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Weyl semimetals are topological quantum materials whose low energy excitations emerges as massles Weyl Fermions. They have a band touching point near the Fermi level called Weyl node.

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What is actually the energy dispersion relation of Weyl Fermions? More specifically, do we have to consider the band as a linear dispersion relation or as a conduction and valence band which takes the shape of two cones?

I came up with this doubt because if we consider the band dispersion relation as linear, then

$$E_k=C\times (k-k_0)$$(For convenience consider 1D. but I know that Weyl semimetals are three dimensional)

then the effective mass is given by $$m*=\frac{\hbar^2}{\frac{\partial^2E_k}{\partial k^2}}$$

that gives

$$m*=\frac{\hbar^2}{0}=\infty$$

Which contradicts the fact that Weyl Fermions are massless.

But If we consider it as the latter one then the derivative is not defined at the Weyl node? Where am I wrong in my calculations and concepts?

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Weyl semimetals (WSMs) are realized when non-degenerate bands cross at isolated points in the Brillouin zone. Consider the simplest example of a time-reversal symmetry broken WSM modeled by the hamiltonian $$H = t' \sin{k_x} \sigma_x + t'' \sin{k_y} \sigma_y + t (\cos{k_x} + \cos{k_y}+ \cos{k_z} - M) \sigma_z.$$ This hamiltonian admits 2 bands ($\sigma_j$ are the Pauli matrices). When $1 < M < 3$ a WSM phase is realized with Weyl points (i.e. points where the two bands cross) at $(k_x, k_y, k_z) = (0, 0, \pm \cos^{-1}(M-2))$.

If you zoom-in close to one of the Weyl points (say the one at $k_z >0$), Taylor expand the terms in the hamiltonian in terms of the deviations $\delta k_j$, and keep only the leading order term in each expansion, then you'll find the following effective hamiltonian $$H_+ = t' \delta k_x~ \sigma_x + t'' \delta k_y~ \sigma_y - t \sin\cos^{-1}(M-2) ~\delta k_z~ \sigma_z.$$

At zero density, the Fermi level will intersect the Weyl points, and the low energy excitations will be described by $H_+$ (here, "low" implies energies at which the quadratic terms that were dropped in the Taylor expansion are small compared to the linear terms). Notice that these excitations disperse linearly, and has 2 branches. In this sense the low energy excitations emulate Weyl fermions.

If one ignores the higher order terms in the Taylor expansion entirely, then the low energy excitations appear to be massless owing to an absence of terms $\propto \delta k_j^2$. In reality there is a finite but small mass. Here, we need to distinguish between the 'band mass' ($m_*$), which is a measure of the local curvature of a band, from the particle-like excitation's mass which is the energy gap in the excitation spectrum: If the excitation spectrum is described by $ E(k) = \sqrt{k^2 + \mu^2}$, then $\mu$ is the mass/energy gap. It is $\mu$ that is usually alluded to when the term massless is invoked in the context of semi-metals or multiband insulators.

It is instructive to check these limiting behaviors with the eigenvalues (bands) of the microscopic hamiltonian, $H$. Make special note of the terms being dropped from the Taylor expansion.

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