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$\underline{\textbf{Model:}}$

Let we have the $t-V$ model for spinless fermions on a 1D lattice, which is defined in second quantization operators as follows:

$$H_1 = -t\sum_i \big(c_i^\dagger c_{i+1} + H.C.\big) + V\sum_i n_i n_{i+1} \qquad (eq.1)$$ here $c_i^\dagger (c_i)$ are creation (annahilation) operators and $n_i$ is number operator.

$\underline{\textbf{JW transformation:}}$

$$c_i^\dagger \rightarrow S_i^+$$ $$c_i \rightarrow S_i^-$$ $$n_i \rightarrow S_i^z + \frac{1}{2}$$ where $S_i$ are spin-1/2 operators. (eq.1) become

$$H_2 = -t\sum_i \big(S_i^+ S_{i+1}^- + H.C.\big) + V\sum_i \big(S_i^z S_{i+1}^z + S_i^z\big) +\frac{NV}{4}\qquad (eq.2)$$ where $N$ is total number of spins.

Ground state energy of $(eq.1)$ can be found at any $N/L$-filling ($N=$ total number of particles, $L=$ total number of sites) using exact diagonalization (as explained here).

$\underline{\textbf{Question:}}$

$(eq.2)$ can also be numerically diagonalized (for small system size). But how can I find its ground state energy at some specific filling, let's say half-filling?

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  • $\begingroup$ That looks like an XXZ model in a magnetic field, for which the solution is known, but hardly trivial - though some limits are rather approachable. The filling only seems to appear in the constant term, so why are you worried about the filling specifically? $\endgroup$
    – Anyon
    Commented Jan 26, 2019 at 21:27
  • $\begingroup$ @Anyon Actually, I am trying to find ground state energy of $t-V$ model at half-filling using traditional DMRG. I approached this problem by converting fermionic operators ($c, c^\dagger)$ into spin operators, as shown above in (eq.2). My results are correct apart from the last term of eq.2 (i verified it against A. Langari's PRB 1998). I am concerned about the filling factor because I can't understand what does filling factor mean in the language of spins. I find ground state energy of [$eq.2 - last\;term$] and then add [$number\;of\;sites*V/4$] in it. But it gives me wrong results. $\endgroup$
    – Luqman Saleem
    Commented Jan 28, 2019 at 12:49

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