So we start with the following hamiltonian describing non-interacting free fermions:

$$ \hat{H}_{\text{free}} = \sum_{i,j,\sigma}\tilde{t}_{ij} \hat{c}_{i\sigma}^\dagger\hat{c}_{j\sigma}.$$

Then we make the following approximation of nearest neighbors:

$$t_{ij}=\begin{cases} -t, & \text{$i$ and $j$ are nearest neighbors} \\ 0, & \text{otherwise} \end{cases}, $$ so we obtain the tight-binding Hamiltonian $$ \hat{H}_{\text{tb}} =-t \sum_{\langle ij\rangle,\sigma} (\hat{c}_{i\sigma}^\dagger\hat{c}_{j\sigma} +\hat{c}_{j\sigma}^\dagger\hat{c}_{i\sigma}). \tag{Bravais lattice}$$

I don't get why we need to add the complex conjugate. If I'm summing over all nearest neighbors.


1 Answer 1


The symbol $<ij>$ means the bond between site $i$ and site $j$. Thus $\sum_{<ij>}$ sums over each bond once. Then one can hop from $i$ to $j$ or from $j$ to $i$. No double counting therefore.

  • $\begingroup$ Thanks! That makes complete sense :) $\endgroup$
    – Caterina
    Jul 31, 2019 at 21:16

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