Moller Scattering. I would like to thank Paul Leonard Große-Bley for pointing out errors in the original version of this document.

Transcription

1 : Moller Scattering Particle Physics Elementary Particle Physics in a Nutshell - M. Tully February 16, 017 I would like to thank Paul Leonard Große-Bley for pointing out errors in the original version of this document. 1 Moller Scattering. Compute the differential cross section for e e e e. Figure 1: The tree-level Feynman diagrams describing the process e + e e + e. (Left) t-channel. (Right) u-channel. The tree-level diagrams for the process e e e e are shown in figure 1, the left diagram representing the t-channel and the right diagram representing the u-channel. 1.1 t-channel. From the diagram, we can write down the matrix element: M t = (eū 3 γ µ u 1 ) ig µν q (eū γ ν u ) = e q [ū 3γ µ u 1 ][ū γ µ u ], (1) where the propagator momentum q = p 1 p 3 = t. The modulus squared is M t t [ū 3γ µ u 1 ][ū γ µ u ][ū 3 γ ν u 1 ] [ū γ ν u ] () t [ū 3γ µ u 1 ][ū γ µ u ][ū 1 γ ν u 3 ][ū γ ν u ] (3) t [ū 3γ µ u 1 ū 1 γ ν u 3 ][ū γ ν u ū γ µ u ], () 1

2 where we ve used the identity: [ūγ ν v] = [u γ 0 γ ν v] = v γ ν γ 0 u = v (γ 0 ) γ ν γ 0 u = v(γ 0 γ ν γ 0 )u = vγ ν u, (5) which holds for u and v spinors. Since the quantities in the square brackets are C-numbers, we can take their traces with no repercussions: M t t Tr[ū 3γ µ u 1 ū 1 γ ν u 3 ] Tr[ū γ ν u ū γ µ u ] t Tr[u 1ū 1 γ ν u 3 ū 3 γ µ ] Tr[u ū γ ν u ū γ µ ]. (6) Now we must average over the initial spins and sum over the final spins: [ ] [ M t t Tr u s 1ū s 1γ ν 3 ū s 3 γ µ Tr u r ū r γ ] ν u r ū r γ µ r s s u s r, (7) we use the completeness relations for Dirac spinors: M t [ ] [ t Tr (/p 1 m e )γ ν (/p 3 m e )γ µ Tr (/p m e )γ ν (/p m e )γ µ], (8) here we will note that any term which is linear in the electron mass will contain an odd number of Dirac matrices inside a trace, which evaluates to zero. Therefore: M t { ] } { [ t Tr [/p 1 γ ν/p 3 γ µ + m e Tr [γ ν γ µ ] Tr { } { t p α 1 p β 3 Tr [γ αγ ν γ β γ µ ] + m e Tr [γ ν γ µ ] /p γ ν /p γ µ] } + m e Tr [γ ν γ µ ] p α p β Tr Using the trace identities of Dirac matrices (Peskin Appendix 3) this is [ γ α γ ν γ β γ µ] } + m e Tr [γ ν γ µ ] M t { } { [ t p α 1 p β 3 Tr [γ αγ ν γ β γ µ ] + g µν m e p α p β Tr γ α γ ν γ β γ µ] } + g µν m e the remaining traces are (9). (10), (11) p α 1 p β 3 Tr [γ αγ ν γ β γ µ ] = p α 1 p β 3 (g ανg βµ g αβ g νµ + g αµ g νβ ) (1) = (p 1ν p 3µ (p 1 p 3 )g νµ + p 1µ p 3ν ) (13) [ p α p β Tr γ α γ ν γ β γ µ] = p α p β Tr [γ α γ ν γ β γ µ] ( = p α p β g αν g βµ g αβ g νµ + g αµ g νβ) (1) = (p ν p µ (p p )g νµ + p µ pν ). (15) Now we must do the multiplication in the matrix element. The first term is the contraction of the above two traces, so using notation (1++3)(A+B+C), the terms are 1A : (p p 1 )(p p 3 ) (16) 1B : (p 1 p 3 )(p p ) (17) 1C : (p p 1 )(p p 3 ) (18) A : (p p )(p 1 p 3 ) (19) B : (p p )(p 1 p 3 ) (0) C : (p p )(p 1 p 3 ) (1) 3A : (p p 3 )(p p 1 ) () 3B : (p 1 p 3 )(p p ) (3) 3C : (p p 1 )(p p 3 ), () Page of 11

3 which we must sum (including an overall factor of ), note the factor of in B comes from g µν g µν =. Let us note the relations: so their sum is 1A = 3C 1C = 3A B + 1B + A + C + 3B = 0, (5) [(p p 1 )(p p 3 ) + (p p 1 )(p p 3 )]. (6) The second term (in the matrix element) is the contraction of the two mass terms: The remaining two terms are the cross-terms : and g µν g µν m e = [ m e]. (7) (p 1ν p 3µ (p 1 p 3 )g νµ + p 1µ p 3ν ) g µν m e = [ m e (p 1 p 3 p 1 p 3 + p 1 p 3 ) ] (8) = [ m ep 1 p 3 ], (9) (p ν p µ (p p )g νµ + p µ pν ) g µν m e = [ m e (p p (p p ) + p p ) ] (30) Gathering these terms, the matrix element is = [ m ep p ]. (31) M t = e t { (p p 1 )(p p 3 ) + (p p 1 )(p p 3 ) + m e m e(p 1 p 3 ) m e(p p ) }. (3) Again we will use our Mandelstam invariants, noting (in the massive case) With these, the matrix element squared is p 1 p = s m e = p 3 p (33) p 1 p 3 = t m e = p p (3) p 1 p = u m e = p p 3. (35) M t = e t { (p1 p )(p p 3 ) + ( )(p p 1 )( )(p p 3 ) + 8m e m e(p 1 p 3 ) m e(p p ) } = e t { (s m e ) + (u m e) + 8m e + m e(t m e) + m e(t m e) } = e { (s m t e ) + (u m e) + 8m e + m e(t m e) } = e { s t + m e sm e + u + m e um e + 8m e + m et 8m } e M t = e { s t + u + m e(t s u) + 8m } e. We recover, in the ultra-relativistic limit (i.e., massless electron): M t = e s + u t. (36) Page 3 of 11

4 1. u-channel. From the diagram, we can write down the matrix element: M u = (eū γ µ u 1 ) ig µν q (eū 3γ ν u ) = e q [ū γ µ u 1 ][ū 3 γ µ u ], (37) the analysis that follows will be abbreviated because the process is nearly identical as the t-channel. The modulus squared of the matrix element is ( ) e M u = [ū γ µ u 1 ][ū 3 γ µ u ][ū 1 γ ν u ][ū γ ν u 3 ] (38) q u [ū γ µ u 1 ][ū 1 γ ν u ][ū γ ν u 3 ][ū 3 γ µ u ] (39) u Tr[u 1ū 1 γ ν u ū γ µ ] Tr[u ū γ ν u 3 ū 3 γ µ ], (0) now we sum and average over final and initial spins, and use the completeness relations for the Dirac spinors: M u u Tr[( /p 1 + m)γ ν (/p + m)γ µ ] Tr[(/p + m)γ ν (/p 3 + m)γ µ ] (1) { ] } { [ u Tr [/p 1 γ ν/p γ µ + m e Tr [γ ν γ µ ] Tr /p γ ν /p 3 γ µ] } + m e Tr [γ ν γ µ ] () { } { [ u p α 1 p β Tr [γ αγ ν γ β γ µ ] + g µν m e p α p 3β Tr γ α γ ν γ β γ µ] } + g µν m e (3) the remaining traces are which contract to give The product of the mass terms is and the cross-terms are so the matrix element squared is p α 1 p β Tr[γ αγ ν γ β γ µ ] = p α 1 p β (g ανg βµ g αβ g νµ + g αµ g νβ ) () = (p 1ν p µ (p 1 p )g νµ + p 1µ p ν ) (5) p α p 3β Tr[γ α γ ν γ β γ µ ] = p α p 3β (g αν g βµ g αβ g νµ + g αµ g νβ) (6) = (p ν p µ 3 (p p 3 )g νµ + p µ pν 3), (7) [(p p 1 )(p 3 p ) + (p 3 p 1 )(p p )]. (8) [ m e], (9) (p 1ν p µ (p 1 p )g νµ + p 1µ p ν ) m eg µν = [ m e(p 1 p ) ] (50) (p ν p µ 3 (p p 3 )g νµ + p µ pν 3) m eg µν = [ m e(p p 3 ) ], (51) M u = e u { (p p 1 )(p 3 p ) + (p 3 p 1 )(p p ) + m e m e(p 1 p ) m e(p p 3 ) }. (5) Page of 11

5 Writing this in terms of Mandelstam invariants, we have M u = e u { (p p 1 )(p 3 p ) + ( )(p 3 p 1 )( )(p p ) + 8m e m e(p 1 p ) m e(p p 3 ) } = e u { (s m e ) + (t m e) + 8m e + m e(u m e) + m e(u m e) } M u = e u { s + t + m e(u s t) + 8m e We recover, in the ultra-relativistic limit (i.e., massless electron): }. M u = e s + t u. (53) It is interesting to note we could have found this result if we had made the substitutions p 3 p and p p 3 (note we do not need to consider a change of sign, because we are not dealing with any antiparticles), so the Mandelstam invariants become s = (p 1 + p ) (p 1 + p ) = s (5) t = (p 3 p 1 ) (p p 1 ) = u (55) u = (p p 1 ) (p 3 p 1 ) = t. (56) Page 5 of 11

6 1.3 Cross Terms. The first cross-term is M t M u = 1 [ ] [ ] e e t [ū 3γ µ u 1 ][ū γ µ u ] u [ū γ ν u 1 ] [ū 3 γ ν u ] tu tu spins (57) [ū 3 γ µ u 1 ][ū 1 γ ν u ][ū γ µ u ][ū γ ν u 3 ] (58) spins Tr[γ µ u 1 ū 1 γ ν u ū γ µ u ū γ ν u 3 ū 3 ] (59) spins tu Tr[γ µ(/p 1 + m)γ ν (/p + m)γ µ (/p + m)γ ν (/p 3 + m)] (60) tu g µαg νβ Tr[γ α (/p 1 + m)γ β (/p + m)γ µ (/p + m)γ ν (/p 3 + m)] (61) We can perform the multiplication inside the trace, and we obtain terms of all powers of m e up to m e. Note that any term that has an odd power of m e will vanish because the trace will contain an odd number of Dirac matrices. We are then left with: { M t M u tu g µαg νβ Tr[γ α /p 1 γ β /p γ µ /p γ ν /p 3 ] + m e Tr[γ α γ β γ µ γ ν ] } + m e Tr[γ α /p 1 γ β (/p γ µ γ ν + γ µ /p γ ν + γ µ γ ν /p 3 ) + γ α γ β (γ µ /p γ ν /p 3 + /p γ µ γ ν /p 3 + /p γ µ /p γ ν )] In the ultrarelativistic limit, where the electron mass vanishes, we retain just the first term: (6) M t M u tu g µαg νβ Tr[/p 1 γ β /p γ µ /p γ ν /p 3 γ α ], (63) where we have permuted the γ α to the end of the trace. For the sake of calculations, let s cast this in a different form, using p 1 p, p k, p 3 p, and p k, we can then write the cross term as M t M u tu g µαg νβ Tr[/pγ β /k γ µ /kγ ν /p γ α ] tu g µαg νβ Tr[/p γ α /pγ β /k γ µ /kγ ν ] (6) now relabel our indeces (swapping β and µ): M t M u tu g αβg µν Tr[/p γ α /pγ µ /k γ β /kγ ν ]. (65) Note the following relations which can be found in Appendix A. of Schwartz Quantum Field Theory and the Standard Model: γ µ γ µ = (66) γ µ γ ν γ µ = γ ν (67) γ µ γ ν γ ρ γ µ = g νρ (68) γ µ γ ν γ ρ γ σ γ µ = γ σ γ ρ γ ν, (69) Page 6 of 11

7 and note the fourth holds for any odd number of gamma matrices between two contracted gamma matrices. Therefore: M t M u tu g αβ Tr[/p γ α /pγ µ /k γ β /kγ µ ] = e tu g αβ Tr[/p γ α /p/k γ β /k] (70) Now in terms of the originally defined momenta: in terms of the Mandelstam invariants this is = e tu Tr[ /p γ α /p/kγ α /k ] = e tu gρσ p ρ k σ Tr[/p /k ] (71) = e tu (p k) Tr[ /p /k ] = 8 e tu (p k)(p k ). (7) Instead of calculating the other cross-term, I will claim M t M u = 8 e tu (p 1 p )(p 3 p ), (73) M t M u = 8 e ( s ( s ) = e tu ) s tu. (7) R{ M t M u } = M t M u + M t M u = R{ M t M u }, (75) so M t M u + M t M u = e s tu. (76) Page 7 of 11

8 1. Differential cross-section. For a two-body final state, the differential cross-section is given by (Peskin eq. 5.1) dω = 1 k Ecm 16π E M. (77) cm In the massless limit, center of mass frame, the differential cross-section simplifies to dω = 1 Ecm E cm / 16π E M = 1 cm Ecm Using the matrix elements from the massless limit sections, we have dω = 1 t dω = 1 u dω = 1 ut E cm E cm E cm 1 { (π) e s t + u } ( e = π 1 (π) e { s u + t } ( e = π 1 e ( ) (π) tu s = ( e π 1 (π) M. (78) ) 1 s + u s t = α s + u s t (79) ) 1 s + t s u = α s + t s u (80) ) 1 s s tu = s α s tu. (81) Now we need to calculate the sum of the differential cross-sections, but we need first determine their sign. Since there are identical fermions in the final state, we have that M = M t M u M = M t + M t M t M u M t M u (8) since under particle interchange, fermionic systems are odd. Then dω = dω + t dω u dω, (83) ut putting in our results: [ dω = α s + u s t + s + t ] u ( ) s = α tu s [ (u t ) + ( t u ) ( 1 + s u + 1 t + ) ], (8) tu which can be written [ (u dω = α ) ( ) t ( s s t u t + 1 ) ] u. (85) If we integrate over the azimuth, we acquire a factor of π, yielding [ (u d cos θ = πα ) ( ) t ( s s t u t + 1 ) ] u. (86) Let s investigate the angular dependence of the differential cross section. In the center of mass frame, we have p 1 = (E cm /, p) p 3 = (E cm /, k) (87) p = (E cm /, p) p = (E cm /, k), (88) Page 8 of 11

9 where p = k = E cm /. Using this, the Mandelstam invariants can be written s = p 1 p = E cm (89) t = p 1 p 3 = (p 0 1p 0 3 (p k)) (90) = E cm (1 cos θ) (91) u = p 1 p = (p 0 1p 0 (p k)) (9) = E cm (1 + cos θ). (93) Inserting these into the differential-cross section yields [ ( ) ( ) d cos θ = πα E cm (1 + cos θ) E cm + (1 cos θ) s E cm (1 cos θ) E cm (1 + cos θ) ( ) ] + Ecm 1 E cm (1 cos θ) + 1, (9) E cm (1 + cos θ) simplification yields [ ( ) d cos θ = πα 1 + cos θ ( ) 1 cos θ ( ) s 1 cos θ 1 + cos θ 1 cos θ + 1 ], (95) 1 + cos θ letting our old pal Mathematica handle the algebra, we obtain { ( dx = πα x + 3 ) } Ecm (x 1), (96) where x = cos θ. We can integrate the angular dependence out, to see the dependence on center of mass energy: { ( πα x + 3 ) } { σ = Ecm (x 1) dx = πα Ecm x 8x } x, (97) 1 which diverges at ±1. If we restrict measurement for particles with a scattering angle greater than 10, we (Mathematica) can evaluate the integral numerically: π { cos(10 ( 180 ) σ = πα x + 3 ) } cos(170 π 180 ) Ecm (x 1) dx = πα Ecm, (98) note the units are GeV, so we pick up a factor: ( ) σ = πα mb Ecm 1 GeV = mb. (99) (E cm /GeV) We now need to include a factor of 1/ due to the identical particles in the final state 1, so σ = 0.03 mb. (100) (E cm /GeV) For example, in colliding electron beams, each with momentum 000 GeV (so a center-of-mass energy of 8000 GeV), the total cross section (for scattering larger than 10 ) is σ = millibarn, or 0.53 pb. 1 In short, a detector can t tell the difference between which electron it detects, so to avoid double counting we need to include a factor of 1/ in the phase space factor. See Peskin pages , in general we include a factor of 1/n! for a final state with n identical particles. Page 9 of 11

10 CalcHep Comparison. We now move to the computational tool CalcHep to verify our result for the process ee ee in the Standard Model option, excluding diagrams propagated with a Z-boson. In our model we want to set the electron charge to e = πα π/137 = After starting n calchep, the IN state was set to give each particle a momentum of 000 GeV. In order to get convergent results, we apply a cut on A(e) with Min bound = 10 and Max bound = 170. The Monte Carlo simulation resulted in a value of σ = 0.53 pb, which is in excellent agreement with our calculated result. A second simulation was run, this time with a cut on the cosine of the scattering angle: 0.95 x The total cross-section calculated using equation 100 is pb. Again, in excellent agreement with the simulation: σ = pb. This time the angular distribution (equation 96) of the electron is plotted with respect to x = cos θ, shown in figure on linear and logarithmic scales. We see excellent agreement between the formula and simulation for the angular distribution (the agreement becomes worse at large angles). For the angular distribution, wee did not need to include the factor of for identical particles, because the 1/n! factor only enters when one integrates over the full phase-space. Thus it has no effect on the differential cross-section. Page 10 of 11

11 Figure : Angular distribution of one of the final state electrons, comparison on CalcHep simulation and analytic calculation. Shown on both linear and logarithmic scales. Page 11 of 11