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Say I have a one-dimensional lattice with lattice constant $a$. With next nearest neighbor hopping (NNN) included, the hopping term that describes such system would be

$$H_{hop} = -t\sum_j(\hat c_{j+1}^{\dagger}\hat c_j + H.c.) -t_2\sum_j(\hat c_{j+2}^{\dagger}\hat c_j +H.c.)$$

When we Fourier transform from lattice site into $k$-space, according to the equation

$\hat c_j= \sum_k \hat c_k\cdot e^{-i\vec k\cdot\vec R_j}$ ,

we get

$H_{hop}$ = $-\sum_k\hat c_{k}^{\dagger}\hat c_k(2t \cos(ka) -2t_2\cos(2ka))$ .

This is very simple for 1D lattice. But what if I want to do it in a 3D cubic lattice? Since electron can be hopping in three dimension, can I just sum them up and write,

for NN hoppping $\rightarrow -2t[\cos(k_xa)+\cos(k_ya)+\cos(k_za)]$,

for NNN hopping $\rightarrow -2t_2[\cos(2k_xa)+\cos(2k_ya)+\cos(2k_za)]$,

It is very straight forward, but I find a different result involving terms like $\cos(k_xa)\cos(k_ya)$ in Table 1.. Am I missing something here?

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1 Answer 1

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If you are at the origin, your nearest neighbors are at $(\pm a,0,0)$, $(0,\pm a,0)$, and $(0,0,\pm a)$. You are essentially claiming that your next nearest neighbors are at $(\pm 2a,0,0)$, $(0,\pm 2a,0)$, and $(0,0,\pm 2a)$. Are you sure you don't have any other neighbors nearer than that?

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  • $\begingroup$ In a simple cubic lattice, I assume that NNN hopping progress would be something like this. Starting at the origin, hops into (a,0,0) and creates a phonon there, then absorbs the phonon and hops into (2a,0,0). (a,a,0) might be the next nearest neighbor but I don't think such hopping would be possible, since the second hopping is triggered by absorbing the phonon created when the electron hops along x-axis in the first place. $\endgroup$
    – Iampotato
    Aug 7, 2019 at 3:27
  • $\begingroup$ The hopping doesn't a priori have anything to do with phonons. Yes, you can have phonon-mediated hopping but you could also ignore phonons entirely and have electrons hopping to nearest, next-nearest, NNN, etc. neighbors. $\endgroup$
    – d_b
    Aug 7, 2019 at 3:38
  • $\begingroup$ @Iampotato As d_b says, there is no concept of lattice vibrations here. Your Hamiltonian allows for hopping between nearest and next-nearest neighbors, which are defined to be quite literally the closest set of points which are further away than the nearest neighbors. $\endgroup$
    – J. Murray
    Aug 7, 2019 at 3:46
  • $\begingroup$ Ahh I get it now. Thank you so much! $\endgroup$
    – Iampotato
    Sep 9, 2019 at 23:24

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