋅5 = 20 bracket 14, so no rung of any representation has those norms. The two norms of a genuine rung are never independent — the algebra ties them together.

Answerm = ½, j = 5⁄2 (six rungs); raising coefficients 2√2ħ then √5ħ, then zero at the top; ⟨Jₓ² + J_y²⟩ = 8.5ħ².