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!


1 Answer 1


From the point of view of Wick's theorem, this should be obvious:

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.

  • $\begingroup$ 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. $\endgroup$ Commented Apr 22, 2021 at 11:00
  • 2
    $\begingroup$ @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. $\endgroup$ Commented Apr 22, 2021 at 11:38
  • $\begingroup$ Thank you so much for your help again! You are a great teacher. $\endgroup$ Commented Apr 22, 2021 at 15:35
  • $\begingroup$ I’m going to add that to your answer, if you don’t mind, in case the comment gets lost in the future for other learners. $\endgroup$ Commented Apr 22, 2021 at 15:36

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