Compton from

Andrew Lawson

17 January, 2014

Abstract

We present a derivation of the differential cross section for the Compton scattering of an and a . 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 [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 . 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 :

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 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 . (3.7)

This definition allows us to rewrite the remaining delta function as

 0 2 2 µ µ 0 2 δ (p + k − k ) − me = δ s − 2 (p + k ) kµ − me 2 √ 0  = δ s − me − 2 s |k | . (3.8)

We can write sin θdθ = d cos θ and use the definition of t to transform the integral further, as

t = (k − k0)2 = −2k0k00 + 2 |k| |k0| cos θ, (3.9) implies that dt = 2 |k| |k0| . (3.10) d cos θ We can substitute these definitions into our expression (3.5) and use the step function to cut off the part of the integral that doesn’t contribute to result in

√ s 2 1 0 dt 2 √ 0  |M1 + M2| σ = d |k | δ s − me − 2 s |k | 2 . (3.11) 8π ˆ ˆ0 2 |k| 2 (s − me)

2 √ √ Because 0 < s − me/2 s < s, we can use the remaining delta function to obtain an expression involving only an integral over t, 1 |M + M |2 √1 2 σ = dt 2 . (3.12) 32π ˆ |k| s (s − me) In our chosen frame we can substitute s − m2 |k| = √ e (3.13) 2 s

5 back into this expression; and thus it follows that

dσ |M + M |2 = 1 2 2 2 dt 16π (s − me) " # e4 −su + m2 (3s + u) + m4 −su + m2 (3u + s) + m4 8m4 + 4m2 (s + u) = e e + e e + e e . (3.14) 2 2 2 2 2 2 2 2 8π (s − me) (s − me) (u − me) (s − me)(u − me)

3.2 Lab Frame

It is of particular use to calculate the quantity

dσ , (3.15) d cos θ in the rest frame of our incident electron (i.e. the lab frame); we can perform this calculation by two different methods. The first and most obvious method would be to return to our Lorentz invariance phase space integral and simplifying using the rest frame of our incident electron rather than the centre of mass frame. We thus return to (3.5) and use the frame specified by the 4-vectors

µ p = (me, 0) , (3.16) kµ = (ω, k) , (3.17)   0µ p 2 0 0 p = me + |p |, p , (3.18) k0µ = (ω0, k0) . (3.19)

Subbing these into (3.5) and changing to polar coordinates we result in the expression

me+|k| 2 1 0 0 0 0 |M1 + M2| σ = |k | d |k | d cos θδ (2me (|k| − |k |) + 2 |k| |k | (cos θ − 1)) 2 . (3.20) 4π ˆ ˆ0 2 (s − me)

This time the delta function simplifies to   0 0 1 0 me |k| δ (2 |k | |k| (cos θ − 1) + 2me (|k| − |k |)) = δ |k | − , (3.21) 2me − 2 |k| (cos θ − 1) me − |k| (cos θ − 1) which allows us to perform the integration over |k0| to yield

0 2 1 ω |M1 + M2| σ = d cos θ 2 , (3.22) 8π ˆ me − ω (cos θ − 1) 2 (s − me) where we have subbed in |k| = ω and |k0| = ω0. Note that although we have kept the variables ω0 in explicitly, the action of the delta function requires that the relation

1 1 1 − 0 = (cos θ − 1) (3.23) ω ω me is satisfied. This is of course the usual formula for the frequency shift that we do indeed observe for Compton scattering [1]. We can now take the derivative of (3.22) with respect to cos θ to find the result we need. We are thus left

6 2.5

2.0

θ 1.5 σ d d cos 2 e 4 e πm 1.0 16

0.5

0.0 1.0 0.5 0.0 0.5 1.0 − − cos θ

2 Figure 3.1: Plot showing the variation of dσ/d cos θ as cos θ varies, at centre of mass energies s & me (blue), 2 2 s ∼ 2me (green) and s  me (red). with the expression

0  0 2 dσ 1 ω 2 1 ω 2 = 2 2 |M1 + M2| = |M1 + M2| , (3.24) d cos θ 32π meω − meω (1 − cos θ) 32π meω where we have used the relation 1 m ω = s − m2 . (3.25) e 2 e However by noting that (3.14) is valid in all reference frames, we can obtain (3.24) by directly converting dt into d cos θ. Recalling the definition of t as

t = (k − k0)2 = 2ωω0 (cos θ − 1) , (3.26) we can simply take the derivative of t with respect to cos θ (while also making use of the relation (3.23)) to obtain dt dω0 2ωω02 = 2ωω0 + 2ω (cos θ − 1) = 2ωω0 + (cos θ − 1) = 2ω02. (3.27) d cos θ d cos θ me If we multiply (3.27) by our expression (3.14) we obtain the desired relation,

 0 2 dσ dt dσ 1 ω 2 = = |M1 + M2| . (3.28) dt d cos θ d cos θ 32π meω

It is easily seen that the results of the two methods agree with each other. Finally it remains to write our expression for our squared matrix element in terms of ω and ω0, which requires use of (3.25) and the relation 1 m ω0 = m2 − u . (3.29) e 2 e Subbing in (3.25) and (3.29) into (3.24), by rearranging and making use of the relation (3.23) we end up with: 4  0 2  0  dσ e ω ω ω 2 = 0 + − sin θ . (3.30) d cos θ 16π meω ω ω

7 We now have enough information to deduce some of the physical properties of Compton scattering from our 2 differential cross sections. Fig. 3.1 shows plots of dσ/d cos θ against cos θ. For the case (a), where s & me (i.e. ω  me, we recover the non-relativistic, elastic version of Compton scattering, known as . In this case we see that are predominantly scattered either in the same direction or in 2 2 the complete reverse direction with relatively equal regularity. At s ∼ 2me (i.e. ω ∼ me/2) we begin to see 2 that forward scattering begins to dominate, and for s  me (i.e. ω  me) forward scattering dominates almost completely.

4 Conclusions

In this paper we have presented a derivation of the cross section for the Compton scattering of a process using QED. In particular we have managed to recover two of the primary relations used to measure Compton scattering; firstly the shift between the incident and outgoing photons and the differential cross section in the electron rest frame. The agreement of these theoretical results with experiment thus show that QED provides an extremely accurate, relativistically invariant description of electromagnetism.

References

[1] A. H. Compton, “A Quantum Theory of the Scattering of X-Rays by Light Elements”, Physical Review 21 (1923) 483.

[2] M. E. Peskin and D. V. Schroeder, “An Introduction to ”, Westview Press, 1995.

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