Hermitian operators and the spectral theorem

The adjoint † is whatever operator lets  cross the inner product: ⟨φ|Âψ⟩ = ⟨†φ|ψ⟩ for every pair of states. Choose an orthonormal basis and the abstract definition becomes a recipe: (†)ᵢⱼ = (Âⱼᵢ)* — transpose, then conjugate every entry. A real symmetric matrix passes; so does σ_y = [[0, −i],[i, 0]], because its imaginary entries mirror into their conjugates; but [[0, i],[i, 0]] fails, since its adjoint is [[0, −i],[−i, 0]], the negative of itself. Two facts fall out before any eigenvalue is computed. Diagonal matrix elements must equal their own conjugates, so they are real. And ⟨ψ|Âψ⟩ = ⟨Âψ|ψ⟩ = ⟨ψ|Âψ⟩* for every state, so every expectation value of a Hermitian operator is real. Keep the physics convention straight: the inner product is conjugate-linear in its first slot, so the star lands on the bra side of every coefficient.

Both classic theorems are two-line computations with the same move: evaluate a matrix element twice, once with  acting rightward and once leftward. Take Â|a⟩ = a|a⟩ with ⟨a|a⟩ = 1. Then ⟨a|Â|a⟩ read rightward is a; read leftward, using † = Â, it is a*. So a = a*: every eigenvalue is real. Now take a second eigenvector with a′ ≠ a and evaluate ⟨a|Â|a′⟩ both ways: rightward gives a′⟨a|a′⟩, leftward gives a⟨a|a′⟩, reality already banked. Subtracting, (a − a′)⟨a|a′⟩ = 0, and since a ≠ a′ the overlap must vanish: outcome states of distinct results are orthogonal. Notice what the proof does not say. Two eigenvectors sharing the same eigenvalue make the bracket a − a = 0, so their overlap is unconstrained — degeneracy is where orthogonality stops being automatic and becomes your job.

A degenerate eigenvalue owns a subspace, not a direction: if Â|u⟩ = a|u⟩ and Â|v⟩ = a|v⟩, then every combination α|u⟩ + β|v⟩ is again an eigenvector with the same a. So "the" eigenvectors of a degenerate level are a choice of basis, and you make that choice orthonormal by running Gram–Schmidt inside each degenerate block. The spectral theorem guarantees the pieces assemble: a Hermitian operator on a finite-dimensional space has a complete orthonormal eigenbasis of the whole space and is rebuilt from it as  = Σ aₙ|aₙ⟩⟨aₙ|, with Σ |aₙ⟩⟨aₙ| = 1. Completeness is the load-bearing word: expand |ψ⟩ = Σ cₙ|aₙ⟩ and Parseval gives Σ|cₙ|² = 1, so the outcome probabilities exhaust certainty. Contrast a matrix that is not Hermitian: [[1, 1],[0, 1]] has eigenvalue 1 twice but only the single eigenvector direction (1, 0). It is defective — no eigenbasis exists — and no Hermitian matrix can ever be.

The physical decode of the eigenvalue equation: measuring  can return only its eigenvalues, and returns aₙ with certainty exactly when the state lies in that eigenvalue's eigenspace: for a nondegenerate aₙ, the state must be |aₙ⟩ up to a phase. For any other state, expand on the eigenbasis and read off probabilities. Take σₓ = [[0, 1],[1, 0]] and prepare |ψ⟩ = (1, 0). The eigenvectors are (1, 1)/√2 and (1, −1)/√2 with eigenvalues ±1; both overlaps are 1/√2, so each outcome appears with probability ½ and ⟨σₓ⟩ = 0 — which matches ⟨ψ|σₓψ⟩ = 0 computed directly, no diagonalisation needed. Guard against the standard misreading: Â|ψ⟩ is not "the state after the measurement", and it is not "a number times the state" unless |ψ⟩ was an eigenvector already. Here σₓ(1, 0) = (0, 1), a different state entirely, because (1, 0) is a basis vector of the representation, not of the observable.

Let [Â, B̂] = 0 and let |a⟩ be an eigenvector of  with a nondegenerate eigenvalue. Then Â(B̂|a⟩) = B̂Â|a⟩ = a(B̂|a⟩): the vector B̂|a⟩ sits in the same one-dimensional eigenspace, so B̂|a⟩ ∝ |a⟩, and the eigenvector of  is automatically an eigenvector of B̂. When the eigenvalue is degenerate, B̂ is only guaranteed to map the block into itself; diagonalise B̂ within the block and the shared basis appears. Physically, commuting Hermitian operators are compatible: one experiment can pin sharp values of both at once, and a complete set of commuting observables labels every basis state by its list of simultaneous eigenvalues — precisely what (n, ℓ, m_ℓ, mₛ) will do for hydrogen later in the course. Non-commuting pairs such as x̂ and p̂ share no eigenbasis at all, which is where Unit 8's uncertainty relation will start.

On the function space L², moving an operator across the inner product is an integration by parts, and a boundary term rides along. For p̂ = −iℏ d/dx, ⟨φ|p̂ψ⟩ − ⟨p̂φ|ψ⟩ = −iℏ[φ*ψ] evaluated at the endpoints, so p̂ is Hermitian only on functions whose boundary conditions kill that term: periodic on a ring, vanishing at ±∞ on the line. Killing the boundary term is necessary, not sufficient — on the half-line x ≥ 0 with ψ(0) = 0 the term vanishes, so p̂ is symmetric, yet it admits no self-adjoint version, and no momentum eigenbasis exists there at all. On the ring the eigenfunctions e^(ikx) are normalisable because k is quantised, but on the line they are not square-integrable; they survive as distributions normalised to ⟨p|p′⟩ = δ(p − p′). The working escape used all course: put the problem on N grid points, build the boundary condition into the finite-difference stencil, and hand the resulting N×N Hermitian matrix to numpy.linalg.eigh, which returns real eigenvalues in ascending order and an orthonormal eigenbasis in the columns of V.