Dirac notation - trace of product of (bipartite) density matrices I'm getting confused by the Dirac notation. Suppose I have the following two objects.
$$\rho = \sum_k p_k (\rho_A \otimes \rho_B) = \sum_k p_k |k \rangle \langle k | \otimes |k\rangle \langle k |  ,$$
$$A = \sum_{ij} |ii \rangle \langle jj | = \sum_{ij} |i \rangle \langle j | \otimes |i \rangle \langle j | .$$
Here $\rho$ is a seperable density matrix, but $A$ isn't quite a density matrix (missing prefactor).
I wanna write down what $A\rho$ is. Do I just write them all "next to" one another?
$$A\rho \rightarrow \sum_{ijk} p_k |i \rangle \langle j | \otimes |i \rangle \langle j |  k \rangle \langle k | \otimes |k\rangle \langle k |$$
This makes no sense, now there's two Kronecker products! So maybe like this?
$$A\rho \rightarrow \sum_{ijk} p_k |i \rangle \langle j |k \rangle \langle k | \otimes  |i \rangle \langle j |k \rangle \langle k |  = \sum_{ij} p_j |i \rangle \langle j | \otimes  |i \rangle  \langle j | $$
This looks suspicious to me, but let's go on. Ultimately I want to take the trace:
$$tr A\rho = \sum_{ij} p_j tr \left(|i \rangle \langle j | \otimes  |i \rangle  \langle j | \right) = \sum_{ij} p_j \delta_{ij} = 1$$
I used the fact that trace of Kronecker product is a product of traces. This seems wrong. I don't like that I got a $1$, because $A$ wasn't a density matrix in the first place. Dividing it by a factor would make it one, but then the trace of the product of two density matrix wouldn't be unity, which it should.
What's the mistake I'm making?
 A: Notation-wise, you can use either of
\begin{align}
A\rho 
& =
\sum_{ijk} p_k \Big(|i \rangle \langle j | \otimes |i \rangle \langle j |\Big) \Big(|  k \rangle \langle k | \otimes |k\rangle \langle k |\Big)
\\ & =
 \sum_{ijk} p_k |i \rangle \langle j |k \rangle \langle k | \otimes  |i \rangle \langle j |k \rangle \langle k | .
\end{align}
As you well note, here the $⟨j|k⟩$ give $\delta_{jk}$'s, so the sum over $k$ goes away:
\begin{align}
A\rho 
& =
 \sum_{ij} p_j |i \rangle \langle j | \otimes  |i \rangle \langle j |.
\end{align}
The rest of your formal manipulations are also fine:
$$\mathrm{Tr} \left(A\rho\right) = \sum_{ij} p_j \mathrm{Tr} \Big(|i \rangle \langle j | \otimes  |i \rangle  \langle j | \Big) = \sum_{ij} p_j \delta_{ij} = 1.$$
This does look vaguely suspicious, but it is reasonable because of the special structure of $A$ and $\rho$. It is not a contradiction because there is nothing that requires the product of density matrices to be anything like a density matrix: if $\rho$ and $\sigma$ don't commute, $\rho\sigma$ is not even hermitian, let alone positive semidefinite; if you want an example of density matrices whose product has trace $≠1$, try $\sigma^2$ for
$$
\sigma=\begin{pmatrix}1/2 & 0 \\ 0 & 1/2\end{pmatrix}.
$$
