Self-adjointness as boundary data

Key claim

A differential expression is not yet an observable. The fracture appears when the same formal Hamiltonian acquires different spectra because its domain changes.

Theme

Self-adjointness is boundary data in disguise.

Guiding question

For a particle on the interval $[0,L]$, suppose we write

What physical information is missing from this line?

Setup

Integration by parts exposes the boundary form

Symmetry requires this expression to vanish on the chosen domain. Self-adjointness further requires that the adjoint have exactly the same domain.

Analysis

Derivation

Build a domain by imposing Dirichlet data, $\psi (0)=\psi (L)=0$. The boundary form vanishes, and the familiar discrete spectrum follows.

Scope and assumptions

Why privilege Dirichlet data? Periodic data,

also makes the boundary form vanish and defines a different self-adjoint operator.

Physical interpretation

The formal differential rule describes local evolution. The domain tells the wavefunction how the ends of configuration space are physically identified.

False claim to diagnose

If two Hamiltonians have the same differential expression, they represent the same observable.

Locate the hidden assumption that makes this statement false.

What follows — and what does not

The Builder treats the boundary condition as a construction choice. The Skeptic reveals a family of equally consistent choices. The Translator identifies the choice as global physical structure rather than a mathematical afterthought.

Exercise

Let $H=-{\mathrm{d}}^2/\mathrm{d}{x}^2$ on $[0,L]$. Show that Robin conditions

with real $\alpha ,\beta$, make the boundary form vanish for every pair of functions satisfying the same conditions.

Hint 1

Evaluate the boundary form separately at $0$ and $L$.

Hint 2

Use the reality of $\alpha$ and $\beta$ when complex-conjugating the condition on $\varphi$.

Oral check 1. Why would complex $\alpha$ generally spoil symmetry?

Oral check 2. Does vanishing of the boundary form alone prove self-adjointness, or only symmetry?

Solution

At $x=0$,

$${\varphi }^{\ast }(0){\psi }^{\prime }(0)-{\varphi }^{\prime \ast }(0)\psi (0)=\alpha {\varphi }^{\ast }(0)\psi (0)-\alpha {\varphi }^{\ast }(0)\psi (0)=0.$$Copy LaTeX

The same cancellation holds at $L$ with $\beta$. Thus the boundary form vanishes. Establishing self-adjointness also requires verifying that the adjoint domain introduces no additional boundary freedom.

Check your understanding

In one sentence: what extra datum turns a formal Hamiltonian into a quantum observable?

Connections and next step

• New pressure point: operator domain versus differential expression.
• Retrieve later: deficiency indices and the $U(2)$ family of interval extensions.
• Unresolved: which extensions can be generated as limits of regular boundary potentials?