Trace of an indefinite operator does not exist I see that there are a few definitions of trace-class operator. They are roughly the same, but some include a clause that the operator should be definite in some flavor. I want to ask why this clause is important. Is it because then the order of the basis elements becomes a question; if the terms in the sum show up in different order the sum is not the same? It appears that on all of the internet, this question hasn't been asked because I have searched all day. If that is indeed the answer, just write "yes", and I will accept it as the answer (provided there isn't some smart admin that interprets the rules to forbid this question or that kind of answer). If it is not the whole story, please provide the full story.
 A: Of course, conventions vary, but I think it is important to realize that "trace class" by itself does not entail any positivity condition. It's just that it is (provably) simpler to (successfully) define trace class for operators that are already known to be positive, etc.
Otherwise, for not-necessarily-positive operators, we have to require that $\sum_n |\langle Te_n,f_n\rangle|<+\infty$ for all pairs of orthonormal bases... rather than just for a single one.
A more robust characterization of "trace class" is as "finite linear combination of composites $S\circ T$ with Hilbert-Schmidt operators $S,T$". The advantage is that $T$ is Hilbert-Schmidt if $\sum_n |Te_n|^2$ for one orthonormal basis... and Parseval's theorem proves that then the same is true for all orthonormal bases.
A: We say that a positive semi-definite, self-adjoint linear operator $T: H \to H$ on some infinite-dimensional Hilbert space $H$ is in trace-class, if for some orthonormal basis $\{e_i\}$ the sum
$$\operatorname{tr}_H(T)=\sum_i \langle Te_i, e_i\rangle$$
is finite. It then follows that this holds true for any orthonormal basis. We call $\operatorname{tr}_H(T)$ the trace of $T$.
If we didn't require $T$ to be positive definite, this definition would not yield a well-defined trace independent of the choice of basis (and not even independent of the order of the basis). Consider on $\ell^2$, the space of square-summable sequences the operator defined by $e_n \mapsto \frac{(-1)1^n}{n} \cdot e_n$, where $\{e_n\}$ is the canonical orthonormal basis. If we compute the trace with respect to the usual ordering, we get
$$ \sum_{n=1}^\infty \langle T e_n, e_n\rangle = \sum_{n=1}^\infty\frac{(-1)1^n}{n} = \ln(2)$$
But by the Riemannian reordering theorem, if we choose a different ordering of our basis, we could make this add up to any $ x\in [-\infty,\infty]$. So yes, in the infinite dimensional case, we only allow for positive semi-definite operators, because otherwise this would not be well-defined.
