# 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)?

• 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 $[s_i,b_j]=0$. – user2723984 May 12 '20 at 13:56
• My first equation is the defining property of fermions. Harmonic oscillator modes are bosons, hence they commute. – loewe May 12 '20 at 14:39
• Is it even clear what a partial trace means? – Norbert Schuch May 14 '20 at 14:52
• 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. – Vadim May 14 '20 at 16:51
• @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)? – 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.