# Commutator of the density operator

1. Sep 18, 2009

### keen23

Hello all!
I hope some of you are more proficient in juggling with bra-kets...
I am wondering if/when the density operator commutes with other operators, especially with unitaries and observables.

1. My guess is, that it commutes with unitaries, but I am not sure if my thinking is correct. It's like this:

$\langle x|\rho U^{\dagger}|x\rangle=\langle x|U^{\dagger}\rho U U^{\dagger}|x\rangle=\langle x|U^{\dagger} \rho|x\rangle$

Is that ok?

2.How do I determine, if a state commutes with its observable? I don't see a physical meaning in this, I'm not sure what to do...

Thanks

2. Sep 18, 2009

### kanato

1. No, you've basically argued that any unitary matrix commutes with any other matrix, which is not true. The problem is that you insert $$U$$ and $$U^\dagger$$ in a weird way, you are assuming that $$\rho = U^\dagger\rho U$$ which is not true. If you want to insert an identity operator by inserting U and its adjoint, you can't split them up around another matrix, unless you assume that U and that other matrix commute.

Now there are rules for computing the trace where this will work, because $$Tr AB = Tr BA$$ regardless of whether A and B commute. So if you're tracing over the product of the density matrix and some other matrix, then something like this can work (be careful about reordering a product of more than two matrices, however). Expressions involving the trace over the density matrix times an operator are quite common, so you may see reordering like this. But this does not mean that the density matrix commutes with that operator.

2. States don't commute with observables. States are vectors, observables are matrices, and you just can't commute them. You can take the Hermitian conjugate, which usually will leave an operator unchanged and might end up looking like you've commuted the state with the operator, but it's not the same operation.

3. Sep 18, 2009

### keen23

Did I get it right, it's in general not possible to swap an operater and $\rho$?

Indeed I want to take the trace, thanks for the hint of swapping "before" :)

I will see later if this works for me (I am afraid, it won't, since $U=U_A U_B$. I was hoping to get rid of $U_B$ first by letting it act on the ket on the right side... So that won't work...)

Thanks :)

4. Sep 18, 2009

### Fredrik

Staff Emeritus
It's not hard to show that $\langle\alpha|XY|\beta\rangle=\langle\beta|Y^\dagger X^\dagger|\alpha\rangle^*$. So if $|\alpha\rangle=|\beta\rangle$ and both operators are hermitian, you can swap them, since the product of two hermitian operators is hermitian, and hermitian operators have real expectation values. In other words, if your U is hermitian instead of unitary, it "commutes" with $\rho$ inside an expectation value, but not in general.