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Hubbard model within mean-field: three different approaches

Physics Asked by Sr Incerteza on January 7, 2021

While reading doi:10.1016/j.carbon.2012.03.009 , the authors mention three types of Hubbard models within mean-field approximation. The first one describes the electron-electron interaction, and to my understanding is the standard way of writting up the model and reads (only interaction term):

$H_{int}=Usum_{i,sigma} n_{isigma}langle{n_{i-sigma}}rangle$

The second version seems to describe electron-hole interaction and reads:

$H_{int}=Usum_{i,sigma}(langle{n_{i-sigma}}rangle-frac{1}{2})n_{isigma}$

And a third one seems to describe moment-moment interaction and reads:

$H_{int}=frac{U}{2}sum_{i} n_{i}langle{n_i}rangle-Usum_i 2m_ilangle{m_i}rangle$

where $n_i=n_{iuparrow}+n_{idownarrow}$ and $m_i=frac{1}{2}(n_{iuparrow}-n_{idownarrow})$.

My question is how can I see that the two last ‘versions’, describe what they are supposed to. To my understanding, all of them are the same version except for a shift on the Fermi level for case 2 which shifts the half-filling below $E=0$, as opposed to $E=U/2$. Case 1 and 3 are the same.

One Answer

To me, the Hubbard interaction per site is defined as $H_{int}=U n_{uparrow} n_{downarrow}$. (I suppressed the $i$, and also there is a factor of two because of your spin sum.)

Then, the mean-field approximation is defined as $n_{uparrow} n_{downarrow}to n_{uparrow} langle n_{downarrow}rangle+n_{downarrow} langle n_{uparrow}rangle-langle n_{uparrow}rangle langle n_{downarrow}rangle$.

So your version (i) lacks the constant term.

Per site, you have two operators, $n_{uparrow}$ and $n_{downarrow}$, i.e. the occupancies for each species of electrons, which, of course, you can trade against total charge, $n=n_{uparrow}+n_{downarrow}$ and moment $m=n_{uparrow}-n_{downarrow}$ (beware, again a factor of two with your definition). This makes you end up with version (iii).

For version (ii), you need a so-called electron-hole transformation, i.e. for one spin species, say $downarrow$, you replace the electron destruction operator by a hole creation operator $c_{downarrow}to a^{+}_{downarrow}$, and vice versa. (The $a$-operators satisfy the same fermionic algebra as the original $c$-operators.) So $n_{downarrow}=c^{+}_{downarrow}c_{downarrow}to a_{downarrow}a^{+}_{downarrow} =1-a^{+}_{downarrow}a_{downarrow}$. This last guy, $a^{+}_{downarrow}a_{downarrow}$, you call it $n_{downarrow}$ again, but remember, it counts holes now. And you end up with the electron-hole interaction in (ii).

Answered by Stesh on January 7, 2021

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