1
$\begingroup$

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

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?

$\endgroup$
  • $\begingroup$ 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 Jan 26 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 Jan 28 at 12:49

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.