Superconductivity in spin-$3/2$ systems: symmetry classification, odd-frequency pairs, and Bogoliubov Fermi surfaces (2106.11983v2)

Abstract: The possible symmetries of the superconducting pair amplitude is a consequence of the fermionic nature of the Cooper pairs. For spin-$1/2$ systems this leads to the $\mathcal{SPOT}=-1$ classification of superconductivity, where $\mathcal{S}$, $\mathcal{P}$, $\mathcal{O}$, and $\mathcal{T}$ refer to the exchange operators for spin, parity, orbital, and time between the paired electrons. However, this classification no longer holds for higher spin fermions, where each electron also possesses a finite orbital angular momentum strongly coupled with the spin degree of freedom, giving instead a conserved total angular moment. For such systems, we here instead introduce the $\mathcal{JPT}=-1$ classification, where $\mathcal{J}$ is the exchange operator for the $z$-component of the total angular momentum quantum numbers. We then specifically focus on spin-$3/2$ fermion systems and several superconducting cubic half-Heusler compounds that have recently been proposed to be spin-$3/2$ superconductors. By using a generic Hamiltonian suitable for these compounds we calculate the superconducting pair amplitudes and find finite pair amplitudes for all possible symmetries obeying the $\mathcal{JPT}=-1$ classification, including all possible odd-frequency (odd-$\omega$) combinations. Moreover, one of the very interesting properties of spin-$3/2$ superconductors is the possibility of them hosting a Bogoliubov Fermi surface (BFS), where the superconducting energy gap is closed across a finite area. We show that a spin-$3/2$ superconductor with a pair potential satisfying an odd-gap time-reversal product and being non-commuting with the normal-state Hamiltonian hosts both a BFS and has finite odd-$\omega$ pair amplitudes. We then reduce the full spin-$3/2$ Hamiltonian to an effective two-band model and show that odd-$\omega$ pairing is inevitably present in superconductors with a BFS and vice versa.