The low-energy effective model for Weyl semimetals (WSM) at a single Weyl point can be written as:

$$H_{w}=\chi \vec{k} \cdot \vec{\sigma}, \tag{1} $$

where $\chi$ is the chirality index, $\vec{k}$ is the crystal momentum, and $\vec{\sigma}$ is the vector of Pauli matrices. The symmetry behavior for $\vec{k}$ and $\vec{\sigma}$ under time-reversal ($\mathcal{T}$) and inversion ($\mathcal{P}$) operations is well-known:

  • $\mathcal{T}\vec{\sigma}=-\vec{\sigma}; \mathcal{T}\vec{k}=-\vec{k}$;
  • $\mathcal{P}\vec{\sigma}=+\vec{\sigma}; \mathcal{P}\vec{k}=-\vec{k}$;

On the other hand, the Weyl semimetal can be classified as nonmagnetic and magnetic ones. Consider the first one, the system respects $\mathcal{T}$ but breaks $\mathcal{P}$ (we assume there is a total of four Weyl points), we can conclude that: $$\mathcal{T}\chi=\chi \tag{2}$$ For the second case, the system respects $\mathcal{P}$ but breaks $\mathcal{T}$ (we assume there is a total of two Weyl points), we can conclude that: $$\mathcal{P}\chi=-\chi \tag{3}$$

Eq.(2) and Eq.(3) are correct? If not, how Eq.(1) can be used to describe the magnetic and nonmagnetic WSMs simultaneously? Or how does the low-energy model by considering all possible Weyl points respect the original symmetry that the system holds?

  • $\begingroup$ Weyl points are subject to the fermion doubling theorem, so they always come in pairs of opposite chiralitly. In Eq.(1) you have a single Weyl point, but time reversal or inversion operations will typically take you to a different Weyl point. $\endgroup$
    – ProfM
    Oct 9, 2022 at 6:01
  • $\begingroup$ @ProfM If I just consider two Weyl points in magnetic WSMs, I need to sum over $\chi$ in Eq.(1). Then how do I understand its transformation under space-time transformation? $\endgroup$
    – Jack
    Oct 9, 2022 at 6:06
  • 2
    $\begingroup$ Eq.(1) is a low-energy expansion around a Weyl point, so an approximation to the true Hamiltonian that is only valid in the vicinity of a single Weyl point. You will have an equation similar to Eq.(1) for every Weyl point in the system. For a full treatment of the symmetry I would recommend looking at the full Hamiltonian, not at the low-energy version in Eq.(1). $\endgroup$
    – ProfM
    Oct 12, 2022 at 6:42
  • 1
    $\begingroup$ It's true that $\chi(\mathbf{k}_0)=\chi(-\mathbf{k}_0)$ for $\mathcal{P}$-breaking WSMs and $\chi(\mathbf{k}_0)=-\chi(-\mathbf{k}_0)$ for $\mathcal{T}$-breaking WSMs, where $\mathbf{k}_0$ is the momentum of a Weyl point, but ProfM is correct that (1) isn't the best starting point to derive this. $\endgroup$
    – Anyon
    Oct 13, 2022 at 13:27
  • $\begingroup$ Did you figure out the answer? Please update us! $\endgroup$ Jun 11 at 17:51

1 Answer 1


I find the answer just by reading this paper:https://www.science.org/doi/epdf/10.1126/science.aav2873.

enter image description here

For the simplest magnetic WSM, we just have two Weyl points, which are related by inversion symmetry, namely: $$\mathcal{P}\chi=-\chi.$$

For the simplest nonmagnetic WSM, we have four Weyl points, in which we have: $$\mathcal{T} \chi = \chi .$$ Note that the different chirality $\chi$ has been represented by different colors in the screenshot.

For each Weyl point, it can be described by $\vec{k}\cdot\vec{\sigma}$, but we need to consider their copies due to the summation of $\chi$ should be zero when we talk about the symmetry, as ProfM told.


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