University Physics IV · The Schrödinger Equation · 6.6

Hermitian Operators & Orthogonality

Every number a measurement can return has to be real, and every expansion you read as a set of probabilities has to be orthogonal. Hermiticity is the one condition that delivers both — and because the proof runs on integration by parts, it is settled at the boundary of the domain, never by the formula you wrote down.

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The measurement postulate hands the operator formalism two bills to pay: the possible outcomes of an observable are the eigenvalues of its operator, so they must be real numbers, and the probability of each outcome is the squared modulus of an expansion coefficient, so the eigenfunctions must form a basis you can project onto. A single algebraic condition pays both: call an operator Hermitian on a space of functions when ∫f*(Ĥg) dx = ∫(Ĥf)*g dx for every admissible pair f, g. Feed one eigenfunction into that identity and reality falls out in a line; feed two, and eigenfunctions belonging to different eigenvalues are forced orthogonal.

What it costs is that the condition is never about the differential expression alone. Getting from one side to the other takes integration by parts, and each pass leaves a boundary term; hermiticity is precisely the claim that those terms cancel, which is a statement about the boundary conditions and the domain. Change the domain and the same symbols can stop being Hermitian, or lose their eigenfunctions entirely.

And the orthogonality proof carries a visible gap: where two eigenvalues coincide it multiplies the overlap by zero and says nothing, so inside a degenerate subspace orthogonality is imposed — by Gram–Schmidt, or by labelling the states with a second, commuting observable.

Simple definition: An operator is Hermitian on a given set of functions when ∫f*(Ĥg) dx equals ∫(Ĥf)*g dx for every pair f, g that the boundary conditions admit — a property of the expression and its domain together, never of the expression alone.
Example: On a ring of circumference L = 1.0 nm, periodicity ψ(L) = ψ(0) makes the leftover bracket −iℏ[f*g]ₐᵇ vanish, so p̂ = −iℏ d/dx is Hermitian there and its eigenvalues pₙ = nh/L are real: p₁ = 6.63 × 10⁻²⁵ kg m s⁻¹.

The Hermitian condition∫ f*(Ĥg) dx = ∫ (Ĥf)* g dx for all f, g in D(Ĥ)

One identity, and both the reality of eigenvalues and the orthogonality of eigenfunctions are corollaries of it.

Both integrals run over the whole domain; D(Ĥ) is the set of functions the boundary conditions admit.

Where it is decided: the boundary term∫[f*(Ĥg) − (Ĥf)*g] dx = −(ℏ²/2m)[f*g′ − (f′)*g]ₐᵇ

Hermiticity is the vanishing of this one bracket — a condition on the boundary, not an algebraic identity.

a and b are the ends of the domain; the two V terms cancel only when V is real.

Eigenvalues come out real(λ − λ*) ∫|ψ|² dx = 0 ⇒ λ = λ*

Guarantees that every possible outcome is a real number, which is what the measurement postulate needs.

ψ is a non-zero eigenfunction, so ∫|ψ|² dx is strictly positive and may be divided out.

Distinct eigenvalues force orthogonality(λₘ − λₙ) ⟨ψₙ|ψₘ⟩ = 0 ⇒ ⟨ψₙ|ψₘ⟩ = 0 for λₘ ≠ λₙ

Projection then works: cₙ = ⟨ψₙ|Ψ⟩ pulls one amplitude out of a superposition, and Σ|cₙ|² = 1.

⟨ψₙ|ψₘ⟩ = ∫ψₙ*ψₘ dx, dimensionless once both are normalised; the step uses the reality already proved.

Momentum: the domain does the choosing∫[f*(p̂g) − (p̂f)*g] dx = −iℏ [f* g]ₐᵇ

Same symbols, two domains, two different physics — the reason hermiticity is never read off a formula.

p̂ = −iℏ d/dx. Periodic f(b) = f(a) gives pₙ = nh/L; hard walls kill the bracket but leave no eigenfunction at all.

Gram–Schmidt inside a degenerate levelχ₂ = ψ₂ − ⟨χ₁|ψ₂⟩ χ₁ , then χ₂ → χ₂/‖χ₂‖

Orthogonality inside a degenerate level is imposed by you; the theorem never supplied it.

χ₁ is already normalised, and ψ₂ shares its eigenvalue, so χ₂ is still an eigenfunction of that same eigenvalue.

Two demands the postulate makes on an operator

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 squares 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.

Where the condition is settled: two integrations by parts

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.

Reality of the eigenvalues, in one line

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.

Orthogonality, in two, and what it buys

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.

Degeneracy: the proof multiplies by zero

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.

Hermitian is a property of the domain, not the symbol

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 = 1.0 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.

Change one variable at a time

Make the relationship visible.

Interactive model

Phase θ of state B0 °

Mode index n of state B1

Hold n at 1 so both curves share one energy, then sweep θ: the overlap runs from 1 to 0 to −1, so orthogonality inside a degenerate level is a choice. Move n to 2 and the overlap is pinned at zero for every θ — the theorem doing the work once the eigenvalues differ.

OVERLAP ⟨A|B⟩1.000

ENERGY RATIO E(B)/E(A)1

ANGLE BETWEEN STATES0.0 °

MEAN OF PRODUCT0.500

Live interpretationOVERLAP ⟨A|B⟩: 1.000. ENERGY RATIO E(B)/E(A): 1. ANGLE BETWEEN STATES: 0.0 °. MEAN OF PRODUCT: 0.500

Catch the common trap

Explain before calculating.

The operator p̂ = −iℏ d/dx is written on the interval 0 ≤ x ≤ L with the infinite well's boundary condition ψ(0) = ψ(L) = 0. Which statement about it is correct?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyAn electron sits in an infinite square well of width L = 0.500 nm, with normalised states ψₙ(x) = √(2/L) sin(nπx/L). Show by direct integration that ψ₁ and ψ₃ are orthogonal, then name the theorem that promised the result in advance.

Write the overlap: ⟨ψ₁|ψ₃⟩ = (2/L)∫ sin(πx/L) sin(3πx/L) dx across the well. Both functions are real, so the conjugation changes nothing.
Use sin A sin B = ½[cos(A − B) − cos(A + B)]. The integrand becomes (1/L)[cos(2πx/L) − cos(4πx/L)].
Integrate: (1/L)[(L/2π) sin(2πx/L) − (L/4π) sin(4πx/L)] between 0 and L. Every sine is taken at an integer multiple of 2π at both limits, so each term contributes 0 − 0, and ⟨ψ₁|ψ₃⟩ = 0.
The theorem had already promised it. ψ(0) = ψ(L) = 0 kills the bracket [f*g′ − (f′)*g]ₐᵇ, so Ĥ is Hermitian on this domain, and E₁ = h²/(8mL²) = 2.41 × 10⁻¹⁹ J = 1.50 eV while E₃ = 9E₁ = 13.5 eV. With a 12.0 eV gap, (E₃ − E₁)⟨ψ₁|ψ₃⟩ = 0 leaves the overlap no choice.

Answer⟨ψ₁|ψ₃⟩ = 0. With E₁ = 1.50 eV and E₃ = 13.5 eV distinct, hermiticity had already forced the overlap to zero; the integral only confirms it.

MediumAn electron moves freely on a ring of circumference L = 2.00 nm. Both u(x) = √(2/L) cos(2πx/L) and w(x) = (1/√L)[cos(2πx/L) + sin(2πx/L)] are normalised eigenfunctions of Ĥ. Find their common energy, compute ⟨u|w⟩, and build a normalised eigenfunction orthogonal to u.

Both functions are combinations of exp(±2πix/L), so both are eigenfunctions of Ĥ = p̂²/2m with E = (h/L)²/2m. Numerically (6.626 × 10⁻³⁴ ÷ 2.00 × 10⁻⁹)² ÷ (2 × 9.109 × 10⁻³¹) = 6.02 × 10⁻²⁰ J = 0.376 eV. One eigenvalue, two independent eigenfunctions: the level is degenerate.
Overlap: ⟨u|w⟩ = (√2/L)∫ cos(2πx/L)[cos(2πx/L) + sin(2πx/L)] dx around the ring. Over one full period ∫cos² dx = L/2 and ∫cos⋅sin dx = 0, so ⟨u|w⟩ = (√2/L)(L/2) = 1/√2 = 0.707.
Nothing is broken. The orthogonality theorem multiplies the overlap by (E − E) = 0 and so asserts nothing about a degenerate pair; 0.707 is an angle of 45° between two perfectly legitimate eigenfunctions of one Hermitian operator.
Gram–Schmidt: χ = w − ⟨u|w⟩u = (1/√L)[cos(2πx/L) + sin(2πx/L)] − (1/√2)⋅√(2/L) cos(2πx/L) = (1/√L) sin(2πx/L).
Normalise and check: ∫(1/L) sin²(2πx/L) dx = 1/2 around the ring, so ‖χ‖ = 1/√2 and the normalised partner is √(2/L) sin(2πx/L). Its overlap with u is (2/L)∫cos⋅sin dx = 0, and it is still an eigenfunction of the same E.

AnswerE = 0.376 eV, doubly degenerate; ⟨u|w⟩ = 1/√2 = 0.707, an angle of 45°; the orthogonal partner built by Gram–Schmidt is √(2/L) sin(2πx/L).

HardA decaying resonance is modelled by giving the potential a constant imaginary part, V = V₀ − iΓ/2 with V₀ and Γ real and positive. Identify the step of the reality proof that fails, find the imaginary part of the energy eigenvalue, and evaluate the lifetime and the surviving norm after 1.00 ps for Γ = 1.00 meV.

Split the mismatch. The kinetic operator is still Hermitian on this domain, so ∫[ψ*(Ĥψ) − (Ĥψ)*ψ] dx = ∫(V − V*)|ψ|² dx = −iΓ∫|ψ|² dx. A real potential was an assumption in the reality proof, not decoration, and here it is gone.
Read the same quantity through the eigenvalue. For Ĥψ = λψ the integral is (λ − λ*)∫|ψ|² dx = 2i Im(λ)∫|ψ|² dx. Equating the two expressions, 2i Im(λ) = −iΓ, so Im(λ) = −Γ/2 and λ = Eᵣ − iΓ/2.
Evolve it. ψ(t) = ψ(0) exp(−iλt/ℏ) = ψ(0) exp(−iEᵣt/ℏ) exp(−Γt/2ℏ), so |ψ(t)|² = |ψ(0)|² exp(−Γt/ℏ). The norm is no longer conserved — the same failure seen from the continuity equation, which acquires a sink once V is complex.
Lifetime: Γ = 1.00 meV = 1.602 × 10⁻²² J, so τ = ℏ/Γ = 1.055 × 10⁻³⁴ ÷ 1.602 × 10⁻²² = 6.58 × 10⁻¹³ s = 0.658 ps.
Surviving norm after t = 1.00 ps: exp(−t/τ) = exp(−1.00/0.658) = exp(−1.52) = 0.219, so 21.9% of the probability is left. The imaginary part is not a bookkeeping trick — Γ is a measured resonance width, and Γτ = ℏ is the lifetime-linewidth statement.

AnswerThe step (λ − λ*)∫|ψ|² dx = 0 fails, since it assumed V real. Here λ = Eᵣ − iΓ/2, the lifetime is τ = ℏ/Γ = 0.658 ps, and 21.9% of the norm survives 1.00 ps.