Trace of operators

Question

Trace of an operator $A$ can be written in its eigenbasis $|a\rangle$ as $${\rm Tr}(A)=\sum_{a} \langle a|A|a\rangle.$$ Since trace is base independent, does it imply that we can also write the expression for trace as $${\rm Tr}(A)=\sum_{b} \langle b|A|b\rangle$$ in some other orthonormal basis (not the eigenbasis of $A$) $|b\rangle$?

Answer 1

In complex Hilbert spaces, one of the various equivalent definitions of trace class operator is a bounded operator $T: H\to H$ such that, for every Hilbert basis $\{u_a\}_{a\in A}$, the sum $$\sum_{a\in A}\langle u_a|Tu_a\rangle$$ absolutely converges to a complex number and this number does not depend on the chosen Hilbert basis. That is the trace of $T$.

Hence, independence of the Hilbert basis is in the definition itself.

In finite dimension, independence is automatic and all operators are trace class.

In the infinite dimensional case that class is a subclass of compact operators. Hence it is quite small. For instance it is not possible to approximate all bounded operators with trace class operators in the natural topology of the space of bounded operators.

Answer 2

Let's check that the two expressions coincide; it can be done by "playing" with the closure relations with respect to each basis, as follows : $$ \begin{array}{rcl} \mathrm{Tr}(A) &=& \displaystyle \sum_a \langle a|A|a \rangle \\ &=& \displaystyle \sum_{a,b} \langle a|A|b \rangle\langle b|a \rangle \\ &=& \displaystyle \sum_{a,b} \langle b|a \rangle\langle a|A|b \rangle \\ &=& \displaystyle \sum_b \langle b|A|b \rangle \end{array} $$