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Can I ask two questions about the Berry curvature? The formula for the berry curvature is written below. $$\Omega_n (k) = -Im \langle \bigtriangledown_k u_{nk} | \times | \bigtriangledown_k u_{nk} \rangle$$ where, $|u_{nk}\rangle$ is the cell-periodic bloch state.

  1. I suppose $|u_{nk}\rangle$ is a matrix; then, $| \bigtriangledown_k u_{nk} \rangle$ is vector with non-zero component along three axes and each component is a matrix. Finally, $\langle \bigtriangledown_k u_{nk} | \times | \bigtriangledown_k u_{nk} \rangle$ is also a vector with non-zero component along three axes and each component is a matrix. In othe words, $\Omega_n (k)$ is a vector with non-zero component along three axes and each component is a matrix. Is my understanding correct or not?

  2. Is possible to convert this formula into the format of Green function? I mean if it is possible to calculate the berry curvature with green function? Thank you.

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  • $\begingroup$ In the second part of the question, I am not sure what kind of Green's function you are referring to. You can already find the Berry Curvature from the Bloch wavefunctions, which are eigenfunctions of the Hamiltonian. What are you trying to achieve? What is the source function such that you would get the Berry curvature as an output, when you integrate the source multiplied by the kernel (the Green's function)? $\endgroup$ May 3, 2021 at 6:46
  • $\begingroup$ Hi Archisman, Thank you for the answer. I am still little bit confused with the definition. Let me take the real system as an example. There are two carbon atoms in pristine graphene unit cell and each carbon atoms have s, px, py and pz orbitals. With spin orbit coupling effect, the Hamiltonian is a 16×16 matrix. After diagonalization, the eigen vector matrix is also a 16×16 matrix. As you said, the |unk⟩ is the eigenfunctions of the Hamiltonian, so it is a 16×16 matrix in the case of pristine Graphene. Is it correct? Thank you again. $\endgroup$
    – Kieran
    May 3, 2021 at 7:16

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$|u_{nk}\rangle$ is a the state corresponding to the periodic part of the Bloch wavefunction. If the position space wavefunction of a Bloch state is $e^{i \vec{k} \cdot \vec{r}} u_{nk}(\vec{r})$, then $|u_{nk}\rangle$ is defined such that $\langle \vec{r}|u_{nk}\rangle = u_{nk}(\vec{r})$, where $\vec{k}$ is the crystal momentum, and $n$ is the band index. $|u_{nk}\rangle$ is certainly not a matrix.

Then,

$\langle \bigtriangledown_k u_{nk} | \times | \bigtriangledown_k u_{nk} \rangle = \int d^n \vec{r} \frac{\partial}{\partial \vec{k}} u_{nk}(\vec{r})^* \times \frac{\partial}{\partial \vec{k}} u_{nk}(\vec{r})$.

In the second part of the question, I am not sure what kind of Green's function you are referring to. You can already find the Berry Curvature from the Bloch wavefunctions, which are eigenfunctions of the Hamiltonian.

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  • $\begingroup$ Hi Archisman, Thank you for the answer. I am still little bit confused with the definition. Let me take the real system as an example. There are two carbon atoms in pristine graphene unit cell and each carbon atoms have s, px, py and pz orbitals. With spin orbit coupling effect, the Hamiltonian is a 16×16 matrix. After diagonalization, the eigen vector matrix is also a 16×16 matrix. As you said, the |unk⟩ is the eigenfunctions of the Hamiltonian, so it is a 16×16 matrix in the case of pristine Graphene. Is it correct? Thank you again. $\endgroup$
    – Kieran
    May 3, 2021 at 7:16
  • $\begingroup$ Does the Hamiltonian not have any hopping elements between the Carbon atoms? In that case, the eigenvector would be a localized wavefunction at a single carbon atom (and it won't have any Berry curvature, which can only appear if there is a crystal lattice). $\endgroup$ May 3, 2021 at 7:18
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    $\begingroup$ In case you are describing a system where there is hopping between the sites, and there are 16 on site orbitals, then $u_k,n (\vec{r}) $ would be 16x1 vector. The 16 components of the vector would stand for the amplitudes of the on-site wavefunctions at position $\vec{r}$, whose linear combinations make up the eigenvector. The complete eigenfunction would still be $e^{i \vec{k} \cdot \vec{r}} |u_{n,k} \rangle$, where $|u_{n,k} \rangle$ is a 1x16 vector, whose components are the amplitudes of the onsite orbitals. $\endgroup$ May 3, 2021 at 7:22
  • $\begingroup$ Hi Archisman, I find a berry curvature calculation formula on the website (phyx.readthedocs.io/en/latest/TI/Lecture%20notes/…). The 2nd formula is $$-Im\sum_{n^{'}\neq n}\frac{\langle n|\bigtriangledown H|n^{'}\rangle \times \langle n^{'}|\bigtriangledown H|n\rangle}{\left( E_n-E_n^{'}\right)^{2}}$$ $\bigtriangledown H$ is a vector, with three non-zero matrices along three axes; then, the Berry curvature should also be a vector with three non-zero matrices. Is it right? $\endgroup$
    – Kieran
    May 3, 2021 at 8:42
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    $\begingroup$ $\langle \bigtriangledown_k u_{nk} | \times | \bigtriangledown_k u_{nk} \rangle$ is a vector with three non-zero values, which are all purely imaginary numbers (it can be shown). As a result, when their imaginary part is taken, we get a real vector having three components. $\endgroup$ May 3, 2021 at 9:26

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