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In celebrated work of Fu and Kane they show appearance of Majorana bound state thanks to presence of superconductor and surface states of topological insulator.

They write Hamiltonian

$H = \tfrac{1}{2} \Psi^\dagger H_{BdG} \Psi$,

where in Nambu notation $\Psi = [(\psi_{\uparrow}, \psi_{\downarrow}), (\psi_{\downarrow}^{\dagger}, -\psi_{\uparrow}^{\dagger}) ] $ in terms of electron field operators for spin up and down.

They then shown Majorana bound state at $E=0$ by solving first quantized Hamiltonian

$H_{BdG} \xi = 0$,

where $\xi$ is four-component column.

Question: Why is $\xi$ four-component column? Since only spin up and down electron appear I naively assume two-component column.

Question: What is basis of $\xi$ and what is physical interpretation of each element?

Any help appreciated.

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    $\begingroup$ Quasiparticle excitations in a superconductor are superpositions of electrons and holes. Since we have spin-$1/2$ electrons, one has a four-component Nambu spinor. If we write $\xi=(u_\uparrow, u_\downarrow, v_\downarrow, -v_\uparrow)^T$, then the corresponding Bogoliugov quasiparticle is $\gamma=u_\sigma \psi_\sigma + v\psi_\sigma^\dagger$. $\endgroup$ – Meng Cheng Mar 23 '16 at 17:35
  • $\begingroup$ @Meng Cheng Thanks. I am confuzed as people in literature treat $\xi$ as first quantized wavefunction. Why is it justified to treat $\xi$ as normal single particle wavefunction in first quantized language? $\endgroup$ – Nigel1 Mar 24 '16 at 8:44

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