Trace class explained

In mathematics, specifically functional analysis, a trace-class operator is a linear operator for which a trace may be defined, such that the trace is a finite number independent of the choice of basis used to compute the trace. This trace of trace-class operators generalizes the trace of matrices studied in linear algebra. All trace-class operators are compact operators.

In quantum mechanics, mixed states are described by density matrices, which are certain trace class operators.

Trace-class operators are essentially the same as nuclear operators, though many authors reserve the term "trace-class operator" for the special case of nuclear operators on Hilbert spaces and use the term "nuclear operator" in more general topological vector spaces (such as Banach spaces).

Note that the trace operator studied in partial differential equations is an unrelated concept.

Definition

Let

H be a separable Hilbert space, \left\{e _{k\right\}} ^{inf ty} ^{k =1} an orthonormal basis and A : H \to H a positive bounded linear operator on H . The trace of A is denoted by \operatorname{Tr} (A) and defined as\operatorname{Tr} (A) = \sum ^{inf ty} _{k =1} \left\langle A e _{k,} e _{k} \right\rangle, independent of the choice of orthonormal basis. A (not necessarily positive) bounded linear operator T:H → H is called trace class if and only if\operatorname{Tr}( |T|) < inf ty, where |T| : = \sqrt{T ^{*} T} denotes the positive-semidefinite Hermitian square root.

The trace-norm of a trace class operator is defined as$$\backslash |T\backslash |\_1\; :=\; \backslash operatorname\; (|T|).$$One can show that the trace-norm is a norm on the space of all trace class operators

B _{1(H)} and that B _{1(H)} , with the trace-norm, becomes a Banach space.

When

H is finite-dimensional, every (positive) operator is trace class and this definition of trace of A coincides with the definition of the trace of a matrix. If H is complex, then A is always self-adjoint (i.e. A = A ^{*=|A|} ) though the converse is not necessarily true.

Equivalent formulations

Given a bounded linear operator

T : H \to H , each of the following statements is equivalent to T being in the trace class:\left(e _{k\right)} _{k} of .

• is a nuclear operator

There exist two orthogonal sequences

\left(x _{i\right)} ^{inf ty} ^{i =1} and \left(y _{i\right)} ^{inf ty} ^{i =1} in H and positive real numbers \left( λ _{i\right)} ^{inf ty} ^{i =1} in \ell ^{1} such that andx \mapsto T(x) = \sum ^{inf ty} _{i =1} λ _{i} \left\langle x, x _{i} \right\rangle y _{i,} \forall x \in H,

where

\left( λ _{i\right)} ^{inf ty} ^{i =1} are the singular values of (or, equivalently, the eigenvalues of |T| ), with each value repeated as often as its multiplicity.

• is a compact operator with

\operatorname{Tr}(|T|)< inf ty.

If is trace class then

\|T\| _{1} = \sup \left\{ |\operatorname{Tr} (C T)| : \|C\| \leq 1 and C : H \to H isacompactoperator \right\}.

• is an integral operator.
• is equal to the composition of two Hilbert-Schmidt operators.
• $\backslash sqrt$

is a Hilbert-Schmidt operator.

Examples

Spectral theorem

Let

T be a bounded self-adjoint operator on a Hilbert space. Then T ^{2} is trace class if and only if T has a pure point spectrum with eigenvalues \left\{ λ _{i(T)\right\}} ^{inf ty} ^{i =1} such that \operatorname{Tr}(T ^{2)} = \sum ^{inf ty} _{i =1} λ ^{2 )} _{i(T} < inf ty.

Mercer's theorem

Mercer's theorem provides another example of a trace class operator. That is, suppose

K is a continuous symmetric positive-definite kernel on L ^{2([a,b])} , defined asK(s,t) = \sum ^{inf ty} _{j =1} λ _{j} e _{j(s)} e _{j(t)} then the associated Hilbert–Schmidt integral operator T _{K} is trace class, i.e.,\operatorname{Tr}(T _{K)} = \int ^{b} _{a} K(t,t)dt = \sum _{i} λ _{i.}

Finite-rank operators

Every finite-rank operator is a trace-class operator. Furthermore, the space of all finite-rank operators is a dense subspace of

B _{1(H)} (when endowed with the trace norm).

Given any

x, y \in H, define the operator x ⊗ y : H \to H by (x ⊗ y)(z) : = \langle z, y \rangle x. Then x ⊗ y is a continuous linear operator of rank 1 and is thus trace class; moreover, for any bounded linear operator A on H (and into H), \operatorname{Tr}(A(x ⊗ y)) = \langle A x, y \rangle.

Properties

1. If
   A : H \to H is a non-negative self-adjoint operator, then A is trace-class if and only if \operatorname{Tr} A < inf ty. Therefore, a self-adjoint operator A is trace-class if and only if its positive part A ^{+} and negative part A ^{-} are both trace-class. (The positive and negative parts of a self-adjoint operator are obtained by the continuous functional calculus.)
2. The trace is a linear functional over the space of trace-class operators, that is, $$\backslash operatorname(aA\; +\; bB)\; =\; a\; \backslash operatorname(A)\; +\; b\; \backslash operatorname(B).$$The bilinear map $$\backslash langle\; A,\; B\; \backslash rangle\; =\; \backslash operatorname(A^*\; B)$$ is an inner product on the trace class; the corresponding norm is called the Hilbert–Schmidt norm. The completion of the trace-class operators in the Hilbert–Schmidt norm are called the Hilbert–Schmidt operators.

3. \operatorname{Tr} : B _{1(H)} \to \Complex is a positive linear functional such that if T is a trace class operator satisfying T \geq 0 and \operatorname{Tr} T = 0 , then T = 0 .
4. If
   T : H \to H is trace-class then so is T ^{*} and \|T\| _{1} = \left\|T ^{*\right\|} _{1 .}
5. If
   A : H \to H is bounded, and T : H \to H is trace-class, then AT and TA are also trace-class (i.e. the space of trace-class operators on H is an ideal in the algebra of bounded linear operators on H), and$$\backslash |A\; T\backslash |\_1\; =\; \backslash operatorname(|A\; T|)\; \backslash leq\; \backslash |A\backslash |\; \backslash |T\backslash |\_1,\; \backslash quad\; \backslash |T\; A\backslash |\_1\; =\; \backslash operatorname(|T\; A|)\; \backslash leq\; \backslash |A\backslash |\; \backslash |T\backslash |\_1.$$Furthermore, under the same hypothesis, $$\backslash operatorname(A\; T)\; =\; \backslash operatorname(T\; A)$$ and |\operatorname{Tr}(A T)| \leq \|A\| \|T\|. The last assertion also holds under the weaker hypothesis that A and T are Hilbert–Schmidt.
6. If
   \left(e _{k\right)} _{k} and \left(f _{k\right)} _{k} are two orthonormal bases of H and if T is trace class then $\backslash sum\_\; \backslash left|\; \backslash left\backslash langle\; T\; e\_k,\; f\_k\; \backslash right\backslash rangle\; \backslash right|\; \backslash leq\; \backslash |T\backslash |\_.$
7. If A is trace-class, then one can define the Fredholm determinant of
   I + A : $$\backslash det(I\; +\; A)\; :=\; \backslash prod\_[1\; +\; \backslash lambda\_n(A)],$$ where \{ λ _{n(A)\}} _{n} is the spectrum of A. The trace class condition on A guarantees that the infinite product is finite: indeed, $$\backslash det(I\; +\; A)\; \backslash leq\; e^.$$It also implies that \det(I + A) ≠ 0 if and only if (I + A) is invertible.
8. If
   A : H \to H is trace class then for any orthonormal basis \left(e _{k\right)} _{k} of H, the sum of positive terms $\backslash sum\_k\; \backslash left|\; \backslash left\backslash langle\; A\; \backslash ,\; e\_k,\; e\_k\; \backslash right\backslash rangle\; \backslash right|$ is finite.
9. If
   A = B ^{*} C for some Hilbert-Schmidt operators B and C then for any normal vector e \in H, $|\backslash langle\; A\; e,\; e\; \backslash rangle|\; =\; \backslash frac\; \backslash left(\backslash |B\; e\backslash |^2\; +\; \backslash |C\; e\backslash |^2\backslash right)$ holds.

Lidskii's theorem

Let

A be a trace-class operator in a separable Hilbert space H, and let \{ λ _{n(A)\}} ^{N\leq inf ty} ^{n =1} be the eigenvalues of A. Let us assume that λ _{n(A)} are enumerated with algebraic multiplicities taken into account (that is, if the algebraic multiplicity of λ is k, then λ is repeated k times in the list λ _{1(A),} λ _{2(A),} ... ). Lidskii's theorem (named after Victor Borisovich Lidskii) states that$$\backslash operatorname(A)=\backslash sum\_^N\; \backslash lambda\_n(A)$$

Note that the series on the right converges absolutely due to Weyl's inequality$$\backslash sum\_^N\; \backslash left|\backslash lambda\_n(A)\backslash right|\; \backslash leq\; \backslash sum\_^M\; s\_m(A)$$between the eigenvalues

\{ λ _{n(A)\}} ^{N} ^{n =1} and the singular values \{s _{m(A)\}} ^{M} ^{m =1} of the compact operator A. ^[1]

Relationship between common classes of operators

\ell ^{1(\N).}

Indeed, it is possible to apply the spectral theorem to show that every normal trace-class operator on a separable Hilbert space can be realized in a certain way as an

\ell ^{1} sequence with respect to some choice of a pair of Hilbert bases. In the same vein, the bounded operators are noncommutative versions of \ell ^{infty} (\N), the compact operators that of c _{0} (the sequences convergent to 0), Hilbert–Schmidt operators correspond to \ell ^{2(\N),} and finite-rank operators to c _{00} (the sequences that have only finitely many non-zero terms). To some extent, the relationships between these classes of operators are similar to the relationships between their commutative counterparts.

Recall that every compact operator

T on a Hilbert space takes the following canonical form: there exist orthonormal bases (u _{i)} _{i} and (v _{i)} _{i} and a sequence \left(\alpha _{i\right)} _{i} of non-negative numbers with \alpha _{i} \to 0 such that $$T\; x\; =\; \backslash sum\_i\; \backslash alpha\_i\; \backslash langle\; x,\; v\_i\backslash rangle\; u\_i\; \backslash quad\; \backslash text\; x\backslash in\; H.$$Making the above heuristic comments more precise, we have that T is trace-class iff the series $\backslash sum\_i\; \backslash alpha\_i$ is convergent, T is Hilbert–Schmidt iff $\backslash sum\_i\; \backslash alpha\_i^2$ is convergent, and T is finite-rank iff the sequence \left(\alpha _{i\right)} _{i} has only finitely many nonzero terms. This allows to relate these classes of operators. The following inclusions hold and are all proper when H is infinite-dimensional:$$\backslash \; \backslash subseteq\; \backslash \; \backslash subseteq\; \backslash \; \backslash subseteq\; \backslash .$$

The trace-class operators are given the trace norm $\backslash |T\backslash |\_1\; =\; \backslash operatorname\; \backslash left[\backslash left(T^*\; T\backslash right)^\{1/2\}\backslash right]\; =\; \backslash sum\_i\; \backslash alpha\_i.$ The norm corresponding to the Hilbert–Schmidt inner product is $$\backslash |T\backslash |\_2\; =\; \backslash left[\backslash operatorname\{Tr\}\; \backslash left(T^*\; T\backslash right)\backslash right]^\; =\; \backslash left(\backslash sum\_i\; \backslash alpha\_i^2\backslash right)^.$$Also, the usual operator norm is $\backslash |\; T\; \backslash |\; =\; \backslash sup\_\; \backslash left(\backslash alpha\_i\backslash right).$ By classical inequalities regarding sequences,$$\backslash |T\backslash |\; \backslash leq\; \backslash |T\backslash |\_2\; \backslash leq\; \backslash |T\backslash |\_1$$for appropriate

T.

It is also clear that finite-rank operators are dense in both trace-class and Hilbert–Schmidt in their respective norms.

Trace class as the dual of compact operators

The dual space of

c _{0} is \ell ^{1(\N).} Similarly, we have that the dual of compact operators, denoted by K(H) ^{*,} is the trace-class operators, denoted by B _{1.} The argument, which we now sketch, is reminiscent of that for the corresponding sequence spaces. Let f \in K(H) ^{*,} we identify f with the operator T _{f} defined by$$\backslash langle\; T\_f\; x,\; y\; \backslash rangle\; =\; f\backslash left(S\_\backslash right),$$where S _{x,y} is the rank-one operator given by$$S\_(h)\; =\; \backslash langle\; h,\; y\; \backslash rangle\; x.$$

This identification works because the finite-rank operators are norm-dense in

K(H). In the event that T _{f} is a positive operator, for any orthonormal basis u _{i,} one has$$\backslash sum\_i\; \backslash langle\; T\_f\; u\_i,\; u\_i\; \backslash rangle\; =\; f(I)\; \backslash leq\; \backslash |f\backslash |,$$where I is the identity operator:$$I\; =\; \backslash sum\_i\; \backslash langle\; \backslash cdot,\; u\_i\; \backslash rangle\; u\_i.$$

But this means that

T _{f} is trace-class. An appeal to polar decomposition extend this to the general case, where T _{f} need not be positive.

A limiting argument using finite-rank operators shows that

\|T _{f\|} _{1} = \|f\|. Thus K(H) ^{*} is isometrically isomorphic to B _{1.}

As the predual of bounded operators

Recall that the dual of

\ell ^{1(\N)} is \ell ^{infty} (\N). In the present context, the dual of trace-class operators B _{1} is the bounded operators B(H). More precisely, the set B _{1} is a two-sided ideal in B(H). So given any operator T \in B(H), we may define a continuous linear functional \varphi _{T} on B _{1} by \varphi _{T(A)} = \operatorname{Tr} (AT). This correspondence between bounded linear operators and elements \varphi _{T} of the dual space of B _{1} is an isometric isomorphism. It follows that B(H) the dual space of B _{1.} This can be used to define the weak-* topology on B(H).

See also

• • • Trace operator

Bibliography

• Book: Conway, John B. . A Course in Operator Theory . American Mathematical Soc. . Providence (R.I.) . 2000 . 978-0-8218-2065-0.
• Book: Conway, John B.. A course in functional analysis. Springer-Verlag. New York. 1990. 978-0-387-97245-9. 21195908.
• Dixmier, J. (1969). Les Algebres d'Operateurs dans l'Espace Hilbertien. Gauthier-Villars.
• Book: Reed . M. . Michael C. Reed . Simon . B. . Barry Simon. Methods of Modern Mathematical Physics: Vol 1: Functional analysis . Academic Press . 1980 . 978-0-12-585050-6.
• Book: Schaefer, Helmut H.. Topological Vector Spaces. Springer New York Imprint Springer. GTM. 3. New York, NY. 1999. 978-1-4612-7155-0. 840278135 .
• Book: Simon, Barry . Barry Simon . Szegő's theorem and its descendants: spectral theory for L² perturbations of orthogonal polynomials. 2010. Princeton University Press. 978-0-691-14704-8.

Notes and References

1. Simon, B. (2005) Trace ideals and their applications, Second Edition, American Mathematical Society.