The Berry curvature of the $n^{\mathrm{th}}$ eigenstate of Hamiltonian $H$ for the vector of external parameters $\vec{R}$ can be derived in part by writing the following two lines:

$$ B^n(\vec{R}) \equiv \nabla \times A^n(\vec{R}) = -\mathrm{Im} \left[ \left< \nabla n(\vec{R}) | \nabla n (\vec{R}) \right> \right] $$

where $A^n(\vec{R})\equiv -\mathrm{Im} \left[ \left< n(\vec{R})| \partial_{\vec{R}} n(\vec{R}) \right> \right]$ is the Berry connection.

I don't know exactly how perform this step in the calculation in detail:

$$ \nabla \times A^n(\vec{R}) = -\mathrm{Im} \left[ \left< \nabla n(\vec{R}) | \nabla n (\vec{R}) \right> \right] $$

Help would be greatly appreciated.

Thank you very much.

  • $\begingroup$ I think you are missing an antisymmetric product between derivatives of states. It should be clearer how to get that expression, but if it's not let me know and I can fill in steps. $\endgroup$
    – d_b
    Feb 13 '14 at 2:10
  • 1
    $\begingroup$ The details of the calculation necessary to explain the above are in Bernevig's textbook. Topological Insulators and Topological Superconductors B. Andrei Bernevig with Taylor L. Hughes press.princeton.edu/titles/10039.html $\endgroup$ Apr 3 '14 at 3:15

The symbols $\partial_{\vec R}$ and $\nabla$ denote the same operation, so I'm going to unify the notation by using $\nabla$ only.

The expression in the OP cannot be correct as written, because $\nabla\times A$ is a vector, while $\langle\nabla n(\vec R)|\nabla n(\vec R)\rangle$ is a tensor of rank two (i.e., a matrix). The problem is that the curvature is not really the curl of the connection, but the exterior derivative instead. This is trivial to solve (cf. Hodge star): we just contract the curl with the Levi-Civita symbol: $\epsilon_{ijk}\partial_i A_j\to \partial_{[i}A_{j]}$. We proceed with $\nabla\times$ for now (because it simplifies the notation), and take its dual at the end.

With this, you only need the basic identity $$\tag 1 \nabla\times(f\vec F)=f(\nabla\times\vec F)+(\nabla f)\times\vec F $$ valid for any scalar function $f$ and any vector function $\vec F$.

First, $\nabla\times$ commutes with $\mathrm{im}$, and therefore $$ \nabla\times A^n(\vec R)=-\mathrm{im}\left(\nabla\times \left\langle n(\vec{R})\bigg| \nabla n(\vec{R}) \right> \right) $$

Next, use the product rule: $$ \nabla\times A^n(\vec R)\overset{\mathrm{(1)}}=-\mathrm{im}\left(\left[\left< n(\vec{R})\bigg|\left[\nabla\times \bigg|\nabla n(\vec{R}) \right>+ \left< \nabla n(\vec{R})\bigg|\right]\times\bigg| \nabla n(\vec{R}) \right>\right] \right) $$

The first term vanishes because $\nabla\times\nabla f=0$ for any $f=f(\vec R)$. We are left with $$ \nabla\times A^n(\vec R)=-\mathrm{im}\left< \nabla n(\vec{R})\bigg|\times\bigg| \nabla n(\vec{R}) \right> $$ which is essentially the result we were after. Taking the dual, $$ \Omega_{ij}=\epsilon_{ijk}(\nabla\times A^n(\vec R))^k=-\mathrm{im}\left< \nabla_{[i} n(\vec{R})\bigg| \nabla_{j]} n(\vec{R}) \right> $$ as required. Note that we may drop the brackets because the $\mathrm{im}$ operation kills the symmetric part anyway.


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