Angular momentum matrices and the uncertainty bound

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.

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.

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.

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.

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.

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 Δ(n′⋅J) Δ(n″⋅J) with (ħ/2)|⟨(n′ × n″)⋅J⟩| for a perpendicular pair of unit vectors n′ and n″. 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.