# Can a general many-body Hamiltonian with quadratic and biquadratic terms be diagonalized?

Can an arbitrary many-body hamiltonian in second quantization form with quadratic and biquadratic terms $$H=\sum_{v_1,v_2} \alpha_{v_1 v_2}\ c_{v_1}^{\dagger}c_{v_2}+ \sum_{v_1,v_2,v_3,v_4}\beta_{v_1 v_2 v_3 v_4}\ c_{v_1}^{\dagger}c_{v_2}^{\dagger}c_{v_3}c_{v_4}$$ be diagonalized into $$H=\sum_{u} \epsilon_u c_{u}^{\dagger}c_{u}$$ ?

No.

I assume you talk about fermions (not that is matters much -- a similar argument would work for bosons).

Consider the 2-fermion Hamiltonian $$H=E_{00}+(E_{10}-E_{00})c_1^\dagger c_1 + (E_{01}-E_{00})c_2^\dagger c_2 + (E_{11}-E_{01}-E_{10}+E_{00}) c_1^\dagger c_1c_2^\dagger c_2\ .$$ This Hamiltonian has eigenstates which are Fock states, with four independent eigenvalues $$E_{00}$$, $$E_{01}$$, $$E_{10}$$, $$E_{11}$$.

On the other hand, the most general Hamiltonian of two non-interacting fermions, $$H'=\epsilon_0 + \epsilon_1 d_1^\dagger d_1 + \epsilon_2 d_1^\dagger d_1 \ ,$$ has only three independent parameters, and thus only three independent eigenvalues. It therefore can by no means reproduce the energy spectrum of a general Hamiltonian $$H$$ above.

(If you want to see this concretely, choose e.g. $$E_{00}=E_{01}=E_{10}=0$$, $$E_{11}=1$$.)

• Get it. Actually I was thinking about that with a fixed particle number N, the many-body Hamiltonian should be diagonalizable. Now I realize being diagonalizable with a fixed N doesn't mean being able to write H into quadratic terms. Because quadratic H cannot necessarily CONNECT N particle's energy spectrum with N+1 particle's energy spectrum. – norman Dec 23 '18 at 1:28
• You can do the same with 3 or more modes, where you project onto a fixed particle number: You will observe the same. The point is that for free fermions, the n-fermion energies are always sums of n single-particle energies, which is a severe constraint. – Norbert Schuch Dec 23 '18 at 1:31