# Calculation problem for Berry phase

The following content comes from David Tong's lecture notes on quantum hall effect, p.34-35.

The resulting Berry curvature in polar coordinates is$$\mathcal{F}_{\theta \phi} = \frac{\partial A_{\phi}}{\partial \theta} - \frac{\partial A_{\theta}}{\partial \phi} = -\sin \theta.\tag{p.33}$$

This is simpler if we translate it back to cartesian coordinates where the rotational symmetry is more manifest. It becomes$$\mathcal{F}_{ij} (\vec{B}) = -\epsilon_{ijk} \frac{B^k}{2|\vec{B}|^3}.\tag{p.33}$$

My question is, how do we get the last equality by starting from the spherical coordinate?

It seems to me that $$\sin \theta$$ would corresponds to $$\sin(\theta) = \frac{\sqrt{x^2 + y^2}}{\sqrt{x^2 + y^2 + z^2}}$$. So we get Berry curvature in the normal coordinate as $$\mathcal{F}_{ij} (\vec{B}) = -\sin(\theta) = \frac{\sqrt{x^2 + y^2}}{\sqrt{x^2 + y^2 + z^2}}$$ . But actually this is not true. So in which step did I get wrong?

• Please don't ask the same question again if the previous one has been closed before. Instead, edit the question and try to re-open. Anyway, it is still not clear for me what your exact question is. What step is unclear? Commented Jul 22 at 13:19
• Hi, just edited it to make it more focused. But it seems to be remain closed, which is also why I delete the same question before. My main confusion is how to transform from the result of spherical coordinate(which is -sin(\theta)) to the result in cartesian coordiante. Commented Jul 22 at 14:04