Common Operator Pitfalls

Operator mistakes often arise from applying a correct finite-dimensional identity where one of its inputs no longer exists. A useful repair starts by identifying the disputed expression’s domain, then checks which limit, inverse, or spectral theorem is actually available. The examples below are diagnostics; the linked pages supply the canonical derivations.

Helpful background. Operators in Quantum Mechanics separates physical roles, and Self-Adjoint Operators states the domain-sensitive observable criterion.

Differential expressions do not select operators

The mistaken step: write $P=-i\hbar\,d/dx$, integrate by parts, and conclude that momentum is self-adjoint.

Diagnosis: integration by parts produces a boundary form. On an interval, that form depends on endpoint traces. Dirichlet conditions at both ends make a first-order momentum symmetric but do not produce its full adjoint domain.

Repair: specify the initial domain; compute the maximal adjoint domain; compare the domains. Quasiperiodic domains give a self-adjoint interval momentum family, while the half-line minimal momentum has unequal deficiency indices and no same-space self-adjoint extension. These are distinct results despite the shared expression. See Adjoints of Unbounded Operators and Deficiency Indices.

A parallel mistake is assuming that two Hamiltonians with the same bulk potential describe the same physics. Robin boundary parameters can change bound-state energies and reflection phases.

Dense tests do not justify every limiting step

The mistaken step: an identity holds on a dense test space, so it holds on the full operator domain.

Diagnosis: density in Hilbert norm controls input vectors but not their unbounded images. On $\ell^2$, $Ae_n=ne_n$ and $\psi_n=e_n/n$ give

$$\psi_n\to0, \qquad \|A\psi_n\|=1.$$

The operator cannot be passed through this limit using continuity in the Hilbert norm.

Repair: identify an operator core and prove graph-norm convergence, or use a bounded extension argument with a uniform estimate. For a quadratic form, use the form norm instead. Compactly supported smooth functions on an interval are Hilbert-dense but are not a core for periodic momentum; see Dense Domains.

Closedness does not repair a missing image limit. It says what happens when both inputs and images converge, not that the second convergence follows from the first.

Moments and operator applications are different requirements

The mistaken step: finite mean energy implies that $H\psi$ exists, or finite variance requires $\psi\in D(H^2)$.

Diagnosis: for self-adjoint $H$, finite absolute mean energy is the condition $\int|\lambda|\,d\mu_\psi<\infty$. Applying $H$ requires $\int\lambda^2\,d\mu_\psi<\infty$. Applying $H^2$ requires a fourth moment.

Repair: write the spectral integral first. A finite second moment gives the second-moment value $\|H\psi\|^2$ on $D(H)$; it does not require the stronger expression $\langle\psi,H^2\psi\rangle$ to be literally defined. A finite first moment can be expressed by the associated form. See Quadratic Forms.

This is why a finite-energy wavefunction can evolve continuously without satisfying the Schrödinger equation as a strong differentiable vector equation.

Products and commutators need intermediate vectors

The mistaken step: cancel an unbounded operator or use $(AB)^*=B^*A^*$ as an unrestricted identity.

Diagnosis: $AB$ requires $B\psi\in D(A)$, while $BA$ requires $A\psi\in D(B)$. Even multiplication by zero shows the difference: $A0$ is defined everywhere, whereas $0A$ is defined only on $D(A)$.

Repair: display the natural product domains and compare the resulting operators. For adjoints, start from the generally valid inclusion $B^*A^*\subseteq(AB)^*$ under the needed density assumptions. Equality needs more information.

Likewise a vanishing commutator on a common dense domain is not automatically strong commutation of the spectral measures. Compatibility of unbounded observables requires the stronger spectral statement; see Compatible Observables.

Eigenvalues do not exhaust spectra

The mistaken step: solve the normalizable eigenvalue equation, find no solutions, and conclude the spectrum is empty.

Diagnosis: position on $L^2(\mathbb R)$ has no normalizable eigenvectors but has spectrum $\mathbb R$. Narrow packets supply approximate eigenvectors at every real number.

Repair: test bounded invertibility or use a justified Weyl criterion. Also distinguish “continuous spectrum” as a range classification from absolutely continuous spectral measures. A diagonal operator can have pure point spectral type while an accumulation point belongs to its continuous spectrum. See Point, Continuous, and Residual Spectrum.

A finite matrix discretization necessarily produces eigenvalues; this alone says nothing about whether the continuum limit has bound states or continuous spectrum.

A small inverse residual needs a conditioning estimate

The mistaken step: $(A-z)\psi_n-f$ is small, so $\psi_n$ must be close to the exact solution.

Diagnosis: this inference needs a bound on the inverse. Near a spectral point the resolvent norm can be large; at a spectral point there may be no bounded inverse at all.

Repair: if $z\in\rho(A)$, then

$$\|\psi_n-R_A(z)f\| \leq\|R_A(z)\|\,\|(A-z)\psi_n-f\|.$$

For self-adjoint $A$, inverse spectral distance supplies the conditioning constant. For nonnormal operators, spectral distance alone need not control it. See Resolvent Operators.

Exercises

A normalized sequence satisfies $\|(A-\lambda)\psi_n\|\to0$. What extra information would turn this into an actual eigenvector proof, assuming $A$ is closed?

Solution

Norm convergence $\psi_n\to\psi$ would suffice. Normalization gives $\|\psi\|=1$, and $A\psi_n=\lambda\psi_n+o(1)\to\lambda\psi$. Closedness then gives $\psi\in D(A)$ and $A\psi=\lambda\psi$. Weak convergence to zero instead provides no eigenvector; in the self-adjoint case it is the pattern used to diagnose essential spectrum.

For a normalized vector with coefficients $c_n=Cn^{-2}$ and $He_n=ne_n$, decide whether $\langle H\rangle$, $\|H\psi\|^2$, and $H^2\psi$ exist.

Solution

The mean involves $\sum n^{-3}$ and is finite. The second moment is $C^2\sum n^{-2}$ and is finite, so $H\psi$ exists. But $H^2\psi$ would have constant coefficients $C$, which are not square-summable. Thus the second moment is valid while the vector $H^2\psi$ is not.

References

• G. Bonneau, J. Faraut, and G. 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, author manuscript — domain errors in elementary quantum examples.
• B. C. Hall, Quantum Theory for Mathematicians, Graduate Texts in Mathematics 267, Springer, 2013, doi:10.1007/978-1-4614-7116-5 — counterexamples and precise operator hypotheses.
• K. Schmüdgen, Unbounded Self-adjoint Operators on Hilbert Space, Graduate Texts in Mathematics 265, Springer, 2012, doi:10.1007/978-94-007-4753-1 — domain algebra, spectra, and functional calculus.