# Dispersion relation in tight binding model with even indices only

Given a tight binding model with Hamiltonian

$H= \sum_{i(even)}t[c_{i+1}^\dagger c_i+h.c]$

containing even indices only, how can we find out the dispersion relation?

Attempt:

My guess is that the dispersion relation is just $\frac{-2tcos(ka)}{2}$ i.e. half the regular tight binding model. I could not find the source for dispersion relation with two different hopping amplitudes. If I had, I would just replace one with zero.

Assuming periodic boundary conditions (PBC):

\begin{align} \hat H =& t\sum_{i(\text{even})}^N \left(a_{i+1}^\dagger a_i + \text{h.c.}\right)= t\sum_{j=1}^{N/2} \left(a_{2j+1}^\dagger a_{2j} + \text{h.c.}\right)=\\ =&t\sum_{j=1}^{N/2}\left[ \left(\frac{1}{\sqrt{N}}\sum_{\bf k}e^{-i{\bf k}(2j+1)}a_{\bf k}^\dagger\right) \left(\frac{1}{\sqrt{N}}\sum_{\bf p}e^{i{\bf p} 2j} a_{\bf p}\right) + \text{h.c.}\right]=\\ =& t \sum_{\bf k,\bf p}\left[ a_{\bf k}^\dagger a_{\bf p} e^{-i\bf k}\frac{1}{N}\sum_{j=1}^{N/2} e^{-i({\bf k-p})2j}+\text{h.c}\right]. \end{align}

Here variables $${\bf k,p}=\frac{2\pi}{N}n$$, where $$n\in\{0,1,\dots,N-1\}$$. Now let's focus on "wierd sum", but this is just a geometric series:

$$\frac{1}{N}\sum_{j=1}^{N/2} e^{-i({\bf k-p})2j} = \frac{1}{N}\sum_{j=1}^{N/2} \left(e^{-i({\bf k-p})2}\right)^j= \begin{cases} \tfrac{1}{2}, & \bf k=p;\\[1ex] \frac{\displaystyle 1-e^{-2i({\bf k-p})N/2}}{\displaystyle 1-e^{-2i({\bf k-p})}}=0, & \bf k\neq p. \end{cases}$$

To summarize:

$$\frac{1}{N}\sum_{j=1}^{N/2} e^{-i({\bf k-p})2j} = \tfrac{1}{2} \delta_{\bf kp},$$

where $$\delta_{\bf kp}$$ is Kronecker delta. Now come back to Hamiltonian:

$$\hat H = t \sum_{\bf k,\bf p}\left[ a_{\bf k}^\dagger a_{\bf p} e^{-i\bf k}\tfrac{1}{2}\delta_{\bf kp}+\text{h.c}\right]=\frac{t}{2}\sum_{\bf k} a_{\bf k}^\dagger a_{\bf k}\left(e^{-i\bf k}+e^{+i\bf k}\right) = t\sum_{\bf k} a_{\bf k}^\dagger a_{\bf k}\cos \bf k.$$