Compact Operators

A compact operator sends bounded sequences to sequences with norm-convergent subsequences. On a Hilbert space this is equivalent to approximation in operator norm by finite-rank operators. Compactness gives the discrete singular values needed to define trace and Hilbert–Schmidt norms; it does not guarantee that their sums converge.

Helpful background. Bounded operators introduces norm estimates; positive operators supplies the square root of $T^*T$.

Compactness and finite-dimensional approximation

For a bounded operator $T:\mathcal H\to\mathcal H$, compactness means that the image of the closed unit ball has compact closure. In this Hilbert-space setting the image is itself compact: the closed unit ball is weakly compact, and a compact operator turns weakly convergent bounded nets into norm-convergent image nets. Every finite-rank operator is compact because bounded subsets of a finite-dimensional range are relatively compact.

Norm limits of compact operators are compact: choose a sufficiently accurate compact approximant to cover the image of the unit ball by finitely many small balls. This shows total boundedness of the limit image, and completeness supplies compact closure.

Conversely, suppose $T$ is compact. Cover its unit-ball image by finitely many balls of radius $\epsilon$ and let $P$ project onto the span of their centers. Then

$$\|(I-P)T\|\leq\epsilon,$$

so the finite-rank operator $PT$ approximates $T$. Thus the compact operators $\mathcal K(\mathcal H)$ form the norm closure of the finite-rank operators. They form a two-sided ideal in $\mathcal B(\mathcal H)$: composing a compact map with a bounded map on either side preserves compactness.

This proof identifies the relevant topology. Strong convergence of finite-rank operators is much weaker. For an orthonormal basis, the projections $P_N$ onto the first $N$ basis vectors satisfy $P_N\psi\to\psi$ for every $\psi$, but $\|I-P_N\|=1$. The identity on an infinite-dimensional space is not compact.

Spectral structure and singular values

A compact self-adjoint operator has an orthonormal family of eigenvectors spanning the orthogonal complement of its kernel. Its nonzero eigenvalues have finite multiplicity and can accumulate only at zero. The kernel can be infinite-dimensional. For a positive compact operator this gives

$$A=\sum_n a_n|u_n\rangle\langle u_n|, \qquad a_n>0,\qquad a_n\to0$$

when infinitely many nonzero eigenvalues occur. The series converges in operator norm, since the norm of the remainder is the largest omitted eigenvalue.

An arbitrary compact operator need not have an orthonormal eigenbasis. Instead apply the positive compact spectral theorem to $|T|=(T^*T)^{1/2}$. Its positive eigenvalues $s_n(T)$ are the singular values. The polar decomposition gives orthonormal singular vectors and

$$T=\sum_n s_n(T)|v_n\rangle\langle u_n|.$$

Order the singular values decreasingly. The rank-$N$ truncation has operator-norm error $s_{N+1}(T)$; no operator of rank at most $N$ can do better. For the lower bound, the span of the first $N+1$ right singular vectors contains a unit vector annihilated by any rank-$N$ approximant. On that vector, $T$ has norm at least $s_{N+1}$. Simon gives the general singular-value and ideal theory.

Three different decay requirements

For $T_\alpha e_n=n^{-\alpha}e_n$ on $\ell^2(\mathbb N)$ with $\alpha>0$, the singular values tend to zero, so $T_\alpha$ is compact. However,

$$\sum_n s_n(T_\alpha)^2<\infty \ \Longleftrightarrow\ \alpha>\tfrac12, \qquad \sum_n s_n(T_\alpha)<\infty \ \Longleftrightarrow\ \alpha>1.$$

The first condition means Hilbert–Schmidt; the second means trace class. These distinct thresholds explain why discretizing a compact operator into a finite matrix does not by itself justify a limiting trace.

A contrasting example is multiplication by $x$ on $L^2([0,1])$. Choose normalized functions with mutually disjoint supports inside $[1/2,1]$. Their images remain mutually orthogonal and have norms at least $1/2$, so no subsequence can converge in norm. The operator is bounded and positive but not compact. Continuous-spectrum observables should therefore not be treated as compact merely because a numerical grid has finitely many points.

Compact resolvents and quantum Hamiltonians

An unbounded Hamiltonian is not a compact bounded operator. It can instead have a compact resolvent: $(H-z)^{-1}$ is compact for one, hence every, $z$ in its resolvent set. For self-adjoint $H$, this entails isolated eigenvalues of finite multiplicity with no finite accumulation point.

This is the structure behind many confined systems. It explains discrete energy levels, but does not alone ensure a finite partition function at every temperature: that requires sufficient eigenvalue growth so that $\sum_ne^{-\beta E_n}<\infty$. Distinguish the spectral compactness statement from the summability statement.

Exercises

1. A density operator has decreasing eigenvalues $p_n$. What is the operator-norm error after keeping its first $N$ eigenvectors?

Solution

For the unnormalized spectral truncation $\rho_N=\sum_{n\leq N}p_n|u_n\rangle\langle u_n|$, the error is $p_{N+1}$. Its missing probability is instead $\sum_{n>N}p_n$. A small largest omitted eigenvalue does not bound that total tail uniformly in the number of omitted levels.

2. Show that an infinite-dimensional orthogonal projection cannot be compact.

Solution

Choose an orthonormal sequence in its range. The projection leaves every member fixed, and the distance between distinct members is $\sqrt2$. The image of this bounded sequence has no norm-convergent subsequence.

References

• Simon, Barry, Trace Ideals and Their Applications, 2nd ed., Mathematical Surveys and Monographs 120, American Mathematical Society, 2005 — compact operators, singular values, and norm ideals. Publisher.