# Converting SSH model into fourier-space

In SSH model hopping between two sites is staggered i.e. hopping strength between site-1 and site-2 is (let's say) $J$ and between site-2 and site-3 is $J'$ and between site-3 and site-4 is again $J$ and so on. SSH Hamiltonian can be written as: $$H=\sum_{n=1}^{L/2}(Jc_{2n-1}^\dagger c_{2n}+J'c_{2n}^\dagger c_{2n+1}+h.c.)$$ Where $c_i^\dagger$ and $c_i$ are creation and annihilation operators respectively and $L$ are total number of sites. I am trying to write this Hamiltonian in k-space. For that one can take fourier transform of creation and annihilation operators as following: $$c_n^\dagger=\frac{1}{\sqrt{L}}\sum_k c_{k}^{\dagger} e^{ikR_n} \\c_n=\frac{1}{\sqrt{L}}\sum_k c_{k}^{} e^{-ikR_n}$$ Where $R_n$ is position of $nth$ particle in 1D discrete real space. Using these transformations 1st term of Hamiltonian in k-space can be written as: $$H(k)_{1st}=J\sum_{k,k'}\sum_{n=1}^{L/2} \frac{e^{ik(R_{2n-1})}}{\sqrt{L}}\frac{e^{-ik'R_{2n}}}{\sqrt{L}}c_k^\dagger c_{k'}$$ As distance between two sites is equal so we can write it as: $$H(k)_{1st}=J\sum_{k,k'}\sum_{n=1}^{L/2}\frac{e^{iR_{2n}(k-k')}}{L}e^{ik}c_k^\dagger c_{k'}$$ Normally when we are taking equal hopping strength between two sites we have Hamiltonian of kind $H_{kin}=t\sum_{n=1}^L c_i^\dagger c_{i+1}$ which after transforming into k-space gives a term of kind $\sum_{n=1}^L e^{iR_n(k-k')}$ that we put equal to $L\delta_{k,k'}$ but here in my $H(k)_{1st}$ I have term which have $R_{2n}$ and summation is over $L/2$ instead of $R_{n}$ and $L$ respectively.
How can I solve this?

• Try thinking in terms of lumping two adjacent sites into a single unit cell. Will lead to simple hopping Hamiltonian in matrix form. – Sunyam Oct 6 '17 at 17:33
• Please define $R_n$. Without a definition, it's really hard to answer this question. – DanielSank Oct 6 '17 at 22:55
• $R_n$ is position of $nth$ particle in 1D discrete real space. – Luqman Saleem Oct 6 '17 at 23:01
• You can look at phyx.readthedocs.io/en/latest/TI/Lecture%20notes/1.html to find precisely same problem being treated. – Sunyam Oct 6 '17 at 23:21
• Are you sure about the second term? Isn't it $J'c_{2n}^{\dagger} c_{2n+1}$ ? – Matteo Oct 7 '17 at 17:20