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?
  • $\begingroup$ I think one needs brackets in the last term, since otherwise the Hamiltonian is non-Hermitian. $\endgroup$
    – Roger V.
    Oct 12, 2020 at 14:05

1 Answer 1


The statement in question is made in a specific context - that of electrons in condensed matter, which interact via Coulomb interaction, which is quartic in fermion operators.

Physically, a quadratic Hamiltonian does not necessarily correspond to non-interacting particles: e.g., bosonization approach reduces Hamiltonians to interactions to quadratic ones. The special thing about quadratic Hamiltonians is that they are exactly diagonalizable, which if the reason why, for the clarity of reasoning, one often talks of them as a non-interacting case, as opposed to more complex/unsolvable interacting Hamiltonian.

Finally, the fact that two electrons cannot occupy the same site is due to the Pauli principle, which is usually not called an interaction.

  • 2
    $\begingroup$ Yes the key is the diagonalization using analytical techniques that can make cross-terms disappear. BTW I don’t think you mean “unsolvable” but rather “exactly solvable analytically in terms of simple functions” or something like this. You can always resort to numerical methods to get arbitrarily good numerical solutions. $\endgroup$ Oct 12, 2020 at 15:19
  • $\begingroup$ @ZeroTheHero agreed. $\endgroup$
    – Roger V.
    Oct 12, 2020 at 16:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.