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?

enter image description here


2 Answers 2


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.

  • $\begingroup$ Could you please be more precise? In which symmetry class would fall a system without spin orbit? As far as I understand the only symmetries involved in this classification are the antiunitary, i.e. particle-hole, time-reversal, and chiral. Also, if $\alpha=0$ the system is still gapped if $\Delta>0$. $\endgroup$
    – sintetico
    Jun 7, 2015 at 12:51
  • $\begingroup$ I've updated the answer to address your comments. $\endgroup$
    – Meng Cheng
    Jun 7, 2015 at 16:26

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)!


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.