I've hit a wall with the Berry's phase.

So here's the thing:

  • I know Berry's phase is defined as \begin{equation} \gamma_C = \int_S \mathcal F d^2\mathbf k \end{equation}
  • I'm considering the 2D $H$ of \begin{equation} H = \left(\begin{matrix}d_z & d_x - id_y \\ d_x + id_y & -d_z\end{matrix}\right) = \hat {\mathbf d}(\mathbf k) \cdot \vec\sigma \end{equation} where $\hat{\mathbf d}$ is essentially a point on $S^2$.

What I don't understand is why it now automatically follows that \begin{equation} \mathcal F = \frac12 \epsilon_{ij}\hat{\mathbf d} \cdot (\partial_{i}\hat{\mathbf d} \times \partial_j\hat{\mathbf d}). \end{equation}

Hopefully somebody can clarify, in particular a geometric argument would be nice. The method I'm reviewing now states,

The Berry curvature is given by the solid angle per unit area in $\mathbf k$ space, which is simply half the solid angle element for the mapping $\mathbf {\hat d}(\mathbf k)$.


Edit: Maybe this additional information is useful: $\mathcal F = \nabla \times \mathbf A$ with \begin{equation} \mathbf A = -i\left< u(\mathbf k)\right| \nabla \left| u(\mathbf k)\right>. \end{equation} with $\left|u(\mathbf k)\right>$ the cell periodic eigenstate of the Bloch Hamiltonian


1 Answer 1


I figured out most of it.

So in the original Berry paper, the following identity is derived: \begin{equation} \mathcal F = \operatorname{Im} \frac{\left<+\right|\nabla\hat H\left|-\right> \times \left<-\right|\nabla\hat H\left|+\right>}{(E_+ - E_-)^2} \end{equation} for the two state system. (Please refer to the paper from '84 for a more detailed overview hereof.)

The original $\hat H$ had a factor $\frac12$ in front of it, such that $\nabla \hat H = \frac12 \vec \sigma$. Using the identity $\left<\pm\right|\vec\sigma\left|\mp\right> = \hat{\mathbf i} \mp i\hat{\mathbf j}$ after rotation of axes rotated alongside $\mathbf d$, the resulting outer product is $2i\hat{\mathbf k}$, of which the imaginary part is obviously $2\hat{\mathbf k}$. This leaves a factor $\frac14$ from the previous expression for $\nabla\hat H$ to be plugged in, and the fact that $E_+ = -E_- = \frac12\lvert\mathbf d\rvert$ (just calculate when the determinant of $\hat H$ above is zero). Apparently cancelling the axis rotation adds in a factor $\hat{\mathbf N} = \frac{\mathbf d}{d}$ (this is the part I can't quite figure out). The result is \begin{equation} \mathcal F = \frac{2\mathbf {\hat k}}{d^2} \cdot \frac14 \cdot \frac{\mathbf d}{d} = \frac{\mathbf d}{2d^3} \end{equation}
Note that the general expression for $\gamma_C = \frac12 \Omega(\mathbf d)$ quickly follows.

The more general expression (using Levi-Civita) can be derived from this, and is in accordance with the result here.

Lastly, why $\gamma_C$ must be a multiple of $2\pi$ from this, is unclear to me...


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.