Compact operator for which a finite trace can be defined

Trace class

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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, quantum 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).

Definition

Let $H$ be a separable Hilbert space, $\left\{e_{k}\right\}_{k=1}^{\infty }$ 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 _{k=1}^{\infty }\left\langle Ae_{k},e_{k}\right\rangle,

independent of the choice of orthonormal basis. A (not necessarily positive) bounded linear operator $T:H\rightarrow H$ is called trace class if and only if

\operatorname {Tr} (|T|)<\infty,

where $|T|:={\sqrt {T^{*}T}}$ denotes the positive-semidefinite Hermitian square root.

The trace-norm of a trace class operator T is defined as $\|T\|_{1}:=\operatorname {Tr} (|T|).$ 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. For $A$ this definition coincides with that 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:

${\textstyle \operatorname {Tr} (|T|)=\sum _{k}\left\langle |T|\,e_{k},e_{k}\right\rangle }$ is finite for every orthonormal basis $\left(e_{k}\right)_{k}$ of H.
T is a nuclear operator.
There exist two orthogonal sequences $\left(x_{i}\right)_{i=1}^{\infty }$ and $\left(y_{i}\right)_{i=1}^{\infty }$ in $H$ and positive real numbers $\left(\lambda _{i}\right)_{i=1}^{\infty }$ in $\ell ^{1}$ such that ${\textstyle \sum _{i=1}^{\infty }\lambda _{i}<\infty }$ and
$x\mapsto T(x)=\sum _{i=1}^{\infty }\lambda _{i}\left\langle x,x_{i}\right\rangle y_{i},\quad \forall x\in H,$

where $\left(\lambda _{i}\right)_{i=1}^{\infty }$ are the singular values of T (or, equivalently, the eigenvalues of $|T|$ ), with each value repeated as often as its multiplicity.
T is a compact operator with $\operatorname {Tr} (|T|)<\infty.$
If T is trace class then
$\|T\|_{1}=\sup \left\{|\operatorname {Tr} (CT)|:\|C\|\leq 1{\text{ and }}C:H\to H{\text{ is a compact operator }}\right\}.$
T is an integral operator.
T is equal to the composition of two Hilbert-Schmidt operators.
${\textstyle {\sqrt {|T|}}}$ 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 $\left\{\lambda _{i}(T)\right\}_{i=1}^{\infty }$ such that

\operatorname {Tr} (T^{2})=\sum _{i=1}^{\infty }\lambda _{i}(T^{2})<\infty.

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 as

K(s,t)=\sum _{j=1}^{\infty }\lambda _{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 _{a}^{b}K(t,t)\,dt=\sum _{i}\lambda _{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 $x\otimes y:H\to H$ by $(x\otimes y)(z):=\langle z,y\rangle x.$ Then $x\otimes y$ 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), $\operatorname {Tr} (A(x\otimes y))=\langle Ax,y\rangle.$

Properties

If $A:H\to H$ is a non-negative self-adjoint operator, then $A$ is trace-class if and only if $\operatorname {Tr} A<\infty.$ 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.)
The trace is a linear functional over the space of trace-class operators, that is, $\operatorname {Tr} (aA+bB)=a\operatorname {Tr} (A)+b\operatorname {Tr} (B).$ The bilinear map $\langle A,B\rangle =\operatorname {Tr} (A^{*}B)$ 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.
$\operatorname {Tr}:B_{1}(H)\to \mathbb {C}$ is a positive linear functional such that if $T$ is a trace class operator satisfying $T\geq 0{\text{ and }}\operatorname {Tr} T=0,$ then $T=0.$
If $T:H\to H$ is trace-class then so is $T^{*}$ and $\|T\|_{1}=\left\|T^{*}\right\|_{1}.$
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 a two-sided ideal in the algebra of bounded linear operators on H), and $\|AT\|_{1}=\operatorname {Tr} (|AT|)\leq \|A\|\|T\|_{1},\quad \|TA\|_{1}=\operatorname {Tr} (|TA|)\leq \|A\|\|T\|_{1}.$ Furthermore, under the same hypothesis, $\operatorname {Tr} (AT)=\operatorname {Tr} (TA)$ and $|\operatorname {Tr} (AT)|\leq \|A\|\|T\|.$ The last assertion also holds under the weaker hypothesis that A and T are Hilbert–Schmidt.
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 ${\textstyle \sum _{k}\left|\left\langle Te_{k},f_{k}\right\rangle \right|\leq \|T\|_{1}.}$
If A is trace-class, then one can define the Fredholm determinant of $I+A$ : $\det(I+A):=\prod _{n\geq 1}[1+\lambda _{n}(A)],$ where $\{\lambda _{n}(A)\}_{n}$ is the spectrum of $A.$ The trace class condition on $A$ guarantees that the infinite product is finite: indeed, $\det(I+A)\leq e^{\|A\|_{1}}.$ It also implies that $\det(I+A)\neq 0$ if and only if $(I+A)$ is invertible.
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 ${\textstyle \sum _{k}\left|\left\langle A\,e_{k},e_{k}\right\rangle \right|}$ is finite.
If $A=B^{*}C$ for some Hilbert-Schmidt operators $B$ and $C$ then for any normal vector $e\in H,$ ${\textstyle |\langle Ae,e\rangle |\leq {\frac {1}{2}}\left(\|Be\|^{2}+\|Ce\|^{2}\right)}$ holds.

Lidskii's theorem

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

Note that the series on the right converges absolutely due to Weyl's inequality $\sum _{n=1}^{N}\left|\lambda _{n}(A)\right|\leq \sum _{m=1}^{M}s_{m}(A)$ between the eigenvalues $\{\lambda _{n}(A)\}_{n=1}^{N}$ and the singular values $\{s_{m}(A)\}_{m=1}^{M}$ of the compact operator $A.$ See also Grothendieck trace theorem.

Relationship between common classes of operators

One can view certain classes of bounded operators as noncommutative analogue of classical sequence spaces, with trace-class operators as the noncommutative analogue of the sequence space $\ell ^{1}(\mathbb {N}).$

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 $\ell ^{\infty }(\mathbb {N}),$ the compact operators that of $c_{0}$ (the sequences convergent to 0), Hilbert–Schmidt operators correspond to $\ell ^{2}(\mathbb {N}),$ 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 $Tx=\sum _{i}\alpha _{i}\langle x,v_{i}\rangle u_{i}\quad {\text{ for all }}x\in H.$ Making the above heuristic comments more precise, we have that $T$ is trace-class iff the series ${\textstyle \sum _{i}\alpha _{i}}$ is convergent, $T$ is Hilbert–Schmidt iff ${\textstyle \sum _{i}\alpha _{i}^{2}}$ 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: $\{{\text{ finite rank }}\}\subseteq \{{\text{ trace class }}\}\subseteq \{{\text{ Hilbert--Schmidt }}\}\subseteq \{{\text{ compact }}\}.$

The trace-class operators are given the trace norm ${\textstyle \|T\|_{1}=\operatorname {Tr} \left[\left(T^{*}T\right)^{1/2}\right]=\sum _{i}\alpha _{i}.}$ The norm corresponding to the Hilbert–Schmidt inner product is $\|T\|_{2}=\left[\operatorname {Tr} \left(T^{*}T\right)\right]^{1/2}=\left(\sum _{i}\alpha _{i}^{2}\right)^{1/2}.$ Also, the usual operator norm is ${\textstyle \|T\|=\sup _{i}\left(\alpha _{i}\right).}$ By classical inequalities regarding sequences, $\|T\|\leq \|T\|_{2}\leq \|T\|_{1}$ 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 $\ell ^{1}(\mathbb {N}).$ 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 $\langle T_{f}x,y\rangle =f\left(S_{x,y}\right),$ where $S_{x,y}$ is the rank-one operator given by $S_{x,y}(h)=\langle h,y\rangle x.$

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 $\sum _{i}\langle T_{f}u_{i},u_{i}\rangle =f(I)\leq \|f\|,$ where $I$ is the identity operator: $I=\sum _{i}\langle \cdot,u_{i}\rangle u_{i}.$

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 $\ell ^{1}(\mathbb {N})$ is $\ell ^{\infty }(\mathbb {N}).$ 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 $B(H)$ is the dual space of $B_{1}.$ This can be used to define the weak-* topology on $B(H).$