
Physics V course sequence · Interactive lesson
Self-adjoint operators and 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.
UPV-03.03 · Lesson 3 of 9 in this unit. Position is not a completion score.Build the model
The formula is only half an observable; the domain is the other half.
The key idea
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 complete physical model and assumptions
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.
1The 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.
2Symmetric 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 = λψ forces ψ ∝ e(iλx/ħ), whose modulus is constant for real λ 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. Each such ψ is orthogonal to the whole range of p − z*, so the spectrum of this perfectly symmetric operator is the closed lower half plane, the mirror image of the adjoint's eigenvalues. Real eigenvalues and a real spectrum are not the same claim, and only self-adjointness delivers the second.
3Integrate 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)]₀L. 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.
4Reality 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.
5The 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.
6What 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, ψN = e(iθ)ψ₀, and the matrix is a circulant twisted by that phase, 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 the θ member of the family above. The central difference also carries spurious doubler modes near n = N/2, which lay a second ladder beside the true one; an FFT derivative with the same twist gives the θ ladder alone. 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, although the operator it imitates owns no eigenvector and has the whole complex plane as its spectrum. Finite dimensions cannot reproduce that, so the grid returns a real ladder that never settles: for L = 2.00 nm the eigenvalues nearest zero approach 0, ±1.657, ±3.313 (×10⁻²⁵ kg m s⁻¹) for odd N but ±0.828, ±2.485 for even N. A Hermitian matrix is not evidence that the operator you meant is self-adjoint.
Key equations
Read the units, assumptions and limits before substituting.
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-adjoint
symmetric: 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)]₀L
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 = a*; then a ≠ b forces ⟨b|a⟩ = 0. Degeneracy fixes a subspace, not a basis.Spectral resolution of the identity
I = 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 extensions
n± = 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.Interactive lab
Predict first. Explore the model. Compare the result.
Explore the model
Try this. 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.
Live results
- LEVEL SPACING h / L
- 3.31 ×10⁻²⁵ kg m/s
- SELECTED pₙ
- 3.31 ×10⁻²⁵ kg m/s
- PHASE ACROSS BOX pₙ L/ħ
- 6.28 rad
- LADDER SHIFT θ / 2π
- 0.00 spacings
Results follow this lesson’s model and assumptions. Displayed values are rounded.
Predict → change one variable → check
Plan a change before running it. Every other input stays at the saved baseline.
Saved baseline. Boundary twist θ: 0 °; Box length L: 2.0 nm; Level index n: 1
This experiment is kept only while this page is open. It does not change saved course progress.
Catch the common trap
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 lower 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.
Practice & worked examples
3 authored examples. Make an attempt, then reveal the reasoning one step at a time.
EasyExample 1
In 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.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 5 steps revealed.
- 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.
Answer. A† = 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.
MediumExample 2
An 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.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 6 steps revealed.
- 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 kg m s⁻¹. For n = 0, p₀ = 8.28 × 10⁻²⁶ kg m s⁻¹; for n = −1, p₋₁ = −0.750 × 3.313 × 10⁻²⁵ = −2.48 × 10⁻²⁵ kg m s⁻¹. The next one up, p₁ = 4.14 × 10⁻²⁵ kg m s⁻¹, 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.
Answer. pₙ = (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.
HardExample 3
Take 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.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 6 steps revealed.
- 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 = λψ forces ψ = C e(iλx/ħ). For real λ 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 ψz = e(izx/ħ) has |ψz|² = e(−2(Im z)x/ħ), whose integral is ħ/(2 Im z) < ∞, and p†ψz = zψz. Every point of the open upper half plane is an eigenvalue of p†. Since ⟨ψz|(p − z*)χ⟩ = ⟨(p† − z)ψz|χ⟩ = 0 for every χ in the domain, the mirror point z* is residual spectrum of p, so the spectrum of p is the closed lower half plane.
- Deficiency indices: solve p†ψ = ±iκψ for a fixed κ > 0 in momentum units. The plus sign gives ψ′ = −(κ/ħ)ψ, so ψ = e(−κx/ħ), which is in L²(0, ∞): n₊ = 1. The minus sign gives ψ′ = +(κ/ħ)ψ, so ψ = e(+κx/ħ), which is not: n₋ = 0.
- Von Neumann's criterion: self-adjoint 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.
Answer. p is symmetric and has no eigenvectors, yet its spectrum is the closed lower 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.
Quick check
Select an answer and explain it before checking.
Review & continue
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.
- Explain the relationship without looking at the equation.
- Check a sign, a unit and a model assumption in your working.
- Return to the lab or worked example if a step is still unclear.