Is the Trace of products of Hilbert-Schmidt Operators stable under cyclic permutations?

Question

It is known, that the product of two hilbert-schmidt operators is in the Trace class. We also know, that for a continuous, linear $S$ and any $T$ in the trace class, $\text{trace}(ST)=\text{trace}(TS)$. If we have two Hilbert-Schmidt Operators $H_{1}$ and $H_{2}$ does $\text{trace}(H_{1}H_{2})=\text{trace}(H_{2}H_{1})$ necessarily hold? What happens to the traces if we choose any two operators $A,B$ auch that $AB$ and $BA$ are both in the Trace class? (If i am not mistaken, that should for example happen for operators lying in Schatten-p-classes for conjugated indexes).

Answer 1

Start with $H_i$ being self-adjoint. Let $x_i$ be an ONB and $p_i=x_i\otimes x_i^*$ the orthogonal projection on the space spanned by $x_i$. One has $$( x_i, H_1 p_j H_2 x_i) = (x_i, H_1 x_j)\,(x_j,H_2,x_i)=(x_j, H_2x_i)(x_i,H_1x_j)=(x_j,H_2p_iH_1x_i).$$ Note that $\sum_{i=0}^n p_i$ converges to $\Bbb 1$ in the strong operator topology. It follows that $\sum_{i=0}^nH_1p_iH_2\to H_1H_2$ in this topology and then $(x,H_1H_2y)=\sum_i(x,H_1p_iH_2y)$. This gives: $$(x_i,H_1H_2x_i)=\sum_j(x_i, H_1p_jH_2x_i)=\sum_j (x_j, H_2p_iH_1x_j) = \mathrm{Tr}(H_2p_iH_1).$$ The trace of $H_1H_2$ is the sum of this over all $i$, and the question is whether $$\mathrm{Tr}(H_1H_2)=\lim_{n\to\infty}\sum_{i=0}^n \mathrm{Tr}(H_2 p_i H_1)\overset?=\mathrm{Tr}(H_2H_1)\tag{1}$$ holds. To do this I want to show that $\sum_i H_1p_i$ converges to $H_1$ in the Hilbert-Schmidt norm, since the trace is the Hilbert-Schmidt scalar product $(1)$ then follows.

$$\|\sum_{i=0}^n H_1p_i- H_1\|^2_{HS}=\sum_{i,j=0}^n\mathrm{Tr}(p_i H_1^*H_1p_j)+\mathrm{Tr}(H_1^*H_1)-\sum_{i=0}^n\mathrm{Tr}(H_1^*H_1p_i + p_iH_1^*H_1).$$

Here the trace will be evaluated in the ONB $x_i$, and if we plug it in the expression simplifies to (using $p_j x_i = \delta_{i,j}x_i$): $$\|\sum_{i=0}^n H_1p_i- H_1\|^2_{HS}=\sum_{i=0}^n (x_i, H_1^*H_1 x_i) +\sum_{i=0}^\infty (x_i, H_1^*H_1,x_i)-2\sum_{i=0}^n(x_i,H_1x_i).$$ Here we retrieve equality in the limit and thus $\mathrm{Tr}(H_1H_2)=\lim_{n\to\infty}\sum_{i=0}^n\mathrm{Tr}(H_2p_iH_1)=\mathrm{Tr}(H_2H_1)$ for self-adjoint Hilbert-Schmidt operators. Since any Hilbert-Schmidt is the a linear combination of 2 self-adjoint Hilbert-Schmidt the general statement follows by linearity of the trace.

What is written here assumes a separable Hilbert space, but it works in non-separable spaces by replacing any countable index with an index of the appropriate cardinality. The sums then make sense in the sense of nets.