Cross-sections. Kevin Cahill for 524 January 24, There are two standard ways of normalizing states. We have been using continuum normalization

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1 Cross-sections Kevin Cahill for 54 January 4, Fermi s Golden Rule There are two standard ways of normalizing states. We have been using continuum normalization p p = δ (3) (p p). (1) The other way is box normalization. We put the universe in a space-time box of duration T and width L. Our possible momenta are p = π L (n 1, n n 3 ) () so that the wave function is single valued when we identify points on opposite sides of the box. So our delta functions become δ (3) (p p) = e i(p p) x d3 x (π) = 3 (π) δ 3 p,p and δ T (E E) = e i(e E) t dt T π. (3) We can normalize our box states p = (π) 3 p (4) 1

2 so that now Similarly, a two-particle box state is and (if the particles are different) p p = δ p,p. (5) p 1, p = (π)3 p 1, p, (6) p 1, p p 1, p = δ p 1,p 1 δ p,p. (7) If they are the same, then p 1, p p 1, p = δ p 1,p 1 δ p,p ± δ p 1,p δ p,p 1 (8) with ± = ( 1) j where j is the spin of the particles. Following Weinberg (section 3.4 of TQTOF1), the probability of a scattering process in which N i incoming particles become N o outgoing particles is P (i o) = S oi (9) in which the S-matrix element S oi in box normalization is related to the one in continuum normalization by ( (π) 3 S oi = ) (No+N i )/ S oi. (10) So ( ) (π) P (i o) = S oi 3 No+Ni = S oi. (11) It follows from our formula () for the possible momenta that the number of outgoing states in do d 3 p 1... d 3 p N o (1) is dn o = ( ) No do. (13) (π) 3

3 So the probability of the process i do is ( ) (π) 3 No+Ni ( ) No ( ) (π) dp (i o) = P (i o) dn o = S oi 3 Ni do = S oi do (14) (π) 3 which is where we stopped on Tuesday. It is conventional to write the S-matrix element as S oi iπ δ 3 (p o p i ) δ T (E o E i ) M oi (15) in which the matrix element M oi is free of delta functions (we are ignoring processes that are disconnected). The squares of our box delta functions are [ δ 3 (p o p i ) = δ 3 (p o p i ) δ 3 (0) = δ 3 (p o p i ) /(π) 3 [δ T (E o E i ) = δ T (E o E i ) δ T (0) = δ T (E o E i ) T/π. (16) So dp (i o) is ( ) (π) 3 Ni ( ) (π) dp (i o) = S oi 3 Ni [ do = π δ 3 (p o p i ) δ T (E o E i ) M oi do ( ) (π) = (π) 3 Ni 1 T π M oi δ 3 (p o p i ) δ T (E o E i ) do. (17) To get the transition rate, we divide by the time T dγ(i o) dp (i o)/t = (π) 3Ni 1 N i M oi δ 4 (p o p i ) do (18) in which we have S oi iπ δ 4 (p o p i ) M oi (19) as we let the box swell to infinity. These last two equations are the keys to cross-section formulas. 3

4 Two cases are of special importance. The case N i = 1 applies to a single particle that decays into N o particles. Here the transition rate is dγ(i o) = π M oi δ 4 (p o p i ) do. (0) The case N i = describes the standard scattering process in which two incoming particles turn into N o outgoing particles. The rate is dγ(i o) = (π) 4 1 M oi δ 4 (p o p i ) do, (1) but we must divide by the flux F of incoming particles. The flux at particle due to a density 1/ of particle 1 coming at speed v is F = v. () The rate divided by the flux is the differential cross-section = dγ(i o)/f = (π) 4 1 M oi δ 4 (p o p i ) do/f = (π) 4 1 M oi δ 4 (p o p i ) do /v = (π) 4 v 1 M oi δ 4 (p o p i ) do. (3) In a general Lorentz frame, the relative speed is v = (p1 p ) m 1m E 1 E. (4) Homework problem 1: Find a simple expression for v when p = (m, 0). Homework problem : Show that a simple expression for v for the case in which p 1 + p = 0 is v = ke E 1 E (5) in which k = p 1 and E = E 1 + E. This speed can exceed unity and can approach. 4

5 The product δ 4 (p o p i ) do is called the phase-space factor. We will find it easiest to compute in the center-of-mass frame p i = 0. There the delta function δ 3 (p o p i ) = δ 3 (p o) lets us drop d 3 p k for one of the N o outgoing particles as long as we remember to set The simplest case is N o =. We set and find that To evaluate this expression, we use the delta-function identities p k = (p p k 1 + p k p N o ). (6) δ(ax) = 1 a p 1 = p (7) δ 4 (p o p i ) do δ(e 1 + E E) d 3 p. (8) and δ(f(x)) = x 1 f (x ) δ(x x ) (9) in which the sum is over the zeros x of f(x). Since p 1 = p k, we must solve the equation E = E 1 + E = k + m 1 + k + m. (30) It has only one root k (E m 1 m ) = 4m 1 m E or if the masses are equal, m 1 = m m, simply (31) k = 1 E 4m. (3) Homework problem 3: Derive this formula (31) for the root k. So the derivative we need is ( ) d k dk + m 1 + k + m = k + k E 1 E = k E E 1E (33) 5

6 and so with d 3 k = k dk dω we have δ 4 (p o p i ) do δ(e 1 + E E) d 3 k = E 1E k E k dω = E 1E E k dω. (34) In these formulas, k is given by (31) or (3). So the two-particle decay rate into solid angle dω is from (0) dγ(i o) = π M oi δ 4 (p o p i ) do = π M oi E 1E E k dω. (35) Similarly, the cross-section for scattering in the center-of-mass frame is from (3) and the result (5) of problem or = (π) 4 v 1 M oi δ 4 (p o p i ) do = (π) 4 E 1E ke M oi E 1E E k dω (36) dω = (π)4 k E 1E E 1 E k E M oi. (37) Muon Pair Production As an example of how to compute M oi, let s recall the amplitude for the process e + + e µ + + µ p, s ; q, t U p, s; q, t = ie (π) u µ(p, s )γ a v µ (q, t )v e (q, t)γ a u e (p, s) δ(p + q p q) (p + q) iɛ = πiδ(p + q e p q) (π) u µ(p, s )γ a v 3 µ (q, t 1 )v e (q, t)γ a u e (p, s) (p + q) iɛ (38) which tells us that M oi = e (π) u µ(p, s )γ a v 3 µ (q, t 1 )v e (q, t)γ a u e (p, s) (p + q) iɛ. (39) 6

7 The x-section is dω = k E 1E (π)4 E 1 E M k E oi. (40) This formula becomes much simpler if we average over the initial spins and sum over the final spins dω = 1 4 s,t,s t (π) 4 k E 1E E 1 E M k E oi. (41) which makes sense because we rarely know the initial spins and rarely measure the final spins. We have carried out the spin sums in section 4 of the notes on perturbation theory. There we found that 1 4 e4 (π) u µ(p, s )γ a v 6 µ (q, t )v e (q, t)γ a u e (p, s) 1 s,t,s t M oi = 1 4 [(p + q) s,t,s t = 1 [ e 4 Tr ( i /q m µ )γ a ( i/p + m µ )γ b Tr [ ( i/q m e )γ a ( i/p + m e )γ b. 4 (π) 6 p 0 q 0 p 0 q 0 (p + q) 4 (4) So combining the last two equations, we get dω = 1 4 s,t,s t (π) 4 k E 1E E 1 E M k E oi = 1 k E 1E 4 (π)4 E 1 E e 4 Tr k E (π) 6 (16π) p p E Tr [ ( i /q m µ )γ a ( i/p + m µ )γ b Tr [ ( i/q m e )γ a ( i/p + m e )γ b p 0 q 0 p 0 q 0 (p + q) [ 4 ( i /q m µ )γ a ( i/p + m µ )γ b Tr [ ( i/q m e )γ a ( i/p + m e )γ b (p + q) 4. (43) Now in the center-of-mass frame p + q = p 0 + q 0 = E, so we have dω = e4 k (16π) k E Tr [ ( i/q m 6 µ )γ a ( i/p + m µ )γ b Tr [ ( i/q m e )γ a ( i/p + m e )γ b. (44) 7

8 3 Traces The gamma matrices in Weinberg s notation are ( ) 0 1 γ 0 = i 1 0 and they satisfy the anticommutation relation and γ = i ( ) 0 σ, (45) σ 0 {γ a, γ b } = η ab. (46) The fifth gamma matrix is γ 5 = γ 5 = iγ 0 γ 1 γ γ 3 = iγ 0 γ 1 γ γ 3 = and it anticommutes with the other four gamma matrices ( ) (47) {γ 5, γ a } = 0. (48) Since γ 5 = 1, we may think of it as a fourth spatial gamma matrix and write {γ α, γ β } = η αβ for α, β = 0, 1,, 3, 4 with η 00 = 1, η kk = 1 for k = 1 4, and η zero otherwise. Homework problem 4: erify equations (46 & 48). We can use these properties of the gamma matrices to show that the trace of an odd number of gamma matrices vanishes Tr [γ a 1... γ a n+1 = 0. (49) To see why, we note that since γ 5 anticommutes with the other four gamma matrices (48), we have γ a = γ 5 γ a γ 5. (50) Thus, if we substitute this expression for every gamma matrix in the trace (49), then we find that all the γ 5 s cancel and so that Tr [γ a 1... γ a n+1 = Tr [( γ 5 γ a 1 γ 5 )... ( γ 5 γ a n+1 γ 5 ) = ( 1) n+1 Tr [γ a 1... γ a n+1 (51) 8

9 which shows that the trace of an odd number of gamma matrices vanishes. Since Trγ a = 0, this result is obviously true for n = 0. Physics would be so much simpler if the trace of an even number of gamma matrices also vanished. Instead, we find Tr ({ γ a, γ b}) = Tr ( γ a γ b) = Tr ( η ab) = 8 η ab, (5) so the trace of two gamma matrices is Tr ( γ a γ b) = 4η ab. (53) The trace of four gamma matrices, all different, is proportional to the trace of γ 5 which vanishes. So the trace of four gamma matrices is nonzero only if two of their indices are the same. But then since Tr ( γ a γ b) = 4η ab, the trace also will vanish unless the other two indices also are equal. Thus, for instance, since (γ c ) = η cc, we have Tr ( γ a γ a γ b γ ) b = Trη aa η bb = 4 η aa η bb. (54) So if a and b are the same, then If instead, a b, then γ a γ b = γ b γ a, and so while So we have Tr (γ a γ a γ a γ a ) = Trη aa η aa = 4 η aa η aa = 4. (55) Tr ( γ a γ b γ a γ b) = Tr ( γ a γ a γ b γ b) = 4 η aa η bb (56) Tr ( γ a γ b γ b γ a) = Tr ( γ a γ b γ a γ b) = Tr ( γ a γ a γ b γ b) = 4 η aa η bb. (57) Tr ( γ a γ b γ c γ d) = 4 ( η ab η cd η ac η bd + η ad η bc). (58) In general, the trace of an even number of gamma matrices will vanish unless all the indices match in pairs in which case it is ±4. Computers programs exist that compute such traces. 9

10 4 Cross-section for Muon Pair Production We must still evaluate the traces T 1 T = Tr [ ( i/q m µ )γ a ( i/p + m µ )γ b Tr [ ( i/q m e )γ a ( i/p + m e )γ b. (59) The first trace is T 1 = Tr [ ( i/q m µ )γ a ( i/p + m µ )γ b = Tr [ /q γ a /p γ b m µtrγ a γ b = q c p d Tr [ γ c γ a γ d γ b ( 4m µη ab = 4q c p d η ca η db η cd η ab + η bc η ad) 4m µη ab = 4 ( q a p b q p η ab + q b p a + m µη ab). (60) And the second one is T = Tr [ ( i/q m e )γ a ( i/p + m e )γ b = Tr [ / qγ a/pγ b m e Trγ a γ b So the product of the traces is = q e p f Tr [γ e γ a γ f γ b 4m eη ab = 4q e p f (η ea η fb η ef η ab + η be η af ) 4m eη ab = 4 ( (61) ) q a p b q p η ab + q b p a + m eη ab. T 1 T = 16 ( q a p b q p η ab + q b p a + m µη ab) ( q a p b q p η ab + q b p a + m eη ab ) = 3 [ (q q) (p p) + (q p) (p q) m µ (q p) m e (q p ) + m µm e. (6) So the differential x-section in the c-o-m frame is dω = e4 k (16π) k E Tr [ ( i/q m 6 µ )γ a ( i/p + m µ )γ b Tr [ ( i/q m e )γ a ( i/p + m e )γ b (16π) k k E 6 3 [ (q q) (p p) + (q p) (p q) m µ (q p) m e (q p ) + m µm e 8π k k E 6 [ (q q) (p p) + (q p) (p q) m µ (q p) m e (q p ) + m µm e (63) 10

11 which has the right dimensions of area because the dimension of energy is inverse length in natural units. In the c-o-m frame, with p = (E 1, p), q = (E, q) = (E 1, p), p = (E 1, p ), q = (E, q ) = (E 1, p ), p p = cos θ kk, q q = cos θ kk, E 1 = E, and E 1 = E, we have q q = kk cos θ E 1 E 1 p p = p p E 1 E 1 = kk cos θ E 1 E 1 = q q q p = q p E E 1 = kk cos θ E 1 E 1 p q = kk cos θ E 1 E 1 = q p q p = k E 1 q p = k E 1. (64) So the differential x-section in the c-o-m frame is dω 8π k k E 6 [ (kk cos θ E 1 E 1) + ( kk cos θ E 1 E 1) m µ ( k E 1 8π k k E 6 [ k k cos θ + E 1E 1 + m µ ( k + E 1 ) ( ) + m e k + E 1 + m µ m e. In the limit in which m e /m µ 1/00 0, we have E 1 = k and E = k, so ) ( ) m e k E 1 + m µ me (65) dω k [ ( ) k k cos θ + E 8π k E 1E m µ k + E1 8π k k (k) 6 [ k k cos θ + k E 1 + m µk 8 π k k 5 [ k cos θ + E 1 + m µ 8 π k k 5 [ k ( 1 + cos θ ) + m µ = e 4. k [ k cos θ + k + m 8 π k 5 µ (66) 11

12 Now E 1 = E 1 = k, and k = k m µ, so dω = e4 k 8 π 8 π 8 π 8 π 8 π 8 π k ) ( 1 + cos θ ) + m µ [( k m k 5 µ k m µ [( ) ( k m k 5 µ 1 + cos θ ) + mµ 1 m µ/k [( k m k 4 µ 1 m µ/k ) ( 1 + cos θ ) + m µ ( 1 cos θ ) [ ( k 1 + cos θ ) + m k 4 µ 1 m µ/k [( 1 + cos θ ) + (m k µ/k ) ( 1 cos θ ) [( ) ( ) 1 m µ 1 + m µ + 1 m µ cos θ. k k k Somewhat more simply, with E E 1 = k and Ecm = k, we have [( ) dω = e 4 1 m µ 1 + m µ + 6 π Ecm E E With α e /4π 1/137.04, this is dω = α ( 1 m µ E [( ) ( ) 1 + m µ + 1 m µ cos θ E E 1 m µ 4Ecm E which is (5.1) of Peskin & Schroeder. Integrating over dω, we get for the total x-section ) which is their (5.13). σ = 4πα 3E cm 1 m µ E 1 ( m µ E (67) ) cos θ. (68) (69) (70)

13 5 Tau Pairs We can immediately deduce from these formulas those that apply to the process e + + e τ + + τ. Since the tau and muon are just heavy copies of the electron, all we need do is to replace m e by m τ : [( ) ( ) dω = α 1 m τ 1 + m τ + 1 m τ cos θ (71) 4Ecm E E E and Measurements of this x-section in 1978 gave m τ = Me. 6 Gauge Invariance ( σ = 4πα 1 m τ ) m τ. (7) 3Ecm E E The reason we could generalize our formulas for muon pair production to tau pair production is that all the charged leptons are coupled to the photon in the same way. Although electrodynamics is an abelian gauge theory, we might as well consider the general case of a nonabelian gauge theory. The action density of a Yang-Mills theory is unchanged when a space-time dependent unitary matrix U(x) changes a vector ψ(x) of matter fields to ψ (x) = U(x)ψ(x). Terms like ψ ψ are invariant because ψ (x)u (x)u(x)ψ(x) = ψ (x)ψ(x), but how can kinetic terms like i ψ i ψ be made invariant? Yang and Mills introduced matrices A i of gauge fields, replaced ordinary derivatives i by covariant derivatives D i i + A i, and required that D iψ = UD i ψ or that Their nonabelian gauge transformation is ( i + A i) U = i U + U i + A iu = U ( i + A i ). (73) ψ (x) = U(x)ψ(x) A i(x) = U(x)A i (x)u (x) ( i U(x)) U (x). (74) 13

14 One often writes the unitary matrix as U(x) = exp( ig θ a (x) t a ) in which g is a coupling constant, the functions θ a (x) parametrize the gauge transformation, and the generators t a belong to the representation that acts on the vector ψ(x) of matter fields. In the case of electrodynamics, the unitary matrix is a member of the group U(1); it is just a phase factor U(x) = exp( ie θ a (x)). The abelian gauge transformation is ψ (x) = U(x)ψ(x) = e ie θ(x) ψ(x) A i(x) = U(x)A i (x)u (x) ( i U(x)) U (x) = A i (x) + ie i θ(x). (75) I have been using a notation in which A i is antihermitian to simplify the algebra. 14