Is the topological phase in a Weyl semimetal is intrinsic or symmetry protected? How can we realize that?
If symmetry protected, which symmetry protects the topological phase of non-centrosymmetric Weyl semimetals and magnetic Weyl semimetals?
$\begingroup$
$\endgroup$
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
0
A Weyl point is a crossing of two bands. A two-band crossing can be described using the general Hamiltonian:
$$ \hat{H}(\mathbf{k})=d_0(\mathbf{k})+d_1(\mathbf{k})\sigma_1+d_2(\mathbf{k})\sigma_2+d_3(\mathbf{k})\sigma_3 $$
where $\sigma_i$ are the Pauli matrices. The eigenvalues are given by $E_{\pm}=d_0\pm\sqrt{d_1^2+d_2^2+d_3^2}$, so that a band crossing occurs when $d_1^2+d_2^2+d_3^2=0$. This equation involves three parameters, and in a three dimensional system, $\mathbf{k}=(k_1,k_2,k_3)$ has three components, so you can generically tune the $\mathbf{k}$ vector to make the two bands degenerate without the need of any symmetry.
Having established this, if you consider a system with both time reversal and inversion symmetry, then every band is doubly degenerate (spin up and spin down electrons have the same energy at every $\mathbf{k}$-point). This means that potential band crossings would involve four (rather than two) bands, and be Dirac rather than Weyl points. So a pre-requisite before you can have a Weyl point is that you need to break the spin degeneracy caused by time reversal and inversion symmetries. This can be accomplished by breaking one of these two symmetries.
