# Proof that the reduced density matrix of free fermions is thermal?

I found this question here but it was partly unanswered. The question remains, namely:

Given a free theory of fermions in a bi-partite system $$S=A\cup B$$ with Hamiltonian $$H = \sum_{ij} t_{ij}a^{\dagger}_ia_j\quad \longrightarrow \quad H = \sum_k E_k c^{\dagger}_kc_k$$ can anybody help me with proving that given any eigenstate $$|\psi\rangle$$ of $$H$$ with its associated density matrix $$\rho=|\psi\rangle\langle\psi|$$ of $$H$$, the reduced density matrix $$\rho_B=\text{tr}_A(\rho)$$ has a "thermal" form: $$\rho_B \sim \exp(-H_B) \quad \text{with} \quad H_B=\sum_i h_{ij}c^{\dagger}_ic_j.$$

The proof I’m reading sketches that since $$|\psi\rangle$$ is a determinant (?), then its correlators factorise $$\langle c_i^{\dagger}c_j^{\dagger}c_kc_l\rangle_{\psi}= \langle c_n^{\dagger}c_l\rangle_{\psi} \langle c_m^{\dagger}c_k\rangle_{\psi} − \langle c^{\dagger}_nc_k \rangle_{\psi}\langle c^{\dagger}_mc_l\rangle _{\psi}$$ ($$\langle \cdot \rangle_{\psi} := \langle \psi | \cdot | \psi \rangle$$), therefore $$C_{ij} := \text{Tr}(\rho_B c_i^{\dagger} c_j)$$ must factorise as well. According to Wick’s theorem, this property holds if (why? is this if an if and only if?) $$\rho$$ is the exponential of a free-fermion operator

$$\rho_B = K\exp(-H_B) \quad \text{with} \quad H_B=\sum_i h_{ij}c^{\dagger}_ic_j.$$ My problem is that this might not be the unique form of $$\rho_B$$. The above to me is a bit meaningless without the if and only if.

• There is a discrepancy in notation between the first and the second equation. Also, the question seems to lack essential information about how the reduced density matrix is defined, what is its relation to $\psi$ etc. The question just doesn't make sense in its current form. – Vadim Feb 19 at 10:56
• The iff part of the argument is what really is dubious. The "if" works according to their reference [8] that proves the variant of the Wick theorem (that represents the products as sum of two-particle correlation functions and smaller product) to compute $\mathrm{Tr}(\rho abc\ldots)$ for the thermal state – OON Feb 19 at 11:53
• If you have Wick's theorem for arbitrary (many-operator) correlations, the "iff" follows trivially since the state is fully specified by all such correlations; if they can be determined from the two-point correlations then this must be the Gaussian state, since the Gaussian state satisfies Wick's theorem and, as stated, the state is determined uniquely by all correlations. Does this answer the unclear direction? – Norbert Schuch Feb 24 at 20:32
• I'm saying: (1) If I know all n-point correlators, for any n, this uniquely determined the state. Is that part ok? If yes, then (2) Gaussian states satisfy Wick's theorem. Is that part ok? Then, from (2): Gaussian states satisfy Wick's theorem, which fixes all n-point correlators, and from (1): A state where I know all n-point correlators is determined uniquely, so => it must be the Gaussian states (as it satisfies the desired properties, and such a state is unique). – Norbert Schuch Feb 25 at 19:07
• Because a state is equivalent to all possible expectation values. States associate an expectation to every possible expectation value - nothing more, and nothing less. Thus, they are fully characterized by all possible expectation values. Anything else which is not specified by those would be entirely unphysical. – Norbert Schuch Feb 26 at 13:56

The answer assumes that you know that for Gaussian states, i.e. states of the form $$\exp(-H_B)$$, with $$H_B$$ quadratric in the fermionic operators, Wick's theorem holds, that is, $$2n$$-point correlators $$\langle f_1\cdots f_{2n}\rangle$$ can be expressed in terms of two-point correlators, where the $$f_i$$ are some fermionic operators (creation or annihiliation).
The question remains why this implies that a state which satisfies Wick's theorem must be a Gaussian state. The reason is that any quantum state is entirely determined by all its expectation values $$\langle O\rangle$$. For fermionic states, all admissible operators must have even parity, and thus, knowing all $$\langle f_1\cdots f_{2n}\rangle$$ means that you know all $$\langle O\rangle$$, and thus you know everything about the state: That is, the state is uniquely determined by all $$\langle f_1\cdots f_{2n}\rangle$$.
So if you have a state which satisfies Wick's theorem, you know it must be Gaussian, since there is a Gaussian state with the same $$\langle f_1\cdots f_{2n}\rangle$$, and those uniquely determine the state.