Trace class

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, $$\left\{e_k\right\}_{k=1}^{\infty}$$ 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
 * $$\operatorname{Tr} (A) = \sum_{k=1}^{\infty} \left\langle A e_k, e_k \right\rangle,$$

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 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:
 * $\operatorname{Tr} (|T|) =\sum_k \left\langle |T| \, e_k, e_k \right\rangle$ 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 $\sum_{i=1}^{\infty} \lambda_i < \infty$ 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
 * $$\|T\|_1 = \sup \left\{ |\operatorname{Tr} (C T)| : \|C\| \leq 1 \text{ and } C : H \to H \text{ is a compact operator } \right\}.$$


 * $T$ is an integral operator.
 * $T$ is equal to the composition of two Hilbert-Schmidt operators.
 * $\sqrt{|T|}$ is a Hilbert-Schmidt operator.

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), $$\operatorname{Tr}(A(x \otimes y)) = \langle A x, y \rangle.$$

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.

$$\operatorname{Tr} : B_1(H) \to \Complex$$ 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 an ideal in the algebra of bounded linear operators on H), and $$\|A T\|_1 = \operatorname{Tr}(|A T|) \leq \|A\| \|T\|_1, \quad \|T A\|_1 = \operatorname{Tr}(|T A|) \leq \|A\| \|T\|_1.$$ Furthermore, under the same hypothesis, $$\operatorname{Tr}(A T) = \operatorname{Tr}(T A)$$ and $$|\operatorname{Tr}(A T)| \leq \|A\| \|T\|.$$ The last assertion also holds under the weaker hypothesis that A and T are Hilbert–Schmidt.

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 $\sum_{k} \left| \left\langle T e_k, f_k \right\rangle \right| \leq \|T\|_{1}.$ 

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 $\sum_k \left| \left\langle A \, e_k, e_k \right\rangle \right|$ is finite.</li>

<li>If $$A = B^* C$$ for some Hilbert-Schmidt operators $$B$$ and $$C$$ then for any normal vector $$e \in H,$$ $|\langle A e, e \rangle| = \frac{1}{2} \left(\|B e\|^2 + \|C e\|^2\right)$ holds.</li> </ol>

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 $$A.$$

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 $$\ell^1(\N).$$

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 $$T x = \sum_i \alpha_i \langle x, v_i\rangle u_i \quad \text{ for all } x\in H.$$ Making the above heuristic comments more precise, we have that $$T$$ is trace-class iff the series $\sum_i \alpha_i$ is convergent, $$T$$ is Hilbert–Schmidt iff $\sum_i \alpha_i^2$  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 $\|T\|_1 = \operatorname{Tr} \left[\left(T^* T\right)^{1/2}\right] = \sum_i \alpha_i.$ 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 $\| T \| = \sup_{i} \left(\alpha_i\right).$ 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 $$\ell^1(\N).$$ 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 $$\ell^1(\N)$$ 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 $$B(H).$$