Operator Theory Minimum

An operator statement in quantum mechanics identifies a rule, a domain, and a claim about that pair. The minimum preparation is the ability to check those three pieces before using spectral or dynamical conclusions. This page gives a compact reading procedure and a worked boundary-condition example; the detailed definitions and proofs live in Operators, Domains, and Spectra.

Helpful background. Domains of Operators introduces formula–domain pairs; Adjoint Operators introduces the inner-product identity defining an adjoint.

Operator expressions have input conditions

For linear operators $A$ and $B$ on a Hilbert space, the natural domains of their sum and product are

$$\begin{aligned} D(A+B)&=D(A)\cap D(B),\\ D(AB)&=\{\psi\in D(B):B\psi\in D(A)\}. \end{aligned}$$

The product acts from right to left. The commutator $[A,B]=AB-BA$ can be evaluated only on $D(AB)\cap D(BA)$ unless a particular extension or quadratic-form interpretation is explicitly supplied. A dense common test space is useful, but equality on that space is not automatically equality of closed operators.

Even if $A=B$ as differential expressions, the operators differ when their domains differ. Conversely, proving equality of two closed operators on a common operator core can identify their closures. The extra word core supplies graph-norm density, which ordinary Hilbert-norm density does not.

The adjoint test contains a quantifier

For densely defined $A$, a vector $\eta$ lies in $D(A^*)$ when there is a vector $\zeta$ satisfying

$$\langle\eta,A\psi\rangle=\langle\zeta,\psi\rangle \quad\text{for all }\psi\in D(A).$$

Then $A^*\eta=\zeta$. Density makes $\zeta$ unique. The test asks whether the functional in $\psi$ is bounded in the Hilbert norm; differentiating a formal expression is not enough to answer it.

A symmetric operator satisfies $A\subseteq A^*$, meaning agreement of actions on $D(A)$ together with domain inclusion. A self-adjoint operator satisfies equality, including $D(A)=D(A^*)$. An essentially self-adjoint operator may start on a smaller domain, but its closure is self-adjoint. These conditions should not be interchanged when invoking the spectral theorem or Stone’s Theorem.

Momentum with endpoint conditions

Take $P_0=-i\hbar\,d/dx$ in $L^2(0,L)$, initially on $H_0^1(0,L)$. This Sobolev domain consists of functions with square-integrable weak derivative and zero endpoint traces. Integration by parts gives

$$\langle\phi,P\psi\rangle-\langle P\phi,\psi\rangle =-i\hbar[\overline{\phi(x)}\psi(x)]_0^L.$$

The boundary term vanishes for $\phi,\psi\in H_0^1$, so $P_0$ is symmetric. But it also vanishes for every $\psi\in H_0^1$ when $\phi$ is any member of $H^1(0,L)$. The weak-derivative adjoint test gives

$$D(P_0^*)=H^1(0,L),$$

with no endpoint restriction. The domains differ, so $P_0$ is not self-adjoint. Real diagonal expectation values did not establish the needed conclusion.

For a fixed phase $\theta$, instead choose the domain $\psi(L)=e^{i\theta}\psi(0)$ within $H^1(0,L)$. Requiring the boundary form to vanish against every such $\psi$ forces the same condition on $\phi$ in the adjoint domain. This operator is self-adjoint. Its eigenvalues are $\hbar(2\pi n+\theta)/L$, $n\in\mathbb Z$. Changing the domain changes the spectral problem even though the derivative formula remains the same. The full extension construction is developed in Self-Adjoint Extensions.

Spectral claims ask about invertibility, not only eigenvectors

For a closed operator, $z$ lies in the resolvent set when $A-zI:D(A)\to\mathcal H$ is bijective with bounded inverse on the Hilbert space. Its complement is the spectrum. Failure of injectivity gives an eigenvalue, but failure of surjectivity or bounded inverse can create spectrum without eigenvectors.

Multiplication by $x$ on $L^2(\mathbb R)$ illustrates this distinction. The equation $(x-a)\psi=0$ has no nonzero $L^2$ solution, yet normalized functions supported near $a$ have arbitrarily small $\|(X-a)\psi\|$. A bounded inverse would rule out these approximate eigenvectors. Thus real $a$ belongs to the spectrum even without a normalizable eigenstate.

Exercise: identify the illegitimate cancellation

For the position and momentum expressions, a calculation on Schwartz functions gives $XP-PX=i\hbar I$. Does the rearrangement $XP\psi=PX\psi+i\hbar\psi$ hold for every $\psi\in L^2(\mathbb R)$?

Solution

No. Both products must first exist, requiring $\psi\in D(XP)\cap D(PX)$. Schwartz functions satisfy this requirement, but a general square-integrable function need not have a square-integrable derivative or a finite position moment. The identity is valid on the specified common domain; extending it requires a separate statement. The bounded Weyl Form of the CCR avoids these product-domain expressions.

References

• Hall, Brian C., Quantum Theory for Mathematicians, Graduate Texts in Mathematics 267, Springer, 2013, doi:10.1007/978-1-4614-7116-5 — unbounded operators and interval examples.
• Teschl, Gerald, Mathematical Methods in Quantum Mechanics: With Applications to Schrödinger Operators, 2nd ed., Graduate Studies in Mathematics 157, American Mathematical Society, 2014, author’s edition — adjoints, self-adjointness, resolvents, and extension theory.