# Trace of the fundamental commutation relation

1. Jul 1, 2010

### daudaudaudau

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?

2. Jul 2, 2010

### 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.

3. Jul 2, 2010

### daudaudaudau

So what's wrong with this "proof":

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

4. Jul 2, 2010

### DrDu

Nothing is wrong with your proof, only that for the operators A=x and B=p Tr(AB) and Tr(BA) are undefined or the value of the trace depends on the basis chosen. As tom.stoer said, the class of operators for which the trace is finite and independent of the basis chosen is only a small subset of the bounded operators (so called trace class operators) while most operators we are interested in in quantum mechanics are unbounded.

5. Jul 2, 2010

### daudaudaudau

Okay I see. I was just considering this question because I was looking at the commutation relations for spin. In that case the trace of the Pauli matrices is actually zero. Just from writing up the commutation relations I obtained a set of equations that I could solve and determine the Pauli matrices... I just wanted to see if a similar thing was possible with x and p.

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