Operator domains and quadratic forms

A differential expression such as $-d^2/dx^2$ is not yet a quantum Hamiltonian. The allowed wavefunctions and their boundary conditions are part of the operator. In field theory this warning also applies to a formal expression built from quantum fields.

Prerequisites: elementary quantum mechanics and the spectral theorem.
Outcomes: identify a domain-dependent claim; distinguish symmetry from self-adjointness; understand how a closed energy form can define an operator.

The same expression, different spectra

Take $K=-d^2/dx^2$ on the interval $[0,L]$, acting on square-integrable wavefunctions. Dirichlet boundary conditions give modes $\sin(n\pi x/L)$, $n=1,2,\ldots$, and eigenvalues $(n\pi/L)^2$. Neumann boundary conditions allow the constant mode, with eigenvalue zero, as well as the cosine modes.

Both use the same differential expression. Their domains encode different physical boundary conditions. An integration by parts produces

$$\langle f,Kg\rangle-\langle Kf,g\rangle =\left[\overline{f'}g-\overline f g'\right]_0^L.$$

One needs appropriate regularity to perform this calculation and conditions that cancel the boundary term. Cancellation proves symmetry on the chosen domain. Self-adjointness additionally means that the adjoint has exactly the same domain. Merely writing $K^\dagger=K$ without checking that domain can miss possible extensions.

Here $K$ has units of inverse length squared. It is not the relativistic field energy itself; for a free relativistic mode that energy is $\sqrt{m^2+K}$.

A simpler infinite-dimensional warning

On $\ell^2(\mathbb N)$, the space of square-summable sequences, consider the diagonal rule $(A\psi)_n=n\psi_n$. Its natural self-adjoint domain is

$$D(A)=\left\{\psi:\sum_{n=1}^\infty n^2|\psi_n|^2<\infty\right\}.$$

The vector with components $\psi_n=1/n$ lies in $\ell^2$, but not in $D(A)$. Applying the rule would produce a constant sequence, which is not square summable. Normalizability alone does not guarantee that applying an unbounded observable is meaningful.

Finite sequences form a useful starting domain. Closing an operator from such a domain, and proving that this closure is the intended self-adjoint operator, is a mathematical step rather than a convention.

Energy can be defined on a larger domain

The energy form associated with a nonnegative self-adjoint operator is

$$q_A[\psi]=\|A^{1/2}\psi\|^2,\qquad Q(A)=D(A^{1/2}).$$

For the diagonal example,

$$q_A[\psi]=\sum_n n|\psi_n|^2,\qquad Q(A)=\left\{\psi:\sum_n n|\psi_n|^2<\infty\right\}.$$

The form domain can be larger than the operator domain. For $\psi_n=n^{-5/4}$, the form sum is $\sum n^{-3/2}$ and converges, while the operator-norm sum is $\sum n^{-1/2}$ and diverges. This state has finite mean energy in the form sense even though $A\psi$ is not a Hilbert-space vector.

A densely defined nonnegative form is closed when its domain is complete in the norm $\bigl(\|\psi\|^2+q[\psi]\bigr)^{1/2}$. A closable form has a well-defined closed extension obtained by completion; arbitrary positive forms need not have this property.

The precise result we use is the representation theorem: a densely defined closed semibounded form determines a unique associated self-adjoint operator. See Teschl, Mathematical Methods in Quantum Mechanics, online version 12 February 2009, §2.3, Theorem 2.13, printed pp. 71–72. The surrounding discussion supplies the closedness assumptions. Positivity by itself is not the theorem.

A gap needs a stronger estimate

A nonnegative Hamiltonian may have energies arbitrarily close to zero. In a sector with a unique normalized vacuum $\Omega$, a spectral-gap estimate is a lower bound on the orthogonal complement:

$$q_H[\psi]\geq\Delta\|\psi\|^2 \quad\text{for }\psi\in Q(H),\ \langle\Omega,\psi\rangle=0, \qquad\Delta>0.$$

The estimate concerns all those vectors, not just a selected trial state. In a continuum construction the domain, vacuum sector, and bound must survive the required limits. Reflection positivity is part of a route to a physical space and Hamiltonian; it does not supply this strict bound automatically.

Exercise: positive energy with no positive gap

Let $H\Omega=0$ and $He_n=n^{-1}e_n$ on an orthonormal basis. Is $H$ bounded and self-adjoint? Does positivity prove a gap?

Solution. The real diagonal entries are bounded between 0 and 1, so the diagonal rule defines a bounded self-adjoint operator on the entire Hilbert space. Each excited basis vector has positive energy, but $\langle e_n,He_n\rangle=1/n\to0$. No positive uniform $\Delta$ works. Zero is an accumulation point of the excited spectrum despite the unique vacuum vector.

Continue with semigroups or return to physical states. The Jaffe–Witten target, §§3–4 requires a physical quantum theory, not just a formal Hamiltonian.