Topological superconductors: what is the role of spin-orbit coupling? Are there topological non-trivial states without spin-orbit?

Question

Let's say I have a one-dimensional system with particle-hole symmetry and with broken time-reversal symmetry. As a consequence, the chiral symmetry is also broken in this case (the chiral symmetry operator is the product of time-reversal and particle hole) and the system is therefore in the Altland-Zirnbauer symmetry class D (see table). This system can be realized by a one-dimensional chain with magnetic field, spin-orbit coupling, and with a $s$-wave superconducting coupling, and it is described by the BdG Hamiltonian with periodic boundary conditions:

$$ \mathcal{H}= \frac12\sum_{i=1}^L \boldsymbol\Psi_{i}^\dagger \cdot \begin{bmatrix} \mathbf{b}\cdot\boldsymbol{\sigma}+\mu\sigma_0&\imath\sigma_y\Delta\\ -\imath\sigma_y\Delta^*&-(\mathbf{b}\cdot\boldsymbol{\sigma}+{\mu}\sigma_0)^* \end{bmatrix} \cdot \boldsymbol\Psi_{i} + \nonumber\\ + %\tfrac12\!\sum_{i=1}^L\! \boldsymbol\Psi_{i}^\dagger \cdot \begin{bmatrix} t\sigma_0-\imath\alpha\sigma_y&0\\ 0&-t\sigma_0+\imath\alpha\sigma_y \end{bmatrix} \cdot \boldsymbol\Psi_{i+1} +\text{h.c.}, $$ where $\boldsymbol\Psi_i=(c_{i\uparrow},c_{i\downarrow},c^\dagger_{i\uparrow},c^\dagger_{i\downarrow})$ is the Nambu spinor (see also answer to this question).

The topological invariant is the $\mathbb{Z}_2$, which can be calculated as the Pfaffian of the BdG Hamiltonian as $(-1)^\nu={\rm sign}\,{\rm Pf}[(\mathcal{H})\imath\tau_x]$, where $\rm Pf$ is the Pfaffian and $\tau_x$ is the Pauli matrix in the particle-hole space (see for example arXiv:1111.6592). If $\nu=0$ the system is in the trivial state, if $\nu=1$ the system is in the topological non-trivial state.

I understand that one needs a finite magnetic field to stay in the symmetry class D and to have a non-trivial topological phase with $\nu=1$. But the question is: what is the role of the spin-orbit coupling? Why we need a finite spin-orbit $\alpha>0$ coupling in this Hamiltonian in order to get a non-trivial topological state? Can I have a topological non-trivial state also with $\alpha=0$? Have I missed something in the reasoning above?

Answer 1

It depends on what kind of pairing you are willing to include in the Hamiltonian. If only s-wave singlet pairing is present, and there is no spin-orbit coupling, the Hamiltonian has an additional $\mathrm{U}(1)$ symmetry (spin rotation around the direction of the Zeeman field), so falls into class A in the table. A bigger issue is that if $\alpha=0$, with s-wave pairing the single-particle excitation gap is zero and the Hamiltonian describes a gapless system. To see this explicitly, consider the continuum version of the Hamiltonian when the spin-orbit coupling vanishes:

$H=\psi_p^\dagger(p^2/2m-\mu)\psi_p+V_z\psi_p^\dagger \sigma_z\psi_p + \Delta \psi_{p\uparrow}^\dagger \psi_{-p,\downarrow}^\dagger+\text{h.c.}$

Diagonalizing the Hamiltonian, the spectrum is given by

$E_p=\pm\sqrt{\Delta^2+(p^2/2m-\mu)^2}\pm V_z$

If one naively applies the Pfaffian criteria for topological superconductivity, one still finds $V_z^2>\Delta^2+\mu^2$. So in this region, we find that if $(p^2/2m-\mu)^2=V_z^2-\Delta^2$, or $p^2=2m(\mu+\sqrt{V_z^2-\Delta^2})$, the gap closes.

On the other hand, if you allow triplet pairing in the Hamiltonian, you do not need spin-orbit coupling at all. In a sense, the role of the spin-orbit coupling is to effectively generate triplet pairing.

Answer 2

If you are interested what happens in the energy spectrum, then this two papers could be very helpful for you: arXiv:1206.1736 and arXiv:1205.7054.

The spin-orbit coupling splits the two spin bands (see in arXiv:1206.1736 Fig. 5a) and the Zeeman term mix them (see in arXiv:1206.1736 Fig. 5b). This looks then in the end like a p-wave pairing in the low-energy regime, but in the basis of Left- and Right-mover you have an s-wave pairing (see. Fig. 3 in arXiv:1205.7054)!