Hermitian operators and orthogonality

The measurement postulate does not treat operators gently. It says the only possible results of measuring an observable are the eigenvalues of its operator, and that if a state is written as Ψ = Σ cₙψₙ over the eigenfunctions, then |cₙ|² is the probability of the result λₙ. Two things have to be true for that to make sense. The eigenvalues must be real, because no instrument has ever read 3 + 2i. And the coefficients must be extractable, with their squared moduli summing to one, which needs ⟨ψₙ|ψₘ⟩ = 0 for n ≠ m — without it, projecting Ψ onto ψₙ picks up contamination from every other term and the probabilities do not add to one. Neither demand is automatic. A general linear operator has complex eigenvalues and non-orthogonal eigenfunctions, so the formalism needs a filter that admits only the operators fit to represent an observable. Hermiticity is that filter, and everything below is the receipt.

Write Ĥ = −(ℏ²/2m) d²/dx² + V(x) and compare ∫f*(Ĥg) dx with ∫(Ĥf)*g dx on a domain a ≤ x ≤ b. The potential contributes ∫(V − V*)f*g dx, which cancels only if V is real. The kinetic term needs two integrations by parts: ∫f*g″ dx = [f*g′]ₐᵇ − ∫(f*)′g′ dx, and ∫(f″)*g dx = [(f′)*g]ₐᵇ − ∫(f*)′g′ dx. The two interior integrals are identical and subtract away, leaving ∫[f*(Ĥg) − (Ĥf)*g] dx = −(ℏ²/2m)[f*g′ − (f′)*g]ₐᵇ. So hermiticity is not an algebraic identity: it is the claim that this one bracket vanishes for every pair of functions the problem admits. Three settings kill it. Square-integrable states on the whole line have f, g → 0 as |x| → ∞. A hard-walled box sets f = g = 0 at both walls. A ring imposes periodicity, so the two ends cancel each other. A Robin condition f′ = αf at an end works too — but only when α is real, since the bracket there collapses to (α − α*)f*g.

Put f = g = ψ, an eigenfunction with Ĥψ = λψ. The left side of the Hermitian condition is ∫ψ*(λψ) dx = λ∫|ψ|² dx; the right side is ∫(λψ)*ψ dx = λ*∫|ψ|² dx. Subtract: (λ − λ*)∫|ψ|² dx = 0. An eigenfunction is by definition not the zero function, so ∫|ψ|² dx is strictly positive and can be divided out, forcing λ = λ*. The proof is three lines, but every clause is load-bearing. It used the conjugation rule (λψ)* = λ*ψ*, which is exactly where a complex eigenvalue would have to appear, and it used the positivity of the norm, which is why the argument says nothing about the zero function. It also shows precisely what breaks the result: give the potential a constant imaginary part, V = V₀ − iΓ/2, and the potential term no longer cancels. The same algebra then returns Im λ = −Γ/2 instead of zero, and the norm of the state decays rather than holding at one.

Now take two eigenfunctions, Ĥψₙ = λₙψₙ and Ĥψₘ = λₘψₘ, and feed them in as f = ψₙ, g = ψₘ. The left side is λₘ⟨ψₙ|ψₘ⟩; the right side is λₙ*⟨ψₙ|ψₘ⟩, which is λₙ⟨ψₙ|ψₘ⟩ because reality has already been proved. Subtracting gives (λₘ − λₙ)⟨ψₙ|ψₘ⟩ = 0, so wherever the two eigenvalues differ the overlap must be zero. Check it on the infinite well, ψₙ = √(2/L) sin(nπx/L): the product-to-sum identity turns ⟨ψ₁|ψ₃⟩ into the integral of (1/L)[cos(2πx/L) − cos(4πx/L)] across the well, and both sines vanish at both limits. What orthogonality buys is the entire probabilistic reading. Expand Ψ = Σ cₙψₙ, multiply by ψₘ* and integrate; every cross term dies and cₘ = ⟨ψₘ|Ψ⟩ survives alone. Normalisation then gives Σ|cₙ|² = 1, so the numbers |cₙ|² are a genuine probability distribution rather than a set of weights that merely resembles one.

The orthogonality proof has a visible hole. If λₘ = λₙ the factor (λₘ − λₙ) is zero, so (λₘ − λₙ)⟨ψₙ|ψₘ⟩ = 0 is satisfied by any overlap whatsoever and makes no claim at all. Degeneracy is not exotic: on a ring of circumference L the states cos(2πx/L) and sin(2πx/L) share the energy h²/(2mL²), and so does every linear combination of them. Take u = √(2/L) cos(2πx/L) and w = (1/√L)[cos(2πx/L) + sin(2πx/L)]. Both are normalised, both are eigenfunctions of the same Ĥ with the same eigenvalue, and ⟨u|w⟩ = 1/√2 = 0.707 — an angle of 45°, not 90°. Orthogonality inside a degenerate subspace is therefore imposed, not derived. Gram–Schmidt does it mechanically: subtract the projection, w − ⟨u|w⟩u, which here returns (1/√L) sin(2πx/L), then renormalise to √(2/L) sin(2πx/L). The physicist's version is to find a second observable that commutes with Ĥ — here p̂ — and label the states by its eigenvalues, which picks out exp(±2πix/L) as the natural orthogonal pair.

Ask whether p̂ = −iℏ d/dx is Hermitian and the honest answer is: on which domain? One integration by parts leaves the bracket −iℏ[f*g]ₐᵇ. On the whole line, with square-integrable states, it vanishes and p̂ is fine. On a ring with f(L) = f(0) it vanishes too, and p̂ then carries a complete orthonormal set exp(2πinx/L)/√L with real eigenvalues pₙ = nh/L — for L = 1.00 nm that is 6.63 × 10⁻²⁵ kg m s⁻¹ per unit of n. Now put the same symbol on the infinite well, 0 ≤ x ≤ L with ψ(0) = ψ(L) = 0. The bracket still vanishes, so p̂ passes the symmetry test — and yet p̂ψ = pψ forces ψ ∝ exp(ipx/ℏ), whose modulus is constant, so nothing in the domain can meet the walls. The operator has no eigenfunctions there at all: it is symmetric but not self-adjoint, and cannot represent a measurable momentum. Ĥ = p̂²/2m is untroubled, which is why the well still has a clean energy spectrum. Same three symbols, three domains, three different answers — which is why the boundary conditions belong in the statement of the operator, not in a footnote.