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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?

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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)? $\endgroup$
    – Archisman Panigrahi
    Commented 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
    Commented 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
    Commented 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). $\endgroup$
    – Archisman Panigrahi
    Commented 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. $\endgroup$
    – Archisman Panigrahi
    Commented 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
    Commented 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. $\endgroup$
    – Archisman Panigrahi
    Commented May 3, 2021 at 9:26
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Regarding the second part of your question there is a series of recent works (in photonics not in condensed matter) where the Chern number (that corresponds to an integral of the Berry curvature) was expressed using Green's function and an integral in the complex plane over a line parallel to the imaginary axis and that is contained in the bandgap.
From the expression of the Chern number, I think that you can quite easily derive an expression for the Berry curvature based on the Green's function formalism. Below are the articles:

  1. original article: Topological classification of Chern-type insulators by means of the photonic Green function
  2. generalization of the formalism to Non-Hermitian systems
  3. application of the formalism in detail to photonic crystals

In the last article, you can find some references for the case of condensed matter. I am not familiar with it but it seems that the formulas are very similar to the photonic case. Maybe the most interesting reference is the following (where you can also find other references)

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    $\begingroup$ While this link may answer the question, it is better to include the essential parts of the answer here and provide the link for reference. Link-only answers can become invalid if the linked page changes. - From Review $\endgroup$
    – Miyase
    Commented Jan 13, 2023 at 11:30
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    $\begingroup$ I have not put the formulas because they are huge formulas that require a lot of extra definitions and also I am afraid of writing mistakes since it is complex. Regarding the links they are DOI so I believe they are quite fixed, no? $\endgroup$
    – lambertmular
    Commented Jan 13, 2023 at 12:08
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    $\begingroup$ The point of answers on this site is to be as self-contained as possible, which is why link-only answers are usually deleted. If you can provide only links, you should post them as comments. $\endgroup$
    – Miyase
    Commented Jan 13, 2023 at 12:29
  • $\begingroup$ I need a reputation of 50 to put a comment, I am not allowed! $\endgroup$
    – lambertmular
    Commented Jan 13, 2023 at 15:41