# Quantum Geometric Tensor and Berry Connection

The Quantum Geometric Tensor is given by $$\begin{split} Q_{\mu\nu}=\langle\partial_{\mu}\psi|\partial_{\nu}\psi\rangle-\langle\partial_{\mu}\psi|\psi\rangle\langle\psi|\partial_{\nu}\psi\rangle \end{split}$$ I wanted to confirm whether for the $$SU(2)$$ coherent state $$\begin{split} |z\rangle=\frac{1}{(1+|z|^2)^j}\sum_{m=-j}^{m=j}\sqrt{\frac{2j!}{(j+m)!(j-m)!}}z^{j+m}|j,m\rangle \end{split}$$ The Berry curvature $$(F)$$ and connection $$(A)$$ are the following $$\begin{split} F=i\frac{dz\wedge d\bar{z}}{(1+|z|^2)^2}~~~~~~ A=i\frac{\bar{z} dz}{(1+|z|^2)} \end{split}$$ I have figured out that the symmetric part viz $$Q_{\mu\nu}dx^{\mu}dx^{\nu}$$ is the metric. But I don't see that how the exterior derivative of the Berry connection $$A=i\langle\psi|d\psi\rangle$$ is proportional to the anti-symmetric part $$Q_{\mu\nu}dx^{\mu}\wedge dx^{\nu}$$. In other words, I wanted to know that the for Kahler manifolds, whether the Berry curvature is proportional to the Kahler two form $$Q_{\mu\nu}dx^{\mu}\wedge dx^{\nu}$$.

• Please note that homework-like questions and check-my-work questions are generally considered off-topic here. We intend our questions to be potentially useful to a broader set of users than just the one asking, and prefer conceptual questions over those just asking for a specific computation. Commented Sep 24, 2020 at 14:57
• @ACuriousMind: I have reframed my question Commented Sep 24, 2020 at 16:12

$$\langle \alpha | \beta \rangle = \langle \beta | \alpha \rangle^*.$$