# A quadratic Hamiltonian is a model of independent particles - why?

I'm reading some notes on the Anderson Hamiltonian:

$$H=\sum h_i c_i^\dagger c_i -q\sum_{\langle i,j\rangle}(c_i^\dagger c_j+c_j^\dagger c_i)$$

Where the $$c_i/c_i^\dagger$$ are fermionic annihilation/creation (ladder) operators. The notes say

Since $$H$$ commutes with $$N=\sum_i c_i^\dagger c_i$$, it conserves the number of fermions, and since it is quadratic in the fermionic operators it is a model for the propagation of independent electrons in a random short ranged potential. We can thus restrict our analysis to the one fermion subspace $$\mathcal H_1$$.

By $$\mathcal H_1$$ the author means the span of the vectors $$|i\rangle:=\bigotimes _{j\neq i}|0\rangle_j\otimes|1\rangle_i$$.

I think the key to see why this is true is that with such a Hamiltonian all the eigenstate will be tensor products of one particle states. My questions are:

• How to see that the above statement is true? It is not obvious to me that any eigenstate will have the form $$\bigotimes_i |\psi_i\rangle$$ where $$|\psi_i\rangle\in \mathcal H_1$$.
• How can I see that this is a model of non interacting fermions? The second terms is a term involving multiple sites, if a fermion is already on a site, another fermion cannot hop on it, why is this not an interaction? Why is it true in general that Hamiltonians which are quadratic in the fermionic operators represent non interacting fermions?
• I think one needs brackets in the last term, since otherwise the Hamiltonian is non-Hermitian. – Vadim Oct 12 '20 at 14:05