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)?

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    $\begingroup$ sorry, I might be naif, but is your first equation that obvious? For example, creation and annihilation operators of two uncoupled harmonic oscillators (for examples of the two dimensions of a 2D oscillator) commute among each others and do not respect the usual bosonic relation $[a,a^\dagger]=1$ if the two act on different spaces. I would instinctively say that $[s_i,b_j]=0$. $\endgroup$ – user2723984 May 12 '20 at 13:56
  • $\begingroup$ My first equation is the defining property of fermions. Harmonic oscillator modes are bosons, hence they commute. $\endgroup$ – loewe May 12 '20 at 14:39
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    $\begingroup$ Is it even clear what a partial trace means? $\endgroup$ – Norbert Schuch May 14 '20 at 14:52
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    $\begingroup$ You might want to read on drone-fermion representations: they have been extensively used for spin systems, where the operators anti-commute on the same sight, but commute one different sites. There might have been similar uses of Schwinger bosons and/or slave-bosons, but I am not aware of them. $\endgroup$ – Vadim May 14 '20 at 16:51
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    $\begingroup$ @loewe The partial trace can be replaced with a more general concept that doesn't rely on any factorization of the Hilbert space: just think of the state as a function from operators $X$ to complex numbers trace$(X(\rho\otimes\omega))$, and simply restrict attention to operators associated with $S$. The partial trace is just a way of enforcing the "restrict attention" rule in the special case of a factorized Hilbert space. Is the question open to that type of answer, or do you have a reason for wanting to use the usual partial trace (e.g., deriving a master equation)? $\endgroup$ – Chiral Anomaly May 14 '20 at 19:29

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.


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