Kraus operator representation of reduced density matrix evolution question [duplicate]

Question

I'm following through a derivation and cannot quite understand a step.

We have a system/bath density matrix (comprised of the system and bath basis states $|s_i\rangle_S$ and $|b_i\rangle_B$ which evolves according to

$$\rho_{SB}(t)=U(t)\rho_{SB}(0)U^\dagger(t)$$

and where

$$\rho_{SB}=\rho_{S}\otimes\rho_B$$ such that $$\rho_{SB}(t)=U(t)(\rho_{S}\otimes\rho_B) U^\dagger(t)$$

Now $U$ is an operator on the full product space so I introduce the embedding of the reduced states/operators into the full space viz.

$$\rho_S^e=\rho_S\otimes I_B$$ $$\rho_B^e=I_S\otimes\rho_B$$ $$|s_i\rangle = |s_i\rangle_S\otimes I_B$$ $$|b_i\rangle = I_S\otimes|b_i\rangle_B$$

so that

\begin{align} \rho_{SB}(t)&=U(t)(\rho^e_{S}\rho^e_B) U^\dagger(t)\\\\ &=\sum_jp_jU(t)\rho^e_{S}|b_j\rangle\langle b_j| U^\dagger(t) \end{align}

We then take a partial trace such that

\begin{align} \rho_{S}&=Tr_B[\rho_{SB}]\\\\ &=\sum_i\sum_jp_j\langle b_i|U(t)\rho^e_{S}|b_j\rangle\langle b_j| U^\dagger(t)|b_i\rangle \end{align}

Now, every single derivation I can find then just makes the next step without a word and just writes

$$\rho_S=\sum_i\sum_jp_j\langle b_i|U(t)|b_j\rangle\rho_{S}\langle b_j| U^\dagger(t)|b_i\rangle$$

And I don't know how they have achieved this. How can they 1) swap the order? and 2) how can they suddenly talk about $\rho_S$ and not $\rho_S^e$?

Or more bluntly, why does $\rho_S^e|b_j\rangle=|b_j\rangle\rho_S$?

Surely, $U(t)$ has to operate on the full Hilbert space, i.e. on $\rho_{S/B}^e$ not $\rho_{S/B}$ so how am I to understand $|b_j\rangle$ in the above.

What am I missing here?

EDIT:

Thanks for the answer. The notation is annoying, I do wish the literature (or at least pedagogy) would distinguish between $\rho_S$ and $\rho_S\otimes I_B$ etc., however it was the last step that was alluding me

I think the most concise way of resolving it is as follows:

There is something called the mixed product property which is

$$(A\otimes B)(C\otimes D)=(AC)\otimes(BD)$$

Consequently

\begin{align} \rho_S^e|b_j\rangle&=(\rho_S\otimes I_B)(I_S\otimes|b_j\rangle_B)\\\\ &=(\rho_SI_S)\otimes (I_B|b_j\rangle_B)\\\\ &=(I_S\rho_S)\otimes (|b_j\rangle_BI_B)\\\\ &=(I_S\rho_S)\otimes (|b_j\rangle_B\cdot 1)\\\\ &=(I_S\otimes |b_j\rangle_B)(\rho_S\otimes 1)\\\\ &=|b_j\rangle\rho_S \end{align}

Answer 1

First of all, formally:

$$ \rho^e_S | b_i \rangle = (\rho_S \otimes I_B) (I_S \otimes |b_i\rangle_B) = (\rho_S I_S) \otimes (I_B |b_I\rangle_B) = \rho_S \otimes |b_i\rangle_B $$

Which is in fact the same as $|b_i\rangle \rho_S$.

The whole problem lies in the notation - I don't really know how to improve it though. Take a look for example here and realize that the partial trace is written as

$$ \sum_j \langle b_j | \rho |b_j\rangle $$

which looks like a scalar quantity at first glance, but is actually an operator acting on $S$ (since the $|b_j \rangle$ are not actually vectors).

So consider $\rho^e_S | b_i \rangle = \rho_S \otimes |b_i\rangle_B$. This acts from the right on objects of the type* $\langle \psi|_S \otimes \langle \phi|_B$, removes the $B$-part and acts with $\rho_S$ on the $S$-part. If you think about this, you see that indeed

$$ (\langle \psi|_S \otimes \langle \phi|_B) (\rho_S \otimes |b_i\rangle_B) = \langle \Phi | b_i \rangle_B \cdot \langle \psi |_S\, \rho_S = (\langle \psi|_S \otimes \langle \phi|_B) |b_i\rangle \rho_S $$

* In general, $\langle b_i | U(t)$ will be a linear combination of such product vectors.