# Commutator of a density matrix and a real symmetric matix

1. Oct 18, 2006

### Einstein Mcfly

Let p1,p2 be two density matrices and M be a real, symmetric matrix.
Now,
<<p1|[M,p2]>>=
<<p1|M*p2>>-<<p1|p2*M>>=
Tr{p1*M*p2}-Tr{p1*p2*M}=
2i*Tr{(Im(p1|M*p2))}.
Why is it that this works out as simply as (x+iy)-(x-iy)?
How is Tr{p1*p2*m}=conjugate(Tr{p1*M*p2})? I can't seem to figure
out why this works out so cleanly. Thanks for any comments.

2. Oct 18, 2006

### Rach3

For the diagonal terms of an operator A, in some matrix representation,

$$(A^\dag)_{ii}=(A_{ii})^*$$

and so

$$Tr(A^\dag) = Tr(A)^*$$.

Of course, we mean $$A\equiv M^\dag \rho_2$$. Since $$Tr(\rho_1)^*=Tr(\rho_1)$$ by self-adjointness,

$$Tr(\rho_1^\dag \rho_2^\dag M)=Tr(\rho_1^\dag) \dot Tr (M^\dag \rho_2)^* = Tr(\rho_1 M^\dag \rho_2)^*$$

Last edited by a moderator: Oct 18, 2006
3. Oct 18, 2006

### Rach3

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