Particle in a Box and Boundary Conditions

A particle confined to an interval does not have a unique kinetic Hamiltonian until boundary conditions are specified. Dirichlet, Neumann, and Robin realizations share the differential expression $-\hbar^2d^2/(2m\,dx^2)$ but have different spectra and energy forms. The Infinite Square Well gives the elementary Dirichlet solution; here the domains explain what that solution assumes.

Required background. Quadratic Forms supplies the representation theorem. Self-Adjoint Extensions distinguishes a boundary realization from closure of a minimal graph.

Dirichlet and Neumann walls use different domains

Let $\alpha=\hbar^2/(2m)>0$ and $\mathcal H=L^2(0,L)$. The Dirichlet Hamiltonian is

$$H_D\psi=-\alpha\psi'',\qquad D(H_D)=H^2(0,L)\cap H_0^1(0,L).$$

Its closed nonnegative form is $\alpha\|\psi'\|^2$ on $H_0^1(0,L)$, whose traces satisfy $\psi(0)=\psi(L)=0$. Integration by parts in the representation identity recovers the displayed operator domain.

For Neumann walls, the form is the same integral but its domain is all $H^1(0,L)$. Arbitrary endpoint test values force

$$D(H_N)=\{\psi\in H^2(0,L):\psi'(0)=\psi'(L)=0\}.$$

Thus boundary conditions can enter either as restrictions on the form domain or as conditions recovered on the smaller operator domain. Treating all boundary conditions as restrictions on arbitrary normalized state vectors would confuse these levels.

The constant normalized state belongs to the Neumann operator domain with energy zero. It is not in the Dirichlet form domain. Both realizations act on the same Hilbert space.

Robin forms determine the signs at each endpoint

For real $\beta_0,\beta_L$, each with units of inverse length, define

$$q_\beta[\psi]=\alpha\left( \|\psi'\|^2+\beta_0|\psi(0)|^2+\beta_L|\psi(L)|^2 \right),\qquad Q=H^1(0,L).$$

The endpoint trace inequalities make both boundary terms infinitesimally form bounded relative to the derivative form. Therefore the form is closed and semibounded for every finite pair of real parameters.

Integration by parts gives the operator domain

$$\psi'(0)=\beta_0\psi(0),\qquad \psi'(L)=-\beta_L\psi(L),$$

with $\psi\in H^2(0,L)$. The opposite signs reflect the two outward endpoint orientations. For each real parameter the probability current vanishes at the corresponding endpoint.

Positive boundary parameters increase the form energy. Negative ones can create negative eigenvalues even though the interior potential is zero. Self-adjointness ensures unitary evolution; it does not enforce nonnegative energy for every extension.

These separated conditions do not exhaust the self-adjoint interval Laplacians. The general family also includes conditions coupling the endpoints, such as periodic or quasiperiodic ones.

Compactness explains why the spectrum is discrete

The form domains above embed compactly into $L^2(0,L)$ by the one-dimensional Rellich theorem. After a constant shift that makes the form positive, its norm is equivalent to the appropriate Sobolev norm. The associated resolvent is therefore compact.

It follows that the spectrum consists of eigenvalues with finite multiplicity tending to positive infinity, and the eigenvectors form a complete orthonormal basis. This conclusion covers Robin conditions even when their eigenvalues are determined by a transcendental equation rather than a simple formula.

For Dirichlet conditions the canonical solution gives $E_n=\alpha(n\pi/L)^2$, $n\geq1$. Spectral coefficients satisfy $\sum E_n|c_n|^2<\infty$ on the form domain and $\sum E_n^2|c_n|^2<\infty$ on the operator domain.

Dirichlet kinetic energy is not a periodic momentum squared

Let $P_{\min}=-i\hbar\,d/dx$ on $H_0^1(0,L)$. This operator is closed and symmetric but not self-adjoint. Its adjoint has domain $H^1(0,L)$, and

$$H_D=\frac1{2m}P_{\min}^*P_{\min}.$$

The product domain gives exactly $H^2\cap H_0^1$. This factorization is a positive-operator construction, not the square of a self-adjoint interval momentum.

By contrast, $P_\theta^2$ requires quasiperiodic conditions on both the function and its derivative, as shown on Momentum on an Interval. Replacing $H_D$ by that square changes the model.

Exercise: a normalized state can have infinite box energy

Expand $\psi(x)=1/\sqrt L$ in the Dirichlet sine basis and determine whether its mean Dirichlet energy is finite.

Solution

The coefficient is $c_n=\sqrt2(1-(-1)^n)/(n\pi)$, so only odd $n$ contribute and $|c_n|^2=8/(n^2\pi^2)$ there. These probabilities sum to one, but $E_n|c_n|^2=8\alpha/L^2$ for every odd $n$, so the mean energy diverges. The state is a valid Hilbert vector and evolves unitarily under $H_D$, but it lies outside its form domain. Its zero interior derivative does not override the missing Dirichlet traces.

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 Hamiltonians and boundary realizations.
• Teschl, Gerald, Mathematical Methods in Quantum Mechanics, 2nd ed., Graduate Studies in Mathematics 157, American Mathematical Society, 2014, author’s text, Chapters 6 and 9 — forms and regular Sturm–Liouville problems.