0
$\begingroup$

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 $$H=\sum_{u} \epsilon_u c_{u}^{\dagger}c_{u}$$ ?

$\endgroup$
0
$\begingroup$

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. $E_{00}=E_{01}=E_{10}=0$, $E_{11}=1$.)

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – norman Dec 23 '18 at 1:28
  • $\begingroup$ 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. $\endgroup$ – Norbert Schuch Dec 23 '18 at 1:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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