Consider a system of free fermions with Hamiltonian $$ H = \sum_{ij} t_{ij}c^{\dagger}_ic_j\quad \longrightarrow \quad H = \sum_k E_k d^{\dagger}_kd_k. $$ An eigenstate $|\psi \rangle$ of $H$ is given by acting on $|0\rangle$ with $d^{\dagger}$s as usual: $$ |\psi^{N_p}\rangle = \prod_{a \in N_p} d_a | 0 \rangle \quad \text{where} \quad d_a = \sum_i\phi^a_ic_i, $$ for some suitable $\phi$s. As shown by Sun Woo Kim in my previous question, any eigenstate is fully characterised (all its correlators can be obtained) from the 2-point correlator matrix $C$ given by $$ C_{ij} = \langle \psi | c^{\dagger}_i c_j |\psi\rangle \equiv \langle c^{\dagger}_i c_j \rangle_{\psi}. $$ If I have a superposition of eigenstates, $$ |\xi\rangle =\sum_n a_n |\psi_n\rangle, $$ does it also have a 2-point correlator that fully characterises $|\xi\rangle $?

I am inclined to think not because if I try the brute force approach $$ C_{ij}\doteq \sum_na_n\sum_{k}\overline{\phi^k_i}\phi^k_j\langle d^{\dagger}_k d_k \rangle_{\psi_n}=\sum_n a_n \langle c^{\dagger}_i c_j \rangle_{\psi_n} $$ then $$ \langle multi \rangle = \sum_n a_n \sum_{\mathcal{i,j,.. \in P}} \langle c^{\dagger}_i c_j\rangle_{\psi_n}\cdots = \sum_{\mathcal{i,j,.. \in P}} \sum_n a_n \langle c^{\dagger}_i c_j\rangle_{\psi_n}\cdots \neq \sum_{\mathcal{i,j,.. \in P}} C_{ij}\cdots $$ as I would need extra $a_n$'s to absorb them into $C_{ij}$'s.

(This is perhaps also related to my previous question of whether or not a sum of gaussian matrices is gaussian)


2 Answers 2


No - coherent states form an (overcomplete) basis for the (infinite-dimensional) Hilbert space, and otherwise, any state could be characterized by its second moments.

More generally, there is a 1-to-1 correspondence between physically admissible 2-point correlation matrices and Gaussian states. Thus, for any non-Gaussian state which can be characterized from its 2-point corrlations, the way in which this is done must be different from what it is for Gaussian states (i.e., Wick's theorem). (Of course, for any given state there is such a relation - the simplest one being "Ignore the 2-point correlations, this is the state.")

  • $\begingroup$ As always, thank you very much for all your help! I guess I will have to explore other routes... $\endgroup$ Commented Apr 21, 2021 at 18:51
  • $\begingroup$ @FriendlyLagrangian Well, what are you trying to achieve? $\endgroup$ Commented Apr 21, 2021 at 18:58
  • $\begingroup$ As you probably know, there is a very effective way of calculating the entanglement negativity of a Gaussian state. But judging from my questions and your answers the state I'm interested in is not gaussian... Tbf that's the only feasable route of computing the negativity of a state that I know. $\endgroup$ Commented Apr 21, 2021 at 20:12
  • $\begingroup$ @FriendlyLagrangian Why don't you ask what you actually want to know? I.e.: I have the state X and want to know its entanglement negativity? $\endgroup$ Commented Apr 21, 2021 at 20:24
  • $\begingroup$ Apologies, you are right, I didn't because I don't believe it has been done before so I assumed that asking the question directly was pointless so I just asked the missing bits of own my ongoing calculation based on close things in the literature (which turned out to be a dead end, of course..). I might ask the question in a new post. $\endgroup$ Commented Apr 21, 2021 at 22:03

You only get everything form the two point correlator when the wavefunction can be written as a single Slater determinant, or equivalently as a single tensor product of one-particle states. The general state is a sum of many tensor product states.

The conditions that an $n$-fermion state can be written a single tensor product are the Plucker relations.

  • $\begingroup$ So are you saying that if one can find some new fermionic operators whose particle states reproduce $|\xi\rangle$ then the state is Gaussian? $\endgroup$ Commented Apr 21, 2021 at 18:48

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