Trace class

Let ${\displaystyle H}$  be a separable Hilbert space, ${\displaystyle \left\{e_{k}\right\}_{k=1}^{\infty }}$  an orthonormal basis and ${\displaystyle A:H\to H}$  a positive bounded linear operator on ${\displaystyle H}$ . The trace of ${\displaystyle A}$  is denoted by ${\displaystyle \operatorname {Tr} (A)}$  and defined as^[2][3]

${\displaystyle \operatorname {Tr} (A)=\sum _{k=1}^{\infty }\left\langle Ae_{k},e_{k}\right\rangle ,}$ 

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

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

where ${\displaystyle |T|:={\sqrt {T^{*}T}}}$  denotes the positive-semidefinite Hermitian square root.^[4]

The trace-norm of a trace class operator T is defined as $${\displaystyle \|T\|_{1}:=\operatorname {Tr} (|T|).}$$  One can show that the trace-norm is a norm on the space of all trace class operators ${\displaystyle B_{1}(H)}$  and that ${\displaystyle B_{1}(H)}$ , with the trace-norm, becomes a Banach space.

When ${\displaystyle H}$  is finite-dimensional, every (positive) operator is trace class and this definition of trace of ${\displaystyle A}$  coincides with the definition of the trace of a matrix. If ${\displaystyle H}$  is complex, then ${\displaystyle A}$  is always self-adjoint (i.e. ${\displaystyle A=A^{*}=|A|}$ ) though the converse is not necessarily true.^[5]

Given a bounded linear operator ${\displaystyle T:H\to H}$ , each of the following statements is equivalent to ${\displaystyle T}$  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 ${\displaystyle \left(e_{k}\right)_{k}}$  of H.^[2]
• T is a nuclear operator^[6][7]

There exist two orthogonal sequences ${\displaystyle \left(x_{i}\right)_{i=1}^{\infty }}$  and ${\displaystyle \left(y_{i}\right)_{i=1}^{\infty }}$  in ${\displaystyle H}$  and positive real numbers ${\displaystyle \left(\lambda _{i}\right)_{i=1}^{\infty }}$  in ${\displaystyle \ell ^{1}}$  such that ${\textstyle \sum _{i=1}^{\infty }\lambda _{i}<\infty }$  and

${\displaystyle x\mapsto T(x)=\sum _{i=1}^{\infty }\lambda _{i}\left\langle x,x_{i}\right\rangle y_{i},\quad \forall x\in H,}$ 

where ${\displaystyle \left(\lambda _{i}\right)_{i=1}^{\infty }}$  are the singular values of T (or, equivalently, the eigenvalues of ${\displaystyle |T|}$ ), with each value repeated as often as its multiplicity.^[8]

• T is a compact operator with ${\displaystyle \operatorname {Tr} (|T|)<\infty .}$ 

If T is trace class then^[9]

${\displaystyle \|T\|_{1}=\sup \left\{|\operatorname {Tr} (CT)|:\|C\|\leq 1{\text{ and }}C:H\to H{\text{ is a compact operator }}\right\}.}$ 

• T is an integral operator.^[10]
• T is equal to the composition of two Hilbert-Schmidt operators.^[11]
• ${\textstyle {\sqrt {|T|}}}$  is a Hilbert-Schmidt operator.^[11]

Let ${\displaystyle T}$  be a bounded self-adjoint operator on a Hilbert space. Then ${\displaystyle T^{2}}$  is trace class if and only if ${\displaystyle T}$  has a pure point spectrum with eigenvalues ${\displaystyle \left\{\lambda _{i}(T)\right\}_{i=1}^{\infty }}$  such that^[12]

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

Mercer's theorem provides another example of a trace class operator. That is, suppose ${\displaystyle K}$  is a continuous symmetric positive-definite kernel on ${\displaystyle L^{2}([a,b])}$ , defined as

${\displaystyle K(s,t)=\sum _{j=1}^{\infty }\lambda _{j}\,e_{j}(s)\,e_{j}(t)}$ 

then the associated Hilbert–Schmidt integral operator ${\displaystyle T_{K}}$  is trace class, i.e.,

${\displaystyle \operatorname {Tr} (T_{K})=\int _{a}^{b}K(t,t)\,dt=\sum _{i}\lambda _{i}.}$ 

Every finite-rank operator is a trace-class operator. Furthermore, the space of all finite-rank operators is a dense subspace of ${\displaystyle B_{1}(H)}$  (when endowed with the trace norm).^[9]

Given any ${\displaystyle x,y\in H,}$  define the operator ${\displaystyle x\otimes y:H\to H}$  by ${\displaystyle (x\otimes y)(z):=\langle z,y\rangle x.}$  Then ${\displaystyle x\otimes y}$  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), ${\displaystyle \operatorname {Tr} (A(x\otimes y))=\langle Ax,y\rangle .}$ ^[9]

1. If ${\displaystyle A:H\to H}$  is a non-negative self-adjoint operator, then ${\displaystyle A}$  is trace-class if and only if ${\displaystyle \operatorname {Tr} A<\infty .}$  Therefore, a self-adjoint operator ${\displaystyle A}$  is trace-class if and only if its positive part ${\displaystyle A^{+}}$  and negative part ${\displaystyle A^{-}}$  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, $${\displaystyle \operatorname {Tr} (aA+bB)=a\operatorname {Tr} (A)+b\operatorname {Tr} (B).}$$  The bilinear map $${\displaystyle \langle A,B\rangle =\operatorname {Tr} (A^{*}B)}$$  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.
3. ${\displaystyle \operatorname {Tr} :B_{1}(H)\to \mathbb {C} }$  is a positive linear functional such that if ${\displaystyle T}$  is a trace class operator satisfying ${\displaystyle T\geq 0{\text{ and }}\operatorname {Tr} T=0,}$  then ${\displaystyle T=0.}$ ^[11]
4. If ${\displaystyle T:H\to H}$  is trace-class then so is ${\displaystyle T^{*}}$  and ${\displaystyle \|T\|_{1}=\left\|T^{*}\right\|_{1}.}$ ^[11]
5. If ${\displaystyle A:H\to H}$  is bounded, and ${\displaystyle T:H\to H}$  is trace-class, then ${\displaystyle AT}$  and ${\displaystyle TA}$  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^[11][13] $${\displaystyle \|AT\|_{1}=\operatorname {Tr} (|AT|)\leq \|A\|\|T\|_{1},\quad \|TA\|_{1}=\operatorname {Tr} (|TA|)\leq \|A\|\|T\|_{1}.}$$  Furthermore, under the same hypothesis,^[11] $${\displaystyle \operatorname {Tr} (AT)=\operatorname {Tr} (TA)}$$  and ${\displaystyle |\operatorname {Tr} (AT)|\leq \|A\|\|T\|.}$  The last assertion also holds under the weaker hypothesis that A and T are Hilbert–Schmidt.
6. If ${\displaystyle \left(e_{k}\right)_{k}}$  and ${\displaystyle \left(f_{k}\right)_{k}}$  are two orthonormal bases of H and if T is trace class then ${\textstyle \sum _{k}\left|\left\langle Te_{k},f_{k}\right\rangle \right|\leq \|T\|_{1}.}$ ^[9]
7. If A is trace-class, then one can define the Fredholm determinant of ${\displaystyle I+A}$ : $${\displaystyle \det(I+A):=\prod _{n\geq 1}[1+\lambda _{n}(A)],}$$  where ${\displaystyle \{\lambda _{n}(A)\}_{n}}$  is the spectrum of ${\displaystyle A.}$  The trace class condition on ${\displaystyle A}$  guarantees that the infinite product is finite: indeed, $${\displaystyle \det(I+A)\leq e^{\|A\|_{1}}.}$$  It also implies that ${\displaystyle \det(I+A)\neq 0}$  if and only if ${\displaystyle (I+A)}$  is invertible.
8. If ${\displaystyle A:H\to H}$  is trace class then for any orthonormal basis ${\displaystyle \left(e_{k}\right)_{k}}$  of ${\displaystyle H,}$  the sum of positive terms ${\textstyle \sum _{k}\left|\left\langle A\,e_{k},e_{k}\right\rangle \right|}$  is finite.^[11]
9. If ${\displaystyle A=B^{*}C}$  for some Hilbert-Schmidt operators ${\displaystyle B}$  and ${\displaystyle C}$  then for any normal vector ${\displaystyle e\in H,}$  ${\textstyle |\langle Ae,e\rangle |={\frac {1}{2}}\left(\|Be\|^{2}+\|Ce\|^{2}\right)}$  holds.^[11]

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

Note that the series on the right converges absolutely due to Weyl's inequality $${\displaystyle \sum _{n=1}^{N}\left|\lambda _{n}(A)\right|\leq \sum _{m=1}^{M}s_{m}(A)}$$  between the eigenvalues ${\displaystyle \{\lambda _{n}(A)\}_{n=1}^{N}}$  and the singular values ${\displaystyle \{s_{m}(A)\}_{m=1}^{M}}$  of the compact operator ${\displaystyle A.}$ ^[14]

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 ${\displaystyle \ell ^{1}(\mathbb {N} ).}$ 

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 ${\displaystyle \ell ^{1}}$  sequence with respect to some choice of a pair of Hilbert bases. In the same vein, the bounded operators are noncommutative versions of ${\displaystyle \ell ^{\infty }(\mathbb {N} ),}$  the compact operators that of ${\displaystyle c_{0}}$  (the sequences convergent to 0), Hilbert–Schmidt operators correspond to ${\displaystyle \ell ^{2}(\mathbb {N} ),}$  and finite-rank operators to ${\displaystyle c_{00}}$  (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 ${\displaystyle T}$  on a Hilbert space takes the following canonical form: there exist orthonormal bases ${\displaystyle (u_{i})_{i}}$  and ${\displaystyle (v_{i})_{i}}$  and a sequence ${\displaystyle \left(\alpha _{i}\right)_{i}}$  of non-negative numbers with ${\displaystyle \alpha _{i}\to 0}$  such that $${\displaystyle Tx=\sum _{i}\alpha _{i}\langle x,v_{i}\rangle u_{i}\quad {\text{ for all }}x\in H.}$$  Making the above heuristic comments more precise, we have that ${\displaystyle T}$  is trace-class iff the series ${\textstyle \sum _{i}\alpha _{i}}$  is convergent, ${\displaystyle T}$  is Hilbert–Schmidt iff ${\textstyle \sum _{i}\alpha _{i}^{2}}$  is convergent, and ${\displaystyle T}$  is finite-rank iff the sequence ${\displaystyle \left(\alpha _{i}\right)_{i}}$  has only finitely many nonzero terms. This allows to relate these classes of operators. The following inclusions hold and are all proper when ${\displaystyle H}$  is infinite-dimensional:$${\displaystyle \{{\text{ finite rank }}\}\subseteq \{{\text{ trace class }}\}\subseteq \{{\text{ Hilbert--Schmidt }}\}\subseteq \{{\text{ compact }}\}.}$$ 

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 $${\displaystyle \|T\|_{2}=\left[\operatorname {Tr} \left(T^{*}T\right)\right]^{1/2}=\left(\sum _{i}\alpha _{i}^{2}\right)^{1/2}.}$$  Also, the usual operator norm is ${\textstyle \|T\|=\sup _{i}\left(\alpha _{i}\right).}$  By classical inequalities regarding sequences, $${\displaystyle \|T\|\leq \|T\|_{2}\leq \|T\|_{1}}$$  for appropriate ${\displaystyle T.}$ 

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

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

This identification works because the finite-rank operators are norm-dense in ${\displaystyle K(H).}$  In the event that ${\displaystyle T_{f}}$  is a positive operator, for any orthonormal basis ${\displaystyle u_{i},}$  one has $${\displaystyle \sum _{i}\langle T_{f}u_{i},u_{i}\rangle =f(I)\leq \|f\|,}$$  where ${\displaystyle I}$  is the identity operator: $${\displaystyle I=\sum _{i}\langle \cdot ,u_{i}\rangle u_{i}.}$$ 

But this means that ${\displaystyle T_{f}}$  is trace-class. An appeal to polar decomposition extend this to the general case, where ${\displaystyle T_{f}}$  need not be positive.

A limiting argument using finite-rank operators shows that ${\displaystyle \|T_{f}\|_{1}=\|f\|.}$  Thus ${\displaystyle K(H)^{*}}$  is isometrically isomorphic to ${\displaystyle B_{1}.}$ 

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

• Nuclear operator – Linear operator related to topological vector spaces
• Nuclear operators between Banach spaces – operators on Banach spaces with properties similar to finite-dimensional operatorsPages displaying wikidata descriptions as a fallback
• Trace operator

• Conway, John B. (2000). A Course in Operator Theory. Providence (R.I.): American Mathematical Soc. ISBN 978-0-8218-2065-0.
• Conway, John B. (1990). A course in functional analysis. New York: Springer-Verlag. ISBN 978-0-387-97245-9. OCLC 21195908.
• Dixmier, J. (1969). Les Algebres d'Operateurs dans l'Espace Hilbertien. Gauthier-Villars.
• Mittelstaedt, Peter (2009). "Mixed State". Compendium of Quantum Physics. Berlin, Heidelberg: Springer Berlin Heidelberg. doi:10.1007/978-3-540-70626-7_120. ISBN 978-3-540-70622-9.
• Reed, M.; Simon, B. (1980). Methods of Modern Mathematical Physics: Vol 1: Functional analysis. Academic Press. ISBN 978-0-12-585050-6.
• Schaefer, Helmut H. (1999). Topological Vector Spaces. GTM. Vol. 3. New York, NY: Springer New York Imprint Springer. ISBN 978-1-4612-7155-0. OCLC 840278135.
• Simon, Barry (2010). Szegő's theorem and its descendants: spectral theory for L² perturbations of orthogonal polynomials. Princeton University Press. ISBN 978-0-691-14704-8.
• Trèves, François (2006) [1967]. Topological Vector Spaces, Distributions and Kernels. Mineola, N.Y.: Dover Publications. ISBN 978-0-486-45352-1. OCLC 853623322.