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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 $[T,U] \neq 0$?

Edit:

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 $$[T+U]|N\rangle \overset{?}{=} H_N|N\rangle.$$

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  • $\begingroup$ What do you mean by "the space of solution of $T$"? $\endgroup$ – Emilio Pisanty Feb 2 '18 at 9:44
  • $\begingroup$ $T$-matrix's dimension. $\endgroup$ – Jack Feb 2 '18 at 9:47
  • $\begingroup$ what is $T$ solution of? $\endgroup$ – ZeroTheHero Feb 2 '18 at 9:48
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    $\begingroup$ 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? $\endgroup$ – Emilio Pisanty Feb 2 '18 at 9:57
  • $\begingroup$ @ ZeroTheHero I have edited my question for clarity. $\endgroup$ – Jack Feb 2 '18 at 9:58
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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.

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  • $\begingroup$ So the difficulty is that $U|\alpha\rangle = U_\alpha |\alpha\rangle?$ $\endgroup$ – Jack Feb 2 '18 at 10:07
  • $\begingroup$ Yes, I suppose you could formulate it like that. $\endgroup$ – Mikael Fremling Feb 3 '18 at 14:17

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