Compatible observables and simultaneous eigenbases

Two Hermitian operators A and B on the same space are compatible when [A, B] = AB − BA = 0. One direction of the theorem is immediate. Suppose an orthonormal basis (|k⟩) exists with A|k⟩ = aₖ|k⟩ and B|k⟩ = bₖ|k⟩. Then AB|k⟩ = aₖ bₖ|k⟩ = BA|k⟩ for every k, and two linear operators that agree on a basis agree on every vector, so [A, B] = 0. Notice what is not required: neither spectrum has to be non-degenerate, and the lists aₖ and bₖ may repeat as often as they like. The contrast is sharp in C². [σ_z, σₓ] = 2iσ_y ≠ 0, and σ_z has eigenvectors |↑⟩ and |↓⟩; but σₓ|↑⟩ = |↓⟩, which is orthogonal to |↑⟩ rather than proportional to it, so that pair shares not one eigenvector, let alone a basis.

Now assume [A, B] = 0 and let A|a⟩ = a|a⟩. Act with A on the vector B|a⟩: A(B|a⟩) = B(A|a⟩) = a(B|a⟩). So B|a⟩ is either the zero vector or another eigenvector of A with the same eigenvalue — B maps the eigenspace Vₐ into itself, and it does so for every a at once. If a is non-degenerate, Vₐ is a line, B|a⟩ has nowhere else to go, and B|a⟩ = b|a⟩: the joint eigenvector was already sitting there, with no choice to make. Keep the corollary. If A has a completely non-degenerate spectrum, then every B commuting with A is diagonal in A's eigenbasis and is therefore a function of A, so A on its own is a complete set. That is why Ĥ alone labels every state of the one-dimensional box, and why in a three-dimensional central potential, where every l ≥ 1 level is (2l + 1)-fold degenerate, it cannot.

If dim Vₐ = d > 1, A has not chosen a basis inside Vₐ and any orthonormal set will do. Because B maps Vₐ into itself, the restriction of B to Vₐ is a Hermitian operator on a d-dimensional space, and the spectral theorem hands you an orthonormal eigenbasis of Vₐ for it. Do that in every eigenspace of A and collect the results: a complete joint basis, built by diagonalising blocks. The matrix statement is the same fact. Write B in any eigenbasis of A; then (a − a′)⟨a|B|a′⟩ = ⟨a|[A, B]|a′⟩ = 0, so every element linking different eigenvalues of A vanishes and B is block diagonal, one block per distinct a. Take A = diag(1, 1, 4) and B with rows (2, 1, 0), (1, 2, 0) and (0, 0, 7). The a = 1 block is [[2, 1], [1, 2]], whose eigenvalues are 3 and 1, so the joint labels are (1, 3), (1, 1) and (4, 7) — three distinct pairs, and no degeneracy left.

np.linalg.eigh(A) returns an orthonormal eigenbasis, but inside a degenerate eigenspace LAPACK's choice is arbitrary: run it on diag(1, 1, 4) and you get e₁ and e₂, neither of which is an eigenvector of the B above. Two recipes fix that. Take a generic combination and diagonalise A + λB, choosing λ so that no two joint labels collide in a + λb; λ = 0.618 sends (1, 3), (1, 1) and (4, 7) to 2.854, 1.618 and 8.326, so the eigenvectors of A + λB are joint eigenvectors, and you read a and b back as ⟨k|A|k⟩ and ⟨k|B|k⟩. Or work block by block: group the eigenvalues of A within a tolerance you state out loud, project with P, diagonalise the small matrix P†BP, and map back. Audit either route with np.linalg.norm(A@B − B@A). At 1e−14 the pair commutes; at 1e−2 you have a physics error, not a rounding one.

Keep adding pairwise-commuting observables until every list of eigenvalues names one ray. For any central potential [Ĥ, L²] = [Ĥ, L_z] = [L², L_z] = 0, so the three can be diagonalised together. Spinless hydrogen at n = 3: E₃ = −1.511 eV is nine-fold degenerate, so energy alone names a nine-dimensional subspace. L² splits it into l = 0, 1, 2 with eigenvalues 0, 2ℏ² and 6ℏ² and multiplicities 1, 3 and 5, summing to 9 but still leaving degeneracy standing; L_z, whose eigenvalues mℏ run from −lℏ to +lℏ, then separates the states inside each block. Every triple (n, l, m) now occurs exactly once, so (Ĥ, L², L_z) is complete — and restoring spin with S_z gives 2n² = 18. Complete does not mean unique: (Ĥ, L², S², J², J_z) is an equally good set, the unitary between the two bases is the Clebsch–Gordan matrix, and which one you want is decided by the next term in the Hamiltonian, not by the states.

First, it produces a basis, not a property of a state: (|a₁, b₁⟩ + |a₂, b₂⟩)/√2 is an ordinary state of a compatible pair with a definite value of neither. Second, non-commuting operators can still share isolated eigenvectors. Lₓ, L_y and L_z pairwise fail to commute, yet all three annihilate |l = 0, m = 0⟩, which is a joint eigenvector of every one of them; what commuting supplies is a complete common basis, not the first shared vector. The Robertson bound carries the same warning: in |l = 1, m = 0⟩ we have ⟨L_z⟩ = 0, so σ(Lₓ)σ(L_y) ≥ ½|⟨[Lₓ, L_y]⟩| = 0, while the true product is ℏ². Third, for unbounded operators AB and BA need not share a domain, and Nelson's example gives two essentially self-adjoint operators commuting on a dense invariant domain whose spectral projections do not commute. The definition that survives is that the unitary groups commute: e^(isA)e^(itB) = e^(itB)e^(isA) for all real s, t.