# If $|\psi\rangle$ is a free fermionic state, why does its reduced density matrix $\text{Tr}_C(|\psi\rangle \langle \psi|)$ also obey Wick's theorem?

I have recently been trying to understand this paper. So far I understand why, given a free fermionic state $$|\psi\rangle$$, it is fully characterised by its 2-point correlation matrix (i.e. obeys Wicks’s theorem). I also understand why if its reduced density matrix $$\rho = \text{Tr}_C(|\psi\rangle\langle \psi|)$$ obeys Wick’s theorem, that is $$C_{ij} = \text{Tr}(\rho c^{\dagger}_ic_j)$$ fully characterises $$\rho$$, then $$\rho$$ must be gaussian (or the exponential of a free fermionic Hamiltonian, i.e. $$\rho \sim \exp(-\sum_{ij}h_{ij}c^{\dagger}_ic_j)$$ ).

But why does $$\rho$$ in the first place need to obey Wick’s theorem?

I was thinking about it and all I could come up with was that $$\rho \sim \sum_n a_n |\tilde\psi_n\rangle \langle \tilde\psi_n|$$ where $$|\tilde \psi_n\rangle$$ is the part of eigenstate $$|\psi_n\rangle$$ not in subsystem $$C$$. However, from my previous questions and specially with the help of @NorbertSchuch, I have gathered that in general that form for $$\rho$$ is not gaussian $$\iff$$ doesn’t obey Wick’s theorem. So I’m certainly missing something!

If you can express any $$N$$-point correlator of operators acting on part A of the system through the corresponding two-point correlators, then this does not depend on the fact whether you trace the other part B of the system or not - that's precisely the point of the partial trace, it describes the same A part of the system (just without requiring to talk about B).
And as we have established previously, a state is fully specified by all its $$N$$-point correlation functions.
• Thank you for your help again :). Wouldn’t that imply that the partial trace and the expectation $\langle \cdot \rangle$ commute in a way? I see that on one hand you can have Wick’s theorem just on $A$ by simply restricting the correlators to be in $A$. So you can have expressions like $\langle \text{ops in$A$} \rangle_{\psi} = \text{Wick in$A$}$. But I dont see how you can infiltrate $\text{Tr}_B$ in the previous expression, they are just $\mathbb{C}$-numbers. Apr 22 at 11:00
• @FriendlyLagrangian It is indeed true that computing expectation values of operators supported on A and tracing B commute. In (non-fermionic) language, $\mathrm{tr}[(O_A\otimes I_B)\ \cdot\ ] = \mathrm{tr}_A[O_A\,\mathrm{tr_B}[\,\cdot\ ]]$. --- But I would argue that this is the definition of the partial trace & reduced density matrices (that's in fact how I introduce it in my lecture): They provide the minimum necessary information to model any measurement on A only. Apr 22 at 11:38