quantum mechanics - Trace of a commutator is zero - but what about the commutator of $x$ and $p$?

Operators can be cyclically interchanged inside a trace:

$${\rm Tr} (AB)~=~{\rm Tr} (BA).$$

This means the trace of a commutator of any two operators is zero:
$${\rm Tr} ([A,B])~=~0.$$

But what about the commutator of the position and momentum operators for a quantum particle?

On the one hand: $${\rm Tr}([x,p])~=~0,$$
while on the other hand: $$[x,p]~=~i\hbar.$$

How does this work out?

Answer

After reading @Peter Morgan's answer, and giving it some more thought, I think this is actually simpler than it seems at first.

For finite-dimensional spaces the trace of a commutator is indeed always zero. For infinite-dimensional spaces the trace is not always defined, since it takes the form of an infinite sum (for countable dimension) or an integral (for continuous dimension) which do not always converge.

When the trace is defined, it obeys the same rules as in finite dimension, specifically the trace of a commutator is zero. For operators such as $x$, $p$ and their products, the trace is simply not defined, so there is no sense in asking questions about it.
When computing thermal averages, the factor $e^{-\beta H}$ makes sure the trace converges, since the energy is always bound from below (otherwise the system is unphysical).

I'm sure the concepts mentioned by @Peter Morgan are important in this context (boundedness, KMS-condition), but I don't know anything about them, and I think the answer I just provided suffices for practical purposes.