Self-adjointness and physical boundaries
Specify an unbounded observable by its action and domain, and see how boundary conditions determine its spectrum and probability flow.
On this page
The question for today
What you’ll understand
- Distinguish a symmetric differential expression from a self-adjoint operator.
- Evaluate boundary forms produced by integration by parts.
- Derive the momentum spectrum for a specified quasiperiodic domain.
An observable written as a derivative is incomplete until the allowed functions and boundary conditions are specified. In a finite matrix, Hermitian conjugation settles self-adjointness on the whole space. In an infinite-dimensional function space, the adjoint can have a larger domain even when integration by parts looks formally Hermitian. Physical boundaries are therefore part of the observable rather than decorative conditions appended after its eigenvalues have been calculated.
The domain is part of the operator
For a densely defined operator A, symmetry means its inner-product boundary form vanishes for pairs of vectors in its chosen domain. Self-adjointness requires more: the adjoint must have exactly that same domain and action. An unbounded operator cannot generally act on every Hilbert-space vector. The adjoint domain consists of vectors for which the relevant inner-product functional is bounded, so it depends on both the expression and the original domain.
Consider momentum on a finite interval. Integrating its first derivative by parts leaves endpoint values. Setting both endpoint values to zero certainly removes this form for vectors in the original domain, making the operator symmetric. But testing against such functions imposes no endpoint restriction on an adjoint vector. The adjoint domain is then larger, so this two-endpoint Dirichlet realization of momentum is not self-adjoint. A real expectation value on selected vectors does not repair that mismatch.
Matching endpoints creates a self-adjoint family
A quasiperiodic condition links the endpoint values by one fixed unit-modulus phase. For the first derivative, the domain of H¹ functions satisfying this relation yields a self-adjoint momentum realization. Both endpoint contributions cancel because bra and ket obey the same phase condition, and the adjoint condition enforces that same relation. Different phases specify different operators, even though their derivative formula is identical.
The eigenfunctions are plane waves. Requiring the phase accumulated across the interval to match the domain phase quantizes the momentum. Periodic boundary conditions allow zero momentum; antiperiodic boundary conditions shift the whole integer ladder by half a step. This model describes a circle with an endpoint phase identification, not a particle bouncing off two impenetrable walls. For the free Hamiltonian , the induced domain also requires . Together the value and derivative conditions make Schrödinger currents agree at the identified endpoints, allowing circulation without loss.
The Hamiltonian has its own boundary problem
The kinetic Hamiltonian uses a second derivative and has a different boundary form involving values and first derivatives. Its Dirichlet realization on an interval can be self-adjoint even though the first derivative with Dirichlet conditions at both ends is not. Squaring a formal expression and assigning unrelated boundary conditions is therefore not the same as taking the operator square with its induced domain. This distinction explains many apparent contradictions about momentum inside an infinite well.
A self-adjoint Hamiltonian generates unitary time evolution through the spectral theorem. Merely demonstrating a formally conserved norm for smooth trial functions does not establish that the full initial-value problem has the required generator. In applications, identify the geometry and boundary physics, specify an appropriate self-adjoint realization, and only then use its eigenbasis or evolution. Not every mathematically allowed extension represents the same apparatus, and a boundary phase is physical input rather than a freely adjustable normalization constant.
Work it through
Antiperiodic momentum on a ring
Use the self-adjoint momentum domain with ψ(L)=−ψ(0). Find its two smallest-magnitude momenta, normalized eigenfunctions, and kinetic energies for mass m using H=P²/(2m) with the induced operator-square domain.
The minus sign is the boundary phase θ=π. The spectral condition gives . The integers n=0 and n=−1 yield the nearest values to zero, +πℏ/L and −πℏ/L.
Their eigenfunctions are and . Each has constant density 1/L and norm one, and each changes sign between x=0 and x=L as required.
Squaring either momentum gives the same kinetic energy . The states remain distinct because their probability currents, , circulate in opposite directions while agreeing at the identified endpoints.
The lowest kinetic level is doubly degenerate with momenta ±πℏ/L and energy π²ℏ²/(2mL²). Zero momentum is excluded by the chosen antiperiodic domain.
The living notebook
Explore the model
Interactive lab · waves
An endpoint sign as a phase condition
Reset and set both controls to Phase difference 0° and Amplitude ratio 1; predict resultant amplitude 2. For a fresh comparison, Reset and set both controls to Phase difference 180° and Amplitude ratio 1; predict zero resultant amplitude. This illustrates a relative sign from a π phase, not the spectrum of the interval derivative.
- Sum at x = 0
- 0.866
- Resultant amplitude
- 1.732
- Phase difference
- 60°
Two waves share the same frequency and wave number. Playback advances t in sin(2x − t) + a sin(2x − t + phase). Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.
Look forPhase conditions can have observable algebraic consequences. The equal-frequency wave lab has no operator domains, quantum boundary conditions, or momentum-eigenvalue computation.
Pause & reason
Check your understanding
Your turn
Try it yourself
Let P=−iℏ d/dx on 0<x<L have domain H¹ functions vanishing at both ends. For the normalized ψ=√(2/L) sin(πx/L), compute ⟨P⟩ and explain why a real answer does not make this P self-adjoint.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
Integrate sin times cos for the expectation. Separately test what endpoint restrictions integration by parts imposes on an adjoint vector.
Compare with the solution
The expectation is proportional to . This real result follows from symmetry for this state. Because every original test function vanishes at both endpoints, the adjoint boundary form places no endpoint restriction on its other argument: D(P*) is H¹(0,L), larger than the original H¹ domain with two zero traces. Thus P is symmetric but not self-adjoint, despite its real expectation. The self-adjoint Dirichlet kinetic Hamiltonian is a separate second-derivative operator.
Keep the idea
Take it with you
- An unbounded observable consists of a differential action and a specified domain.
- Symmetry is weaker than equality with the adjoint including its domain.
- Boundary conditions determine spectra and can differ for momentum and kinetic energy.
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