Gauss law with staggered fermions

Question

I was wondering if someone could explain how to derive the discrete version of Gauss law in 1+1 QED using staggered Fermions. The result I am trying to reproduce is found in multiple references [see for instance this paper, Appendix F, paragraph below Eq. (F2)]. Gauss' law in the staggered representation takes the form $$E_{j, j+1} - E_{j-1, j} = \Phi^\dagger_j \Phi_j - \frac{1 - (-1)^j}{2} \ .$$ The LHS seems clear; one simply approximates the derivative with a central finite difference, $\partial E(x) \rightarrow \frac{1}{a} (E_{j, j+1} - E_{j-1, j})$.

If I take the staggered fields defined on even and odd lattice sites such that $\Phi_{2j} := \sqrt{a}\psi_{e^-}(x_{2j})$ and $\Phi_{2j+1} := \sqrt{a}\psi^\dagger_{e^+}(x_{2j+1})$, respectively, then, using anticommutation relations $\{\psi_i, \psi_j^\dagger\}=\delta_{ij}$, I get for the RHS of Gauss' law $$\psi^\dagger_j \psi_j = \begin{cases} \Phi^\dagger_j \Phi_j & j \ \text{even} \\ \Phi_j \Phi^\dagger_j & j \ \text{odd} \end{cases} = \begin{cases} \Phi^\dagger_j \Phi_j \\ - \Phi^\dagger_j \Phi_j + 1 \end{cases} = (-1)^j \Phi^\dagger_j \Phi_j + \frac{1 - (-1)^j}{2} \ .$$

Can anyone tell me what I do wrong?

Answer 1

I think that in this article it is well explained why the Gauss law has that expression.

Let's start from a state with no particles, and consider for simplicity a $1$ dimensional lattice. If you add a fermion to the system at the site $j$ (that will be even since we consider a fermion), the difference in electric field values between the links connected to this site will be $$ E_{j,j+1} - E_{j-1,j} = 1 $$

On the other hand, if you consider adding to the system an anti-fermion, and so an odd site, the difference will be $$ E_{j,j+1} - E_{j-1,j} = -1 $$

These relations come from the fact that if the incoming electric field value is $E_{j-1,j}$, the outgoing value will be $E_{j-1,j}$ plus the charge on the site, as in the classical analog of the Gauss law. Putting everything together we have $$ E_{j,j+1} - E_{j-1,j} = (-1)^{j}\psi_j^{\dagger}\psi_j $$

Now, following your derivation, you have $$ \psi_j^{\dagger}\psi_j = (-1)^j \phi^{\dagger}_j\phi_j + \frac{1}{2}[1-(-1^j)] $$

and substituting you get the result you were looking for: $$ E_{j,j+1} - E_{j-1,j} = \phi^{\dagger}_j\phi_j - \frac{1}{2}[1-(-1^j)] $$