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?


1 Answer 1


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?

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.