# How is a Majorana fermion created when a s-wave superconductors is in proximity to a topological insulator (e.g. via an antidot)

Kane and Fu proposed a few geometries how to create Majorana zero modes using a s-wave superconductor in proximity to a 3D topological insulator (TI).

I understand that we need the superconductor to induce the particle-hole symmetry and we need the topological insulator to get a protected edge state at zero energy. But overall I must admit that I have no intuitive description for this process. Especially the importance of the Berry phase, the vortex or the magnetic flux quantum is not clear to me. Especially in the so called Antidot-experiment, where there's a hole in a s-wave superconductor on top of a TI and a magnetic flux quantum through this hole creates a Majorana zero mode.

It seems like a big question, but maybe someone can give a simple intuitive answer why we need all these ingredients and end up with a Majorana zero mode.

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For a moment let's assume you subscribe to the Hamiltonian used by Fu-Kane to describe the TI surface with an induced superconducting proximity effect, $$\mathcal{H}(\mathbf{k})=\left( \begin{array}{cc} H(\mathbf{k}) & i\sigma_y \Delta \\ -i\sigma_y \Delta^* & - H^*(-\mathbf{k}) \end{array} \right)=v_F(\sigma_x k_y-\tau_z\sigma_y k_x) - \tau_z\mu +\tau_y\sigma_y\Delta,$$

where $H(k)=v_F(\sigma_x k_y-\sigma_y k_x)$ and $\Delta=\Delta(\bf{x})$ some spatially dependent single valued function. The energy eigenvalues for constant $\Delta(x)=\Delta_0$,

$$E=\pm\sqrt{|\Delta|^2 +(v_F |\bf{k}|\pm \mu)^2}.$$ Let's assume for a moment that $\mu=0$, for without this condition you get flat-band Majoranas (ArXiv: http://arxiv.org/abs/1207.5534 PRB: http://link.aps.org/doi/10.1103/PhysRevB.86.161108 shameless self-publicity). The energy eigenvalues will be written as $$E=\pm\sqrt{|\Delta|^2 +(v_F \,k)^2}.$$ This energy dispersion does not pass through $E=0$ and is not linear, $E\propto k$, but if we did set $\Delta=0$ the energy dispersion does fall into the form of a linear Dirac-like spectrum, $E=\pm v_F\,k$. But, how can we set $\Delta=0$ in a superconductor? The answer first provided by Jackiw-Rebbi (http://link.aps.org/doi/10.1103/PhysRevLett.37.172) and then by Read-Green (http://link.aps.org/doi/10.1103/PhysRevB.61.10267).

The "mass term", in this case $\Delta$ must go through a transition spatially from $+|\Delta(\bf{x})|\rightarrow-|\Delta(\bf{x})|$, which forces $\Delta(\bf{x})=0$ at exactly one point. One way to do this is by having the superconductor be in a vortex where $\Delta(r,\phi)=\Delta(r) e^{i\phi}$. Right at the center of the vortex, $\Delta(r,\phi)$ must go to 0, because there can be no value satisfies the vortex. One one point on the side of the vortex $\Delta(r_0,\phi_0)=|\Delta|$ while the opposite side, $\Delta(-r_0,\phi_0)=\Delta(r_0,\phi_0+\pi)=-|\Delta|$ because of the $e^{i\pi}$ phase difference between the two locations. This allows one location to host the Majorana fermion where the energy dispersion is linear. There are other requirements for the Majorana, but this is what I can offer at the moment. Once my thesis is in complete form (~ two weeks as of this writing), I'll post a link here because I attempt to explain your question in its introduction.

Another perspective to understanding this story is through a simpler model, a Josephson \pi junction. See the first PRB paper referenced above for my attempt at explaining this Majorana through Andreev bound states.