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


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?

  • $\begingroup$ Consider to spell out acronyms. $\endgroup$
    – Qmechanic
    Commented Jun 9, 2018 at 6:32

1 Answer 1


Your notation is slightly different from: Asboth, where in Chapter 1 they cover the winding number of the SSH model. You can look there for help. But I think I know the answer to your problem.

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.

  • $\begingroup$ @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? $\endgroup$ Commented Jun 9, 2018 at 19:40

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