Trace-Class Operators

Trace-class operators are the infinite-dimensional operators for which the trace is absolutely convergent and independent of an orthonormal basis. They form a Banach ideal whose positive trace-one elements are density operators. Absolute summability is essential: a finite or conditionally convergent diagonal sum in one basis is not an adequate definition.

Helpful background. Compact operators supplies singular values; Hilbert–Schmidt operators supplies the product estimate used below.

Trace class and its norm

For a compact operator $T$ on a separable complex Hilbert space, let $s_n(T)$ be its singular values, counted with multiplicity. Define

$$\mathcal S_1(\mathcal H) =\left\{T:\sum_ns_n(T)<\infty\right\}, \qquad \|T\|_1=\sum_ns_n(T)=\operatorname{Tr}|T|.$$

For a positive bounded operator $A$, its possibly infinite trace can first be defined as $\sum_n\langle e_n,Ae_n\rangle$. Basis independence follows by expressing each term as $\|A^{1/2}e_n\|^2$ and applying Parseval with nonnegative sums. Finiteness forces $A^{1/2}$ to be Hilbert–Schmidt, hence compact, and therefore forces $A$ to be compact. This reconciles the positive-trace definition with the singular-value definition.

The singular-value decomposition gives

$$T=\sum_ns_n|u_n\rangle\langle v_n|$$

with trace-norm convergence. In particular finite-rank operators are dense in $\mathcal S_1$. Its norm is stronger than the operator and Hilbert–Schmidt norms:

$$\|T\|\leq\|T\|_2\leq\|T\|_1.$$

Completeness and ideal inequalities are part of the standard trace-ideal theory developed by Simon.

Why the trace is basis independent

For a rank-one operator, $\operatorname{Tr}(|u\rangle\langle v|)=\langle v,u\rangle$. The singular-value expansion therefore defines

$$\operatorname{Tr}T =\sum_ns_n\langle v_n,u_n\rangle, \qquad |\operatorname{Tr}T|\leq\|T\|_1.$$

This agrees with the diagonal sum in every orthonormal basis. To justify exchanging the two sums, apply Cauchy–Schwarz to each rank-one term:

$$\sum_j |\langle e_j,u_n\rangle\langle v_n,e_j\rangle| \leq\|u_n\|\,\|v_n\|=1.$$

Multiplication by $s_n$ and summation gives a bound by $\sum_ns_n$. Thus the double series is absolutely convergent, and the exchange is legitimate. The proof identifies precisely what finite-matrix manipulations conceal.

For positive $T$, the trace equals $\|T\|_1$. For nonpositive $T$, the trace can vanish even though its trace norm is large. The trace measures a signed or complex sum; the trace norm measures total singular-value weight.

Products and legitimate cyclicity

If $T\in\mathcal S_1$ and $A,B$ are bounded, then

$$\|ATB\|_1\leq\|A\|\,\|T\|_1\,\|B\|, \qquad \operatorname{Tr}(AT)=\operatorname{Tr}(TA).$$

Prove cyclicity first for rank-one operators and then pass to finite-rank approximants in trace norm. Both products converge in trace norm by the ideal bound, and the trace is continuous in that norm. More generally, if $S,T$ are Hilbert–Schmidt, then both $ST$ and $TS$ are trace class and their traces agree.

An unbounded operator is not covered by these statements. Expressions such as $\operatorname{Tr}(\rho H)$ require a separate integrability or product-domain argument. The density-operator treatment defines unbounded expectations through spectral measures.

Even for bounded operators, having a trace-class commutator does not imply its trace is zero. Let $S$ be the unilateral shift, $Se_n=e_{n+1}$. Then $S^*S=I$ and $SS^*=I-P_1$, so

$$[S^*,S]=P_1, \qquad \operatorname{Tr}[S^*,S]=1.$$

Neither product is trace class. Subtracting their divergent traces would be invalid.

A summability test for a candidate state

Consider $Te_n=n^{-\alpha}e_n$ for $\alpha>0$. It is positive and compact, but its trace is finite exactly for $\alpha>1$. In that regime

$$\rho_\alpha =\frac{1}{\zeta(\alpha)} \sum_{n=1}^{\infty}n^{-\alpha}|e_n\rangle\langle e_n|$$

is a density operator. At $\alpha=1$, no finite normalizing constant exists. Every finite cutoff can be normalized, yet that does not produce a trace-norm limit of states as the cutoff grows. Normalization at finite dimension cannot replace a summability proof.

For an arbitrary bounded observable $A$, the trace estimate now ensures $|\operatorname{Tr}(\rho_\alpha A)|\leq\|A\|$. The physical expectation is finite without any reference to a coordinate representation.

Exercises

1. For the above $\rho_\alpha$, let $R_N$ be its unnormalized spectral truncation. Compute $\|\rho_\alpha-R_N\|_1$ and the error bound for a bounded measurement effect.

Solution

The difference is positive, so its norm is the tail probability $\zeta(\alpha)^{-1}\sum_{n>N}n^{-\alpha}$. For $0\leq E\leq I$, the prediction error is at most this tail by trace duality. This is the rigorous content of truncating a density-matrix calculation.

2. Why does a diagonal operator with entries $(-1)^n/n$ fail to be trace class despite convergence of its displayed diagonal sum?

Solution

Its singular values are $1/n$, whose sum diverges. The alternating sum depends on ordering and cannot be the basis-independent trace-class trace. A conditionally convergent sum is insufficient.

References

• Simon, Barry, Trace Ideals and Their Applications, 2nd ed., Mathematical Surveys and Monographs 120, American Mathematical Society, 2005 — trace ideals, basis independence, and cyclicity. Publisher.