Self-adjoint domains in one dimension

Consider $H=-C\,d^2/dx^2$ on $L^2(0,\infty)$, where $C=\hbar^2/(2m)>0$. Two integrations by parts express the difference between $\langle\phi,H\psi\rangle$ and $\langle H\phi,\psi\rangle$ as endpoint terms. To obtain a symmetric operator, these terms must vanish for every pair of functions in its domain, not merely for one selected eigenfunction. At infinity the usual square-integrable regular domain supplies the required vanishing behaviour; the endpoint at zero still needs a condition.

A real Robin condition $\psi'(0)=\gamma\psi(0)$ makes the zero-endpoint form vanish for every pair obeying the same condition. With the usual Sobolev domain of square-integrable functions having square-integrable weak derivatives through order two, this gives a self-adjoint half-line realization. The real parameter γ has inverse-length units. Dirichlet condition $\psi(0)=0$ is another realization, treated separately rather than obtained by inserting an infinite number into an ordinary calculation.

Self-adjointness requires the operator and its adjoint to have equal domains as well as matching actions. Imposing both value and derivative zero at the regular endpoint makes the boundary form vanish, but restricts the domain too far: the adjoint allows more functions. This produces a symmetric minimal operator rather than a self-adjoint Hamiltonian. Extra boundary constraints are therefore not automatically a safer or more precise choice; they can remove the structure needed for a unique unitary evolution.

For the half-line Laplacian the deficiency indices are $(1,1)$, which underlies the one-parameter family of endpoint choices. Different choices need a physical endpoint model or experimental input. The Dirichlet condition of an ordinary impenetrable wall is a particular choice, not a theorem that every boundary interaction must use it. A Robin endpoint can represent an effective short-distance interaction while maintaining zero probability flux into the excluded half-line. Mathematical freedom does not mean the choices describe the same experiment.

Seek a negative-energy normalizable solution $\psi=Ae^{-\kappa x}$ with $\kappa>0$. Its endpoint derivative imposes $-\kappa=\gamma$. Consequently a negative Robin parameter supports one bound state, while nonnegative γ supports none. For γ = 0, a constant zero-energy solution is not square-integrable on the half-line. The bound energy is supplied by the endpoint model; calling the differential expression free does not make every self-adjoint realization nonnegative.

At positive energy an incoming wave and reflected wave must satisfy the same endpoint relation. For real γ their reflection amplitude has unit magnitude, but its phase depends on γ. Thus absence of probability leakage does not imply identical scattering. The endpoint current $j(0)=(\hbar/m)\operatorname{Im}[\psi^*(0)\psi'(0)]$ vanishes because γ is real. Reflection states use continuum normalization, whereas the boundary-bound exponential has an ordinary finite norm; the two should not be conflated.