# Winding number of SSH model 3

SSH model can be written as $$H=-\sum_n\big[Jc_n^\dagger d_n + J'd_n^\dagger c_{n+1}\big]+h.c.$$ in Fourier space $$H(k)= \begin{bmatrix} c_k^\dagger && d_{k}^\dagger \end{bmatrix} \begin{bmatrix} 0 && -Je^{ik}-J'e^{-ik}\\ -Je^{-ik}-J'e^{ik} && 0 \end{bmatrix} \begin{bmatrix} c_k \\ d_{k} \end{bmatrix}\\$$ $$h(k)=h_1\sigma_1+h_2\sigma_2+h_3\sigma_3$$ with $h_1=-(J+J')\cos{k}$,$h_2=-(J-J')\sin{k}$ and $h_3=0$

Winding number is defined as: $$\nu=\frac{1}{2\pi}\int_{-\pi}^\pi dk \frac{h_1\partial_kh_2-h_2\partial_kh_1}{h_1^2+h_2^2}$$

Question:

It is known that winding number is $1$ for $J'>J$ and $0$ for $J>J'$ but when I calculated using above equation I found $-1$ and $1$ respectively. What am I missing?

• Consider to spell out acronyms. – Qmechanic Jun 9 '18 at 6:32

I believe your fourier transformation is incorrect. It does not agree with what is found in Asboth on page 4. The thing is that the on-site potential term $$Jc_n^\dagger d_n$$ should not produce an exponential when you Fourier transform. Therefore I think that your Bloch Hamiltonian $h(k)$ whould look like this $$\left(\begin{array}{cc} 0 & -J-J'e^{-ik}\\ -J-J'e^{ik} & 0 \end{array}\right).$$ Try calcuating the winding number with this matrix and see if you get the right answer.
• @LuqmanSaleem I don't really understand what you mean, maybe you can expand. At any rate, if you insist that your $h(k)$ matrix is correct then answer me this. Does $n$ label different unit cells? – Jens Roderus Jun 9 '18 at 19:40