Superconducting state in the Kondo-Heisenberg model on a triangular lattice

Question

In this paper, "Fractionalized Fermi Liquids" by T. Senthil, Subir Sachdev and Matthias Vojta, the authors state in the last paragraph on page 2, "the pairing of the spinons and the condensation of $B_{1,2}$ implies that the resulting phase also has pairing of the conduction electrons, and is a superconductor." Can anyone prove it?

They are considering the Kondo-Heisenberg model on a triangular lattice and the Hamiltonian is given by eq(1) \begin{equation} H = -\sum_{j,j'} t(j,j')c^{\dagger}_{j\sigma}c_{j'\sigma} + \frac{1}{2}\sum_j J_k(j) S_j\cdot c^{\dagger}_{j\sigma}\tau_{\sigma\sigma'}c_{j\sigma'} +\sum_{j<j'}j_H(j,j')S_j\cdot S_{j'}. \end{equation} $t(j,j')$ is for hopping, $J_K$ for Kondo interaction, and $J_H$ for Heisenberg interaction. The order parameters are $B_1 = f^{\dagger}_{\sigma}c_{\sigma}$,$B_2 = \epsilon^{\sigma \sigma'} f_{\sigma}c_{\sigma'}$, and $ D = \epsilon^{\sigma \sigma'} f_{\sigma}f_{\sigma'}$. Here, $f_{j\sigma}$ comes from the pseudo-fermion representation of the spin operator $S_j = f^{\dagger}_{j\sigma}\tau_{\sigma\sigma'}f_{j\sigma'}/2$.

In fact, can anyone show $\langle B_{1i} \rangle \langle B_{1j} \rangle D_{ij} \propto \langle \epsilon^{\sigma \sigma'} c_{i\sigma}c_{j\sigma'} \rangle$ ?

Answer 1

Well, there have several papers on this issue, e.g. PRB 86 4526. I have also studied this Hamiltonian in my paper arXiv:1410.6261. In my opinion, the RVB pairing induces the pairing of conduction electrons via the nonvanishing Kondo screening. To my surprise, such SC state is able to fit the experimental data of the heavy fermion superconductor CeCoIn5.[The fermi surface measured by ARPES, the entropy and London penetration depth.]