# Diagonalization of the Fermionic Hopping Model

I am studying the fermionic hopping model on a one-dimensional chain with $N$ sites and lattice constant $1$: $$H=\sum_\ell c_{\ell}^\dagger c_{\ell+1}+c_{\ell+1}^\dagger c_\ell.$$ This is usually done by diagonalizing $H$ with the help of a Fourier transform. In the course of the calculation, the identity $$\frac{1}{N}\sum_\ell e^{i\ell(k-k^\prime)}=\delta_{k,k^\prime}$$ is used. I was now wondering where this identity comes from. I followed the computation of Scalettar where the identity is listed on page 11.

The identity you are talking about is : $$\frac{1}{N}\sum_{l=1}^{N}e_{}^{il(k-k')}=\delta_{kk'}^{}$$ with $k=\frac{2\pi n}{N}$ and $k'=\frac{2\pi n'}{N}$for some $n,n' \in \{1,\dots,N\}$. To see it note : $$\sum_{k=0}^{N-1}x^{k}_{}=\frac{1-x_{}^{N}}{1-x}.$$ Use it to proove.
$\textbf{Proof :}$
For $k=k'$ : Clearly $\frac{1}{N}\sum_{l=1}^{N}e_{}^{il(k-k')}=\frac{1}{N}\sum_{l=1}^{N} 1 =1.$
For $k \neq k'$ : $\frac{1}{N}\sum_{l=1}^{N}e_{}^{il(k-k')}=\frac{e^{i\frac{2\pi}{N}(n-n')}}{N}\sum_{l=0}^{N-1} e^{il\frac{2\pi}{N}(n-n')} = \frac{e^{i\frac{2\pi}{N}(n-n')}}{N}\frac{1-e^{i2\pi(n-n')}}{1-e^{i\frac{2\pi}{N}(n-n')}}=\frac{e^{i\frac{2\pi}{N}(n-n')}}{N}\frac{1-1}{1-e^{i\frac{2\pi}{N}(n-n')}}=0.$
Hence :$$\mathbf{\frac{1}{N}\sum_{l=1}^{N}e_{}^{il(k-k')}=\delta_{kk'}^{}}$$