Consider a one-dimensional non-interacting Hubbard Hamiltonian:

$$ H=-t\sum_{\langle i, j \rangle}\sum_{\sigma=\uparrow,\downarrow} \left(c_{i,\sigma}^\dagger c_{j,\sigma} +\mathrm{h.c.} \right) $$ where the number of electrons is equal to the number of sites.

This Hamiltonian can be easily diagonalized thanks to momentum modes. One thus obtains two degenerate bands (one for spin-up electrons, the other for spin-down electrons). Filling the bands from the lowest possible energy, one obtains the following band diagram:

Band diagram

corresponding to a metallic solution. The local charge fluctuation $$ \langle n_i ^2 \rangle - \langle n_i \rangle^2 $$ is non-zero.

Now let us assume that Hamiltonian $H$ is supplemented by a magnetic-field term $+h\sum_i \left(n_{i,\uparrow}-n_{i,\downarrow}\right)$. In this case the two degenerate bands are no longer degenerate. The band corresponding to spin-up electrons rises, while the band corresponding to spin-down electrons is pushed downward. Suppose that the magnetic field $h$ is so strong that the lower band is completely filled while the upper band is completely empty (this is termed a band insulator). My question is: in the band-insulating state, are local charge fluctuations (see definition above) zero (as in a Mott insulator) or non-zero (as in a metal)?

Is this a general property of all band insulators?


1 Answer 1


When the band is completely filled, there is one electron on each site, i.e., $\langle n_{i,\sigma}\rangle=\langle n_{i,\sigma}^2\rangle=1$, and the charge fluctuations vanish. Same is true for the empty band. Note that this would also be true, if we include Coulomb interaction (e.g., as in the Hubbard model), although the condition under which the bands can be considered as completely empty and filled might be altered.


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