University Physics V · Angular Momentum in Quantum Mechanics · 9.4

Angular Momentum Matrices & the Uncertainty Bound

The ladder argument handed you a spectrum and no object to compute with. This topic hands you the object: one (2j+1)-square array per component, filled from a single square root, that you can multiply, commute and diagonalise in ten lines of NumPy — and then read the uncertainty bound straight off the commutator.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Angular momentum was quantised by algebra alone, and the same algebra is what makes it computable. Fix the basis |j, m⟩ of simultaneous eigenvectors of J² and J_z, ordered m = j down to −j, and every component becomes a finite Hermitian matrix whose entries are the inner products ⟨j, m′|Jᵢ|j, m⟩. J_z is diagonal with mħ; J₊ is a single stripe above the diagonal carrying ħ√(j(j+1) − m(m+1)); J₋ is its adjoint; Jₓ and J_y are the Hermitian combinations (J₊ + J₋)/2 and (J₊ − J₋)/2i, each entry half a ladder element.

Nothing else is needed. The commutators, the Casimir J² = ħ² j(j+1) I, the rotation exp(−iθ n⋅J/ħ) and the spectrum of J along any axis all fall out of matrix arithmetic, and a commutator residual near machine ε is the proof that your arrays are the representation and not a transcription error. The price is written in the same algebra: [Jₓ, J_y] = iħJ_z fed into the Robertson inequality returns ΔJₓ ΔJ_y ≥ (ħ/2)|⟨J_z⟩|, a floor that moves with the state and forbids sharp Jₓ and J_y whenever ⟨J_z⟩ ≠ 0.

In |j, m⟩ itself the transverse spreads are each ħ√((j(j+1) − m²)/2), never zero, so the familiar cone of half-angle cos θ = m/√(j(j+1)) is a diagram of variances: it has no azimuth, nothing on it moves, and it is not a trajectory.

Simple definition: The matrix representation of angular momentum writes Jₓ, J_y and J_z as (2j+1)×(2j+1) Hermitian matrices whose entries are ⟨j, m′|Jᵢ|j, m⟩ in the eigenbasis of J² and J_z, so operator identities become checks on matrix products.
Example: For j = 1 the entries are J_z = ħ diag(1, 0, −1) and Jₓ = (ħ/√2) times the tridiagonal matrix with 1s beside the diagonal; multiplying out JₓJ_y − J_yJₓ returns iħ diag(1, 0, −1) entry by entry, which is iħJ_z.

Ladder matrix element⟨j, m+1|J₊|j, m⟩ = ⟨j, m|J₋|j, m+1⟩ = ħ √(j(j+1) − m(m+1))

The only input. J₊ is the stripe just above the diagonal in the m-descending ordering, J₋ = J₊† the stripe just below, and every entry of Jₓ and J_y is half of one of these numbers.

j, m dimensionless; ħ = 1.055 × 10⁻³⁴ J s carries the unit; the Condon–Shortley phase makes every entry real and ≥ 0

Cartesian components from the laddersJₓ = (J₊ + J₋)/2, J_y = (J₊ − J₋)/2i, J_z = ħ diag(j, j−1, …, −j)

Hermiticity is built in because J₋ = J₊†. The 1/2 and 1/2i are where hand-built matrices most often go wrong, and the error shows up later as every variance four times too large.

ordering m = j down to −j; Jₓ real symmetric, J_y imaginary antisymmetric, J_z real diagonal — all three Hermitian

Algebra check[Jₓ, J_y] = iħ J_z, Jₓ² + J_y² + J_z² = ħ² j(j+1) I

Two lines of NumPy that catch a wrong sign, a dropped half or a mis-ordered basis before anything downstream inherits the mistake.

I the (2j+1)-square identity; in float64 the Frobenius norm of JₓJ_y − J_yJₓ − iħJ_z should sit near 10⁻¹⁵ ħ²

Robertson bound for Jₓ and J_yΔJₓ ΔJ_y ≥ (ħ/2) |⟨J_z⟩|

Read straight off σ_A σ_B ≥ ½|⟨[A, B]⟩| with [Jₓ, J_y] = iħJ_z. Cyclic permutation gives the other two pairs, and all three go empty wherever the mean on the right vanishes.

ΔJᵢ = √(⟨Jᵢ²⟩ − ⟨Jᵢ⟩²), in units of ħ; the right side is an expectation in the same state, so the floor moves with the state

Spreads in an eigenstate |j, m⟩⟨Jₓ⟩ = ⟨J_y⟩ = 0, ΔJₓ = ΔJ_y = ħ √((j(j+1) − m²)/2)

Never zero, even at m = j, where product and bound both equal j ħ²/2: the stretched state saturates the inequality and every other m sits strictly above it.

from Jₓ² + J_y² = J² − J_z² and the x–y symmetry of the state; the product is (j(j+1) − m²) ħ²/2 against a bound of |m| ħ²/2

Spectrum along any axiseigenvalues of n⋅J = nₓ Jₓ + n_y J_y + n_z J_z are mħ, m = −j, …, j

Rotational invariance made concrete: diagonalise n⋅J numerically for a random n and the same 2j+1 numbers return, each with a new eigenvector.

n a unit vector; the eigenvectors are U|j, m⟩, U the rotation carrying z into n — they change while the eigenvalues do not

Choose the basis before you write a single entry

A matrix element is meaningless until the basis is named. Take the 2j+1 simultaneous eigenvectors |j, m⟩ of J² and J_z, orthonormal, and order them m = j, j−1, …, −j from the top row down. That ordering is a convention, but it is the convention every Clebsch–Gordan table and every NumPy script assumes, and changing it permutes the rows and columns of everything that follows. In this basis J_z is already finished: ⟨j, m′|J_z|j, m⟩ is mħ when m′ = m and zero otherwise, a real diagonal matrix reading ħ diag(j, …, −j). J² is finished too, and it is invisible: ħ² j(j+1) times the identity, so within one j it commutes with everything and cannot help you tell states apart. All the content sits in the two components that fail to commute with J_z, and those you have to build.

Assemble J₊ from one square root, then take halves

The norm of J₊|j, m⟩ is fixed by the algebra: ⟨j, m|J₋J₊|j, m⟩ = ⟨J² − J_z² − ħJ_z⟩ = ħ²(j(j+1) − m(m+1)), and with the Condon–Shortley choice of phases its root is real and non-negative, so ⟨j, m+1|J₊|j, m⟩ = ħ√(j(j+1) − m(m+1)). For j = 3/2 the three entries are ħ√3, 2ħ and ħ√3, for the steps from m = 1/2 to 3/2, from −1/2 to 1/2, and from −3/2 to −1/2. In the m-descending ordering they sit on the stripe just above the diagonal, J₋ = J₊† puts the same numbers just below, and Jₓ = (J₊ + J₋)/2 is the real symmetric tridiagonal matrix with (√3, 2, √3)ħ/2 beside the diagonal. J_y = (J₊ − J₋)/2i carries the same magnitudes, as −i times them above the diagonal and +i times them below. The halves are not decoration: leave them out and every variance you compute later comes out four times too big.

Verify the algebra numerically before trusting anything else

In NumPy, with ħ = 1: m = np.arange(j, −j - 1, −1); off = np.√(j*(j+1) - m[:-1]*m[1:]); Jp = np.diag(off, 1); Jm = Jp.conj().T; Jx = (Jp + Jm)/2; Jy = (Jp - Jm)/(2*1j); Jz = np.diag(m). The product m[:-1]*m[1:] equals m(m+1) for the lower member of each adjacent pair, which is what the ladder formula wants. Then test: np.linalg.norm(Jx@Jy - Jy@Jx - 1j*Jz) should return something like 3 × 10⁻¹⁶, and Jx@Jx + Jy@Jy + Jz@Jz should equal j(j+1) times the identity to the same precision. By hand at j = 1, JₓJ_y is (ħ²/2) times the matrix with rows (i, 0, −i), (0, 0, 0), (i, 0, −i), while J_yJₓ has rows (−i, 0, −i), (0, 0, 0), (i, 0, i); subtracting gives (ħ²/2) diag(2i, 0, −2i) = iħ · ħ diag(1, 0, −1) = iħJ_z. A residual of order 1 rather than 10⁻¹⁵ almost always means a dropped half, a conjugate missed on J₋, or m ordered upward while the stripe was placed above the diagonal.

Read the uncertainty bound off the commutator

The Robertson inequality σ_A σ_B ≥ ½|⟨[A, B]⟩| needs one commutator and one state. Feed it [Jₓ, J_y] = iħJ_z and it returns ΔJₓ ΔJ_y ≥ (ħ/2)|⟨J_z⟩| — a floor that depends on the state through ⟨J_z⟩, unlike the constant ħ/2 for x and p. Evaluate it in |j, m⟩. Jₓ only connects m to m ± 1, so ⟨Jₓ⟩ = 0, and likewise ⟨J_y⟩ = 0. For the second moments use Jₓ² + J_y² = J² − J_z², whose expectation is ħ²(j(j+1) − m²); the state is symmetric under rotations about z, so the two shares are equal and ⟨Jₓ²⟩ = ⟨J_y²⟩ = ħ²(j(j+1) − m²)/2. The product ΔJₓ ΔJ_y is therefore (j(j+1) − m²)ħ²/2 against a bound of |m|ħ²/2. At j = 1, m = 1 both sides equal ħ²/2: the stretched state saturates the inequality. At j = 1, m = 0 the product is ħ² while the bound is zero — satisfied, but the theorem has gone silent, and only the direct calculation says how spread Jₓ actually is.

The cone is a plot of variances, not a path

Draw |J| = ħ√(j(j+1)) at the angle cos θ = m/√(j(j+1)) to the z-axis and sweep it round: that is the textbook cone, and it encodes exactly two numbers. Its height is ⟨J_z⟩ = mħ; its rim radius is √(⟨Jₓ²⟩ + ⟨J_y²⟩) = ħ√(j(j+1) − m²), the root-mean-square transverse spread. For j = 1, m = 1 the half-angle is 45° and the rim radius is ħ — as large as the height. Nothing in the picture is a position on the rim, because the state has no azimuth: ⟨Jₓ⟩ and ⟨J_y⟩ are zero, and a standard deviation has no direction to point in. Nothing in it moves either. Without a Hamiltonian |j, m⟩ is stationary, and with H = ωJ_z it acquires the phase e^(−imωt), which changes no expectation value at all. Precession is real only for a superposition of different m, and even then it is ⟨J⟩ that rotates, not a tip riding a circle. The reason the arrow cannot lie along z even at m = j is the same bookkeeping: √(j(j+1)) exceeds j because ΔJₓ = ΔJ_y = ħ√(j/2) is not zero in the top state.

Rotate the axis and the spectrum stays put

The eigenvalues of n⋅J = nₓJₓ + n_yJ_y + n_zJ_z are the same 2j+1 numbers mħ for every unit vector n, because n⋅J is unitarily equivalent to J_z through the rotation that carries z into n. At j = 1 the eigenvector of Jₓ with eigenvalue +ħ is (1, √2, 1)/2 in the (|1, 1⟩, |1, 0⟩, |1, −1⟩) ordering — check it: Jₓ on that vector gives (ħ/√2)(√2, 2, √2)/2, which is ħ times the vector back. In that state ⟨J_z⟩ = ħ(1/4 − 1/4) = 0, so the bound on ΔJₓ ΔJ_y is zero and is met with ΔJₓ = 0 exactly; but the cyclic partner ΔJ_y ΔJ_z ≥ (ħ/2)|⟨Jₓ⟩| = ħ²/2 is live, and ⟨J_y²⟩ = ⟨J_z²⟩ = ħ²/2 makes it saturate. Numerically this is the whole lab: build n⋅J for a random n, call np.linalg.eigh, confirm the spectrum, then compute variances in each eigenvector and compare with (ħ/2)|⟨n″⋅J⟩| for a perpendicular pair. Within a single j the spectrum of n⋅J is non-degenerate, so eigh returns the eigenvectors unambiguously; stack several j and the m-values repeat, eigh hands back an arbitrary basis of each eigenspace, and a variance computed there is a property of that choice, not of the physics.

Change one variable at a time

Make the relationship visible.

Interactive model

Angular momentum j1.0

Steps k below the top state0

At j = 1 move k from 0 to 1: m drops to 0, the open ring falls to the axis as the bound vanishes while the filled dot climbs to the peak — then return k to 0 and raise j to 3: dot and ring coincide again at the right-hand end, yet the cone still opens by √3 ħ in the stretched state.

m = j − k1.0

ΔJx = ΔJy0.707 ħ

PRODUCT ΔJx⋅ΔJy0.500 ħ²

BOUND ħ|⟨Jz⟩|/20.500 ħ²

Live interpretationm = j − k: 1.0. ΔJx = ΔJy: 0.707 ħ. PRODUCT ΔJx⋅ΔJy: 0.500 ħ². BOUND ħ|⟨Jz⟩|/2: 0.500 ħ²

Catch the common trap

Explain before calculating.

Misconception check

The error is to read the cone as a real vector that has a definite length ħ√(j(j+1)), sits at a definite tilt, and sweeps round the z-axis so fast that its x and y components average to zero. Nothing in the state supports any of that. Measure Jₓ in |j, m⟩ and you get one of the same 2j+1 values m′ħ, never the projection of a tilted arrow; the transverse quantity the cone draws is ΔJₓ = ΔJ_y = ħ√((j(j+1) − m²)/2), a standard deviation, and a standard deviation has no azimuth to precess through. Nor is there any motion to average away: |j, m⟩ is stationary, and even under H = ωJ_z its only change is a global phase. The cone is variance bookkeeping — height ⟨J_z⟩, rim radius √(⟨Jₓ²⟩ + ⟨J_y²⟩) — and it is the picture of why ⟨Jₓ²⟩ can never be zero while [Jₓ, J_y] = iħJ_z has a non-zero mean.

A spin-1 system (j = 1) is prepared in |1, 1⟩, the m = +1 eigenstate of J_z. What is ΔJₓ ΔJ_y in that state, and how does it compare with the Robertson bound?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyUse the ladder matrix element ⟨j, m+1|J₊|j, m⟩ = ħ√(j(j+1) − m(m+1)) to build J₊, J₋, Jₓ and J_y for j = 1/2 in the basis (|½, ½⟩, |½, −½⟩), and identify the Pauli matrices.

Only one raising step exists, from m = −½ to m = +½: the element is ħ√(¾ − (−½)(½)) = ħ√(¾ + ¼) = ħ.
In the m-descending ordering that entry sits in row 1, column 2: J₊ = ħ [[0, 1], [0, 0]], and J₋ = J₊† = ħ [[0, 0], [1, 0]].
Jₓ = (J₊ + J₋)/2 = (ħ/2) [[0, 1], [1, 0]] = (ħ/2) σₓ, and J_y = (J₊ − J₋)/2i = (ħ/2) [[0, −i], [i, 0]] = (ħ/2) σ_y.
J_z = ħ diag(½, −½) = (ħ/2) σ_z. Check: [Jₓ, J_y] = (ħ²/4)[σₓ, σ_y] = (ħ²/4)(2iσ_z) = iħ (ħ/2) σ_z = iħJ_z.

AnswerJₓ = (ħ/2)σₓ, J_y = (ħ/2)σ_y, J_z = (ħ/2)σ_z — the j = ½ representation is the Pauli matrices scaled by ħ/2, and the commutator closes.

MediumFor j = 3/2, write the non-zero entries of J₊, build Jₓ, and find ΔJₓ in the state |3/2, 1/2⟩. Compare ΔJₓ ΔJ_y with the Robertson bound (ħ/2)|⟨J_z⟩|.

Ladder entries ħ√(15/4 − m(m+1)) for the lower member m of each pair: m = ½ gives ħ√(15/4 − 3/4) = ħ√3; m = −½ gives ħ√(15/4 + 1/4) = 2ħ; m = −3/2 gives ħ√(15/4 − 3/4) = ħ√3.
In the ordering (3/2, 1/2, −1/2, −3/2) these fill the stripe above the diagonal of J₊; Jₓ = (J₊ + J₊†)/2 is symmetric tridiagonal with (√3, 2, √3)ħ/2 on both off-diagonals.
⟨Jₓ⟩ = 0 because Jₓ has no diagonal entries. The diagonal entry of Jₓ² in the m = ½ row is the sum of the squares of the two off-diagonal entries touching it: (ħ²/4)(3 + 4) = 7ħ²/4.
So ΔJₓ = ħ√7/2 ≈ 1.32ħ; by the x–y symmetry of the state ΔJ_y is the same, and the shortcut ħ²(j(j+1) − m²)/2 = ħ²(15/4 − 1/4)/2 = 7ħ²/4 agrees.
Product ΔJₓ ΔJ_y = 7ħ²/4 = 1.75ħ². Bound: (ħ/2)|⟨J_z⟩| = (ħ/2)(ħ/2) = ħ²/4 = 0.25ħ². The inequality holds by a factor of 7 — far from saturated, as every m below j is.

AnswerΔJₓ = ΔJ_y = (√7/2)ħ ≈ 1.32ħ; ΔJₓ ΔJ_y = 1.75ħ² against a bound of 0.25ħ².

HardFor j = 1, find the normalised eigenvector of Jₓ with eigenvalue +ħ in the basis (|1, 1⟩, |1, 0⟩, |1, −1⟩). In that state compute ⟨J_z⟩, ΔJ_y and ΔJ_z, and test the cyclic bound ΔJ_y ΔJ_z ≥ (ħ/2)|⟨Jₓ⟩|.

Jₓ = (ħ/√2) times the tridiagonal matrix with 1s beside the diagonal. Writing v = (a, b, c), the eigen-equation Jₓ v = ħv reads b/√2 = a, (a + c)/√2 = b and b/√2 = c.
So b = √2 a and c = a; normalising, |a|² + 2|a|² + |a|² = 1 gives a = ½ and v = (1, √2, 1)/2. Check: Jₓ v = (ħ/√2)(√2, 2, √2)/2 = ħ(1, √2, 1)/2.
⟨J_z⟩ = ħ(¼ · 1 + ½ · 0 + ¼ · (−1)) = 0 and ⟨J_z²⟩ = ħ²(¼ + 0 + ¼) = ħ²/2, so ΔJ_z = ħ/√2.
⟨J_y⟩: J_y v = (ħ/√2)(−i√2, i − i, i√2)/2 = (ħ/2)(−i, 0, i), and v · (J_y v) = (ħ/4)(−i + i) = 0. From Jₓ² + J_y² + J_z² = 2ħ²I with ⟨Jₓ²⟩ = ħ²: ⟨J_y²⟩ = 2ħ² − ħ² − ħ²/2 = ħ²/2, so ΔJ_y = ħ/√2.
ΔJ_y ΔJ_z = ħ²/2, and the bound is (ħ/2)|⟨Jₓ⟩| = (ħ/2)(ħ) = ħ²/2: saturated, as for a stretched state along any axis. Meanwhile ΔJₓ = 0 and ⟨J_z⟩ = 0, so ΔJₓ ΔJ_y ≥ 0 holds trivially.

Answerv = (1, √2, 1)/2; ⟨J_z⟩ = 0; ΔJ_y = ΔJ_z = ħ/√2 ≈ 0.707ħ; ΔJ_y ΔJ_z = ħ²/2 equals the bound ħ²/2.