Inmathematics, specificallyfunctional analysis, atrace-class operator is a linear operator for which atrace 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 inlinear algebra. All trace-class operators arecompact operators.

Inquantum mechanics,quantum states are described bydensity matrices, which are certain trace class operators.^[1]

Trace-class operators are essentially the same asnuclear operators, though many authors reserve the term "trace-class operator" for the special case of nuclear operators onHilbert spaces and use the term "nuclear operator" in more generaltopological vector spaces (such asBanach spaces).

Let${\displaystyle H}$ be aseparableHilbert space,${\displaystyle {\left\{e_k\right\}}_{k=1}^{\mathrm{\infty }}}$ anorthonormal basis and${\displaystyle A:H\to H}$ apositivebounded linear operator on${\displaystyle H}$. Thetrace of${\displaystyle A}$ is denoted by${\displaystyle \mathrm{Tr}(A)}$ and defined as^[2][3]

${\displaystyle \mathrm{Tr}(A)=\sum _{k=1}^{\mathrm{\infty }}⟨A{e}_k,e_k⟩,}$

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

where${\displaystyle |T|:=\sqrt{T^{\ast }T}}$ denotes the positive-semidefiniteHermitiansquare root.^[4]

Thetrace-norm of a trace class operatorT is defined as$${\displaystyle \Vert T{\Vert }_1:=\mathrm{Tr}(|T|).}$$One can show that the trace-norm is anorm on the space of all trace class operators${\displaystyle B_1(H)}$ and that${\displaystyle B_1(H)}$, with the trace-norm, becomes aBanach space.

When${\displaystyle H}$ is finite-dimensional, every (positive) operator is trace class. For${\displaystyle A}$ this definition coincides with that of thetrace of a matrix. If${\displaystyle H}$ is complex, then${\displaystyle A}$ is alwaysself-adjoint (i.e.${\displaystyle A=A^{\ast }=|A|}$) 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:

• $\mathrm{Tr}(|T|)=\sum _k⟨|T|e_k,e_k⟩$ is finite for everyorthonormal basis${\displaystyle {\left(e_k\right)}_k}$ ofH.^[2]
• T is anuclear operator.^[6][7]

  There exist twoorthogonal sequences${\displaystyle {\left(x_i\right)}_{i=1}^{\mathrm{\infty }}}$ and${\displaystyle {\left(y_i\right)}_{i=1}^{\mathrm{\infty }}}$ in${\displaystyle H}$ and positivereal numbers${\displaystyle {\left({\lambda }_i\right)}_{i=1}^{\mathrm{\infty }}}$ in${\displaystyle {\ell }^1}$ such that and

  ${\displaystyle x\mapsto T(x)=\sum _{i=1}^{\mathrm{\infty }}{\lambda }_i⟨x,x_i⟩y_i,\mathrm{\forall }x\in H,}$

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

• T is acompact operator with

  IfT is trace class then^[9]

  ${\displaystyle \Vert T{\Vert }_1=sup\left\{|\mathrm{Tr}(CT)|:\Vert C\Vert \le 1\text{ and }C:H\to H\text{ is a compact operator }\right\}.}$

• T is anintegral operator.^[10]
• T is equal to the composition of twoHilbert-Schmidt operators.^[11]
• $\sqrt{|T|}$ 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 classif and only if${\displaystyle T}$ has apure point spectrum with eigenvalues${\displaystyle {\left\{{\lambda }_i(T)\right\}}_{i=1}^{\mathrm{\infty }}}$ such that^[12]

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

${\displaystyle K(s,t)=\sum _{j=1}^{\mathrm{\infty }}{\lambda }_j{e}_j(s)e_j(t)}$

then the associatedHilbert–Schmidt integral operator${\displaystyle T_K}$ is trace class, i.e.,

${\displaystyle \mathrm{Tr}(T_K)={\int }_a^{b}K(t,t)dt=\sum _i{\lambda }_i.}$

Everyfinite-rank operator is a trace-class operator. Furthermore, the space of all finite-rank operators is adense 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${\displaystyle (x\otimes y)(z):=⟨z,y⟩x.}$ Then${\displaystyle x\otimes y}$ is a continuous linear operator of rank 1 and is thus trace class; moreover, for any bounded linear operatorA onH (and intoH),${\displaystyle \mathrm{Tr}(A(x\otimes y))=⟨Ax,y⟩.}$^[9]

1. If${\displaystyle A:H\to H}$ is a non-negativeself-adjoint operator, then${\displaystyle A}$ is trace-class if and only if Therefore, a self-adjoint operator${\displaystyle A}$ is trace-classif 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 thecontinuous functional calculus.)
2. The trace is alinear functional over the space of trace-class operators, that is,$${\displaystyle \mathrm{Tr}(aA+bB)=a\mathrm{Tr}(A)+b\mathrm{Tr}(B).}$$The bilinear map$${\displaystyle ⟨A,B⟩=\mathrm{Tr}(A^{\ast }B)}$$ is aninner product on the trace class; the corresponding norm is called theHilbert–Schmidt norm. The completion of the trace-class operators in the Hilbert–Schmidt norm are called the Hilbert–Schmidt operators.
3. ${\displaystyle \mathrm{Tr}:B_1(H)\to \mathbb{C}}$ is a positive linear functional such that if${\displaystyle T}$ is a trace class operator satisfying${\displaystyle T\ge 0\text{ and }\mathrm{Tr}T=0,}$ then${\displaystyle T=0.}$^[11]
4. If${\displaystyle T:H\to H}$ is trace-class then so is${\displaystyle T^{\ast }}$ and${\displaystyle \Vert T{\Vert }_1={\Vert T^{\ast }\Vert }_1.}$^[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 onH is a two-sidedideal in the algebra of bounded linear operators onH), and^[11][13]$${\displaystyle \Vert AT{\Vert }_1=\mathrm{Tr}(|AT|)\le \Vert A\Vert \Vert T{\Vert }_1,\Vert TA{\Vert }_1=\mathrm{Tr}(|TA|)\le \Vert A\Vert \Vert T{\Vert }_1.}$$Furthermore, under the same hypothesis,^[11]$${\displaystyle \mathrm{Tr}(AT)=\mathrm{Tr}(TA)}$$ and${\displaystyle |\mathrm{Tr}(AT)|\le \Vert A\Vert \Vert T\Vert .}$ The last assertion also holds under the weaker hypothesis thatA andT are Hilbert–Schmidt.
6. If${\displaystyle {\left(e_k\right)}_k}$ and${\displaystyle {\left(f_k\right)}_k}$ are two orthonormal bases ofH and ifT is trace class then$\sum _k\left|⟨T{e}_k,f_k⟩\right|\le \Vert T{\Vert }_1.$^[9]
7. IfA is trace-class, then one can define theFredholm determinant of${\displaystyle I+A}$:$${\displaystyle det(I+A):=\prod _{n\ge 1}[1+{\lambda }_n(A)],}$$ 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,$${\displaystyle det(I+A)\le e^{\Vert A{\Vert }_1}.}$$It also implies that${\displaystyle det(I+A)\ne 0}$ if and only if${\displaystyle (I+A)}$ is invertible.
8. If${\displaystyle A:H\to H}$ is trace class then for anyorthonormal basis${\displaystyle {\left(e_k\right)}_k}$ of${\displaystyle H,}$ the sum of positive terms$\sum _k\left|⟨A{e}_k,e_k⟩\right|$ is finite.^[11]
9. If${\displaystyle A=B^{\ast }C}$ for someHilbert-Schmidt operators${\displaystyle B}$ and${\displaystyle C}$ then for any normal vector${\displaystyle e\in H,}$$|⟨Ae,e⟩|=\frac{1}{2}\left(\Vert Be{\Vert }^2+\Vert Ce{\Vert }^2\right)$ holds.^[11]

Let${\displaystyle A}$ be a trace-class operator in a separable Hilbert space${\displaystyle H,}$ and let${\displaystyle \{{\lambda }_n(A){\}}_{n=1}^{N\le \mathrm{\infty }}}$ 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 afterVictor Borisovich Lidskii) states that$${\displaystyle \mathrm{Tr}(A)=\sum _{n=1}^N{\lambda }_n(A)}$$

Note that the series on the right converges absolutely due toWeyl's inequality$${\displaystyle \sum _{n=1}^N\left|{\lambda }_n(A)\right|\le \sum _{m=1}^M{s}_m(A)}$$between the eigenvalues${\displaystyle \{{\lambda }_n(A){\}}_{n=1}^N}$ and thesingular 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 classicalsequence spaces, with trace-class operators as the noncommutative analogue of thesequence space${\displaystyle {\ell }^1(\mathbb{N}).}$

Indeed, it is possible to apply thespectral 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${\displaystyle {\ell }^{\mathrm{\infty }}(\mathbb{N}),}$ thecompact operators that of${\displaystyle c_0}$ (the sequences convergent to 0), Hilbert–Schmidt operators correspond to${\displaystyle {\ell }^2(\mathbb{N}),}$ andfinite-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$${\displaystyle Tx=\sum _i{\alpha }_i⟨x,v_i⟩u_i\text{ for all }x\in H.}$$Making the above heuristic comments more precise, we have that${\displaystyle T}$ is trace-class iff the series$\sum _i{\alpha }_i$ is convergent,${\displaystyle T}$ is Hilbert–Schmidt iff$\sum _i{\alpha }_i^2$ 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$\Vert T{\Vert }_1=\mathrm{Tr}\left[{\left(T^{\ast }T\right)}^{1/2}\right]=\sum _i{\alpha }_i.$ The norm corresponding to the Hilbert–Schmidt inner product is$${\displaystyle \Vert T{\Vert }_2={\left[\mathrm{Tr}\left(T^{\ast }T\right)\right]}^{1/2}={\left(\sum _i{\alpha }_i^2\right)}^{1/2}.}$$Also, the usualoperator norm is$\Vert T\Vert =\underset{i}{sup}\left({\alpha }_i\right).$ By classical inequalities regarding sequences,$${\displaystyle \Vert T\Vert \le \Vert T{\Vert }_2\le \Vert T{\Vert }_1}$$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.

Thedual space of${\displaystyle c_0}$ is${\displaystyle {\ell }^1(\mathbb{N}).}$ Similarly, we have that the dual of compact operators, denoted by${\displaystyle K(H{)}^{\ast },}$ 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${\displaystyle f\in K(H{)}^{\ast },}$ we identify${\displaystyle f}$ with the operator${\displaystyle T_f}$ defined by$${\displaystyle ⟨T_{f}x,y⟩=f\left(S_{x,y}\right),}$$where${\displaystyle S_{x,y}}$ is the rank-one operator given by$${\displaystyle S_{x,y}(h)=⟨h,y⟩x.}$$

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$${\displaystyle \sum _i⟨T_f{u}_i,u_i⟩=f(I)\le \Vert f\Vert ,}$$where${\displaystyle I}$ is the identity operator:$${\displaystyle I=\sum _i⟨\cdot ,u_i⟩u_i.}$$

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

A limiting argument using finite-rank operators shows that${\displaystyle \Vert T_f{\Vert }_1=\Vert f\Vert .}$ Thus${\displaystyle K(H{)}^{\ast }}$ isisometrically isomorphic to${\displaystyle B_1.}$

Recall that the dual of${\displaystyle {\ell }^1(\mathbb{N})}$ is${\displaystyle {\ell }^{\mathrm{\infty }}(\mathbb{N}).}$ 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-sidedideal in${\displaystyle B(H).}$ So given any operator${\displaystyle T\in B(H),}$ we may define acontinuouslinear functional${\displaystyle {\phi }_T}$ on${\displaystyle B_1}$ by${\displaystyle {\phi }_T(A)=\mathrm{Tr}(AT).}$ This correspondence between bounded linear operators and elements${\displaystyle {\phi }_T}$ of thedual space of${\displaystyle B_1}$ is anisometric isomorphism. It follows that${\displaystyle B(H)}$is the dual space of${\displaystyle B_1.}$ This can be used to define theweak-* topology on${\displaystyle B(H).}$

• Nuclear operators between Banach spaces

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