revealing form. As a first step towards this, we note that the spin vector may be expressed as s µ = λ m$_{ν}$ ( k, E$_{ν}$ ˆ k ) . (11) Using this and Eq. (10), we write K$_{1}$ = η$_{1}$ (1 , ˆ k ); K$_{2}$ = η$_{2}$ (1 , − ˆ k ); (12) with η$_{1}$ = E$_{ν}$ + ( E 2 ν − m 2 $_{ν}$) 1 / $^{2}$; η$_{2}$ = E$_{ν}$ − ( E 2 ν − m 2 $_{ν}$) 1 / $^{2}$. (13) Note that for m$_{ν}$ << E$_{ν}$ we have η$_{1}$ ≈ 2 E$_{ν}$ ; η$_{2}$ ≈ m 2 ν 2 E$_{ν}$ . (14) We now evaluate the traces given in Eqs. (5) and (6) and the contraction N $^{µν}$E$_{µν}$ is given by N $^{µν}$E$_{µν}$ = 8 ( a 2 + b $^{2}$) {[ a $^{2}$( g e $_{V}$) 2 + b $^{2}$( g e $_{A}$) + 4 a $^{2}$b2 ( a 2 + b $^{2}$) g e $_{V}$g e A ] ( p$_{1}$ · K$_{1}$ )( p$_{2}$ · K$_{2}$ ) + [ a $^{2}$( g e $_{V}$) 2 + b $^{2}$( g e $_{A}$) − 4 a $^{2}$b2 ( a 2 + b $^{2}$) g e $_{V}$g e A ] ( p$_{1}$ · K$_{2}$ )( p$_{2}$ · K$_{1}$ ) + [ a $^{2}$( g e $_{V}$) 2 − b $^{2}$( g e $_{A}$) ] m 2 $_{e}$( K$_{1}$ · K$_{2}$ ) } , (15) where g e V = − 1 2 + 2 sin $^{2}$θ$_{W}$ and g e A = − 1 $_{2}$. From this expression and Eqs. (12) and (14) above, we see that the amplitude of transition vanishes for massless neutrino, as expected. Furthermore, Eq. (15) is taken as the usual weak pair production amplitude with the replacement K$_{i}$ → k$_{i}$ . From Eqs. (4) and (15) the explicit form for the squared transition amplitude is 1 2 × 1 2 |M| 2 = G 2 F ( a 2 + b $^{2}$) {[ a $^{2}$( g e $_{V}$) 2 + b $^{2}$( g e $_{A}$) + 4 a $^{2}$b2 ( a 2 + b $^{2}$) g e $_{V}$g e A ] ( p$_{1}$ · k$_{1}$ )( p$_{2}$ · k$_{2}$ ) 6