Standard Examples Table

A good standard example isolates one mathematical distinction with as little extra physics as possible. Multiplication operators expose spectral measures; diagonal operators expose domain and summability conditions; shifts expose the difference between isometries and unitaries. Use the table to choose a test model, then follow the link for its full construction.

Choose an example for the claim being tested

Example and space	What it tests	Development
$Qf(x)=xf(x)$ on $L^2(\mathbb R)$, maximal domain	A self-adjoint observable can have continuous spectrum with no Hilbert-space eigenvectors	Continuous spectrum without eigenvectors
$-i\,d/dx$ on an interval with a specified endpoint relation	One differential expression has different self-adjoint realizations and spectra	Self-adjoint extensions
$De_n=n^{-1}e_n$ on $\ell^2(\mathbb N)$	Compactness does not imply trace class; zero can be a spectral limit point	Compact operators
$Ne_n=ne_n$ on its maximal diagonal domain	Domain membership is a weighted square-summability condition	Unbounded operators
$Se_n=e_{n+1}$ on $\ell^2(\mathbb N_0)$	An isometry can fail to be surjective	Stinespring dilation
Matrix transpose on a qubit	Positivity of a map is weaker than complete positivity	Completely positive maps
$\rho=\sum_np_n\lvert n\rangle\langle n\rvert$	Trace, moments, and entropy require different summability tests	Entropy and trace
$\psi=\sum_ns_n\lvert n,n\rangle$	Schmidt coefficients control reductions and finite-rank approximation	Schmidt decomposition
$\Phi_t=e^{-\gamma t}\mathrm{id}+(1-e^{-\gamma t})R_\tau$	A bounded generator can act on an infinite-dimensional state space	Quantum dynamical semigroups
Pure birth rates $\lambda_n=2^n$	Formal local probability balance need not imply conservation	Bounded and unbounded generators

The examples are diagnostic rather than interchangeable. A finite matrix cannot test a genuinely unbounded domain problem. A multiplication operator with continuous spectrum does not establish anything about complete positivity of a map on states. Matching the mathematical category is the first step in selecting a counterexample.

For positive diagonal operators, ask separately whether the entries are bounded, tend to zero, sum to a finite value, or have finite entropy after normalization. For an unbounded diagonal observable, replace these questions by weighted square summability on vectors and weighted integrability in the state being measured.

Worked example: two spectral ladders and a disappearing energy penalty

On $\ell^2(\mathbb N_0^2)$ define the self-adjoint diagonal operator

$$H_a e_{m,n}=(m+an)e_{m,n},\qquad a\ge0,$$

on its maximal domain. Units are chosen so that the first ladder spacing is one. For $a>0$, only finitely many pairs satisfy $m+an\le E$ at a fixed finite $E$. Its resolvent is compact: the diagonal entries of $(H_a+I)^{-1}$ tend to zero outside finite sets.

When $a$ is irrational, equality $m+an=m'+an'$ forces $m=m'$ and $n=n'$, so every energy eigenvalue is simple. When $a$ is a positive rational number, degeneracies occur, but each energy level still has finite multiplicity. Compact resolvent is compatible with those finite degeneracies.

At $a=0$, every energy $m$ has infinitely many orthogonal eigenvectors $e_{m,n}$. The resolvent is no longer compact: it acts by the same nonzero number $(m+1)^{-1}$ on an infinite-dimensional subspace. The spectral set is $\mathbb N_0$, and every eigenvalue now has infinite multiplicity.

The thermal trace sees this distinction as well. For $\beta>0$ and $a>0$,

$$Z_a(\beta)=\operatorname{Tr}e^{-\beta H_a} =\frac1{(1-e^{-\beta})(1-e^{-\beta a})}.$$

At $a=0$, the sum over $n$ diverges at every temperature. Each finite-dimensional cutoff has a Gibbs state, but there is no normalized Gibbs density operator for the full zero-penalty model. The issue is an uncontrolled multiplicity, not an energy unbounded below.

This example connects the compactness tests in the table to quantum-statistical state construction without requiring an interacting many-body Hamiltonian.

Exercises

The domain of the two-ladder Hamiltonian. State $D(H_a)$ for $a>0$ and compare it with the form domain.

Solution

For $\psi=\sum c_{m,n}e_{m,n}$, $D(H_a)$ requires $\sum(m+an)^2|c_{m,n}|^2<\infty$. The nonnegative form domain $D(H_a^{1/2})$ instead requires $\sum(m+an)|c_{m,n}|^2<\infty$. The form can assign a finite mean energy even when the vector is outside the operator domain, as explained by quadratic forms.

Thermal weight escaping through a soft ladder. For fixed $\beta>0$, calculate the Gibbs probability that $n\le N$ and send $a\downarrow0$ with $N$ fixed.

Solution

The second ladder has geometric probabilities $(1-e^{-\beta a})e^{-\beta an}$, so $\Pr(n\le N)=1-e^{-\beta a(N+1)}\to0$. Mass escapes every fixed finite set of second-ladder levels. This diagnoses why the normalized finite-$a$ Gibbs states cannot converge in trace norm to a density operator at $a=0$.

References

• Hall, Brian C. Quantum Theory for Mathematicians. Graduate Texts in Mathematics 267, Springer, 2013. doi:10.1007/978-1-4614-7116-5. Multiplication, differentiation, and diagonal spectral models.
• Holevo, Alexander S. Quantum Systems, Channels, Information: A Mathematical Introduction, 2nd ed. De Gruyter, 2019. doi:10.1515/9783110642490. State and channel 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. Spectral multiplicity, compact resolvents, and domains.