University Physics V · The Schrödinger Equation · 4.1

The Hamiltonian as a Self-Adjoint Operator

Before you solve a single Schrödinger equation you must build its left-hand side, and that is two jobs, not one: promote the classical energy to p̂²/2m + V(x̂), choosing an ordering wherever x and p mix, then pin down the domain — because hermiticity is decided at the boundary, not by the formula.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Classical mechanics hands you an energy function H(x, p) = p²/2m + V(x); quantum mechanics needs the generator of time evolution, an operator on the Hilbert space L². Canonical quantisation is the bridge: x becomes multiplication by x, p becomes −iℏ d/dx, and the energy becomes Ĥ = −(ℏ²/2m) d²/dx² + V(x̂). The recipe has two costs, and this topic is about both.

First, it is ambiguous wherever a classical product mixes x and p — x̂p̂ and p̂x̂ differ by iℏ, so classical xp has several quantum images, and hermiticity forces at least the symmetrised (x̂p̂ + p̂x̂)/2. Second, and deeper: what the recipe delivers is a differential expression, not yet an operator. An unbounded operator exists only together with its domain — which states it may act on, and what they do at the boundary — and the familiar check ⟨φ|Ĥψ⟩ = ⟨Ĥφ|ψ⟩ (symmetry) is strictly weaker than self-adjointness, Ĥ = Ĥ† with equal domains.

The gap is not pedantry: only a self-adjoint Ĥ carries a complete real spectrum, and only a self-adjoint Ĥ exponentiates, by Stone's theorem, into the unitary evolution e^{(−iĤt/ℏ)} that conserves probability. The boundary conditions of every wall, well and delta spike in this unit are exactly the data that finish the operator.

Simple definition: The Hamiltonian is the self-adjoint energy operator built by substituting x̂ and p̂ = −iℏ d/dx into the classical energy, together with the domain of states on which its boundary terms vanish and D(Ĥ†) = D(Ĥ).
Example: For an electron on [0, 1 nm] with Dirichlet walls ψ(0) = ψ(L) = 0, Ĥ = −(ℏ²/2m) d²/dx², and its lowest eigenvalue is π²ℏ²/2mL² = 6.02 × 10⁻²⁰ J = 0.376 eV.

Canonical quantisationĤ = p̂²/2m + V(x̂), with p̂ = −iℏ d/dx

Turns the classical energy function H(x, p) into the operator that will generate the dynamics; [x̂, p̂] = iℏ is what calibrates the pair.

x̂ multiplies by x (m); p̂ carries kg m s⁻¹; ℏ = 1.055 × 10⁻³⁴ J s

Kinetic operator in the position basisp̂²/2m = −(ℏ²/2m) d²/dx² → −(ℏ²/2m)∇² in 3D

Curvature is kinetic energy: the faster ψ bends, the more the state costs. 3.81 eV Å² is why atoms, bonded on ångström scales, trade in electron-volts.

ℏ²/2mₑ = 6.11 × 10⁻³⁹ J m² = 3.81 eV Å² for an electron

Ordering ambiguityx̂p̂ − p̂x̂ = iℏ → classical xp ↦ (x̂p̂ + p̂x̂)/2

Classical mechanics cannot order what commutes; hermiticity forces the symmetrised product wherever a Hamiltonian mixes x with p.

x̂p̂ alone is not Hermitian: (x̂p̂)† = p̂x̂ = x̂p̂ − iℏ

Symmetry lives at the boundary⟨φ|Ĥψ⟩ − ⟨Ĥφ|ψ⟩ = (ℏ²/2m)[φ̄′ψ − φ̄ψ′]ₐᵇ

For real V everything interior cancels: hermiticity of Ĥ is decided entirely by four boundary numbers, so it is a condition on the domain.

φ̄ is the conjugate; the bracket is evaluated at the interval's endpoints a, b

Self-adjoint, not just symmetricĤ = Ĥ† requires D(Ĥ†) = D(Ĥ)

Only self-adjointness delivers the spectral theorem and, via Stone's theorem, the unitary e^{(−iĤt/ℏ)}; a merely symmetric Ĥ can generate no dynamics at all.

symmetric means Ĥ ⊂ Ĥ†: the adjoint's domain may still be strictly bigger

One expression, two spectraψ(0) = ψ(L) = 0 → Eₙ = n²π²ℏ²/2mL²periodic → 4n² × π²ℏ²/2mL²

Changing only the domain changes the measured energies: the boundary condition is physical data, not decoration.

Dirichlet levels 1, 4, 9, …; periodic gives 0, 4, 16, …, each nonzero one doubly degenerate

Promote x and p, and the energy comes along

Canonical quantisation is a substitution with rules. Keep the classical Hamiltonian H(x, p) = p²/2m + V(x), then promote: x̂ acts as multiplication by x, p̂ as −iℏ d/dx, both on the position-basis wavefunction ψ(x) in L²(ℝ). Kinetic energy becomes −(ℏ²/2m) d²/dx² — in three dimensions the Laplacian −(ℏ²/2m)∇² — and the potential becomes multiplication by V(x). The pair is calibrated by the commutator: [x̂, p̂] = iℏ mirrors the Poisson bracket {x, p} = 1 with the factor iℏ, which is the entire content of the word 'quantisation'. Numbers anchor it: for an electron ℏ²/2mₑ = 3.81 eV Å², so a wavefunction forced to bend on the scale of an ångström carries electron-volt kinetic energies — atomic physics fixed by one coefficient.

Where the recipe stalls: operator ordering

The substitution is only well defined because p²/2m + V(x) never multiplies x by p. The classical observable xp has at least three quantum images — x̂p̂, p̂x̂ and (x̂p̂ + p̂x̂)/2 — differing by multiples of iℏ, since x̂p̂ − p̂x̂ = iℏ. The first two are not even Hermitian: (x̂p̂)† = p̂x̂ = x̂p̂ − iℏ. Hermiticity therefore forces the symmetrised product, and Weyl ordering extends the rule to any polynomial in x and p. Nor is the ambiguity exotic: a charged particle in a magnetic field carries (p̂ − qA(x̂))²/2m, whose expansion produces the cross term p̂⋅A + A⋅p̂, collapsing to 2A⋅p̂ only in Coulomb gauge ∇⋅A = 0, where the factors commute. Every ordering choice shifts the operator at order ℏ — invisible classically, measurable quantum mechanically.

An operator is an expression plus a domain

In ℂⁿ every matrix acts on every vector. In L² that fails: Ĥ is unbounded and cannot act on the whole space. The function ψ = 1/(1 + |x|) is square-integrable, yet x̂ψ is not, because ∫ x²/(1 + |x|)² dx diverges; the kink e^(−|x|) is normalisable, but its second derivative contains −2δ(x), which the kinetic term cannot digest. So defining Ĥ means naming its domain D(Ĥ): the dense set of states smooth enough, decaying fast enough, and — on an interval — obeying stated boundary conditions. Change the domain and you change the operator, spectrum included: −(ℏ²/2m) d²/dx² on [0, L] with Dirichlet walls has levels 1, 4, 9, … in units of π²ℏ²/2mL², while the periodic domain ψ(0) = ψ(L), ψ′(0) = ψ′(L) delivers 0, 4, 16, … in the same unit, each nonzero level twice. Same expression, different physics.

Hermiticity is decided at the boundary

Test the kinetic term against two states φ and ψ on [0, L]. Two integrations by parts give ⟨φ|T̂ψ⟩ − ⟨T̂φ|ψ⟩ = (ℏ²/2m)[φ̄′ψ − φ̄ψ′]₀ᴸ: the interior cancels identically, and the entire question of hermiticity condenses onto four boundary numbers. The figure below computes this live: φ = sin(2πx/L) respects the Dirichlet walls, while ψ is given a leak ψ(0) = β, and the defect comes out as −4πβ in units of ℏ²/2mL — growing linearly with the leak and utterly indifferent to the interior amplitude. Killing the bracket for every pair of states in the domain is a condition on the domain, and its solutions form a four-parameter family: Dirichlet ψ = 0, Neumann ψ′ = 0, periodic identification, and the Robin interpolations ψ′ = λψ — every one a legitimate, distinct, self-adjoint Hamiltonian.

Symmetric is not yet self-adjoint

Symmetry — ⟨φ|Ĥψ⟩ = ⟨Ĥφ|ψ⟩ on D(Ĥ) — can hold while the adjoint Ĥ†, defined on every φ for which ⟨φ|Ĥψ⟩ = ⟨η|ψ⟩ has a solution η, lives on a strictly larger domain. The clean counterexample is momentum on the half-line: p̂ = −iℏ d/dx on x ≥ 0 with ψ(0) = 0. The boundary term iℏφ̄(0)ψ(0) dies through ψ(0) = 0 alone, so the adjoint imposes no condition on φ — and it owns the normalisable eigenfunction e^(−x/a) with the imaginary eigenvalue iℏ/a. Von Neumann's deficiency indices count the L² solutions of Ĥ†χ = ±iλχ: equal indices mean self-adjoint extensions exist and form a family; here they are (1, 0), so no boundary condition whatever can repair p̂. The physics is honest — momentum generates translations, and translating a half-line state pushes it off its space. A confined particle keeps a kinetic energy, since −d²/dx² there has indices (1, 1), but it has no momentum observable.

What self-adjointness buys, and how a grid hides it

Self-adjointness is the hypothesis of the two theorems this unit runs on. The spectral theorem gives Ĥ a complete orthonormal eigenbasis with real eigenvalues — the levels Eₙ the coming topics compute. Stone's theorem exponentiates Ĥ into the unitary U(t) = e^{(−iĤt/ℏ)}, and unitarity is probability conservation; a merely symmetric operator generates no such group. Numerically the issue seems to evaporate: discretise on N points and Ĥ becomes a finite real symmetric matrix, and in finite dimensions symmetric and self-adjoint coincide. But the domain has not vanished — it has moved into the stencil. Ending the three-point Laplacian at the first and last rows silently sets ψ = 0 one step beyond each wall, so scipy.linalg.eigh_tridiagonal is solving the Dirichlet extension; modelling periodic or Robin physics means editing the corner entries of the matrix, which is the boundary condition wearing numerical clothes.

Change one variable at a time

Make the relationship visible.

Interactive model

Wall leak β = ψ(0)0.30

Interior amplitude α1.00

Drive β to 0 and the two matrix elements snap together as the defect bar collapses — then slide α anywhere: the interior amplitude reshapes the curve but never moves the defect, because hermiticity lives at the walls, not in the bulk.

ψ AT LEFT WALL β0.30

⟨φ|T̂ψ⟩1.26 ℏ²/2mL

⟨T̂φ|ψ⟩5.03 ℏ²/2mL

DEFECT (DIFFERENCE)-3.77 ℏ²/2mL

Live interpretationψ AT LEFT WALL β: 0.30. ⟨φ|T̂ψ⟩: 1.26 ℏ²/2mL. ⟨T̂φ|ψ⟩: 5.03 ℏ²/2mL. DEFECT (DIFFERENCE): −3.77 ℏ²/2mL

Catch the common trap

Explain before calculating.

Misconception check

Students import the finite-dimensional theorem — a matrix equal to its conjugate transpose is Hermitian, done — into L², and conclude that any Ĥ with real V is self-adjoint because a formal integration by parts works. But the integration by parts is exactly where the claim leaks: it leaves the bracket (ℏ²/2m)[φ̄′ψ − φ̄ψ′] at the endpoints, which vanishes only on a chosen domain, and different choices are different operators — Dirichlet walls on [0, L] give levels 1, 4, 9 × π²ℏ²/2mL², periodic identification gives 0, 4, 16. Worse, a domain can make an operator symmetric while its adjoint keeps a strictly bigger domain: −iℏ d/dx on the half-line with ψ(0) = 0 passes every formal check yet has no self-adjoint version at all. Symmetric is a property of the formula; self-adjoint is the formula plus a domain, and only the second is an observable.

The kinetic operator T̂ = −(ℏ²/2m) d²/dx² is to act on wavefunctions on [0, L]. What, if anything, must be added before T̂ counts as an observable?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyQuantise the classical energy H = p²/2m + ½mω²x² for an electron in a trap with ω = 2.00 × 10¹⁵ s⁻¹. Write Ĥ in the position basis and evaluate both coefficients, the kinetic one in eV Å² and the potential one in eV Å⁻².

Promote: x → x̂ (multiplication by x) and p → p̂ = −iℏ d/dx. No term mixes x with p, so no ordering choice arises and Ĥ = −(ℏ²/2m) d²/dx² + ½mω²x̂² is unambiguous.
Kinetic coefficient: ℏ²/2mₑ = (1.055 × 10⁻³⁴)²/(2 × 9.109 × 10⁻³¹) = 6.11 × 10⁻³⁹ J m². Dividing by 1.602 × 10⁻¹⁹ J eV⁻¹ and multiplying by 10²⁰ Å² m⁻² gives 3.81 eV Å².
Potential coefficient: ½mω² = 0.5 × 9.109 × 10⁻³¹ × (2.00 × 10¹⁵)² = 1.82 J m⁻², which is 1.82 × 10⁻²⁰ J Å⁻² = 0.114 eV Å⁻².
Check the operator by balance: the kinetic scale 3.81/d² eV equals the potential scale 0.114 d² eV at d = (3.81/0.114)^¼ = 2.4 Å — precisely √(ℏ/mω), the trap's natural width, read off before solving anything.

AnswerĤ = −(3.81 eV Å²) d²/dx² + (0.114 eV Å⁻²) x², with x in Å; balancing the two coefficients gives the natural width (3.81/0.114)^¼ ≈ 2.4 Å = √(ℏ/mω).

MediumThe classical observable A = xp is proposed as the operator x̂p̂. Show that in every normalised state its expectation value has imaginary part ℏ/2, and write the operator that actually represents A.

Conjugate the expectation value: ⟨x̂p̂⟩* = ⟨ψ|(x̂p̂)†|ψ⟩ = ⟨ψ|p̂x̂|ψ⟩, using x̂† = x̂ and p̂† = p̂ and the reversal of order under the adjoint.
Subtract: ⟨x̂p̂⟩ − ⟨x̂p̂⟩* = ⟨ψ|[x̂, p̂]|ψ⟩ = ⟨ψ|iℏ|ψ⟩ = iℏ for any normalised ψ — the commutator is a multiple of the identity, so the state drops out.
Since z − z* = 2i Im z, the imaginary part is ℏ/2 = 5.27 × 10⁻³⁵ J s, state-independent. A measured value must be real, so x̂p̂ represents no observable.
Symmetrise: Â = (x̂p̂ + p̂x̂)/2 is Hermitian by construction. Writing p̂x̂ = x̂p̂ − iℏ shows Â = x̂p̂ − iℏ/2 — exactly the offending imaginary part subtracted off.

AnswerIm⟨x̂p̂⟩ = ℏ/2 ≈ 5.27 × 10⁻³⁵ J s in every normalised state; the observable is the symmetrised (x̂p̂ + p̂x̂)/2, which equals x̂p̂ − iℏ/2.

HardA particle lives on the half-line x ≥ 0 with ψ(0) = 0. Show that p̂ = −iℏ d/dx is symmetric on this domain but admits no self-adjoint version, using the trial function χ(x) = √(2/a) e^(−x/a) with a = 1.0 nm.

Symmetry: ⟨φ|p̂ψ⟩ − ⟨p̂φ|ψ⟩ = −iℏ[φ̄ψ]₀^∞ = iℏ φ̄(0)ψ(0) = 0, because every ψ in the domain has ψ(0) = 0. The formal hermiticity check passes.
But it passes through ψ(0) = 0 alone: φ needed no condition, so the adjoint's domain D(p̂†) carries no boundary condition and is strictly larger than D(p̂).
χ is normalised and admissible for the adjoint: ∫₀^∞ (2/a) e^(−2x/a) dx = (2/a)(a/2) = 1, and χ(0) = √(2/a) ≠ 0 is now allowed.
Apply the adjoint: p̂†χ = −iℏ(−1/a)χ = (iℏ/a)χ — a normalisable eigenfunction with the purely imaginary eigenvalue iℏ/a = i × 1.05 × 10⁻²⁵ kg m s⁻¹, impossible for any self-adjoint operator.
Von Neumann's test makes it terminal: p̂†χ = +iλχ has the one L² solution e^(−x/a), while p̂†χ = −iλχ gives e^(+x/a), which diverges. Deficiency indices (1, 0) are unequal, so no self-adjoint extension exists.
Physics: momentum generates translations, and a translation pushes a half-line state off its space. Kinetic energy survives confinement — −d²/dx² there has equal indices (1, 1) and the family ψ′(0) = λψ(0) — but momentum itself does not.

AnswerSymmetric, yes; self-adjoint, never: p̂† owns the normalisable eigenfunction χ with imaginary eigenvalue iℏ/a ≈ i × 1.05 × 10⁻²⁵ kg m s⁻¹, and deficiency indices (1, 0) forbid every extension. A half-line particle has no momentum observable.