Probably a dumb question, but, given that $\mathcal{H}$ is a separable Hilbert space, is there anything wrong with the following definition of trace-class operator?
Definition. A linear map $A: \mathcal{D}_{A} \rightarrow \mathcal{H}$ is said to be of trace-class if $\mathcal{D}_{A}=\mathcal{H}$ and, for any
orthonormal basis $\left\{e_{n}\right\}$ of $\mathcal{H},$ the sum/series
$$
\sum_{n}\left\langle e_{n} | A e_{n}\right\rangle<\infty
$$
If $A: \mathcal{H} \rightarrow \mathcal{H}$ is of trace-class, one can show that the value of $\sum_{n}\left\langle e_{n} | A e_{n}\right\rangle$ does not
depend on the choice of orthonormal basis $\left\{e_{n}\right\} .$
Definition. Let $A: \mathcal{H} \rightarrow \mathcal{H}$ be of trace-class. Then the trace of $A$ is
$$
\operatorname{Tr} A:=\sum_{n}\left\langle e_{n} | A e_{n}\right\rangle
$$
where $\left\{e_{n}\right\}$ is any orthonormal basis of $\mathcal{H} .$
I took it from Dr. Schuller's lectures on Quantum Theory. I want to know if this is right, because what I see everywhere is that the series $\sum \left\langle e_{n}|\sqrt{A^{\dagger}A}e_{n}\right\rangle$ is the one which needs to converge, probably to guarantee that the series above converges uniformly, but I do not really know why.
Best Answer
It's not a dumb question; actually, it's been asked on this site more than once, and never answered. And I cannot give you a full answer because I don't know it, but here are some thoughts.
The condition works when $A$ is positive, and in that case convergence for only one orthonormal basis is sufficient, and it agrees with the "good" definition. Maybe that's what your source is doing?
But when $A$ is not positive, you cannot initially say that the value of $\sum_n\langle Ae_n,e_n\rangle$ does not depend on the orthonormal basis, because to prove that you need to exchange series, and for that you need absolute convergence.
Absolute convergence does apply, though, because you are allowed to reorder any fixed basis. And a series converges under all permutations if and only if it is absolutely convergent.
The problem is that you still don't know that $\sum_n\langle Ae_n,e_n\rangle$ convergent (note that there is no need for the value to be positive or even real), implies that the sum is the same for all bases. The usual argument goes by showing that $\operatorname{Tr}(AB)=\operatorname{Tr}(BA)$ and then using it to show that $\operatorname{Tr}(V^*AV)=\operatorname{Tr}(A)$ for any unitary, but this requires $AB$ and $BA$ to be trace-class. With the usual definition that $\sum_n\langle (A^*A)^{1/2}e_n,e_n\rangle<\infty$, one has inequalities available that allow you to show that the space of trace-class operators is an ideal and so if $A$ is trace-class so are $AB$ and $BA$. But with your definition, I don't see how you could show this.
In summary, I cannot prove that your definition is wrong (that would require finding an operator such that those sums are finite for all orthonormal bases while the operator is not trace-class), but at the very least it is not useful unless some smart calculation allows you to use it to show that the trace-class operators, as you define them, form an ideal.
For more information on this, here you can see how a stronger requirement than yours does imply trace-class.