Trace-class operator (${\mathcal{S}}_1$)

A bounded operator $T$ on a separable Hilbert space is trace class when $|T{|}^{1/2}$ is Hilbert--Schmidt. Its trace norm is

The dual of the trace-class operators, and also of the normed space of finite-rank operators with the trace norm, is isometrically the space of bounded operators. The pairing is $F_B(T)=\mathrm{Tr}(BT)$. For $R_{x,y}z=⟨z,y⟩x$, one has $\Vert R_{x,y}{\Vert }_1=\Vert x\Vert \Vert y\Vert$ and $F_B(R_{x,y})=⟨Bx,y⟩$, using an inner product linear in its first argument. A bounded functional on these rank-one operators gives a bounded sesquilinear form and hence a unique $B$ by the Riesz representation theorem. The singular value decomposition proves $\Vert F_B\Vert =\Vert B\Vert$ and the triangle inequality for the trace norm.

For a trace-class operator $K$ on a Hilbert space, the Fredholm determinant is the convergent product over its eigenvalues, counted with algebraic multiplicity. For self-adjoint $K$, spectral theorem for compact self-adjoint operators reduces the product to a diagonal calculation. For $\Vert K\Vert <1$, $\log \det (I+K)={\sum }_{n\ge 1}(-1{)}^{n+1}\mathrm{tr}(K^n)/n$. This identity exposes the successive cumulants in a Gaussian quadratic exponential moment.

For bounded operators with trace-class commutator, the multiplicative commutator differs from identity by a trace-class operator. Along $F(t)=e^{tA}{e}^B{e}^{-tA}{e}^{-B}$, the trace of its logarithmic derivative is the constant $\mathrm{Tr}[A,B]$. This follows by writing $e^{B}A{e}^{-B}-A$ as the integral of conjugates of $[B,A]$ and using trace invariance under bounded similarities. Integrating the Fredholm determinant derivative from $F(0)=I$ proves the formula. Individual traces of $A$ and $B$ need not exist.

An operator is trace class exactly when it factors as $T=AB$ with $A$ and $B$ Hilbert--Schmidt. The factors can be chosen so that $\Vert T{\Vert }_1=\Vert A{\Vert }_{\mathrm{HS}}\Vert B{\Vert }_{\mathrm{HS}}$ by using the polar decomposition of a bounded operator.

The Schatten norm Hölder inequality gives $\Vert AB{\Vert }_r\le \Vert A{\Vert }_p\Vert B{\Vert }_q$ when $1/r=1/p+1/q$. In particular, the product of two Hilbert-Schmidt operators is trace class and

For a trace-class operator, the series $\mathrm{tr}T={\sum }_n⟨T{e}_n,e_n⟩$ converges absolutely and is independent of the orthonormal basis. It satisfies $|\mathrm{tr}T|\le \Vert T{\Vert }_1$.

The trace-class operator condition makes this operator trace well-defined; the class of eligible operators and the trace functional are distinct concepts.

Every bounded operator $S$ defines a functional $T\mapsto \mathrm{tr}(ST)$ on the trace-class operators, andFinite-rank density in ${\mathcal{S}}_1$ makes every continuous functional arise uniquely this way.

For vectors $x,y$ in a Hilbert space, the rank-one operator $(x\otimes y)z=⟨z,y⟩x$ has $\Vert x\otimes y{\Vert }_1=\Vert x\Vert \Vert y\Vert$ and $\mathrm{tr}(x\otimes y)=⟨x,y⟩$.