3 the latter case the solvability condition Eq. (20) becomes bi-quadratic and yields the energy spectrum The spinors ˆ a$_{↑}$ and ˆ a$_{↓}$ are eigenstates of excitations in the normal graphene. We also introduce vectors in the Nambu space, In this limit, the Bogoliubov–de Gennes equations can also be solved analytically (see Appendix B 2). One sees that the energy, Eq. (21), for µ = 0 does not have the usual Doppler term proportional to the vector k . This may lead to a confusion 15 when calculating the supercur- rent. Each component here is a pseudo-spinor. We find for the upper sign in Eq. (11) E 2 ± = | Δ | 2 + v 2 $_{F}$( p 2 + k $^{2}$/ 4) ± √ | Δ | $^{2}$v 2 $_{F}$k 2 + v 4 $_{F}$( p · k ) 2 . (21) ˇ ψ = ( ˆ u ˆ v ) , ˇ ψ + = ( ˆ u † , ˆ v $^{†}$) . Let us consider the linear correction to the energy and to the wave functions due to superconducting momentum k assuming v$_{F}$ | k | ≪ µ . We put E = E (0) + E ′ where E ′ ≪ E (0) and E (0) is the energy of one of the states with k = 0 determined by Eqs. (12), (14). Within the linear approximation in k we find from Eq. (20) for any finite µ = 0 A. Linear response For k = 0 the order parameter is real Δ = | Δ | . Therefore, E (0) 1 , 2 = ± E$_{↑}$ , E$_{↑}$ = √ ( v$_{F}$ p − µ ) 2 + | Δ | 2 . (12) ( ˆ u (0) 1 ˆ v (0) 1 ) = ( u$_{↑}$ v$_{↑}$ ) ˆ a$_{↑}$e i p · $^{r}$, ( ˆ u (0) 2 ˆ v (0) 2 ) = ( v$_{↑}$ − u$_{↑}$ ) ˆ a$_{↑}$e i p · $^{r}$. (13) For the lower sign in Eq. (11) we have and E (0) 3 , 4 = ± E$_{↓}$ , E$_{↓}$ = √ ( v$_{F}$ p + µ ) 2 + | Δ | 2 , (14) ( ˆ u (0) 3 ˆ v (0) 3 ) = ( u$_{↓}$ v$_{↓}$ ) ˆ a$_{↓}$e i p · $^{r}$, ( ˆ u (0) 4 ˆ v (0) 4 ) = ( v$_{↓}$ − u$_{↓}$ ) ˆ a$_{↓}$e i p · $^{r}$. (15) Here for the upper (lower) sign in Eq. (11). Therefore, correc- tions to the energies are E ′ = ± v$_{F}$ ( p · k ) / 2 p ≡ ± E$_{D}$ E (1) 1 , 2 = − E (1) 3 , 4 = E$_{D}$ . (22) The energy E$_{D}$ = ( dξ$_{p}$/dp )( k / 2) is the usual Doppler shift for the normal-state energy ξ$_{p}$ = v$_{F}$ p . Equation (22) coincides with the result of Ref. 10 obtained in the linear approximation in k . At the same time, it differs from the linear in k term obtained from Eq. (21) for µ = 0. This means that the undoped case µ = 0 requires a special consideration. This will be done later in Section III B (see also Appendix B). First-order corrections to the wave functions can be found by expanding the total functions in terms of the zero-order functions ˆ u (0) β , ˆ v (0) β given by Eqs. (13), (15): The different wave functions are orthogonal, ˇ ψ + α ˇ ψ$_{β}$ = δ$_{αβ}$ . Equation (13) goes over into Eq. (15) under the transformation E → − E and µ → − µ . Using Eq. (10) one can check that where z$_{0}$ is the unit vector in the z direction perpendic- ular to the graphene layer plane. Inserting this into Eqs. (1), (2) we find ˇ ψ$_{α}$ = ˇ ψ (0) α + ∑ β = α B$_{αβ}$ ˇ ψ (0) β . (23) u$_{↑}$ = 1 √$_{2}$ √ 1 + v$_{F}$ p − µ E$_{↑}$ , v$_{↑}$ = 1 √$_{2}$ √ 1 − v$_{F}$ p − µ E$_{↑}$ , (16) u$_{↓}$ = 1 √$_{2}$ √ 1 − v$_{F}$ p + µ E$_{↓}$ , v$_{↓}$ = 1 √$_{2}$ √ 1 + v$_{F}$ p + µ E$_{↓}$ . (17) ˆ a † $_{↑}$ˆ σ ˆ a$_{↑}$ = − ˆ a † $_{↓}$ˆ σ ˆ a$_{↓}$ = p /p , (18) ˆ a † $_{↑}$ˆ σ ˆ a$_{↓}$ = − ˆ a † $_{↓}$ˆ σ ˆ a$_{↑}$ = i [ z$_{0}$ × p ] /p . (19) III. CURRENT-CARRYING STATE For the current-carrying state the solvability condition of the Bogoliubov–de Gennes equations (8), (9) takes the form where p$_{±}$ = p ± k / 2. Equation (20) cannot be solved analytically for nonzero k , except for the zero doping limit µ = 0. In One can check that B$_{βα}$ = − B ∗ $_{αβ}$. We find B$_{12}$ = B$_{21}$ = B$_{34}$ = B$_{43}$ = 0 while B$_{αβ}$ = v$_{F}$ ˇ ψ (0)+ β (ˆ σ · k ) ˇ ψ (0) α 2( E (0) α − E (0) β ) . B$_{13}$ = − B$_{24}$ = − iv$_{F}$ ([ p × k ] · z ) 2 p ( u ∗ $_{↓}$u$_{↑}$ + v ∗ $_{↓}$v$_{↑}$ ) E$_{↑}$ − E$_{↓}$ , (24) B$_{23}$ = B$_{14}$ = iv$_{F}$ ([ p × k ] · z ) 2 p ( u $_{↓}$v$_{↑}$ − v $_{↓}$u$_{↑}$ ) E$_{↑}$ + E$_{↓}$ . (25) ( E 2 − µ $^{2}$)2 − 2 | Δ | $^{2}$( E 2 − µ $^{2}$) + | Δ | 4 + 2 | Δ | $^{2}$v2 $_{F}$p$_{+}$ p$_{−}$ − ( E + µ ) $^{2}$v2 $_{F}$p 2 − − ( E µ ) $^{2}$v2 $_{F}$p 2 + + v 4 $_{F}$p 2 $_{+}$p 2 − = 0 , (20)