# 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, $$$$\{s_i , b_j \} = 0$$$$ and hence the $$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 $$$$\textrm{tr}_B [ s_i b_i s_j b_j (\rho \otimes \omega ) ],\ \textrm{tr}_B [ s_i b_i (\rho \otimes \omega ) s_j b_j ].\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$$$$ The first expression I know how to evaluate since $$$$\textrm{tr}_B [ s_i b_i s_j b_j (\rho \otimes \omega ) ] = - \textrm{tr}_B [ (s_i s_j) (b_i b_j) (\rho \otimes \omega ) ] = - (s_i s_j) \rho\ \textrm{tr}[ (b_i b_j) \omega ],$$$$ where in the second equality I used that $$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 $[s_i,b_j]=0$. Commented May 12, 2020 at 13:56
• My first equation is the defining property of fermions. Harmonic oscillator modes are bosons, hence they commute. Commented May 12, 2020 at 14:39
• Is it even clear what a partial trace means? Commented May 14, 2020 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. Commented May 14, 2020 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)? Commented May 14, 2020 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.

• Interesting trick. Is this the same Klein operator that is used in bosonization? Roughly there one constructs a boson-like operator out of the fermionic particle density operator (which has thererefore two fermions). But then one needs an operator to change the number of bosons. This is achieved by the Klein operator.
– lcv
Commented Oct 14, 2022 at 17:10
• @lcv I'm not sure. The name Klein might be associated with some related-but-different things (physics.stackexchange.com/q/442145), and I haven't spent enough time studying bosonization to know which differences are real and which are only superficial. Commented Oct 16, 2022 at 1:06