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I'm new to Hubbard model and I have a few questions about it. From the sources I can find on the internet, Fermionic-Hubbard model is often written as (please correct me if I'm wrong!) \begin{align} H &= H_{hop} + H_{int}+ H_{chem} \\ &=\sum_{i.j\sigma} t_{ij}\, (c_{i,\sigma}^{\dagger}c_{j,\sigma}+H.c.) \,+\, \sum_i U_i \,n_{i,\uparrow}n_{i,\downarrow} \,-\, \sum_i\mu\,(n_{i,\uparrow}+n_{i,\downarrow}). \end{align}

My first question is , why is the hermitian conjugate term in $$H_{hop}$$ sometimes omitted?

The second question is, why do we need the chemical potential term in this model but not in others let's say Hartree-Fock Hamiltonian?

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    $\begingroup$ I would suppose that the H.C term is the same as the other term in the first bracket upon the exchange of i and j which is allowed since t_ij is symmetric (probability of hopping from i to j is same as hopping from j to i) and so you could omit the H.C term and add a factor of 2 which can then be absorbed into the coefficient t_ij $\endgroup$
    – ravjotsk
    Jul 14, 2018 at 13:52

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The Hermitian conjugate term is never omitted! It might be accounted within $t_{ij}$ to avoid double counting. The chemical potential is needed to set the electron filling over the lattice. For a translation invariant system at half-filling, the chemical potential($\mu$) and the Hartree correction($\phi$) cancel out. As a result, the effective chemical potential, $\phi - \mu$ goes to zero, and you can omit that term. The simplest example of this is the mean-field treatment of the square lattice Hubbard model at half-filling.

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  • $\begingroup$ What is the meaning of the h.c. part? why do we need take into account the hopping from j to i AND i to j.. isn't it like double counting? $\endgroup$
    – relaxon
    Jan 7, 2022 at 19:38

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