Quantum Field Theory Example Sheet 4 Michelmas Term 2011

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1 Quantum Field Theory Example Sheet 4 Michelmas Term 0 Solutions by: Johannes Hofmann Laurence Perreault Levasseur Dave M. Morris Marcel Schmittfull jbh38@cam.ac.uk L.Perreault-Levasseur@damtp.cam.ac.uk dmm49@cam.ac.uk M.Schmittfull@damtp.cam.ac.uk Exercise In the previous example sheet we considered -particle to -particle scattering in φ 4 theory, L = µφ µ m φ λ φ4. In this exercise, we shall compute the leading order contribution to the amplitude for the scattering of 3-particles to produce 3-particles. Giving labels to the momenta of these particles, our initial and final states therefore look like i = 8E p E p E p3 a p a p a p 3 0, f = 8E p E p E p 3 a p a p a p. 3 0 As in Exercise of example sheet 3, we determine the S-matrix element [ lim f U(t, t) i = f T exp iλ ] φ(x) 4 i. () t As before there are a number of diagrams which the prescription for calculating this amplitude tells us we must throw away: these are, first, the O() term as this contains no interactions. Second, there being 6 external lines but only 4 fields φ appearing in () at O(λ), not all the external lines in i and f can be contracted with fields in the time order product. The diagrams below correspond to such terms and, as they are disconnected, are consequently ignored. The quantity of interest is thus the term ( ) iλ d 4 y f T { φ 4 (x)φ 4 (y) } i Via Wick s theorem, this contains further contractions some of which again correspond to disconnected diagrams. These are those containing two or more propagators, as such terms require at least four contracted fields in the time ordered product, leaving at most four to be contracted with external states, and, consequently, at least one in-going line is matched with an out-going line. Discarding these, we are left with those of the form giving the diagram φ(k )φ(k )φ(k 3) : φ(x)φ(x)φ(x)φ(x)φ(y)φ(y)φ(y)φ(y) : φ(k )φ(k )φ(k 3 ), ()

2 k k k k + k + k 3 k k k and three similar contractions in which one in-going leg is instead contracted with a φ(x) and one out-going leg with a φ(y). These latter contractions have diagrams of the form The contractions () may easily be calculated via the methods detailed in the solutions to Exercise of the previous example sheet. After performing the two integrals, this short calculation gives ( ) iλ d 4 y φ(k )φ(k )φ(k 3) : φ(x)φ(x)φ(x)φ(x)φ(y)φ(y)φ(y)φ(y) : φ(k )φ(k )φ(k 3 ) = ( ) iλ i (p + p + p 3 ) m (π)4 δ (4) (p + p + p 3 p p p 3), and we claim that an overall combinatorial factor of () arises when we sum over every contraction of the form (). To see this, first note that a factor of arises from interchanging the integration variables x and y in (), since two contractions differing in only this respect give identical expressions when these integrals are evaluated. Next, consider the number of ways of obtaining a propagator from contractions φ(x)φ(y) This is the number of ways of pairing one φ(x) in φ(x) 4 with one φ(y) in φ(y) 4. There are 4 ways of doing this, so that an overall factor of 4 is obtained. Now, having extracted the factor of discussed above, x and y may be seen as fixed in such a way that each φ(y) contracts with an operator in the initial state, while each φ(x) contracts with an operator in the final state. Moreover, having performed one pair of contractions in evaluating a propagator, the overall factor reflecting the remaining freedom is the number of ways to contract the remaining three φ(y) s with the operators in i multiplied by the number of ways to contract the remaining three φ(x) s with the operators in f. This gives a factor of (3!), which, together with the factors previously removed, gives () as claimed. In conclusion then, the amplitude is Exercise M(φ(p )φ(p )φ(p 3 ) φ(p )φ(p )φ(p 3)) = i( iλ) i (p + p + p 3 ) m + O(λ3 ) In φ 4 theory we evaluate order by order up to O(λ ) the vacuum to vacuum amplitude [ ] 0 S 0 = 0 T exp i dth I (t) 0 = iλ 0 φ 4 I(x) 0 + ( ) iλ d 4 y 0 T { φ 4 I(x)φ 4 I(y) } 0 + O(λ 3 ).

3 Trivially, the O(λ 0 ) term is. The O(λ) contribution is by, Wick s theorem, ( i) λ 0 φ 4 I(x) 0 = i λ ( ) combinatorial 0 φ(x)φ(x)φ(x)φ(x) 0 factor = ( i) λ ( ) combinatorial φ(x)φ(x)φ(x)φ(x) factor There are three ways to contract a chosen φ(x) with another in φ(x) 4 and, as the only remaining contraction can be between the remaining two φ(x) s, this is the total combinatorial factor. The O(λ) vacuum to vacuum amplitude is thus i λ φ(x)φ(x)φ(x)φ(x). 8 This contraction corresponds to the diagram Let us now proceed to O(λ ), where three distinct kinds of diagrams arise from the contractions in namely the following ( ) iλ d 4 y 0 T { φ 4 I(x)φ 4 I(y) } 0, The first of these is the sum of all contractions of the form = ( ) iλ d 4 y φ(x)φ(x)φ(x)φ(x)φ(y)φ(y)φ(y)φ(y), of which we calim there to be. To see this, first observe that interchanging x and y in any such expression gives an identical result, and therefore a factor of. The remaining freedom consists in pairing a φ(y) with a φ(x); there are such choices. Thus ( ) iλ The second diagram is the sum of all contractions of the form ( ) iλ d 4 y φ(x)φ(x)φ(x)φ(x)φ(y)φ(y)φ(y)φ(y) d 4 y φ(x)φ(x)φ(x)φ(x)φ(y)φ(y)φ(y)φ(y), 3

4 of which there are 3, there being 3 means of contracting two φ(x) s in φ(x) 4 and likewise for contractions between φ(y) s. Thus = 3 ( ) iλ = ( iλ 8 = ) φ(x)φ(x)φ(x)φ(x) d 4 y φ(x)φ(x)φ(x)φ(x)φ(y)φ(y)φ(y)φ(y) The final diagram is that corresponding to contractions of the form ( ) iλ d 4 y φ(x)φ(x)φ(x)φ(x)φ(y)φ(y)φ(y)φ(y) To find the symmetry factor, we may first, as before, fix the position of x and y by extracting a factor of. There are then 3 ways of contracting two φ(x) s and of contracting two φ(y) s. Next, of the remaning fields, there are two ways of contracting a φ(x) with a φ(y) and the remaining φ(x) and φ(y) are necessarily contracted. The overall symmetry factor is thus (3!), so that In conclusion, = (3!) ( ) iλ d 4 y φ(x)φ(x)φ(x)φ(x)φ(y)φ(y)φ(y)φ(y). 0 S 0 = O(λ 3 ) = exp +++ Exercise 3 Consider the following Lagrangian, L = = = ( 3 ( µ φ i ) ( µ φ i ) 3 3 m φ i λ 8 3 ( µ φ i ) ( µ φ i ) 3 m φ i λ 8 3 ( µ φ i ) ( µ φ i ) 3 m φ i 3 λ 4 φ i ) 3 φ 4 i λ φ i φ j 8 i j ( φ 4 + φ 4 + φ 4 λ ( 3) φ 4 φ + φ φ 3 + φ 3φ )

5 We make the following identifications L 0 = 3 L Int = 3λ [ ( µφ i ) ( µ φ i ) ] φ i ( φ 4 + φ 4 + φ 4 λ ( 3) φ 4 φ + φ φ 3 + φ 3φ ) The free (λ = 0) theory is exactly solved by 3 mutually commuting (or non-interacting ) scalar fields of mass m. Therefore, the propagator is the familiar expression 0 φ i (x)φ i (y) 0 = D F (x y), Since they commute, φ i fields may not propagate to φ j fields, if i j, 0 φ i (x)φ j (y) 0 = 0. (3) More concisely, These give the Feynman rule 0 φ i (x)φ i (y) 0 = δ ij D F (x y). iδ φ i φ ij j p p m It remains now to compute the various possible vertex terms. The allowed vertices are described by either the first or second terms in parentheses in (3). The amplitude associated to the former kind of vertex is precisely the lowest order term in the amplitude for φ i φ i φ i φ i in the φ 4 i theory with coupling 3λ, i.e. φ i φ i 3iλ φ i φ i The second allowed vertex terms occur in φ i φ j φ i φ j or φ i φ i φ j φ j scattering for i j. From Exercise of Example Sheet 3 with φ = φ i, = φ j, the single vertex amplitude in the former process is simply iλ, so that φ j φ j iλ φ i φ i The vertex in φ i φ i φ j φ j has the same amplitude, as is reflected in the symmetry of the above diagram. As these are the only vertices allowed by the interaction (3), the amplitude for φ i φ j φ k φ l scattering is zero if neither {i, j} = {k, l} nor i = j, k = l hold. We may therefore write the above two rules more concisely as follows φ j φ l iλ (δ ij δ kl + δ ik δ jl + δ il δ jk ), φ i φ k which expression is the leading order contribution to φ i φ j φ k φ l scattering. 5

6 Exercise 4 We summarize the Feynman rules for the Yukawa theory (cf. Peskin and Schroeder, Ch. 4.7) of a single Dirac field and a single scalar field φ interacting via the following term in the Lagrangian L Int = gφ. The propagators for scalar and fermion fields are, respectively, p i p µ, p i(/p + m) p m Vertex amplitudes are assigned as φ ig The amplitudes for external fermionic and anti-fermionic lines are given by the contractions (x) s (p) = u s (p), s (p) (x) = ū s (p) (x) s (p) = v s (p), s (p) (x) = v s (p) Finally, one imposes four-momentum conservation at each vertex and includes the necessary minus signs given by statistics. The first amplitude we compute using these rules is that for scattering. Here the initial and final states are and we employ Dyson s formula i = E p Eq a s p a r q 0, f = E p Eq a s p a r q 0, to compute the leading order contribution to this process. [ ] f S i = f T exp i dt H I (t ) i, As there are two in-going and two out-going fermions, the first non-zero term in Dyson s formula must contain two fermion creation opertors and two fermion annihilation operators. This is the order O(g ) term f S I i = ( ig) d 4 y f T { φ(x) (x)(x)φ(y) (y)(y) } i, in which the fermion operators are contracted with the inital and final states. The only nonzero contractions are q, s p, r q, s p p p, r = ( ig) = ( ig) [ūs (p ) u s (p)][ū r (q ) u r (q)] (p p) µ d 4 y r (q ) s (p ) : φ(x) (x)(x)φ(y) (y)(y) : s (p) r (q) 6

7 and q, s p, r p, s p q q, r = ( ig) = ( ig) [ūs (p ) u r (q)][ū r (q ) u s (p)] (q p) µ. d 4 y r (q ) s (p ) : φ(x) (x)(x)φ(y) (y)(y) : s (p) r (q) Let us explain both the combinatoric factors and minus signs arising from statistics. First, each term comes with a factor of since we obtain equivalent contributions by interchanging the vertex positions x and y. Second, to evaluate the first set of contractions we must anti-commute the (x) past (y) and then anti-commute (y) through (x). These give a total factor of ( ) = +. In the second contraction only one anti-commutation is performed, namely that between (y) and (x). The final amplitude is the sum of the above, [ ] M( ) = ( ig) [ū s (p ) u s (p)][ū r (q ) u r (q)] (p p) µ [ūs (p ) u r (q)][ū r (q ) u s (p)] (q p) µ + O(g 3 ) We next compute the amplitude for scattering at leading order, i.e. at order O(g ), as before. The only contractions that contribute to this order are q, s q, s p, r p + q p, r = ( ig) d 4 y r (q ) s (p ) : φ(x) (x)(x)φ(y) (y)(y) : s (p) r (q) and q, s p, r p q p, s q, r = ( ig) d 4 y r (q ) s (p ) : φ(x) (x)(x)φ(y) (y)(y) : s (p) r (q), where the combinatoric factor of again arises from interchanging x and y and the minus signs may be accounted for by keeping careful track of all the necessary anti-commutations. The total amplitude is then [ ] M( ) = ( ig) [ v r (q) u s (p)][ū s (p ) v r (q )] (p + q) µ [ūs (p ) u s (p)][ v r (q) v r (q )] (p p ) µ + O(g 3 ). Exercise 5 Consider now the Yukawa theory of the previous exercise but with the modified pseudo-scalar interaction term L Int = gφ γ 5. 7

8 This only affects the Feynman rules by altering the vertex amplitude, φ ig γ 5 Identical computations to those of the previous exercise find that, to order g in pseudo-scalar Yukawa theory, ( ) M( ) = ( ig) [ū r (q )γ 5 u r (q)][ū s (p )γ 5 u s (p)] (q q) µ [ūr (q )γ 5 u s (p)][ū s (p )γ 5 u r (q)] (q p) µ ( ) M( ) = ( ig) [ū s (p )γ 5 v r (q )][ v r (q)γ 5 u s (p)] (p + q ) µ [ūs (p )γ 5 u s (p)][ v r (q)γ 5 u r (q )] (p p) µ Exercise 6 Suppose that any 3-vector field f may be decomposed into a transverse ( Φ = 0) part and a longitudinal ( Φ = 0) part, i.e. for some three-vector field g with g = 0 and function h, we may write f = f T + f L, f T = g, f L = h. Modulo some assumptions about the regularity of f, by finding expressions for g and h we demonstrate that such a decomposition is indeed possible. To find h, take the divergence of f, Formally, this gives f = ( g) + h = h. }{{} =0 h = ( ) f. In fact, if we restrict ourselves to suitable classes of functions, the Poisson equation h = φ, will possess a unique solution for a given φ. For example, for functions compactly supported on R 3, ( ) is none other than the linear operator given by integrating against the correct Green s function, ( ) φ(x) = d 3 x φ(x ) 4π R x. (4) 3 To find g, take the curl of f, Formally then, f = ( g) + ( h) = ( g) g = g. }{{}}{{} =0 =0 g = ( ) ( f). We can write the transversal component of f as f T = f f L = f h = f ( ) ( f), or The linear operator f T i = [δ ij i ( ) j ]f j. P ij = δ ij i ( ) j 8

9 is a projection since P ij P jk = [δ ij i ( ) j ][δ jk j ( ) k ] = δ ij i ( ) k + i ( ) ( ) k = P ik. We can also show that P ij = P ji as follows. First, since δ ij and i j are symmetric, it will suffice to show that i ( ) = ( ) i. Now, suppose that φ solves the Poisson equation φ = f, i.e. φ = ( ) f. Then, by commutativity of partial derivatives, i.e. ( ) ( i f) = i φ. In other words, for any f, ( i φ) = i f, ( ) ( i f) = i φ = i ( ( ) f ) and so i ( ) = ( ) i, as required. This can also be seen less formally from (4). Now suppose that we are interested in the dynamics of a gauge field A: by the above, we may decompose A into a longitudinal and a transverse part, A = A T + A L. In Coulomb gauge, where A = 0, we must project out the longitudinal component of the gauge field and consider the dynamics of A T i = P ij A j. In addition, recall that Π = 0 is implemented as a constraint on the dynamics, so that we must further project onto transverse momentum configurations also. The canonical commutation relations [A i (x), Π j (y)] = iδ ij δ (3) (x y) are now to be projected onto the configuration space of transverse gauge fields and their momenta. superscript x in P ij to keep track of the variable of differentiation, we have Including a [A T i (x), Π T j (y)] = [P x ika k (x), P y jl Π l(y)] = P x ikp y jl [A k(x), Π l (y)] = ip x ikp y jk δ(3) (x y) = ip x ikp x jkδ (3) (x y) = ip ij δ (3) (x y), where in the last line we use that P ij is a projection and that P ij = P ji. These are the commutation relations for the gauge-fixed theory. Exercise 7 The momentum space Feynman rules relevant to the leading order perturbative contribution to Compton scattering γ γ are the external lines In-going Out-going Electron lines p, s u e s(p) p, s ū e s(p) Photon Lines γ µ ɛ(p) µ µ γ ɛ(p) µ ɛ(p) ɛ(p) 9

10 together with the electron propagator, i(/p + m) p m Imposing momentum conservation at each vertex, we may compute the following amplitude contributing to Compton scattering at leading order ɛ in (q) ɛ out (q ) p, s p + q p, s = (ū s (p ) ɛ out (q ) µ ) }{{} γ in final state final vertex {}}{ ( ieγ µ ) i(/p + /q + m) (p + q) m }{{ } propagator first vertex {}}{ ( ieγ ν ) (ɛ in (q) ν u s (p)) }{{} γ in initial state = ( ie) ū s (p ) /ɛ out (q ) i( /p + /q + m) (p + q) m /ɛ in (q) u s(p) The remaining order e contribution comes from permuting the contraction of the vertices with the external photon lines, ɛ in (q) p, s p q = ( ie) ū s (p ) /ɛ in (q) i( /p /q + m) (p q) m /ɛ out (q ) u s (p) p, s ɛ out (q ) The total amplitude at leading order is the sum of these two contributions, { M( γ γ) = ( ie) ū s (p ) /ɛ out (q ) i( /p + /q + m) (p + q) m /ɛ in (q) u s(p) + ū s (p ) /ɛ in (q) i( /p } /q + m) (p q) m /ɛ out (q ) u s (p) We now ask what happens if we make the change in photon polarization ɛ in (q) ɛ in (q) + q, i.e. change the polarization by a direction transverse to the incoming photon. We will see that the amplitude is unaltered to leading order by this replacement, as one would expect. Explicitly, the change in the amplitude is { ( ie) ū s (p ) /ɛ out (q ) i( /p + /q + m) (p + q) m / q u s (p) + ū s (p ) /q i( /p } /q + m) (p q) m /ɛ out (q ) u s (p) (5) To show that this vanishes, first take the adjoint of and multiply on the right by γ 0 to find that From Exercise of Example Sheet 3, (/p m)u s (p) = 0 u s (p) γ 0 ( γ 0 (γ µ ) γ 0 p µ m ) = 0, γ 0 (γ µ ) γ 0 = γ µ, (6) so that the Dirac conjugate of u s (p) satisfies ū s (p)(/p m) = 0. 0

11 From these identities we find, so that the firt term in the expression (5) is /qu s (p) = (/p + /q m)u s (p), ū s (p )/q = ū s (p )( /p /q m), ū s (p ) /ɛ out (q ) i( /p + /q + m) (p + q) m / q u s (p) = ū s (p ) /ɛ out (q i ) (p + q) m ( /p + /q + m)(/p + /q m)u s (p) whereas the second is = ū s (p ) /ɛ out (q i ) (p + q) m ((p + q) m )u s (p) = iū s (p ) /ɛ out (q )u s (p), ū s (p ) /q i( /p /q + m) (p q) m /ɛ out (q ) u s (p) = ū s (p )( /p /q m)(/p i /q + m) (p q) m /ɛ out (q ) u s (p) which two expressions precisely cancel as required. If instead we make the replacement = iū s (p )/ɛ out (q ) u s (p), ɛ out (q ) ɛ out (q ) + q, we find M( γ γ) changes as { ( ie) ū s (p ) /q i( /p + /q + m) (p + q) m /ɛ in (q) u s(p) + ū s (p ) /ɛ in (q) i( /p } /q + m) (p q) m / q u s (p). (7) This time, /q u s (p) = (/p /q m)u s (p), ū s (p )/q = ū s (p )( /p + /q m), but by 4-momentum conservation p q = p q and p + q = p + q, and we can therefore now compute that (7) vanishes using the same trick as before. Exercise 8 We show that the identity [ γ µ χ ] = χγ µ, (8) holds for any Dirac spinors, χ. From the previous example sheet, [ γ µ χ ] = [ α[γ 0 ] αβ [γ µ ] βγ χ γ ] = α [(γ 0 ) ] βα [(γ µ ) ] γβ χ γ (γ 0 ) = γ 0, γ 0 (γ µ ) γ 0 = γ µ, and therefore γ µ χ = α [(γ µ ) ] γβ [γ 0 ] βα χ γ = = [ α (γ µ ) γ 0] γα χ γ [ α γ 0 γ µ] γα χ γ = χ γ[γ 0 ] γβ [γ µ ] βα α = χγ µ With = u s (p ), χ = u s (p) in (8), and using the spin sum u s (p) α ū s (p) β = (/p m) αβ s

12 from Example Sheet, we find s,s [ū s (p )γ ν u s (p)] [ū s (p )γ µ u s (p)] = = s s,s [ū s (p)γ ν u s (p )] [ū s (p )γ µ u s (p)] ū s (p)γ ν ( s u s (p )ū s (p ) ) γ µ u s (p) = s ū s (p)γ ν (/p + m)γ µ u s (p) = s ū s (p) α [γ ν ] αβ [/p + m] βγ [γ µ ] γδ u s (p) δ ( ) = [γ ν ] αβ [/p + m] βγ [γ µ ] γδ u s (p) δ ū s (p) α s = [γ ν ] αβ [/p + m] βγ [γ µ ] γδ [/p + m] δα = Tr [ γ ν (/p + m)γ µ (/p + m) ] = Tr [ (/p + m)γ µ (/p + m)γ ν], where in the final line we use Tr(ABC) = Tr(BCA) to put the trace in the desired form. If we now set = v s (q) and χ = u s (p) in (8), then using the sum v s (p) α v s (p) β = (/p m) αβ, in addition to that used previously, we obtain s,s [ v s (q)γ ν u s (p)] [ v s (q)γ µ u s (p)] = s = s s,s [ū s (p)γ ν v s (q)] [ v s (q)γ µ u s (p)] ū s (p)γ ν ( s v s (q) v s (q) ) γ µ u s (p) = s ū s (p)γ ν (/q m)γ µ u s (p) = s ū s (p) α [γ ν ] αβ [/q m] βγ [γ µ ] γδ u s (p) δ ( ) = [γ ν ] αβ [/q m] βγ [γ µ ] γδ u s (p) δ ū s (p) α s = [γ ν ] αβ [/q m] βγ [γ µ ] γδ [/p + m] δα = Tr [ γ ν (/q m)γ µ (/p + m) ] = Tr [ (/q m)γ µ (/p + m)γ ν] From Exercise 0 of Example Sheet, we have the following identities = Tr [ /qγ µ /pγ ν mγ µ /pγ ν + m/qγ µ γ ν m γ µ γ ν] (9) Tr /p /p /p 3 /p 4 = 4 [(p p )(p 3 p 4 ) (p p 3 )(p p 4 ) + (p p 4 )(p p 3 )] (0) Tr γ µ γ ν = 4 g µν Tr γ µ γ ν γ ρ = 0 so that the middle two terms in (9) vanish and, with p = q, p 3 = p, (p ) ρ = δ ρ µ and (p 4 ) ρ = δρ ν in (0), [ v s (q)γ ν u s (p)] [ v s (q)γ ν [ u s (p)] = 4 p ν q µ + p µ q ν (p q)g µν m g µν] s,s Exercise 9 = 4 [ p ν q µ + p µ q ν (p q + m )g µν] () The interaction term in the Lagrangian for a theory describing the electrodynamics of both electron fields e and muon fields m is taken to be e m /A m + e e /A e,

13 Assigning momenta and spins to the initial and final states, it is clear from the form of the interaction that the first non-zero term in the perturbative expansion of the amplitude for e + µ µ + scattering is the order e term ( ie) d 3 x d 3 y µ (p, s ), µ + (q, r ) T { m (x) /A(x) m (x) e (y) /A(y) e (y) } (p, s), e + (q, r). There is only one set of contractions that contribute to this amplitude, namely those from the diagram e + q, r p + q q, r µ + p, s p, s µ The QED momentum space Feynman rules tell us to assign amplitudes to each particular component of this diagram as follows In-going Electron Lines p, s u e s(p) e + p, s v e s(p) Out-going Muon Lines p, s µ ū m s (p) p, s µ + v m s (p) Vertices Photon Propagator e + ie γ ν γ µ ν i g µν k k µ + ie γ ν γ µ the whole amplitude then being computed by multiplying each contribution, contracting indices for each vertex and enforcing 4-momentum conservation. Whence, the amplitude M( e + µ µ + ) at leading order is e + p, s q, r p + q q, r p, s µ + µ = ( ie) [ ve r(q)γ µ u e s(p)] [ū m s (p )γ µ v m r (q )] (p + q) () Employing the identity () proved earlier to evaluate the sums, the spin-averaged probability is therefore given by r,s,r,s M( e + µ µ + ) = = ( ie) (p + q) 4 r,s,r,s [ v e r(q)γ µ u e s(p)] [ v e r(q)γ ν u e s(p)] [ū m s (p )γ µ v m r (q )] [ū m s (p )γ ν v m r (q )] 6e 4 (p + q) 4 [ p ν q µ + p µ q ν (p q + m )g µν] [ p νq µ + p µq ν (p q + M )g µν ], where m = p = q and M = p = q define the rest masses of the electron and muon, respectively. In the limit that m and M are negligible, we find r,s,r,s M( e + µ µ + ) = 3e4 (p + q) 4 ((q q)(p p) + (p q)(q p)). 3

14 In the C.O.M. frame q = p and q = p, and, since m = M = 0, q 0 = p 0 = p and q 0 = p 0 = p. By conservation of energy, p = p. The collision is confined to the plane Π spanned by p and p, defining the scattering angle as cos θ = p p p p. As M is a scalar and thus rotationaly invariant, we may without loss of generality take p to be aligned along the 3-axis and Π to be the plane x = 0, so that p µ = ( p, 0, 0, p ), q µ = ( p, 0, 0, p ) p µ = ( p, 0, p sin θ, p cos θ), q µ = ( p, 0, p sin θ, p cos θ) It is then a simple matter to compute that (p + q) = 4 p and p p = p ( cos θ), p q = p ( + cos θ) q p = p ( + cos θ), q q = p ( cos θ) and so r,s,r,s M( e + µ µ + ) = 3e4 6 p 4 ( p 4 ( cos θ) + p 4 ( + cos θ) ) = 4e 4 ( + cos θ). 4