# Symmetry for the low-energy effective model of Weyl semimetal

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?

• 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. Oct 9, 2022 at 6:01
• @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?
– Jack
Oct 9, 2022 at 6:06
• 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). Oct 12, 2022 at 6:42
• 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. Oct 13, 2022 at 13:27