University Physics V · The Quantum Wavefunction · 3.3

Self-Adjoint Operators & Their Spectra

Hermiticity is the promise physics needs: real outcomes, mutually exclusive alternatives, a complete set to expand in, and a clock that conserves probability. This lesson is where you learn to check that promise properly — by naming the domain, integrating by parts, and reading the boundary term.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Quantum mechanics needs its observables to deliver four things at once: outcomes that are real numbers, alternatives that exclude one another, a complete set of alternatives to expand any state in, and a time evolution that conserves probability. Self-adjointness is the single condition that delivers all four, and it is strictly stronger than the A = A† you write in a finite basis. The adjoint is defined by the inner product, ⟨φ|Aψ⟩ = ⟨A†φ|ψ⟩, and in infinite dimensions that equation fixes a domain as well as a rule, because unbounded operators such as x and p are only defined on part of L².

Symmetric means the two sides agree on D(A); self-adjoint means D(A†) has not quietly grown larger than D(A). The gap is not pedantry: −iħ d/dx on a half-line is symmetric, owns no eigenvector, carries a whole half plane of complex numbers in its spectrum, and no boundary condition repairs it. What self-adjointness costs is that you must state the boundary condition before you have finished naming the operator, so different conditions are genuinely different observables with different spectra.

What it buys is the spectral theorem — a resolution of the identity split into a discrete sum and a continuous integral — and Stone's theorem, which converts the operator into the unitary group that actually moves the state.

Simple definition: An observable is a self-adjoint operator: a rule A together with a dense domain D(A) on which ⟨φ|Aψ⟩ = ⟨Aφ|ψ⟩ for every pair, and for which the adjoint's domain D(A†) is no larger than D(A) itself.
Example: On 0 ≤ x ≤ 2.00 nm, −iħ d/dx paired with ψ(L) = e^(iθ)ψ(0) is self-adjoint, with spectrum pₙ = (n + θ/2π)h/L spaced 3.31 × 10⁻²⁵ kg m s⁻¹; the same formula paired with ψ(0) = ψ(L) = 0 is only symmetric, and owns no eigenvector at all.

The adjoint, from the inner product⟨φ|Aψ⟩ = ⟨A†φ|ψ⟩ for all ψ ∈ D(A)

The adjoint is built from the geometry of the space, so shrinking D(A) makes D(A†) grow.

In a finite orthonormal basis (A†)ᵢⱼ = (Aⱼᵢ)*; in L² the same equation defines D(A†) as well as the rule.

Symmetric versus self-adjointsymmetric: D(A) ⊆ D(A†) · self-adjoint: D(A) = D(A†)

Only the equality yields the spectral theorem and a unitary group exp(−iAt/ħ), by Stone's theorem.

Both demand ⟨φ|Aψ⟩ = ⟨Aφ|ψ⟩ on D(A). Textbook "Hermitian" usually means only the first.

Boundary term for p = −iħ d/dx⟨φ|pψ⟩ − ⟨pφ|ψ⟩ = −iħ [φ*(x) ψ(x)]₀^L

The one-line test: choose the domain that kills this bracket for every pair drawn from inside it.

ħ in J s and ψ in m^(−1/2), so the bracket carries kg m s⁻¹, the units of p itself.

Real spectrum, orthogonal eigenvectors(a − b) ⟨b|a⟩ = 0 for A|a⟩ = a|a⟩, A|b⟩ = b|b⟩

Real outcomes and exclusive alternatives — but only for eigenvectors that exist inside the space.

b = a with ⟨a|a⟩ > 0 forces a real; a ≠ b forces ⟨b|a⟩ = 0. Degeneracy fixes a subspace, not a basis.

Spectral resolution of the identityI = Σₙ |n⟩⟨n| + ∫ da |a⟩⟨a| , A = Σₙ aₙ |n⟩⟨n| + ∫ da a |a⟩⟨a|

The continuous piece is the whole story for x and p, and it supplies no basis vectors in L².

⟨n|m⟩ = δₙₘ is dimensionless; ⟨a|a′⟩ = δ(a − a′) carries units of 1/a.

Deficiency indices and the extensionsn± = dim ker(A† ∓ iκ)p on [0, L]: (1,1) → ψ(L) = e^(iθ)ψ(0)

Equal indices give a whole family of observables from one formula; on (0, ∞) they are (1,0) and none exists.

κ > 0 only fixes units. Then pₙ = (2πn + θ)ħ/L, n ∈ ℤ, spaced h/L = 3.31 × 10⁻²⁵ kg m s⁻¹ at L = 2.00 nm.

The adjoint is fixed by the inner product, not by transposing

In a finite orthonormal basis the adjoint is the conjugate transpose, (A†)ᵢⱼ = (Aⱼᵢ)*, and ⟨φ|A|ψ⟩* = ⟨ψ|A†|φ⟩ says the same thing without naming a basis. Nothing more is needed there. In L²(ℝ) it is not that easy, because the operators physics cares about are unbounded: there is no constant c with ‖pψ‖ ≤ c‖ψ‖, and applying −iħ d/dx to a merely square-integrable ψ can throw you straight out of the space. So an operator is a pair — a rule together with a dense domain D(A) — and the adjoint has to be constructed rather than written down. A vector φ belongs to D(A†) exactly when some χ satisfies ⟨φ|Aψ⟩ = ⟨χ|ψ⟩ for every ψ ∈ D(A), and only then is A†φ defined, as χ. Read that definition twice: the smaller you make D(A), the fewer conditions φ must satisfy, so the larger D(A†) becomes. Every difficulty in this topic is a consequence of that inverse relation.

Symmetric is cheap; self-adjoint is what physics needs

Symmetric asks only that ⟨φ|Aψ⟩ = ⟨Aφ|ψ⟩ whenever both vectors sit in D(A) — equivalently D(A) ⊆ D(A†), with A† agreeing with A there. Self-adjoint asks for the two domains to be equal. Most textbooks say "Hermitian" and mean the first. The example that separates them is p = −iħ d/dx on L²(0, ∞) with the domain ψ(0) = 0. It is symmetric. It owns no eigenvector: pψ = pψ forces ψ ∝ e^(ipx/ħ), whose modulus is constant for real p and therefore not square-integrable on a half-line. Its adjoint meanwhile carries no boundary condition at all, so for every complex z with Im z > 0 the function e^(izx/ħ) decays, lies in L²(0, ∞) with squared norm ħ/(2 Im z), and satisfies p†ψ = zψ. The spectrum of this perfectly symmetric operator is the closed upper half plane. Real eigenvalues and a real spectrum are not the same claim, and only self-adjointness delivers the second.

Integrate by parts and look at what is left over

The working test is one line. For p = −iħ d/dx on an interval [0, L], integration by parts gives ⟨φ|pψ⟩ − ⟨pφ|ψ⟩ = −iħ [φ*(x)ψ(x)]₀^L. The factor of i is what makes this possible at all: bare d/dx is antisymmetric, and multiplying by −i flips the sign back. Self-adjointness is now the demand that the bracket vanish for every pair drawn from one and the same domain. Dirichlet, ψ(0) = ψ(L) = 0, kills it — but only by constraining ψ, leaving φ entirely free, so D(p†) is strictly larger and the operator is symmetric only. The repair is a condition that ties the two ends together: ψ(L) = e^(iθ)ψ(0). Then φ*(L)ψ(L) = e^(−iθ)φ*(0) · e^(iθ)ψ(0) = φ*(0)ψ(0), the bracket cancels identically, and D(p†) = D(p). Its eigenfunctions e^(ipx/ħ)/√L are honest members of L²(0, L), with pₙ = (2πn + θ)ħ/L spaced h/L apart — 3.31 × 10⁻²⁵ kg m s⁻¹ for L = 2.00 nm.

Reality and orthogonality in two lines, and their limits

Let A|a⟩ = a|a⟩ with |a⟩ ∈ D(A) and A symmetric. Then a⟨a|a⟩ = ⟨a|Aa⟩ = ⟨Aa|a⟩ = a*⟨a|a⟩, and since ⟨a|a⟩ > 0 the eigenvalue satisfies a = a*. The identical computation on two eigenvectors gives (a − b)⟨b|a⟩ = 0, so distinct eigenvalues carry orthogonal eigenvectors. Degeneracy weakens that: a repeated eigenvalue fixes a subspace, orthogonality is guaranteed between eigenspaces but not automatically inside one, and you pick a basis there yourself, usually by diagonalising a second commuting observable. And every line of the argument is conditional on the eigenvector existing in the space. The position operator on L²(ℝ) is self-adjoint on D(x) = (ψ : xψ ∈ L²), yet (x − a)ψ(x) = 0 forces ψ = 0 almost everywhere: not one eigenvector, anywhere in the spectrum.

The spectral theorem resolves the identity in two pieces

For a self-adjoint A the theorem supplies a projection-valued measure E, with A = ∫ λ dE(λ), I = ∫ dE(λ), and P(outcome lies in Δ) = ⟨ψ|E(Δ)|ψ⟩. Where the spectrum is discrete, dE is a sum of projectors and you recover the familiar orthonormal basis with Σₙ |n⟩⟨n| = I. Where it is continuous there is no basis of the space at all: x and p on L²(ℝ) both have spectrum ℝ, and their so-called eigenvectors δ(x − a) and e^(ipx/ħ)/√(2πħ) are δ-normalised objects living in a rigged space, not in L². The completeness statements ∫|x⟩⟨x| dx = I and ∫|p⟩⟨p| dp = I still hold as operator identities, and they carry dimensions, since ⟨x|x′⟩ = δ(x − x′) has units of inverse length. So "Hermitian implies an eigenbasis" is a finite-dimensional theorem. Its infinite-dimensional replacement is the spectral measure, and what you compute from it is the probability of an interval.

What the Python grid quietly decides on your behalf

Sample ψ on N points spaced a = L/N and build p as −iħ times the central-difference matrix. What you write in the first and last rows is a choice of self-adjoint extension, made by your array indexing. Wrap the ends with a phase, ψ_N = e^(iθ)ψ₀, and the matrix is circulant with eigenvectors e^(i(2πn+θ)j/N) and eigenvalues ħ sin((2πn + θ)/N)/a, real by construction and converging to (2πn + θ)ħ/L for n ≪ N: that is exactly the θ member of the family above. Drop the corner entries instead — the naive Dirichlet grid — and the matrix stays exactly antisymmetric, so −iħD is still Hermitian and every eigenvalue is still real. Finite dimensions cannot reproduce a half plane. What they do instead is converge to the wrong thing: for L = 2.00 nm the eigenvalues nearest zero come out at 0, ±1.657, ±3.313 (×10⁻²⁵ kg m s⁻¹), spaced πħ/L = h/2L, half the spacing of any genuine extension. A Hermitian matrix is not evidence that the operator you meant is self-adjoint.

Change one variable at a time

Make the relationship visible.

Interactive model

Boundary twist θ0 °

Box length L2.0 nm

Level index n1

Turn θ from 0° to 360° and the whole ladder slides by exactly one spacing: a different self-adjoint operator at every angle, all from one differential formula. Then stretch L and watch Δp = h/L close up towards the continuous spectrum p has on the whole line.

LEVEL SPACING h / L3.31 ×10⁻²⁵ kg m/s

SELECTED pₙ3.31 ×10⁻²⁵ kg m/s

ITS WAVELENGTH h/|p|2.00 nm

LADDER SHIFT θ / 2π0.00 spacings

Live interpretationLEVEL SPACING h / L: 3.31 ×10⁻²⁵ kg m/s. SELECTED pₙ: 3.31 ×10⁻²⁵ kg m/s. ITS WAVELENGTH h/|p|: 2.00 nm. LADDER SHIFT θ / 2π: 0.00 spacings

Catch the common trap

Explain before calculating.

Misconception check

The error is to treat Hermiticity as a property of the formula. Check that p = −iħ d/dx has the i in the right place, or that a matrix looks symmetric, and declare the thing an observable. But ⟨φ|Aψ⟩ = ⟨Aφ|ψ⟩ is a statement about a set of vectors, and the two sets need not match. Momentum on the half-line with ψ(0) = 0 passes the formula test and the symmetry test, owns no eigenvector, and has the whole closed upper half plane as its spectrum — and with deficiency indices (1,0), no boundary condition anywhere repairs it. Two corollaries are worth memorising. Real eigenvalues do not imply a real spectrum, because an operator can have no eigenvalues at all. And self-adjoint does not imply an eigenbasis: x and p on L²(ℝ) are as self-adjoint as operators get, and neither owns a single eigenvector inside the space.

Take p = −iħ d/dx on the interval 0 ≤ x ≤ L, with the candidate domain being the smooth functions satisfying ψ(0) = ψ(L) = 0. Which statement about this operator is correct?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyIn an orthonormal basis of a two-dimensional Hilbert space an operator has the matrix A = [[2, 3 − i], [3 + i, −1]]. Show that A is Hermitian, find its spectrum, and verify that its eigenvectors are orthogonal.

A† is the conjugate transpose. Swapping the off-diagonal entries and conjugating sends 3 + i to (3 + i)* = 3 − i in the (1,2) slot and 3 − i to 3 + i in the (2,1) slot; the diagonal entries 2 and −1 are already real. So A† = A. In finite dimensions there is no domain question, so Hermitian and self-adjoint coincide here.
Characteristic polynomial from the trace and determinant: tr A = 2 + (−1) = 1, and det A = (2)(−1) − (3 − i)(3 + i) = −2 − (9 + 1) = −12. So λ² − λ − 12 = 0.
λ = (1 ± √(1 + 48))/2 = (1 ± 7)/2, giving λ = 4 and λ = −3. Both real, exactly as (a − a*)⟨a|a⟩ = 0 demands.
For λ = 4 the first row gives −2v₁ + (3 − i)v₂ = 0, so |4⟩ ∝ (3 − i, 2). For λ = −3 it gives 5v₁ + (3 − i)v₂ = 0, so |−3⟩ ∝ (−3 + i, 5).
Overlap: ⟨4|−3⟩ = (3 − i)*(−3 + i) + 2* · 5 = (3 + i)(−3 + i) + 10 = −10 + 10 = 0. The norms are √14 and √35, so the normalised kets are orthonormal and |4⟩⟨4| + |−3⟩⟨−3| = I.

AnswerA† = A, spectrum (4, −3), both real. The eigenvectors (3 − i, 2)/√14 and (−3 + i, 5)/√35 are orthonormal, and their projectors sum to the identity.

MediumAn electron is confined to 0 ≤ x ≤ L with L = 2.00 nm, and the domain of p = −iħ d/dx is fixed by ψ(L) = e^(iθ)ψ(0) with θ = π/2. Show that the boundary term vanishes, find the momentum eigenvalues, and evaluate the two smallest in magnitude.

The defect is ⟨φ|pψ⟩ − ⟨pφ|ψ⟩ = −iħ(φ*(L)ψ(L) − φ*(0)ψ(0)). Both φ and ψ are drawn from the same domain, so both carry the twist: φ*(L)ψ(L) = e^(−iθ)φ*(0) · e^(iθ)ψ(0) = φ*(0)ψ(0). The bracket cancels for every pair, so D(p) = D(p†) and p is self-adjoint.
Eigenfunctions: −iħψ′ = pψ gives ψ(x) = C e^(ipx/ħ), normalised by C = 1/√L on the interval. Imposing the twist: e^(ipL/ħ) = e^(iθ), so pL/ħ = θ + 2πn with n ∈ ℤ.
Hence pₙ = (2πn + θ)ħ/L = (n + θ/2π) h/L, a ladder of real, non-degenerate eigenvalues with spacing Δp = h/L.
Δp = 6.626 × 10⁻³⁴ J s ÷ 2.00 × 10⁻⁹ m = 3.313 × 10⁻²⁵ kg m s⁻¹.
With θ = π/2 the offset is θ/2π = 0.250, so pₙ = (n + 0.250) × 3.313 × 10⁻²⁵. For n = 0, p₀ = 8.28 × 10⁻²⁶; for n = −1, p₋₁ = −0.750 × 3.313 × 10⁻²⁵ = −2.48 × 10⁻²⁵. The next one up, p₁ = 4.14 × 10⁻²⁵, is larger than both in magnitude.
These are genuine vectors in L²(0, L) with ⟨pₙ|pₘ⟩ = δₙₘ, unlike the plane waves on the whole line, which are only δ-normalisable.

Answerpₙ = (n + 1/4) h/L. With L = 2.00 nm, Δp = h/L = 3.31 × 10⁻²⁵ kg m s⁻¹, and the two smallest in magnitude are p₀ = +8.28 × 10⁻²⁶ and p₋₁ = −2.48 × 10⁻²⁵ kg m s⁻¹.

HardTake p = −iħ d/dx on L²(0, ∞) with the domain of smooth functions obeying ψ(0) = 0 and decaying at infinity. Show that p is symmetric, that it has no eigenvectors, and that its spectrum nevertheless contains complex numbers. Compute its deficiency indices and say what they rule out.

Symmetry: ⟨φ|pψ⟩ − ⟨pφ|ψ⟩ = −iħ[φ*ψ]₀^∞ = −iħ(0 − φ*(0)ψ(0)) = 0, since ψ(0) = 0 for every ψ in the domain and both functions decay at infinity. So p is symmetric.
No eigenvectors: pψ = pψ forces ψ = C e^(ipx/ħ). For real p the modulus |ψ| = |C| is constant, so ∫₀^∞ |ψ|² dx diverges unless C = 0. The point spectrum is empty.
The adjoint is bigger. The bracket already died using only the ψ side, so p† acts by −iħ d/dx on every ψ ∈ H¹(0, ∞) with no condition at x = 0.
For any z with Im z > 0 the function ψ_z = e^(izx/ħ) has |ψ_z|² = e^{(−2(Im z)}x/ħ), whose integral is ħ/(2 Im z) < ∞, and p†ψ_z = zψ_z. Every point of the open upper half plane is an eigenvalue of p†, that is, residual spectrum of p.
Deficiency indices: solve p†ψ = ±iħκ ψ for a fixed κ > 0. The plus sign gives ψ′ = −κψ, so ψ = e^(−κx), which is in L²(0, ∞): n₊ = 1. The minus sign gives ψ′ = +κψ, so ψ = e^(+κx), which is not: n₋ = 0.
Von Neumann's criterion: extensions exist if and only if n₊ = n₋. Here (1,0), so no boundary condition at the origin rescues this operator, and by Stone's theorem there is no unitary translation group exp(−ipa/ħ) on the half-line. On L²(ℝ) neither exponential is square-integrable, the indices are (0,0), and p is essentially self-adjoint with purely continuous spectrum ℝ.

Answerp is symmetric and has no eigenvectors, yet its spectrum is the closed upper half plane. The deficiency indices are (1,0), so no self-adjoint extension exists and momentum is simply not an observable on the half-line.