Compton Scattering from Quantum Electrodynamics Andrew Lawson 17 January, 2014 Abstract We present a derivation of the differential cross section for the Compton scattering of an electron and a photon. We explicitly calculate the squared matrix element for this process in Feynman gauge before recovering an expression for the cross-section for this process by explicitly performing an integration in Lorentz-invariant phase space. We derive two expressions for the differential cross section of this process, dσ/dt and dσ/d cos θ, which we then analyse to deduce some of the physical properties of this scattering process. 1 Introduction Compton scattering is the inelastic scattering of a photon with an electically charged particle, first discovered in 1923 by Arthur Compton [1]. This scattering process is of particular historical importance as classical electromagnetism is insufficient to describe the process; a successful description requires us to take into account the particle-like properties of light. Furthermore, the Compton scattering of an electron and a photon is a process that can be described to a high level of precision by the theory of quantum electrodynamics (QED). The cross-section is an important quantity in particle physics, as it is directly measurable quantity that can be used to test the validity of a theory [2]. In this report we therefore present a derivation of the differential cross section for this process using QED. Our derivation will proceed as follows: in section 2 we use QED to derive the squared matrix element for this scattering process. In section 3 we will then derive two expressions for the differential cross section: the general Lorentz-invariant expression dσ/dt and then the directly measurable quantity dσ/d cos θ (in the rest frame of the incident electron). We also discuss the physical relevance of these derived quantities. Finally we present our conclusions in section 4. 2 Scattering Matrix Elements At tree level, two diagrams contribute to the Compton scattering process. These diagrams are shown in Fig. 2.1. We have indicated the respective 4-momenta of the incident electron and photon to be pµ = p0, p , (2.1) kµ = k0, k , (2.2) 1 (a)(b) Figure 2.1: The two tree-level Feynman diagrams contributing to the matrix element of the Compton scat- tering process. as well as 4-momenta for the outgoing electron and photon respectively: pµ = p00, p0 , (2.3) kµ = k00, k0 . (2.4) Using the Feynman rules for QED, we can calculate the matrix elements corresponding to the diagrams Fig. 2.1(a) and Fig. 2.1(b) rather simply. We can thus write down the respective matrix elements rather simply: " # 2 (s0) 0 0∗ ν p/ + k/ + m µ (s) M1 = e u¯ (p ) ν γ µγ u (p) , (2.5) (p + k)2 − m2 " # 2 (s0) 0 ν p/ − k/ + m 0∗ µ (s) M2 = e u¯ (p ) ν γ µ γ u (p) , (2.6) (p − k0)2 − m2 where the spinor u(s) (p) represents the incoming electron and µ is the polarisation vector describing our incident photon.The total matrix element is given by the sum of the matrix elements corresponding to these two diagrams, i.e. M = M1 + M2. In order to write our expressions in a manifestly Lorentz-invariant way, we will make use of the Mandelstam variables: s = (pµ + kµ)2 = (p0µ + k0µ)2 , (2.7) t = (kµ − k0µ)2 = (pµ − p0µ)2 , (2.8) u = (pµ − k0µ)2 = (kµ − p0µ)2 . (2.9) 2 µ 0µ 0 2 µ 0µ 0 As the variables satisfy s + t + u = 2me, along with p pµ = p pµ = me and k kµ = k kµ = 0 we can use 2 the following relations: 1 pµk = s − m2 , (2.10) µ 2 1 pµp0 = (s + u) , (2.11) µ 2 1 kµp0 = m2 − u . (2.12) µ 2 We now proceed with the calculation of the total squared matrix element. The calculation can be split up using the formula 2 2 2 ∗ ∗ |M1 + M2| = |M1| + |M2| + M1M2 + M2M1. (2.13) We will illustrate the calculation in detail for the first squared element; the rest follow by similar working. We start with the full expression including a sum over the spins and polarisations of all the incoming and outgoing electrons: 4 2 e X X r0∗ r r r0∗ h (s0) ν µ (s)i h (s) ρ σ (s0)i |M1| = u¯ γ p/ + k/ + me γ u u¯ γ p/ + k/ + me γ u . (2.14) 2 2 ν µ ρ σ 4 (s − me) s,s0 r,r0 Making use of the polarisation sum X r r∗ µν = −gµν , (2.15) r and the spinor completeness relation X (s) (s) u (p)u ¯ (p) = p + m , (2.16) j k / e jk s=1,2 we can write the amplitude as a trace, leaving us with 4 2 e 0 ν µ |M1| = Tr p/ + me γ p/ + k/ + me γ p/ + me γµ p/ + k/ + me γν 2 2 4 (s − me) 4 e 0 = Tr −2p/ + 4me p/ + k/ + me −2p/ + 4me p/ + k/ + me 2 2 4 (s − me) 4 e 0 2 0 2 Tr p/ p/ + k/ p/ p/ + k/ + m p/ p/ + 4m p/ + k/ p/ + k/ 2 2 e e = (s − me) 0 4 −4p/ p/ + k/ − 4 p/ + k/ p/ + 4me 4 e h 0µ ν 2 0ν 2 0ν 2 2 8p (pµ + kµ) p (pν + kν ) − 4 (pµ + kµ) p pν + 4m p pν + 16m (pµ + kµ) − 2 2 e e = (s − me) 0µ ν 4 16p (pµ + kµ) − 16p (pν + kν ) + 16me 2e4 = −su + m2 (3s + u) + m4 . (2.17) 2 2 e e (s − me) µ µ1 µn In line 4 we used cyclicity of trace and γ pγ/ µ = −2p/. As Tr [γ . γ ] = 0 for n odd, we have only 3 considered terms with even numbers of γµ matrices. Next we have 4 2 e h 0 ν 0 µ 0 i |M2| = Tr p/ + me γ p/ − k/ + me γ p/ + me γµ p/ − k/ + me γν 2 2 4 (u − me) 2e4 = ... = −su + (3u + s) m2 + m4 . (2.18) 2 2 e e (u − me) Finally we can calculate 4 −e n h 0 ν µ 0 i Tr p/ + me γ p/ + k/ + me γ p/ + me γµ p/ − k/ + me γν + ∗ ∗ 4 (s − m2)(u − m2) M1M2 + M2M1 = e e h 0 ν 0 µ io Tr p/ + me γ p/ − k/ + me γ p/ + me γµ p/ + k/ + me γν 4 4e 4 2 = ... = 2 2 2me + me (s + u) . (2.19) (s − me)(u − me) The full squared matrix element for the Compton scattering process is therefore " 2 4 2 4 4 2 # 2 4 −su + me (3s + u) + me −su + me (3u + s) + me 8me + 4me (s + u) |M1 + M2| = 2e + + . (2.20) 2 2 2 2 2 2 (s − me) (u − me) (s − me)(u − me) 3 Cross Section In general, the cross section of a 2 → n scattering process is given by n 4 n ! 2 Y d qi 4 X |M| σ = δ q2 − m2 Θ q0 (2π) δ q − p − k , (3.1) ˆ 3 i i i i F i=1 (2π) i=1 where F is the Møller flux factor, given by 2 F = E1E2v12 = 2 s − me . (3.2) The energies E1 and E2 are the energies of the incident particles, and v12 is their relative velocity [2]. Of course here we will take n = 2 and use the 4-vectors we introduced in the previous section. 3.1 Centre of Mass Frame We can plug our expression for the amplitude (2.20) into the formula for the cross-section to yield 2 1 4 0 4 0 02 2 02 00 00 4 0 0 |M1 + M2| σ = d p d k δ p − me δ k θ p θ k δ (p + k − p − k) (2π)2 ˆ 2 (s − m2) 2 1 4 0 0 2 2 02 00 0 0 00 |M1 + M2| = 2 d k δ (p + k − k ) − me δ k θ k θ p + k − k 2 (2π) ˆ 2 (s − me) 2 1 0 3 0 0 2 2 002 0 2 00 0 0 00 |M1 + M2| = 2 dk d k δ (p + k − k ) − me δ k − |k | θ k θ p + k − k 2 , (2π) ˆ ˆ 2 (s − me) (3.3) To simplify the integral further we must consider this scattering process in an arbitrary frame with the incoming and outgoing 4-vector described by (2.1), (2.2), (2.3) and (2.4). We can now simplify the second 4 delta function in our expression using the definition 2 1 δ k02 = δ k00 − |k0|2 = δ k00 + |k0| + δ k00 − |k0| . (3.4) 2 |k0| However the δ k00 + |k0| expression can be neglected as our step function θ k00 ensures that we only have contributions from postive k00, so we can perform the integration over k00 to yield 3 0 2 1 d k 0 2 2 0 0 0 |M1 + M2| σ = 2 0 δ (p + k − k ) − me θ p + k − |k | 2 . (3.5) (2π) ˆ 2 |k | 2 (s − me) This integral can be simplified by converting to spherical polar coordinates and replacing d3k0 = |k0|2 d |k0| sin θdθdφ. (3.6) To simplify things further, we choose our frame to be the centre of mass frame. We can therefore define the 4-vector √ pµ + kµ = s, 0 .