# Winding number in Su-Schrieffer-Heeger (SSH) model

I'm studying the SSH model from this review and on page 14, equation (1.38), they give a formula from evaluating the winding number saying it's easy to check it. Now, I've done the math and came up with: $$\begin{equation} \left(\boldsymbol{\tilde{d}}(k)\times\frac{d}{dk}\boldsymbol{\tilde{d}}(k)\right)_z=\frac{\omega^2+v\omega\cos k}{v^2+\omega^2+2v\omega\cos k} \end{equation}$$ and to me this is not a function I'd say is easy to integrate to check if the definition of the winding number is correct.

Is there a simpler way to check that $$\frac{1}{2\pi}\int \left(\boldsymbol{\tilde{d}}(k)\times\frac{d}{dk}\boldsymbol{\tilde{d}}(k)\right)_z dk = \nu, \,\, \nu\in \mathbb{Z}$$

is true (eq 1.38) ?

The definition of $$\boldsymbol{d}$$ is given by (1.18)

$$d_x(k) = \nu + w \cos k; \qquad \qquad d_y(k) = w \sin k; \qquad \qquad d_z(k)=0.$$

• It's not clear what you want to derive. What definition of winding number do you want to start with to get 1.38? Because what Asboth is saying is that equation 1.38 is the mathematical definition of a winding number, and you should be able to check it matches your intuitive definition of a winding number. – Jahan Claes Jun 22 at 15:55
• That's the point, I'm not able to intuitively check that it is correct. – Karim Chahine Jun 22 at 16:06
• You will definitely make this too complicated if you plug in a specific function for $\tilde d$, because this equation holds for ANY $\tilde d (k)$. – Jahan Claes Jun 22 at 16:34
• Yes, I suspect so. By plugging in the specific function I was trying to gain some insight but couldn't. Could you maybe show me the intuitive explanation of that formula or link something? All I can think of is that the cross product will be parallel to $\boldsymbol{\hat{z}}$ for all $k$ and I expect it to be a unitary contribution to the integral for all $k$ because it's the area of a 1x1 square. If that's all it is then I can see why it holds. – Karim Chahine Jun 22 at 16:49

\begin{align} w &= \frac{1}{2\pi}\int \frac{\vec{d}}{d}\times \partial_k \frac{\vec{d}}{d} dk\\ &= \frac{1}{2\pi}\int \frac{\vec{d}}{d}\times \left( \frac{\partial_k\vec{d}}{d}-\frac{\vec{d}\partial_k d}{d^2}\right) dk &\qquad\text{(product rule)}\\ &= \frac{1}{2\pi}\int \frac{\vec{d}\times \partial_k\vec{d}}{d^2} dk&\qquad\text{(\vec d\times \vec d=0)}\\ \end{align}
Now, draw a picture to convince yourself of the following: $$\frac{\vec{d}\times \partial_k\vec{d}dk}{d^2}=\frac{\vec{d}\times\delta \vec{d}}{d^2}$$ is the angle, in radians, between $$\vec{d}$$ and $$\vec{d}+\delta\vec d$$. Then by integrating this, you find the total number of radians that the vector $$\vec{d}$$ rotates through. This (divided by $$2\pi$$) is exactly the winding number.