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As much I understand, in a crystal Berry connection and Berry curvature is defined for a particular band. The Hall conductance is given by the total Chern number (integrated Berry curvature over the full Brillouin zone) of all the bands below Fermi level. In this paper [ arXiv:cond-mat/0307337], the anomalous Hall coefficient for Fe is calculated by integrating the Berry curvature. It also shows a density plot of Berry curvature over the full Brillouin zone. I don't understand

  1. How to calculate/define Berry curvature for a metallic system?

In a metal, one or multiple bands can cross the Fermi level. Above Fermi level, the occupation is zero. How would that affect the derivation?

  1. What data is shown in this figure [Fig.4 of the above paper]?

Is it the sum of Berry curvature of all the bands below Fermi level? Since the main calculation is done for bulk (3D), how the 3rd momentum component is fixed?

enter image description here

caption: Fermi surface in (010) plane (solid lines) and Berry curvature
Ωz(k) in atomic units (color map).

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  1. How to calculate/define Berry curvature for a metallic system?

We know that, in the Bloch theorem, eigenfunction of a periodic function can be written as $\psi_{k}(r) =r^{ik.r} u_{k}(r)$, where $u_{k}(r)$ is a periodic function $u_{k}(r+R)= u_{k}(r)$, and $R$ is defined on the Bravaius lattice.

Knowing the band structure of your system, namely $\psi_{nk}$ which is the Bloch state where $n$ is the band index, you can employ Eq.(4) of the referenced paper and can determine the Berry curvature of your system. The band structure is plotted in Fig. 1 and upper panel of Fig. 3.

2.What data is shown in this figure [Fig.4 of the above paper]?

Taking advantages of lattice symmetries, we can reduce the amount of calculation inside the first Brillouin zone to only calculations along the high-symmetry points of the lattice, in this system consisting of $\{\Gamma, H, P, N \}$. An intuitive 3D plot can be found in this link.

In Fig. 4, authors focused on a particular cross-section of the system along the (010)-plane, determined the associated Berry curvature and plotted that in the momentum space along $\Gamma -H$ points of the lattice. As you may also be noticed from the caption, the black solid lines are the Fermi surface of the system. Some 3D plots of the fermi-surfaces can be found here.

In a metal, one or multiple bands can cross the Fermi level. Above Fermi level, the occupation is zero. How would that affect the derivation?Is it the sum of Berry curvature of all the bands below Fermi level? Since the main calculation is done for bulk (3D), how the 3rd momentum component is fixed?

It has been statedd in the paper, Eq. (5), $$\Omega(\mathbf{k})=\sum\limits_{n} f_{n} \Omega_{n}(\mathbf{k}),$$ where $f_{n}$ is the Fermi-Dirac distribution. Thereby, only occupied states have a contribution in determining the Berry curvature of the system. Subsequently, the conductance of the system is obtained from the Berry curvature which has been summed over the occupied states.

Regarding the third component of momentum, please recall that the presented calculations in Fig. (4) are performed on a particular plane of the system, namely (010), imposing that due to the plane equation (F(k_{x}, k_{y}, k_{z})=0) only two of the components are free and the third component can be expressed as a function the others.

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  • $\begingroup$ Thanks, @Shasa. I am aware of the methods. What I don't understand is how to handle the situation when some bands cross the Fermi level and how to fix the 3rd momentum component when calculating Berry curvature on a plane. I modified my question accordingly. $\endgroup$ – Sumit Jul 4 '17 at 6:52
  • $\begingroup$ I have modified the answer accordingly to address your new questions. $\endgroup$ – Shasa Jul 4 '17 at 8:34
  • $\begingroup$ So in this case, the simplest assumption would be $k_z=0$. In that case, would it be right to consider that it is the Berry curvature of a 2D system with a width of one unit cell? $\endgroup$ – Sumit Jul 4 '17 at 9:10
  • $\begingroup$ Despite the 2D facade of the plot, you should remember that the band structure is obtained from the three-dimensional lattice. As a result, the 3D nature of the problem has been penetrated into the projected 2D response. Nevertheless, this particular plane is associated with a fixed $k_{y}$, and due to the location of $\Gamma$ and $H$ points in the 1st BZ, this plane is confined in this zone as well. $\endgroup$ – Shasa Jul 4 '17 at 9:20

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