Momentum on an Interval: Transport and Boundary Phase

The self-adjoint interval momenta form a family indexed by a boundary phase. The canonical interval theorem derives that family and its spectrum; Self-Adjoint Operators supplies the domain framework. Here the phase is interpreted through wrapped translations and gauge transport, and the resolvent shows how it enters measurable spectral data.

Required background. The linked interval theorem supplies $P_\theta=-i\hbar\,d/dx$ on $D(P_\theta)=\{\psi\in H^1(0,L):\psi(L)=e^{i\theta}\psi(0)\}$. This page uses $L>0$ and $\theta$ modulo $2\pi$.

Wrapped translation accumulates the boundary phase

Extend a function on $[0,L)$ quasiperiodically to the line by

$$\psi_{\rm ext}(x+nL)=e^{in\theta}\psi(x).$$

The unitary group generated by $P_\theta$ is

$$(T_\theta(a)\psi)(x)=\psi_{\rm ext}(x-a).$$

The formula is well defined for almost-everywhere classes and preserves the norm on one period. It gives a strongly continuous group, with derivative $-iP_\theta/\hbar$ on the specified domain.

One full circuit has the action

$$T_\theta(L)=e^{-i\theta}I.$$

The phase is global after a complete circuit, but it shifts momentum eigenvalues and affects interference between paths with different windings. This is a circle-type transport model, not translation within a box with reflecting hard walls.

For $0<a<L$, values crossing the left endpoint re-enter from the right with the factor $e^{-i\theta}$. Reversing the direction of transport conjugates that factor. The sign follows from the convention $\psi(x-a)$.

A gauge map transfers the phase into the operator

Let $(V_\theta f)(x)=e^{i\theta x/L}f(x)$. It maps the periodic domain $D(P_0)$ onto $D(P_\theta)$, and differentiation gives

$$V_\theta^*P_\theta V_\theta =P_0+\frac{\hbar\theta}{L}I.$$

Thus all boundary conditions can be compared on one periodic domain at the cost of a constant shift in the operator. A gauge transformation transports both action and domain; changing only the differential expression would be incomplete.

Increasing $\theta$ by $2\pi$ returns the same boundary condition and relabels the integer momentum index. The phase family is periodic as a family of operators, even though an individual continuously tracked eigenvalue branch need not return with the same index.

This is the mathematical mechanism used in ring and flux models. Relating $\theta$ to an electromagnetic flux additionally requires a charge and a specified minimal-coupling convention; the boundary theorem alone fixes neither.

The boundary condition determines resolvent poles

For nonreal $z$, solving $(P_\theta-z)\psi=f$ gives

$$\psi(x)=e^{izx/\hbar} \left[C+\frac{i}{\hbar}\int_0^x e^{-izy/\hbar}f(y)\,dy\right].$$

Writing $I_f(z)$ for the integral with upper limit $L$, the endpoint condition determines

$$C=\frac{i\,e^{izL/\hbar}I_f(z)} {\hbar\bigl(e^{i\theta}-e^{izL/\hbar}\bigr)}.$$

The denominator cannot vanish for nonreal $z$. Its real zeros occur at $z=\hbar(2\pi n+\theta)/L$, exactly the eigenvalues in the canonical theorem.

The resolvent maps $L^2$ into $H^1(0,L)$. Rellich compactness of the inclusion $H^1(0,L)\hookrightarrow L^2(0,L)$ then makes it compact. Thus this realization has a discrete spectrum with a complete orthonormal eigenbasis; the discreteness follows from the bounded configuration and domain control, not from the derivative expression alone.

Squaring momentum retains both endpoint conditions

The operator square has domain

$$\begin{aligned} D(P_\theta^2)=\{\psi\in H^2(0,L):\;& \psi(L)=e^{i\theta}\psi(0),\\ &\psi'(L)=e^{i\theta}\psi'(0)\}. \end{aligned}$$

The second condition follows because $P_\theta\psi$ must itself lie in $D(P_\theta)$. Therefore $P_\theta^2/(2m)$ is the quasiperiodic kinetic energy.

It is not the Dirichlet box Hamiltonian, whose domain requires $\psi(0)=\psi(L)=0$ without identifying the endpoint derivatives. An identity between differential expressions is insufficient to identify their self-adjoint realizations. The box boundary page develops that distinction.

Exercise: compare two familiar kinetic spectra

For $\theta=0$ and $\theta=\pi$, determine whether $P_\theta^2/(2m)$ has a zero-energy state and whether its lowest energy is degenerate.

Solution

At $\theta=0$, momentum zero occurs once, so the ground energy is zero and is nondegenerate. At $\theta=\pi$, the momenta nearest zero are $\pm\pi\hbar/L$, giving a twofold lowest energy $\pi^2\hbar^2/(2mL^2)$. Neither statement is the Dirichlet-box spectrum. The degeneracy at $\theta=\pi$ comes from the paired momenta; no spin or Kramers assumption is involved.

References

• Bonneau, Guy, Jacques Faraut, and Galliano Valent, “Self-adjoint extensions of operators and the teaching of quantum mechanics,” American Journal of Physics 69, 322–331 (2001), doi:10.1119/1.1328351, arXiv:quant-ph/0103153 — interval momentum, boundary phases, and kinetic-energy domains.
• Teschl, Gerald, Mathematical Methods in Quantum Mechanics, 2nd ed., Graduate Studies in Mathematics 157, American Mathematical Society, 2014, author’s text — resolvents, compactness, and boundary realizations.