7 hospitality the Racah Institute of Physics of the Hebrew University of Jerusalem, where a part of this work has been performed. tions in the second line We start with Eq. (29). Using Eqs. (13), (15), and (24), (25) we find after averaging over momentum direc- Appendix A: Current in the linear approximation j = − ev$_{F}$ ∑ p p p [ tanh E$_{↑}$ + E$_{D}$ 2 T − tanh E$_{↑}$ − E$_{D}$ 2 T + tanh E$_{↓}$ + E$_{D}$ 2 T − tanh E$_{↓}$ − E$_{D}$ 2 T ] − ev 2 $_{F}$k ∑ p    ∣ u ∗ $_{↑}$u$_{↓}$ + v ∗ $_{↑}$v$_{↓}$ ∣ ∣$^{2}$ E$_{↑}$ − E$_{↓}$ ( tanh E$_{↑}$ 2 T − tanh E$_{↓}$ 2 T ) + ∣ v ∗ $_{↑}$u$_{↓}$ − u ∗ $_{↑}$v$_{↓}$ ∣ ∣$^{2}$ E$_{↑}$ + E$_{↓}$ ( tanh E$_{↑}$ 2 T + tanh E$_{↓}$ 2 T )    . (A1) Appendix B: Current in the degenerate case 1. Wave functions for weak doping µ ≪ Δ are small, in Eqs. (B1) and (B2) we can take the coeffi- cients in the zero order approximation. As a result C$_{12}$ an C$_{14}$ are small (i.e., are proportional to the perturba- tion), while C$_{11}$ and C$_{13}$ are of the order unity. We put C$_{11}$ = C (0) 11 , C$_{13}$ = C (0) 13 while C$_{12}$ = C (1) 12 , C$_{14}$ = C (1) 14 , and find up to the first order terms Since and The contribution from the zero-order term has the form of Eq. (30) of surface integral over a remote sphere in the momentum space. Using Eq. (A2) we find This expression formally diverges for large v$_{F}$p due to contribution from regions far from the Dirac points. As we discussed already in Section III we remove this spu- rious divergence by transforming the zero-order terms. With Eqs. (16), (17) we have in the zero-order approxi- mation Consider Eq. (38) for the state α = 1. We have within the linear approximation in H (1) When added to Eq. (A1), this compensates the diverg- ing terms there. As a result, we obtain the converging expression, Eq. (31). The same result can be obtained if we subtract the normal current, i.e., Eq. (A1) for Δ = 0. For T ≪ | Δ | the terms cosh − $^{2}$( E$_{↑}$$_{,}$$_{↓}$/ 2 T ) are small while tanh( E$_{↑}$$_{,}$$_{↓}$/ 2 T ) = 1. Therefore, the first line in Eq. (31) vanishes. The current becomes j = ev 2 $_{F}$k ∫ ∞ 0 pdp 2 π    1 v$_{F}$p − 2 ∣ v ∗ $_{↑}$u$_{↓}$ − u ∗ $_{↑}$v$_{↓}$ ∣ ∣$^{2}$ E$_{↑}$ + E$_{↓}$    . (A4) Calculating the current with help of Eqs. (16), (17) we obtain Eq. (32). while C (0) 11 δE (1) 1 = C (0) 11 H$_{11}$ + C (0) 13 H$_{13}$ , (B5) C (0) 13 δE (1) 1 = C (0) 11 H$_{31}$ + C (0) 13 H$_{33}$ , (B6) 2 E$_{0}$C (1) 14 = C (0) 11 H$_{41}$ + C (0) 13 H$_{43}$ . (B8) 2 E$_{0}$C (1) 12 = C (0) 11 H$_{21}$ + C (0) 13 H$_{23}$ , (B7) E$_{1}$ − E (0) 1 = E$_{1}$ − E (0) 3 ≡ δE$_{1}$ j (0) = − ∫ dφ (2 π ) $_{2}$[( p · k ) j (0) K ( p )]$_{p}$$_{→∞}$ = ev 2 $_{F}$k ∫ ∞ 0 pdp 2 π [ 1 v$_{F}$p ] . (A3) j (0) K ( p ) = − j (0) − $_{K}$( p ) = − ev$_{F}$ [ v$_{F}$p − µ E$_{↑}$ tanh E$_{↑}$ 2 T + v$_{F}$p + µ E$_{↓}$ tanh E$_{↓}$ 2 T ] p p . (A2) C$_{11}$ [ E$_{1}$ − E (0) 1 ] = ∑ β C$_{1}$$_{β}$H$_{1}$$_{β}$ , (B1) C$_{13}$ E$_{1}$ − E (0) 3 = β C$_{1}$$_{β}$H$_{3}$$_{β}$ , (B2) C$_{12}$ [ E$_{1}$ − E (0) 2 ] = ∑ β C$_{1}$$_{β}$H$_{2}$$_{β}$ , (B3) C$_{14}$ E$_{1}$ − E (0) 4 = β C$_{1}$$_{β}$H$_{4}$$_{β}$ . (B4)