A Calculation of the Cross Section for Compton Scattering

A Calculation of the Differential Cross Section for Compton Scattering in Tree-Level Quantum Electrodynamics

Declan Millar

School of Physics and Astronomy, University of Southampton, Highﬁeld, Southampton, SO17 1BJ, UK

January 2014

Abstract The differential cross section for unpolarised Compton scattering is calculated with respect to both t and cos θ. (e γ e γ) 2 is evaluated to ﬁrst order in α using pertur- |M − → − | em bative QED. The Klein-Nishina formula is derived using two different approaches, working dσ dσ both from the centre of mass and lab frames. Plots of dt and d cos θ resemble classical Thom- dσ = son scattering at low s. At higher energy d cos θ is largest at θ 0, decreasing rapidly towards π. The overall cross section decreases as s increases. 1 Introduction

Compton scattering occurs when electromagnetic radiation is scattered by free electrons at rest in the lab reference frame. The initial and ﬁnal states are an electron and a photon: e γ − → e−γ. The cross section of this interaction is intrinsic to the colliding particles and allows us to calculate the probability of this ﬁnal state, independently of the luminosity L of any particular experiment. The number of corresponding scattering events N is related to the cross section by:

N = L σ. (1.1) ·

The most general formula for the inﬁnitesimal cross section of a two particle collision is given by [1]

1 dσ = (p , p p ) 2dΠ . (1.2) 2E 2E v v |M A B → { f } | n A B| A − B| And the phase space integral over the ﬁnal states has the form

3 ! Z d p f 1 dΠ = (2π)4δ(4)( p p p ). (1.3) n ∏ ( )3 ∑ f A B f 2π 2Ef − −

The last two factors in (1.2) are Lorentz invariant, while the first is invariant under co-linear boosts. The square matrix element 2 may be calculated using the Feynman rules of quantum electrodynamics (QED) without|M| any reference to a particular frame of reference. We will undertake this calculation first in section 2, before returning, in section 3, to evaluate the remaining factors in both the centre of mass and lab frames. This will allows us to calculate the dσ differential cross section with respect to both: the scattering angle in the lab frame d cos θ , and the dσ square momentum transfer between initial-state and final-state photons dt . In section 4 we plot dσ dσ dt and d cos θ for three different of centre of mass energies s. Conclusions are made in section 5.

2 Calculating the square S-matrix element

2.1 Applying the Feynman Rules

At zero loop (tree) level, there are two possible Feynman diagrams with initial and ﬁnal states representative of Compton scattering. These are shown in ﬁgure 2.1.

2 k k′ k p + k k′

p p′ p p′ p k′ −

Figure 2.1: Feynman diagrams for Compton scattering. Time runs left to right.

Following the reverse fermion ﬂow and applying the QED Feynman rules [2] to each diagram, we may immediately write down the corresponding transition amplitudes and 1. M1 M2 s µ r i(/p + /k + m) ν r s i = u¯ 0 (p0)( ieγ )e∗ 0 (k0) ( ieγ )e (k)u (p) (2.1) M1 − µ (p + k)2 m2 − ν − s ν r i(/p /k + m) µ r s i = u¯ 0 (p0)( ieγ )e (k) − ( ieγ )e∗ (k0)u (p) (2.2) M2 − ν (p k)2 m2 − µ − − In the canonical quantisation of QFT these amplitudes are terms in the Wick expansion of the scattering matrix element f S i [3]. Therefore, the total transition amplitude is equal to the sum of these expressions. h | | i

= + (2.3) M M1 M2 In accordance with Fermi-Dirac statistics, there is no relative minus sign between the two terms due to the identical fermion ﬂow in both diagrams. In addition, we are uninterested in terms corresponding to processes where no scattering occurs and so we have

f S 1 i = i (2π)4δ(4)(P P ). (2.4) h | − | i M F − I The momentum conserving delta function is shared by all S-matrix elements, thus it is absorbed in to the general formula for a cross section (1.2). In order to calculate this formula we require 2. This is given by2: |M| 2 2 2 = ( + )( ∗ + ∗) = + + 2 ( ∗) (2.5) |M| M1 M2 M1 M2 |M1| |M2| R M1M2 The total square matrix element is therefore equal to the sum of three terms which may be calculated separately: two terms simply equal to the square of each amplitude and a third 2 ‘interference’ term between the diagrams. We shall begin by evaluating 1 in full detail before returning to the later two terms. |M |

1Note that spin notation is coupled to momentum notation and so spin superscripts are henceforth left implicit. In addition, we will implement the notation u = u(p), u0 = u(p0), e = e(k) and e0 = e(k0). 2As the matrix elements are commutative, the second equality follows simply from:

∗ + ∗ = (a + ib)(c id) + (c + id)(a ib) = 2(ac + bd) = 2 ( ∗) M1M2 M1 M2 − − R M1M2

3 2.2 Calculating 2 |M1| We may simplify the expression for (2.1). First, by rearranging commuting terms: M1 µ ν 2 γ (/p + /k + m)γ i = ie e0∗e u¯0 u. (2.6) M1 − µ ν (p + k)2 m2 − The denominator may then be rewritten as follows:

(p + k)2 m2 = p2 + p k + k p + k2 m2 − · · − = 2p k. (2.7) ·

Where the second equality follows from p2 = m2 and k2 = 0. The numerator may also be simpliﬁed using

ν µ ν ν (/p + m)γ u = (γ γ pµ + γ m)u = (2pν γν p + γνm)u − / = 2pνu. (2.8)

Where the second equality follows from the Dirac algebra [γµ, γν] = 2ηµν and the third from the equation of motion for positive frequency solutions of the free Dirac field (p m)u = 0. / − With simplified, we may find the complex conjugate .3 M1 M1∗ µ ν µ ν 2 2γ p + γ /kγ i = ie e0∗e u¯0 u (2.9) M1 − µ ν 2p k · (2.10) ρ σ σ ρ 2 2γ p + γ /kγ i ∗ = +ie e0 e∗u¯ u0 (2.11) − M1 ρ σ 2p k · Hence, the square matrix element is

µ ν µ ν σ ρ σ ρ 2 4 2γ p + γ /kγ 2γ p + γ /kγ = e e0∗e0 e e∗u¯0 uu¯ u0. (2.12) |M1| µ ρ ν σ 2p k 2p k · · If we make the spinor indices explicit we are free to rearrange this.

µ µ ν σ ρ σ ρ 2 4 2γ p + γ /kγ 2γ p + γ /kγ 1 = e eµ0∗eρ0 eνeσ∗u¯0a ubu¯c ud0 |M | 2p k ab 2p k cd µ · µ ν σ ρ· σ ρ 4 2γ p + γ /kγ 2γ p + γ /kγ = e e0∗e0 e e∗u0 u¯0 u u¯ (2.13) µ µ ν ν d a 2p k b c 2p k · ab · cd 3The complex conjugate of a bi-spinor product v¯γµu may be calculated as:

µ † µ † 0 † † µ † 0 † 0 µ 0 0 † 0 µ (v¯γ u)∗ = u (γ ) (γ ) v = u (γ ) γ v = u γ γ γ γ v = (u γ )γ v = u¯γµv

4 2.2.1 Summing over Polarisation States

Currently the expression retains freely speciﬁed spin and polarization states for the electrons and photons. Experimental coulomb scattering, however, typically involves unpolarised photons colliding with electrons in an unpolarised medium and so we must average over these states. Likewise, photon and electron detectors are also commonly blind to polarisation and so we must sum over the ﬁnal spin states. Therefore, we are interested in the unpolarised square matrix element

1 1 2 1 2 2 ∑ ∑ ∑ ∑ (s, r s0, r0) = ∑ = . (2.14) 2 s 2 r |M → | 4 |M| |M| s0 r0 spins

The spin summation for the electron states can be performed using the completeness relation

s s ∑ = u (p)u¯ (p) = /p + m. (2.15) s=1,2

While the polarisation summation for the photon polarizations can be performed by making the replacement:

r r ∑ = eµ∗ (k)eν(k) gµν (2.16) r=1,2 → −

Which is valid here as both Lorentz indices are contracted within the expression for . Sum- ming over spin states in (2.13) by using these techniques we arrive at the expression M

4 µ ν + µ ν σ ρ + σ ρ 2 e 2γ p γ /kγ 2γ p γ /kγ = g g (p0 + m) (p + m) . (2.17) |M1| 4 µρ νσ / da 2p k / bc 2p k · ab · cd

2.2.2 Trace Algebra

An examination of (2.17) reveals that it has the form of a trace of a product of four 4-dimensional matrices: 3 3 tr(MNOP) = ∑ (mnop)dd = ∑ mdanabobc pcd (2.18) d=0 c,b,a,d=0

Therefore, we have: e4 1 2 = (2.19) |M1| 4 (2p k)2 A ·

µ ν µ ν = tr[(p0 + m) (γ /kγ + 2γ p ) (p + m) γ /kγ + 2γ p ] (2.20) A / / ν µ µ ν

On expansion consists of 24 = 16 terms: A

= + + ... + . (2.21) A I II XVI

5 These traces may be evaluated individually by hand using the suitable contraction identities [1]. If, in multiplying out, we move from the end of the expression to the start while cycling within brackets from left to right, we ﬁnd that terms , , , and contain an odd number of gamma matrices and, consequently,III IV immediatelyVII − X vanish.XIII AppendixXIV A contains the evaluations of the other 8 traces. Adding all these terms together returns

4 2 2 2 = 32[2m + 2m (p k) m (p p0) m (p0 k) + (p k)(p0 k)]. (2.22) A · − · − · · ·

2.2.3 Mandelstam Variables

It is helpful at this point to utilise the Lorentz invariant Mandelstam variables, which follow directly from 4-momentum conservation p + k = p0 + k0.

s = (p + k)2 = 2p k + m2 · 2 2 = (p0 + k0) = 2p0 k0 + m − · 2 2 t = (p0 p) = 2p p0 + 2m − − · 2 = (k0 k) = 2k k0 − − · 2 2 u = (p0 k) = 2k p0 + m − − · 2 2 = (k0 p) = 2p k0 + m (2.23) − − ·

These also yield the relation s + t + u = 2m2. Substituting these in to gives A

= 8[2m2(2s + t + u) + (s m2)(u m2)] A − − = 8[4m4 + 2m2(s m2) (s m2)(u m2)] A − − − − 4m4 2m2 (u m2) 2 = 2e4 + − (2.24) |M1| (s m2)2 (s m2) − (s m2) − − − Now we may return to calculate the other terms in 2. |M|

2.3 Calculating the full square matrix element

We may simplify in the same way as and similarly ﬁnd its complex conjugate. M2 M1

ν µ ν µ 2 γ /k 0γ + 2γ p i = ie e0∗e u¯0 − u (2.25) M2 − ν µ 2p k − · 0 ρ σ σ ρ 2 γ /k 0γ + 2γ p i ∗ = +ie e0 e∗u¯ − u0 (2.26) − M2 ρ σ 2p k − · 0 With these in hand we may ﬁnd

ν µ ν µ ρ σ σ ρ 2 4 γ /k 0γ + 2γ p γ /k 0γ + 2γ p = e e0∗e0 e e∗u¯0 − uu¯ − u0. (2.27) |M2| µ ρ ν σ 2p k 2p k − · 0 − · 0

6 2 2 Comparing this with (2.12) we see that 1 is identical to 2 under the replacement k k (s u). This means that the evaluation|M | is symmetric and|M immediately| allows us to write:→ − 0 →

4m4 2m2 (s m2) 2 = 2e4 + − (2.28) |M2| (u m2)2 (u m2) − (u m2) − − − Now, all that is left is to calculate the interference term, which from (2.10) and (2.26) is:

µ ν µ ν σ ρ σ ρ 4 γ /kγ + 2γ p γ /k 0γ + 2γ p 2 ∗ = 2e e0∗e0 e e∗u¯0 uu¯ − u0 (2.29) M1M2 µ ρ ν σ 2p k 2p k · − · 0 Due to the symmetry between this expression and (2.12) we see that its evaluation will mirror sections 2.2.1 and 2.2.2 with expressions equivalent to (2.19) and (2.20) of

e4 1 2 = 0 (2.30) M1M2∗ 2 (2p k)( 2p k ) A · − · 0 µ ν µ ν 0 = tr[(p0 + m) (γ /kγ + 2γ p ) (p + m) γ /k 0γ + 2γ p ] (2.31) A / / − µ ν ν µ When expanded out likewise comprises 16 terms A0

0 = 0 + 0 + ... + 0 (2.32) A I II XVI Multiplying out identically to (2.22), the equivalent terms immediately vanish. Appendix A contains the evaluations of the other 8 traces. Adding all these terms together:

2 0 = 32(k k0)(p p0) 32(p k)(p p0) + 16m (p0 k) + 32(p k0)(p p0) A · · − · · · · · 2 2 2 2 4 16m (p0 k0) + 16m (p p0) 16m (k k0) + 32m (p k) 32(p k0) + 16m (2.33) − · · − · · − · Substituting the Mandelstam variables gives us:

h 4 2 2 2 2 i 0 = 8 4m + m (u m ) + m (s m ) (2.34) A − − 8m4 2m2 2m2 2 = 2e2 + . (2.35) M1M2∗ (s m2)(u m2) (u m2) − (s m2) − − − − Bringing all the terms together ﬁnally gives us:

" 1 1 2 2 = 2e4 4m4 + |M| s m2 u m2 − − 1 1 u m2 s m2 +4m2 + − − (2.36) s m2 u m2 − s m2 − u m2 − − − − To calculate the cross section from this we must now specify the frame of reference.

7 3 Frame speciﬁc Evaluation

3.1 Centre of mass frame

In the centre of mass (CM) frame the 4-momenta of the particles may be written

k =(ω∗, 0, 0, ω∗) k0 = (ω∗, ω∗ sin θ∗, 0, ω∗ cos θ∗) (3.1) p =(E, 0, 0, ω∗) p0 = (E, ω∗ sin θ∗, 0, ω∗ cos θ∗) − − −

Where ω∗ and θ∗ are the photon energy and scattering angle in the CM frame. This makes the Mandelstam variables:

2 s = 2ω∗(E + ω∗) + m 2 t = 2ω∗ (cos θ∗ 1) − 2 u = 2ω∗(E + ω∗ cos θ∗) + m . (3.2) −

Therefore, (1.2) becomes

1 2 dσ = dΠ2 (3.3) 2E2ω 1 ω∗ |M| ∗| − E |

In terms of frame invariant variables:

1 dσ = 2dΠ . (3.4) 2(s m2) |M| 2 −

3.2 Evaluating the phase space integral in the CM frame

We may also evaluate the ﬁnal state phase space integral in the CM frame4.

Z ZZZ 3 ZZZ 3 d p0 1 d k0 1 4 (4) dΠ = (2π) δ (p + k p0 k0). (3.5) 2 ( π)3 2E ( π)3 2E 2 p0 2 k0 − −

We integrate over all 3 components of p0 which, given p = k, sets p0 = k0. And so with a change of variable to spherical polar coordinates in phase space− 5we have −

Z ZZZ 2 dk0k0 d cos θ∗dφ∗ dΠ = δ(E + ω∗ E (k0) E (k0)). (3.6) 2 (2π)24E E p0 k0 p0 k0 − −

4We temporarily forget manual conservation of energy and momentum in order for the Dirac delta function to take over. 5 µ When working in spherical polar coordinates k0 is one dimensional momentum, not k0 .

8 However, the collision is symmetric about φ so we pick up another factor of 2π. We now have an integral over momentum, but a Dirac delta function in terms of energies, which are functions of momentum via the dispersion relation. So, if f (k ) = E + ω E (k ) E (k ), we must 0 ∗ p0 0 k0 0 evaluate − −

Z ZZ dk k 2d cos θ δ(k k ) = 0 0 ∗ 0 0 dΠ2 (− ) 8πE E ∂ f k0 p0 k0 ∂k0 ! 1 Z k 2d cos θ k k − = 0 ∗ 0 + 0 8πE E E E p0 k0 p0 k0 ! Z k 2d cos θ k = 0 ∗ 0 (3.7) 8π E + E p0 k0

Where f (k0 ) = 0. The exact value of k0 is unnecessary, but integrating over this distribution enforces energy conservation E + w = E + E . Hence, ∗ p0 k0

Z Z ω∗ dΠ2 = d cos θ∗ . (3.8) 8π(E + ω∗)

2 We may also calculate that in the CM frame dt = 2w∗ d cos θ∗. Thus,

Z Z dt dΠ2 = 16πω∗(E + ω∗) Z dt = . (3.9) 8π(s m2) − This gives us a manifestly Lorentz invariant formula for the cross section:

dσ 1 = 2. (3.10) dt 16π(s m2)2 |M| −

Where 2 is given by (2.36). To express this in terms of more physical variables we now move in to the|M| lab frame.

3.3 Lab frame

In the lab frame the electron is at rest and the 4-momenta of the particles may be written

k =(ω, 0, 0, ω) k0 = (ω, ω sin θ, 0, ω cos θ) (3.11) p =(m, 0) p0 = (E, p0) −

This makes the Mandelstam variables in the lab frame:

9 s = 2mw + m2

t = 2ww0(cos θ 1) − 2 u = 2mw0 + m (3.12) −

Note that in this frame w0 and θ are not independent variables. To ﬁnd this dependence we derive the Compton formula. Conserving 4-momentum p + k = p0 + k0 once again:

2 2 p0 = (p + k k0) (3.13) − 2 2 2 = p + 2p (k k0) 2k k0 + k + k0 (3.14) · − − · 2 2 m = m + 2m(w w0) + 2ww0(cos θ 1) (3.15) − − Thus,

1 1 1 = (1 cos θ). (3.16) w0 − w m −

Rearranging for w0:

w w0 = . (3.17) 1 + w (1 cos θ) m − Which allows us to express t only in terms of cos θ:

2w2(1 cos θ) t = − (3.18) 1 + w (1 cos θ) m − dt 2w2 = = 2 2 2w0 (3.19) d cos θ 1 + w (1 cos θ) m − So we ﬁnd the differential cross section with respect to the angle in the lab frame:

dσ dσ dt 2 1 2 = = 2w0 . d cos θ dt d cos θ 16π(2mw)2 |M|

To express 2 in terms of lab frame variables we substitute (3.12) and (3.17) in to (2.36) to get |M| " # cos θ 1 2 cos θ 1 w w 2 = 4 4 + 2 + 0 + 2e 4m −2 4m −2 |M| 2m 2m w w0 w w = 2e4 0 + sin2 θ . (3.20) w w0 −

Which ﬁnally produces the Klein-Nishina formula [4]:

dσ πα2 w 2 w w = 0 0 + 2 2 sin θ . (3.21) d cos θ m w w w0 −

10 3.4 Evaluating the phase space integral in the lab frame

In order to verify this result it is possible to return to the phase space integral and carry it out directly in the rest frame of the incident electron.

Z ZZZ d3 p 1 ZZZ d3k 1 = 0 0 ( )4 (4)( + ) dΠ2 3 3 2π δ p k p0 k0 . (3.22) (2π) 2w0 (2π) 2E0 − −

0 0 Integrating once again over all three components of p sets p0 = k k and leaves the remaining delta function over the energies. Changing to spherical polar coordinates− once again:

Z ZZ 2 dω0ω0 d cos θ dΠ2 = δ(g(ω0)). (3.23) 8πE0ω0 Where:

g(ω0) = E0 + ω0 m ω (3.24) q − − 2 2 = (k k ) + m + ω0 m ω − 0 − − p 2 2 = ω + ω0 2ωω0 cos θ + ω0 m ω (3.25) − − − ∂g(ω0) 1 2 2 1 = 1 + 2(ω0 ω cos θ) (m + ω + ω0 2ωω0 cos θ)− 2 ∂ω0 − 2 − ω ω cos θ = 1 + 0 − . (3.26) E0 (3.22) then becomes Z ZZ 1 dω0ω0d cos θ ω0 ω cos θ − dΠ2 = 1 + − δ(ω0 ω0) 8πE0 E0 − Z d cos θ ω = 0 8π E + ω ω cos θ 0 0 − Z d cos θ ω = 0 8π m 1 + ω (1 cos θ) m − Z d cos θ ω 2 = 0 (3.27) 8π mω

Where for the 3rd equality, integrating over the delta function enforces that E0 + ω0 = ω + m. Substituting (3.27) in to (3.4) produces

dσ 1 ω 2 1 = 0 2 d cos θ 8π mω 2(2mω) |M| 1 ω 2 = 0 2 32πm2 ω |M| πα2 w 2 w w = 0 0 + 2 2 sin θ . (3.28) m w w w0 −

So, (3.21) is equal to (3.28), we have once again produced the Klein-Nishina formula.

11 4 Plots of Differential cross section

We now have two expressions for the differential cross section. The ﬁrst is with respect to t and in terms of Mandelstam variables (3.10). It may rearranged as:

" dσ 2πα2 1 1 2 = 4m4 dt (s m2)2 s m2 − s m2 + t − − − 1 1 s m2 + t (s m2) +4m2 + − + − . (4.1) s m2 − s m2 + t s m2 s m2 + t − − − − The other is with respect to cos θ and in terms of lab variables (3.21). It may be written:

dσ πα2 1 2 1 w = + (1 cos θ) + cos2 θ . (4.2) d cos θ m2 1 + w (1 cos θ) 1 + w (1 cos θ) m − m − m −

ω is related to s by

s m2 ω = − (4.3) 2m

In these forms it is simple to plot the differential cross sections for different centre of mass energies in the following limits: 1. s m2 << m2 (s > m2) − 2. s m2 2m2 − ≈ 3. s m2 >> m2. − Note that s and t and not independent and so the limits on t are not the same in each case. (3.2) relates t and ω∗, therefore, the limits on t in terms of s are:

(s m2)2 − t 0. (4.4) − 4s ≤ ≤

dσ Note that t is always negative. Figure 4.1 shows the plots for d cos θ while ﬁgure 4.2 shows those dσ for dt .

12 ×10-3

1.2 s = 1.001m2 s = 2m2 2 1 s = 10m ]

-2 0.8 [MeV θ

σ 0.6 d dcos

0.4

0.2

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 cos θ

Figure 4.1: Plot of against cos θ for three different centre of mass energies.

×103 ×10-3 10 2 s = 1.001m2 12 s = 2m

9 11

10 8 ] ] -4 -4 9 [MeV [MeV

σ 7 σ dt dt d d 8

6 7

6 5 × -6 × -3 10 5 10 -0.25 -0.2 -0.15 -0.1 -0.05 0 -140 -120 -100 -80 -60 -40 -20 t [MeV2] t [MeV2]

(a) s m2 << m2 (b) s m2 2m2 − − ≈

×10-3 0.6 s = 10m2

0.5

] 0.4 -4 [MeV σ dt d 0.3

0.2

0.1 -2 -1.5 -1 -0.5 0 t [MeV2]

13 4.1 Plot Discussion

In the low energy limit of the lab reference frame, the graph is approximately symmetric about cos θ = 0: the likelihood of forward and backward scattering is the same. In this limit ω 0, → ω0 1. Hence, (3.21) becomes: ω → dσ πα2 = 1 + cos2 θ) . (4.5) d cos θ m2 This is the formula for Thomson scattering and may be derived within the conﬁnes of classical electromagnetism. At higher energy the probability of scattering at large angles decreases while the probability of a θ = 0 scattering event is unchanged. Therefore, the total cross section decreases as s increases. In the high s limit the probability of backward scattering is low and fairly constant with angle, while likelihood of forward scattering increases rapidly as θ becomes smaller. Figure 4.2 also resembles classical scattering of electromagnetic radiation for low centre of mass energy, and shows that as the centre of mass energy increases, the total cross section decreases. At high s the differential cross section is larger for higher t , deceasing rapidly as t approaches 0. In the centre of mass this corresponds to a greater cross| section| for backwards scattering.

5 Conclusion

We have calculated the unpolarised differential cross section with respect to both t and cos θ. Where θ is the scattering angle in the lab frame and t is the square momentum transfer between 2 the initial-state and ﬁnal-state photon. The square matrix element (e−γ e−γ) was evaluated using perturbative quantum electrodynamics at the tree level.|M → | The Klein-Nishina formula was reproduced using two different approaches. The ﬁrst by sub- dσ stituting dt , as calculated in the centre of mass frame, with lab frame variables by direct con- version of dt to d cos θ. The second by carrying out the calculation, including the phase space integral, directly in the lab frame. dσ dσ Plots of dt and d cos θ reproduce classical scattering of electromagnetic radiation by free electrons in the low energy limit. In the lab frame, at high s, events with a small angle are more probable, peaking sharply at θ = 0. The cross section decreases as the centre of mass energy increases.

References

[1] M. E. P. . D. V. Schroeder. An Introduction to Quantum Field Theory. Westview Press, 1995. [2] F. J. Dyson. “The S Matrix in Quantum Electrodynamics”. In: Physical Review 75 (June 1949), pp. 1736–1755. DOI: 10.1103/PhysRev.75.1736. [3] D. Tong. “Quantum Field Theory”. University of Cambridge Part III Mathematical Tripos. 2007. [4] O. Klein and Y. Nishina. “Uber¨ die Streuung von Strahlung durch freie Elektronen nach der neuen relativistischen Quantendynamik von Dirac”. German. In: Zeitschrift f¨urPhysik 52.11-12 (1929), pp. 853–868. ISSN: 0044-3328. DOI: 10.1007/BF01366453.

14 Appendices

A Trace evaluations for and A A0 µ ν µ ν = tr(p0γ /kγ pγν/kγµ) 0 = tr(/pγ /kγ /pγµk/0γν) I / / I − µ µ = tr(p0γ /k( 2p)/kγµ) = tr(/pγ /k( 2/kγµ/p)) / − / − − = 2tr(p0( 2/k p/k)) = 8tr(/p0(k k0)/p) − / − / · = 4tr(p0/k(2p k /k p)) = 32(k k0)(p p0) / · − / · · = 32(p k)(p0 k) · · µ 0 = tr(p0γ p2γ p ) II / / ν µ µ ν = tr(p0γ /kγ p2γµ pν) = 2tr(/p0/p/k( 2/p)) II / / − = 2tr(p0( 2p p/k)pν) = 4tr(/p0(2(p k) /k/p)/p) / − // · − 2 2 = 4tr(p0m /k) = 32(p k)(p p0) + 16m (p k) − / − · · · 2 = 16m (p0 k) − · µ ν 0 = tr(p02γ p pγ /k 0γ ) V − / / µ ν µ ν = tr(p02γ p pγµ/kγµ) = 2tr(/p0( 2/p)/k 0/p) V / / − − µ = 2tr(p γ p p/kγ ) = 4tr(p0(2(p k0) k/0 p)p) /0 // µ / · − / / 2 2 = 2m tr(p0( 2/k)) = 32(p k0)(p p0) 16m (p0 k0) / − · · − · 2 = 16m (p0 k) − · µ ν 0 = tr(p02γ p p2γ p ) VI / / ν µ µ ν = tr(p02γ p p2γ p ) = 4tr(/p0/p/p/p) VI / / µ ν µ 2 = 4m2tr(pγ pγ ) = 16m (p p0) / / µ · 2 = 32m (p p0) − · µ ν 0 = tr(mγ /kγ mγ /k 0γ ) XI − µ ν µ ν 2 = tr(mγ /kγ pγ /kγ ) = 4m tr(/k 0/k) XI / ν µ − 2 µ 2 = 4m tr(γ /k/kγ ) = 0 = 16m (k k0) µ − ·

µ ν µ ν = tr(mγ /kγ p2γµ pν) 0 = tr(mγ /kγ m2γ up ) XII / XII n µ 2 µ 2 = 2m tr(γ /k pγµ) = 32m (p k) / · = 32m2(k p) · µ 0 = tr(m2γ p mγ /k 0γ ) XV − ν µ ν µ ν 2 = tr(mγ /kγ mγν/kγµ) = 32m (p k) XV 2 µ · = 2m tr(γ /p/kγµ) 2 µ ν = 32m (p k) 0 = tr(m2γ p m2γν pµ) · XVI = 16m4 = tr(m2γµ pνm2γ p ) XVI µ ν 4 µ = 4m tr(γ γµ) = 64m4 15