Fermionic operators, bipartite systems and partial traces Consider a composite system (e.g. a system $S$ and a bath $B$) and fermionic operators $s_i$ and $b_j$, where $s_i$ annihilate fermions in $S$ and $b_j$ annihilate fermions in $B$. Obviously, 
\begin{equation}
\{s_i , b_j \} = 0
\end{equation}
and hence the $s_i$ do not decompose as $\textrm{stuff}_S \otimes \textrm{id}_B$ and also the $b_i$ do not decompose as $\textrm{id}_S \otimes \textrm{stuff}_B$ because if they did, the $s_i$ and the $b_j$ would commute. So the fermionic operators somehow also "act" on the part of the world they do not "belong to". 
However, products $s_i s_k$ or $s^{\dagger}_i s_k$ commute with $b_j$ and act nontrivially only on the system so they should be of the form $\textrm{stuff}_S \otimes \textrm{id}_B$ and similarly even powers of $b_i$ should be of the form $\textrm{id}_S \otimes \textrm{stuff}_B$. 
How does this constrain the matrix representation of the fermion operators? 
Can one find a simple expression for the fermion operators similar to $\textrm{id}_S \otimes \textrm{stuff}_B$ for bosonic operators?
In particular, I would like to evaluate partial traces of the type
\begin{equation}
\textrm{tr}_B [ s_i b_i s_j b_j (\rho \otimes \omega )  ],\ 
\textrm{tr}_B [ s_i b_i (\rho \otimes \omega ) s_j b_j ].\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  (1)
\end{equation}
The first expression I know how to evaluate since
\begin{equation}
\textrm{tr}_B [ s_i b_i s_j b_j (\rho \otimes \omega )  ] = - \textrm{tr}_B [ (s_i s_j) (b_i b_j) (\rho \otimes \omega )  ] = - (s_i s_j) \rho\ \textrm{tr}[ (b_i b_j) \omega ],
\end{equation}
where in the second equality I used that $s_i s_j$ $(b_i b_j)$ act only in the system (bath) Hilbert space. Note the extra minus sign. But I don't know how to deal with the second expression in $(1)$ since the partial trace would only allow to move $b_j$ to the front if it were of the form $\textrm{id}_S \otimes \textrm{stuff}_B$. Is there a trick that allows to perform these partial traces (without writing down a representation for the fermion operators)? 
 A:  The magic transformation 
One way to approach this problem is to use a Klein transformation. A Klein transformation can be used to make $S$ and $B$ commute with each other without changing the (anti)commutation relations within $S$ or $B$ individually. This is done by using the $S$-operators to construct an operator $K$ that commutes with everything in $B$ and anticommutes with the creation/annihilation operators in $S$, and shown in detail below. Then the original pair $(S,B)$ can be replaced by the new pair $(S,KB)$. The operators in $S$ and $KB$ commute with each other, so we can represent them on a factorized Hilbert space ${\cal H}_S\otimes{\cal H}_{KB}$, and then the partial trace can be defined as usual.
That might feel illegal, because now the "bath" $KB$ involves an operator $K$ from the "system" $S$. This is relatively harmless, though, because $K$ has only two eigenspaces, and those eigenspaces are not mixed with each other by any observables in $S$. In other words, as far as $S$ is concerned, the eigenspaces of $K$ are superselection sectors. Thus allowing $K$ to be regarded as part of the "bath" won't affect any predictions about other observables in $S$, as long as the state doesn't involve any correlations between superselection sectors.
We can have observables that are not "localized" in either $S$ or $B$, and those observables can mix the eigenspaces of $K$. In particular, the Hamiltonian of the combined system may mix the two eigenspaces of $K$. However, since $K$ is constructed from operators in $S$, we can diagnose such mixtures just using the reduced density matrix for $S$.
 Single-mode example 
To illustrate the idea in the simplest possible setting, suppose that the observables in $S$ and $B$ are each generated by a single creation-annihilation operator pair:
\begin{gather}
\newcommand{\db}{b^\dagger}
\newcommand{\dd}{d^\dagger}
\newcommand{\ds}{s^\dagger}
 \{s,\ds\}=1
\hskip2cm
 \{s,s\}=0
\tag{1}
\\
 \{b,\db\}=1
\hskip2cm
 \{b,b\}=0
\tag{2}
\\
 \{s,\db\}=0
\hskip2cm
 \{s,b\}=0.
\tag{3}
\end{gather}
I'm using the standard notation
$$
 \{A,B\} := AB+BA
\hskip2cm
 [A,B] := AB-BA.
$$
Using a Klein transformation, we can change the anticommutation relations (3) to commutation relations without affecting (1) and (2). Define
$$
 K = [s,\ds].
\tag{4}
$$
This operator has the properties
\begin{gather}
 K^2 = 1
\hskip2cm
 \{K,s\}=0
\\
 K^\dagger = K
\hskip2cm
 [K,b]=0.
\tag{5}
\end{gather}
Use these to see that the operator $d := Kb$ satisfies
\begin{gather}
\\
 \{d,\dd\}=1
\hskip2cm
 \{d,d\}=0
\tag{6}
\\
 [s,\dd]=0
\hskip2cm
 [s,d]=0.
\tag{7}
\end{gather}
This is the desired result.
 Multi-mode case 
Define 
$$
 K=\prod_j [s_j,\ds_j].
\tag{8}
$$
This operator has the properties
\begin{gather}
 K^2 = 1
\hskip2cm
 \{K,s_j\}=0
\\
 K^\dagger = K
\hskip2cm
 [K,b_j]=0.
\tag{9}
\end{gather}
Define $d_j := Kb_j$ to get the desired result.
