See, for example, equation (5) of:
http://arxiv.org/abs/nucl-th/9508008
No reason to argue over terminology. And \rho(x,x') is a lot more complicated than just a number.
Carl, I know one or two things about the density matrix, so you don't need to point a "reference" to me
A physically acceptable (unique, hermitian & non-negative) density matrix must satisfy;
1) <\hat{A}> = Tr(\rho\hat{A})
2) the quantum Liouville equation;
i\partial_{t} \rho (x,x';t) = [H,\rho (x,x';t)]
So, Carl, be kind and show us how YOU prove that the talked about expession satisfies (1) & (2) ?
I am aware that some people who work on many-particle systems, non-relativistic field theory, DO ,
occasionally, call;
<\psi^{\dagger}(\vec{x},t)\psi(\vec{x'},t)>_{0}
(in Heisenberg picture)
or,
<\psi^{\dagger}(\vec{x})\psi(\vec{x'})>_{t}
(in Schrodinger picture)
a one-body density matrix!
Well, they are wrong, because these expressions are C-numbers. As such they do not satisfy (1) & (2).
Indeed, from (1) we see that these expressions are equal to;
Tr(\rho_{0}\psi^{\dagger}(x,t)\psi(x',t))=Tr(\rho_{t}\psi^{\dagger}(x)\psi(x'))
where the density matrix;
\rho(t) = exp(-iHt) \rho(0) exp(iHt)
does satisfy the quantum Liouville equation (2).
Don't ask me why, but even the number-density operator
n(\vec{x})=\psi^{\dagger}(\vec{x})\psi(\vec{x})
sometimes is called (by similar people) a density matrix!
I believe, it is a simple case of misuse of language, because these people do a fine work and make use of
<n(\vec{x})> = Tr \Left(\rho\psi^{\dagger}\psi\Right)
in their work!
regards
sam
