Operator Classes Cheat Sheet

Operator classes answer different questions: whether an action is continuous, whether finite-rank approximation works, whether singular values are summable, or whether an unbounded observable has the correct adjoint domain. The comparisons below keep these questions separate. The linked articles develop the definitions and proofs; this page supplies a compact selection guide and a borderline classification exercise.

Bounded and ideal classes at a glance

For operators between Hilbert spaces, the norm and summability criteria below refer to their everywhere-defined bounded extensions. Singular values $s_n$ are arranged in nonincreasing order with multiplicity.

Class	Test or distinguishing property	Canonical development
Bounded	$\lVert A\psi\rVert\le C\lVert\psi\rVert$; an everywhere-defined continuous action	Bounded Operators
Compact	Images of bounded sets have compact closure; norm limits of finite-rank operators on Hilbert space	Compact Operators
Hilbert–Schmidt	$\sum_n s_n^2<\infty$; $\lVert A\rVert_2^2=\operatorname{Tr}(A^\dagger A)$	Hilbert–Schmidt Operators
Trace class	$\sum_n s_n<\infty$; $\lVert A\rVert_1=\operatorname{Tr}\lvert A\rvert$	Trace Class Operators
Positive bounded	$\langle\psi,A\psi\rangle\ge0$ for every vector; admits a bounded positive square root	Positive Operators
Density operator	Positive trace class with $\operatorname{Tr}\rho=1$	Density Operators

On an infinite-dimensional Hilbert space,

$$\mathcal S_1\subsetneq\mathcal S_2 \subsetneq\mathcal K\subsetneq B(\mathcal H),\qquad \|A\|\le\|A\|_2\le\|A\|_1$$

where the norms on the right are used only when finite. Positivity is not another step in this chain: a positive operator can be the identity, a compact operator with divergent trace, or a density operator.

For quick calculations, bounded multiplication preserves an operator ideal, and two Hilbert–Schmidt factors have trace-class product:

$$\|BAC\|_1\le\|B\|\|A\|_1\|C\|,\qquad \|XY\|_1\le\|X\|_2\|Y\|_2.$$

The trace-class treatment proves the cyclic trace identities and their hypotheses. An arbitrary product of unbounded operators is outside these estimates until its domain and ideal membership are established.

Domain classes answer a separate question

Property	What must be checked	Canonical development
Densely defined	$\overline{D(A)}=\mathcal H$; needed for a single-valued adjoint	Dense Domains
Closed	The graph is closed in $\mathcal H\oplus\mathcal H$	Closed and Closable Operators
Symmetric	$A\subseteq A^\dagger$, including the domain inclusion	Symmetric Operators
Self-adjoint	$A=A^\dagger$, including equality of domains	Self-Adjoint Operators
Essentially self-adjoint	The closure of the symmetric operator is self-adjoint	Essential Self-Adjointness

A closed unbounded operator need not be symmetric. A symmetric operator need not be closed, and a closed symmetric operator need not be self-adjoint. For a bounded operator defined on all of $\mathcal H$, symmetry does imply self-adjointness; this finite-matrix habit is exactly what becomes unsafe when domains are suppressed.

An unbounded positive self-adjoint operator can be discussed through its quadratic form and spectral square root. The bounded positive-operator row should not be read as imposing an everywhere-defined action on such an observable.

Worked classification with a logarithmic boundary

On $\ell^2(\mathbb N)$, let

$$A_{\alpha,\beta}e_n =n^{-\alpha}[\log(n+1)]^{-\beta}e_n, \qquad \alpha\ge0,\quad \beta\in\mathbb R.$$

For $\alpha>0$ the diagonal tends to zero for every fixed $\beta$, so the operator is bounded, positive, and compact. Schatten membership is determined by

$$\sum_n n^{-\alpha p}[\log(n+1)]^{-\beta p}<\infty.$$

The integral test gives convergence when $\alpha p>1$, or when $\alpha p=1$ and $\beta p>1$. Thus $A_{1/2,3/4}$ is Hilbert–Schmidt because the squared sequence is $1/[n\log(n+1)^{3/2}]$, but is not trace class. It cannot be normalized into a density operator by multiplying by a positive constant.

At $\alpha=0$, the operator is compact exactly when $\beta>0$, bounded but noncompact for $\beta=0$, and unbounded for $\beta<0$ on its maximal diagonal domain. A logarithm can decide compactness or a borderline Schatten test, but it cannot overcome a strictly divergent power.

Exercises

A state from an operator square. For which $\beta$ does $A_{1/2,\beta}^\dagger A_{1/2,\beta}$ have finite nonzero trace, so that its normalization is a density operator?

Solution

Its eigenvalues are $1/[n\log(n+1)^{2\beta}]$. The trace converges exactly when $2\beta>1$. Thus an operator that is not itself trace class can still produce a density operator through $A^\dagger A$ if it is Hilbert–Schmidt. This is the operator-ideal condition behind square-root representations of mixed states.

A bounded noncompact limit. Let $P_N$ project onto the first $N$ basis vectors. Explain how compact operators can converge strongly to a noncompact operator.

Solution

$P_N\psi\to\psi$ for every fixed $\psi$, so the strong limit is $I$. But $\|I-P_N\|=1$ for every $N$; there is no operator-norm convergence. Compact operators are closed under operator-norm limits, not under arbitrary strong limits.

References

• Hall, Brian C. Quantum Theory for Mathematicians. Graduate Texts in Mathematics 267, Springer, 2013. doi:10.1007/978-1-4614-7116-5. Bounded and unbounded operator distinctions.
• Simon, Barry. Trace Ideals and Their Applications, 2nd ed. Mathematical Surveys and Monographs 120, American Mathematical Society, 2005. Publisher record. Schatten ideals and product estimates.