Nambu notation and the Majorana bound state

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.

• 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 $\gamma=u_\sigma \psi_\sigma + v\psi_\sigma^\dagger$. – Meng Cheng Mar 23 '16 at 17:35
• @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? – Nigel1 Mar 24 '16 at 8:44