Let's take a 1D Hubbard model $$H=-J\sum\limits_{i} \sum_\sigma (c_{i,\sigma}^{\dagger}c_{i+1,\sigma} +c_{i+1,\sigma}^{\dagger}c_{i,\sigma})+U\sum\limits_{i} n_{i,\uparrow} n_{i,\downarrow}$$ with $L$ lattice sites at quarter filling, i.e. the number of electrons is $N_e=L/2$.

If I calculate the two lowest energy eigenvalues of this system, I find, that their is a finite difference $\Delta = E_1-E_0$ which decreases with increasing $U$. For example, in a system with $L=100$ I get $\Delta = 0.04$ for $U=0$ and $\Delta = 0.01$ for $U=10$. The calculations were done using the DMRG.

Two questions:

1.) Why is there this small gap?

2.) Why does this gap close for large $U$?


The gap is actually zero :)

Indeed, for $U = 0$ the model can be easily diagonalized by a Fourier transform, in which case we see we are just partially filling a band spectrum, which is of course gapless. The $\Delta = 0.04$ you see is simply a finite size effect. For this special point ($U=0$) you can in fact analytically calculate this finite-size gap, but more generally it is useful (if not essential) to numerically actually compute the energy for difference system sizes and see how the gap scales in order to make a statement about the thermodynamic limit (in this case you would see $\Delta \sim \frac{1}{L}$).

This non-interacting point can then be perturbed by the Hubbard interactions. It is well-known that at half-filling this creates a charge gap (which is intuitive: $U$ demands that every site has exactly one fermion in it), but leaves the spin degrees of freedom to be gapless (indeed if $U$ is large enough, the low-energy effective theory is given by the Heisenberg anti-ferromagnet). At less-than-half-filling, no charge gap is created and so both the charge and spin sectors remain gapless! For more details, see for example the phase diagram on page 198 (pdf page 218) of The One-Dimensional Hubbard Model (links to Korepin's e-copy).


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