Majorana fermions in s wave superconductor

I need some help to understand the majorana fermions in $s-$ wave superconductor and to check whether following method is correct

For $s-$ wave superconductor we can write the Hamiltonian as

$$H=-t\sum_{i,\sigma} c_{i,\sigma} ^{\dagger} c_{i+1,\sigma} +h.c +u c_{i,\sigma} ^{\dagger}c_{i,\sigma} +\Delta( c_{i,\uparrow} ^{\dagger}c_{i,\downarrow} ^{\dagger}+c_{i,\uparrow} c_{i,\downarrow})$$

following the procedure of kitaev model for $p-$wave superconductor the hamiltonian in majorana fermions form using following two equations

$$c_{j,\uparrow}=\frac{1}{2}(\gamma_{A_{j\uparrow}}+i \gamma_{B_{j\uparrow}})$$ $$c_j^\dagger=\frac{1}{2}(\gamma_{A_{j,\uparrow}}^\dagger-i \gamma_{B_{j,\uparrow}}^\dagger)$$ 1. Does my following transformation of hamiltonian in majorana fermion form is correct ?

\begin{equation} H=\sum_i [\frac{-it}{2} (\gamma_{A_j \uparrow} \gamma_{B_{j+1} \uparrow}-\gamma_{A_j \uparrow}\gamma_{A_{j} \uparrow}) +\frac{i u}{2}( \gamma_{A_j \uparrow} \gamma_{B_j \uparrow}+\gamma_{A_j \downarrow}\gamma_{B_j \downarrow})-i \Delta( \gamma_{B_j \uparrow} \gamma_{A_j \downarrow}+\gamma_{A_j \uparrow}\gamma_{B_j \downarrow})] \end{equation}

Condition for observing uncoupled Majorana fermion at edge $u=0$ ,$\Delta=0$

the above equation reduce to \begin{equation} H=\sum_j \frac{-it}{2}[ (\gamma_{A_j \uparrow} \gamma_{B_{j+1} \uparrow}-\gamma_{B_j \uparrow}\gamma_{A_{j+1} \uparrow}+ (\gamma_{A_j \downarrow} \gamma_{B_{j+1} \downarrow}-\gamma_{B_j \downarrow}\gamma_{A_{j+1} \downarrow}))] \end{equation} which shows that there are two uncoupled majorana at each end site $\gamma_{A_j \uparrow}$ and $\gamma_{A_j \downarrow}$

1. But I have a problem of understanding how we can understand physically $\Delta=0$ term because at the boundary of trivial and non-trivial superconductor $\Delta\neq 0$
• Even with $u=0$ and $\Delta=0$, $\gamma_{Aj\uparrow}$ and $\gamma_{Aj\downarrow}$ are still coupled to other Majoranas. Of course, we know that the $s$-wave superconductor is topologically trivial so there should not be any protected zero modes. Turning off $\Delta$ just means you make that part of the system a metal. – Meng Cheng Dec 9 '15 at 3:42
• @MengCheng how does the $\gamma_{Aj\uparrow}$ and $\gamma_{Aj\downarrow}$ are coupled if $\Delta=0$ and $u=0$ – user48826 Dec 9 '15 at 10:49
• Aren't they coupled to $\gamma_{Bj+1\uparrow}$ and $\gamma_{Bj+1\downarrow}$? – Meng Cheng Dec 9 '15 at 16:41
• How do you get read off the sum over $\sigma$ going to the Majorana basis ? The transformation should reads $c_{i,\sigma}=\left(\gamma_{A,i,\sigma}+i\gamma_{B,i,\sigma}\right)/2$ and its complex conjugate with self-Hermitian $\gamma$ I hardly suspect your Hamiltonian in point 1 to be completely wrong – FraSchelle Dec 9 '15 at 17:10