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The particle hole transformation for a bipartite lattice $\Lambda$ (with sublattices $A$ and $B$) can be written as $$U^\dagger c_{i,\uparrow} U = \epsilon(i) c^\dagger_{i\uparrow} \\ U^\dagger c_{i,\downarrow} U = \epsilon(i) c^\dagger_{i\downarrow}$$
The factor $\epsilon(i)$ is $+1$ for the lattice index $i\in A$ and $-1$ for $i \in B$. The $c$ operators are the usual creation and annihilation operators for electrons. Now, if I have a Hamiltonian which has a term like $$H_s = \kappa \sum_i (n_{i,\uparrow} - n_{i,\downarrow})$$ how do I know how this transforms? My guess was that I only have to substitute all the $c_{i\downarrow}$ with $c^\dagger_{i\downarrow}$ and all the $c_{i\uparrow}$ with $c^\dagger_{i\uparrow}$ (and vice versa). This would give me an overall minus sign in this part of the Hamiltonian, so that $$H_s' = -\kappa \sum_i (n_{i,\uparrow} - n_{i,\downarrow})$$ However, this question made be believe that I also have to flip the spins, so that $c_{i\downarrow} \rightarrow c^\dagger_{i\uparrow}$, $c_{i\uparrow} \rightarrow c^\dagger_{i\downarrow}$, $c^\dagger_{i\downarrow} \rightarrow c_{i\uparrow}$ and $c^\dagger_{i\uparrow} \rightarrow c_{i\downarrow}$.

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  • $\begingroup$ The Hamiltonian $H_s$ you considered is not particle-hole symmetric, that is why it is not invariant under the transformation. You need to redefine the particle-hole transformation with spin flipping in order to make the transformation a symmetry of the Hamiltonian. $\endgroup$ – Everett You Oct 29 '15 at 23:22
  • $\begingroup$ Is there more than one particle-hole transformation and why am I allowed to just define one with spin flip? Is this still a unitary transformation? $\endgroup$ – Merlin1896 Oct 30 '15 at 0:41
  • $\begingroup$ There are infinitely many particle-hole transformations, all related by unitary transforms. Each particle-hole transformation followed by a nontrivial unitary transform is another particle-hole. Now spin flip itself is a unitary transform. Because a unitary transform followed by another unitary transform will still be unitary, spin flip does not change the unitary nature of particle-hole transform. $\endgroup$ – Everett You Oct 30 '15 at 19:49

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