# Trace of the fundamental commutation relation

Hi.

So I have learned that this holds for the trace if A and B are two operators: $\text{Tr}(AB)=\text{Tr}(BA)$. Now I take the trace of the commutator between $x$ and $p$: $\text{Tr}(xp)-\text{Tr}(px)=\text{Tr}(xp)-\text{Tr}(xp)=0$. But the commutator of x and p is $i\hbar$. Certainly the trace of $i\hbar$ is not zero. What is wrong here?

tom.stoer
The problem is that x and p are unbounded operators on an infinite-dimensional Hilbert space; therefore certain relations known from finite-dimensional matrices do no-longer hold.

So what's wrong with this "proof":

$$\text{Tr}(AB)=\sum_n \langle n|AB|n\rangle = \sum_n\sum_m\langle n|A|m\rangle\langle m|B|n\rangle=\sum_n\sum_m\langle m|B|n\rangle\langle n|A|m\rangle=\sum_m \langle m|BA|m\rangle=\text{Tr}(BA)$$

DrDu