1 Spinor-Scalar Scattering in Yukawa Theory

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1 Physics 610 Homework 9 Solutions 1 Spinor-Scalar Scattering in Yukawa Theory Consider Yukawa theory, with one Dirac fermion ψ and one real scalar field φ, with Lagrangian L = ψ(i/ m)ψ 1 ( µφ)( µ φ) M φ λ 4 φ4 φ ψ(y iγ 5 y )ψ. (1) Assume that y and y are small but comparable to each other. Consider theprocessinwhichafermionofmomentumpscattersoffascalarofmomentum k, to the final state containing a fermion of momentum p and scalar of momentum k. 1.1 Matrix element At lowest order in the Yukawa couplings (which will be second order), the scattering arises from two diagrams. Draw the diagrams, and use them to write an expression for the matrix element for the scattering process to this order Solution The two diagrams are The contribution of the first diagram to the matrix element is M 1 = iū(p,σ )( iy γ 5 y i( /p /k +m) ) (p+k) +m ( iy γ 5y )u(p,σ) () and the contribution of the second diagram is M = iū(p,σ )( iy γ 5 y ) i( /p+/k +m) (p k ) +m ( iy γ 5y )u(p,σ). (3) 1. Nonrelativistic case Consider the case where p,k are both close to rest, p µ = (E,0,0,p z ) with p z E m and k µ = (E,0,0, p z ) with E M p z. For this case, at lowest (zero) order in p z, use the explicit expressions for the spinors u(p,σ) and ū(p,σ) and for the gamma matrices to simplify your expression for the matrix element for each of the four cases: 1

2 initial spin and final spin, initial spin and final spin, initial spin and final spin, initial spin and final spin. Your result should depend only on y, y, and the two masses Solution In this case (p + k) = (m + M) and (p k ) = (M m) so the denominators become mm M and +mm M respectively. /p = +mγ 0 and /k = +Mγ 0 while +/k = Mγ 0. The spinors are explicitly 1 0 u(p, ) = 0 m, u(p, ) = 1 m, ū(p, ) = m [ 1010 ], ū(p, ) = m [ 0101 ] (4) The explicit expression for the case is M 1, = iy y m [ ] 0 iy y mm M 0 0 iy +y iy +y m 0 m+m 0 0 m 0 m+m m+m 0 m 0 0 m+m 0 m iy y iy y iy +y iy +y 0 = 4y m +(y +y )mm mm +M = y m M +y m m+m. (5) The other three cases can be evaluated by changing which row and column matrices we put on the ends. The result for is identical, while for and for we find zero.

3 The result for M is the same but with M M; that is, M, = 4y m (y +y )mm mm +M = y m M +y m m M. (6) Again the case of is the same and the other two cases are zero. Summing the contributions, M = M = 8m y Only y contributes in the nonrelativistic limit. 4m M. (7) 1.3 Spin averaging The book presents a derivation for the case where y = 0 but y 0. So consider instead the case y = 0 but y 0. For this case, consider generic p,k (make no assumption about the relative size compared to the two masses). Compute the squared matrix element, averaged over the initial spin and summed over the final spin of the fermion. Call it M. Express your result in terms of the Mandelstamm variables and the particle masses Solution Write (p+k) +m = m s and (p k ) +m = m u. So our expression for the total matrix element is M = y ū(p,σ )γ 5 ( /p /k +m m s ) + /p+/k +m γ 5 u(p,σ). (8) m u Move the γ 5 across the quantity in large parenthesis; it reverses the sign of the gamma matrices, and γ 5 = 1, so M = y ū(p,σ ) ( /p+/k +m m s + /p ) /k +m u(p,σ). (9) m u Now (/p+m)u(p,σ) = 0 by the Dirac equation, which simplifies the expression: ( ) M = y ū(p /k,σ ) m s + /k u(p,σ). (10) m u its conjugate is ( ) M = y ū(p,σ) /k m s + /k u(p,σ ) (11) m u and the spin summed and averaged product is M = y 4 ( ) Tr /k m s + /k ( /p+m) m u ( ) /k m s + /k ( /p +m). (1) m u 3

4 There is a term from m and a term with no numerator m factors; the term with one m factor is odd in gamma matrices and gives zero. The m term is ( k (y ) 4 m (m s) + k k ) (m s)(m u) + k (m u) but k = M and k = M while k k = t M. So these terms are ( (y ) 4 m M (m s) + M (m u) + t M ). (14) (m s)(m u) The term without m is ( k pk p (y ) 4 k p p + k pk p k p p k pk p +k p k p k k p p ) (m s) (m u) (m s)(m u) (15) which using p p = t m (13), (16) k k = t M, (17) p k = p k = m +M s, (18) p k = p k = u m M, (19) and k = k = M, and using t = s u+m +M, becomes 1.4 Limits M = y 4(s u) ( su+m(s+u)+m m ) (s m ) (m u). (0) Find the limit of M from the last subsection, in each of the following limits: the nonrelativistic limit the ultra-relativistic limit in which one neglects both masses: m,m s, t, u. Verify that the nonrelativistic limit coincides with the spin averaged and summed square of what you found directly two subsections ago. 4

5 1.4.1 Solution In the nonrelativistic limit s = (m + M) and u = (M m), so (s u) = 4mM and su = (M m ) while s+u = (m +M ). So our expression simplifies to M = 64m4 y 4 (4m M ). (1) This is the square of M ; since M is zero it does not contribute to the spin summed, squared matrix element. In the ultra-relativistic limit we set m = 0 and M = 0, finding M = y 4(s u) su Since u < 0 and s > 0 these terms are each positive. = (y ) 4( s u u s +). () 1.5 Total cross-section Consider the ultra-relativistic limit. Work in the center of mass frame. Express the squared matrix element as a function of s and of the cosine of the angle between p and p, cosθ pp. (It might help to first find expressions for t,u in terms of this angle.) Express the cross section as an integral over the angle θ pp, of a function of cosθ pp. Don t do the integral (it may result in a log divergence!) Solution As we saw in class, the ultra-relativistic limit has so the squared matrix element is The spin sum-averaged cross section is σ = 1 v 1 v p 0 k 0 u s = 1+cosθ pp, (3) M = y 4(3+cosθ pp ) (1+cosθ pp ). (4) d 3 p d 3 k (π) 4 δ 4 (p+k p k ) M. (5) (π) 6 p 0k 0 Now v 1 v = and p 0 k 0 = s, likewise there is an s in the denominator of the integral. We use the spatial delta function to perform the k integrals, and express the p integral in spherical coordinates: σ = 1 s π 0 π p dφ sinθ pp dθ dp pp 0 0 (π) s δ(p0 p 0 )y 4(3+cosθ pp ) (1+cosθ pp ) (6) 5

6 The φ integral gives π and the p integral gives (p 0 ) / = s/8. Rewrite the θ integral as an integral over cosθ pp : σ = y 4 +1 s π 8 1 (3+cosθ pp ) dcosθ pp (1+cosθ pp ) (7) which is the form we are seeking. Note that this integral diverges as cosθ pp 1. 6