Chiral anomaly in Weyl semimetal

Question

In the presence of electromagnetic fields $E$ and $B$, four current is not conserved in a Weyl semimetal i.e. $\partial_{\mu} j^{\mu}\propto E\cdot B \neq 0$. There are some proofs in the literature where this is proved with the machinery of Lagrangian and Action, but I am looking for a physical and more intuitive explanation of this phenomena as to why this happens, in a Weyl semimetal where say for example we have two Weyl nodes.

Answer 1

Your statement itself is not quite right. What is not conserved is the chiral current, namely the current of fermions at one of the Weyl nodes. The physics can be understood essentially in one-dimensional version of the Weyl metal: consider a 1D electron gas. There are two Fermi points, and the low-energy theory is given by two "Weyl fermions" in 1D with opposite chiralities. If we apply an electric field, obviously it drives a current. This current can be understood as fermions near one of the Fermi points moving to the other, through the bottom of the bands which are not present in the low-energy theory and connect the two "Weyl nodes". Therefore if you just look at fermions near one of the Fermi point, what you see is exactly the chiral anomaly.

The three-dimensional chiral anomaly is quite similar. First we can solve the Landau levels in the presence of a uniform magnetic field $\vec{B}=B\hat{z}$, and one sees that for each node and momentum $k_z$, one essentially has a 2D Dirac fermions with an out-of-plane magnetic field. Landau levels of Dirac fermions are well-understood, and in particular there is a $n=0$ level with zero energy at $k_z=0$. With finite $k_z$, the spectrum starts to disperse, but in a "chiral" way, with the chirality set by the Weyl node. So the situation is very much the same as the 1+1-dimensional chiral anomaly.

You can find more details in http://www.sciencedirect.com/science/article/pii/0370269383915290.