The Hubbard model is a model to describe electrons in a lattice. In general, the Hubbard model Hamiltonian $H$ contains two terms:

  1. The kinetic term:$$T=-t\sum_{\langle ij\rangle\sigma} [c_{i\sigma}^\dagger c_{j\sigma} + h.c.] $$
  2. The onsite Coulomb interaction term:

$$U=u\sum_{i=1}^N n_{i\uparrow}n_{i\downarrow}$$

So my question is: why can we not diagonalize the Hamiltonian: $H=T+U$? Some books attribute the reason that $T$ doesn't commute with $U$, therefore we need to formulate a perturbation theory. In particular, I want to know whether the space of solution of $T$ has the same dimension compared to the space of solution of $U$ due to $[T,U] \neq 0$?


For $T$ operator we have the following eigenequation: $$T|n\rangle=T_n|n\rangle$$ For $U$ operator we have another one eigenequation: $$U|\alpha\rangle=U_\alpha|\alpha\rangle$$ Due to $[T,U] \neq 0$, I am wandering whether $|n\rangle$ has the same dimension compared to $|\alpha\rangle$? And why $$[T+U]|N\rangle \overset{?}{=} H_N|N\rangle.$$

  • $\begingroup$ What do you mean by "the space of solution of $T$"? $\endgroup$ Feb 2, 2018 at 9:44
  • $\begingroup$ $T$-matrix's dimension. $\endgroup$
    – Jack
    Feb 2, 2018 at 9:47
  • $\begingroup$ what is $T$ solution of? $\endgroup$ Feb 2, 2018 at 9:48
  • 1
    $\begingroup$ For you to be able to add $T$ and $U$ (as in $H=T+U$) or multiply them together (as in $[T,U]$) they need to have the same dimension. That much is obvious - why not remove that part and focus on the interesting question at the start? $\endgroup$ Feb 2, 2018 at 9:57
  • $\begingroup$ @ ZeroTheHero I have edited my question for clarity. $\endgroup$
    – Jack
    Feb 2, 2018 at 9:58

1 Answer 1


As for the question of dimensionality: Yes, the two sets of solutions for $T$ and $H$ have the same dimensions (Because they act on the same Hilbert space).

The pedestrian reason why you cannot solve the Hubbard model analytically is that the $U$ term contains quartic interactions ($c^\dagger c^\dagger c c$), and it is only in special cases that these can be diagonalized exactly.

  • $\begingroup$ So the difficulty is that $U|\alpha\rangle = U_\alpha |\alpha\rangle?$ $\endgroup$
    – Jack
    Feb 2, 2018 at 10:07
  • $\begingroup$ Yes, I suppose you could formulate it like that. $\endgroup$ Feb 3, 2018 at 14:17

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.