8 Appendix B: Derivation of Eqs. 7a and 7b 0 1 2 3 x 10 −11 −0.4 −0.2 0 0.2 0.4 0.6  [eV−m] V X  (norm.) Numerical Integral Analytic expression 0 1 2 3 x 10 −11 −0.4 −0.2 0 0.2 0.4 0.6  [eV−m] V Y  (norm.) Numerical Integral Analytic expression FIG. 3: Left : Numerical calculation (squares) of Eqs. 6, 5a versus analytical expression (solid) in Eq. 7a as a function of α . Right : Numerical calculation (squares) of Eqs. 6, 5b versus analytical expression (solid) in Eq. 7b as a function of α . From Eqs. 6 and 5a, Noting that the phase has a stationary point at s = 0 [11], we expand it in Taylor’s series around s = 0 to obtain as stated in Eq. 7a. Similarly from Eqs. 6 and 5b, as stated in Eq. 7b. V$_{Y}$ = B π ∫ 1 0 d s (1 − s $^{2}$) + B π Re {∫ 1 0 d s s $^{2}$exp ( iθ$_{L}$ √$_{1}$ − s 2 )} ≃ 2 B 3 π V$_{X}$ ≃ B 3 π + B π Re [∫ 0+ ϵ 0 d s (1 − s $^{2}$) exp { iθ$_{L}$ ( 1 + s 2 2 )}] ≃ B 3 π + B π Re { exp( iθ$_{L}$ ) ∫ ∞ 0 d s exp ( iθ$_{L}$ s 2 2 )} = B 3 π + B π Re exp( iθ$_{L}$ ) exp( i π $_{4}$) √$_{2}$$_{θ}$$_{L}$ Γ ( 1 2 )} = B 3 π + B √$_{2}$$_{πθ}$$_{L}$ cos( θ$_{L}$ + π 4 ) V$_{X}$ = 1 π ∫ 1 0 d s V$_{X}$$_{0}$ ( s ) = B π 1 0 d s s 2 + B π Re {∫ 1 0 d s (1 − s $^{2}$) exp ( iθ$_{L}$ √$_{1}$ − s 2 )}