# Dispersion Relation and Eigenvectors of SSH Model in Tight Binding

Consider a one-dimensional chain of atoms as shown in the figure. Let the spacing between the atoms be $$a$$. Assume that the onsite energy is the same at each point and is equal to $$0$$ (without any loss of generality), but the hopping terms are of two types: $$w$$ denoted by a single bond and $$v$$ denoted by the double bond.

So, the hamiltonian would be like $$H = \sum_n^Nw(c_{An}^{\dagger}c_{Bn} + c_{Bn}^{\dagger}c_{An})+v(c_{Bn}^{\dagger}c_{A(n+1)} + c_{A(n+1)}^{\dagger}c_{Bn})$$ where $$A$$ and $$B$$ are the two interstitial sites and $$c^{\dagger}$$ and $$c$$ are the creation and annihilation operators respectively. If we transform them into their momentum space analogues $$a_k$$ and $$b_k$$ and take the spinor $$\psi_k = \begin{pmatrix} a_k\\ b_k \end{pmatrix}$$ we get $$H = \sum_k \psi_k^{\dagger} \begin{pmatrix} 0 & w+ve^{-ika} \\ w+ve^{ika} & 0 \end{pmatrix}\psi_k$$ We can diagonalize the Hamiltonian to get the dispersion relation as $$E(k) = \sqrt{v^2+w^2+2vwcos(ka)}$$.

I have a doubt in this. In this diagonalization, we consider only the matrix $$\begin{pmatrix} 0 & w+ve^{-ika} \\ w+ve^{ika} & 0 \end{pmatrix}$$ for our purpose, but the Hamiltonian involves a sum over $$\psi_k^{\dagger}$$ and $$\psi_k$$ as well. Also, if I go on to find the eigenvectors in order to get the Bloch spinors $$|u_{\pm}(k)>$$, do I use this matrix again? Can someone provide an answer as to why we don't use the entire explicit form of the Hamiltonian to get the dispersion relation or the Bloch spinors?

Well, what you are actually doing is a unitary transformation on the spinor $$\psi_k$$, i.e. a transformation that preserves the anticommutation relations of the fermionic operators. More precisely, you are mapping $$\psi_k \to \phi_k = P_k \psi_k$$, where $$P_k$$ is a unitary $$2\times 2$$ matrix such that $$P_k P^{\dagger}_k = 1$$, and this preserves the fermionic algebra, namely if the only non trivial anticommutator of the original fermions is $$\{\psi_{k,a} , \psi^{\dagger}_{k,b} \} = \delta_{a,b},$$ then the only non trivial anticommutator of the transformed fermions is $$\{\phi_{k,a} , \phi^{\dagger}_{k,b} \} = \delta_{a,b}$$ as a direct consequence of the unitarity of $$P_k$$. Now if you choose $$P_k$$ such that it diagonalizes the $$2\times 2$$ matrix $$h_k = \left( \begin{array}{cc} 0 & w+ve^{-ika}\\ w+ve^{ika} & 0 \end{array} \right),$$ i.e. such that $$P_k h_k P^{\dagger}_k = \lambda_k$$, where $$\lambda_k$$ is a diagonal matrix with entries $$\pm E(k)$$, you get a more transparent Hamiltonian for the transformed fermions: $$H = \sum_k \phi^{\dagger}_k \left( \begin{array}{cc} -E(k) & 0 \\ 0 & E(k) \end{array} \right) \phi_k.$$ Now you see that this Hamiltonian describes independent fermions for every momentum $$k$$ with two available energy levels $$\pm E(k)$$.
Finally yes, the Bloch spinors are just the two eigenvectors $$u_{\pm}(k)$$ corresponding to the two eigenvalues $$\pm E(k)$$.