ScatteringCross-sectionsin Quantum Electrodynamics

Final expression after simplifying equation 2.10 T r (γ σ p′σ γ µ γ ρ pρ γ ν ) T r m2e γ µ γ ν 4T r p′µ pν g µν p′ρ pρ p′ν pµ 4m2e g µνT r [(γ σ p′σ me )γ µ (γ ρ pρ me )γ ν ] Similarly, we can get the expression for T r (γ σ kσ mµ )γµ (γ ρ kρ′ mµ )γν . T r (γ σ kσ mµ )γµ (γ ρ kρ′ mµ )γν 4T r kµ′ kν g µν k ′ρ kρ kν′ kµ 4m2µ g µνNow, we can write (1/4)Pspin M 2 in terms of above simplified formulaiih1X4e4 h M 2 4 p′µ pν g µν p′ρ pρ p′ν pµ 4m2e g µν kµ′ kν g µν k ′ρ kρ kν′ kµ 4m2µ g µν4 spinqHere, we are taking me 0 because incoming particles require high energy (Kinetic Energy) toproduce larger mass particles. Then, we will get"# e4 ′µ ν1X2′ν µµν ′′′′2 M 4 p p p p g [p · p] kµ kν kν kµ gµν [k · k mµ ]4 spinq(2.11)By solving above equation, We will geti8e4 h1X M 2 4 (p′ · k)(p · k ′ ) (p′ · k ′ )(p · k) (p · p′ )m2µ4 spinqSince, we know thatPspin(2.12) M 2 is proportional to a differential cross-section which is a physicalquantity. So, we must need to write above expression regarding energy and angle. The vectors p,p′ , k, k ′ and q concerning the basic kinematic variables-energies and angles. We need to take aparticular frame of reference for which our equation become accessible. Now, we are making thesimplest choice to evaluating cross-section in center-of-mass frame.k (E, Ek)θ′p (E, E ẑ)p (E, E ẑ)k ′ (E, Ek)Figure 2.3: e e µ µ interaction in center-of-mass frame7

To compute the modulus square matrix element, we need to write equation (2.12) in terms ofenergy and angle.q2 (p p′ )2 4E 2p · p′ 2E 2p·k p′ · k ′ E 2 E k cos θp · k′ p′ · k E 2 E k cos θNow, putting the values of p · k , p · k ′ etc in the (1/4)result1X M 24 spinTheP2.3spin Pspin M 2 expression. It gives the followingie4 h(E k cos θ)2 (E k cos θ)2 2m2µ22E(2.13) M 2 expression is written in the center-of-mass energy form. Here E (Ecm /2)Total scattering cross-sectionWhen a beam of particles strikes a target consisting of particles of a different type, some of theparticles pass directly through the target while other deflected. Those deflected particles are said tointeract when they collide with the target particles. The cross section is a measure of the effectivenessof the incident and target particles interaction. Larger the cross section, the more likely it is thatthe incident particles deflected.In the center of mass frame, our differential cross-section is given by dσdΩ cm P M 22EA 2EB vA vB (2π)2 4Ecm(2.14)Above formula comes when we take the interaction part of S-matrix. For our given interactionrelative velocity and energy of the incoming particle in center of mass frame defined as vA vB EA EB dσdΩ cm 2Ecm 2 k 11 X M 2 222Ecm 16π Ecm 4 spinsIntegrating differential scattering cross-section over solid angle (dΩ) for total cross-section8(2.15)(2.16)(2.17)

2σT α24Ecms1 m2µE2Z2πrm2µ1 2Eπm2µ1 2E00α2 24EcmZm2µ1 2E!!sin θdθdφR 2π R 2π cos(2θ) 1 002sin θdθdφBy solving integration, our total scattering cross-section formula isσT4π 2α3 s"#m2µ1 m2µ1 2 1 E2 E2(2.18)In the above expression when the energy of the incoming particle is less than the rest mass energyof the muon then the total cross-section will become zero. This information gives us that the energyof the incoming particle should always be greater than or equal to the rest mass energy of muonsto produce them. The energy for which our total cross-section becomes finite called as thresholdenergy.2.4ResultIn the high energy limit whereE mµ , these formula reduce to dσdΩ cm E mµα2(1 cos2 θ)24EcmσT E mµ4πα223Ecm(2.19)(2.20)Here, we can see the angular dependence on the differential cross-section in the high energy limit.But the total scattering cross-section has energy dependence and strength of the electromagneticinteraction dependence. But as we can see from the above formula, differential cross-section hasmaximum value for (θ 0, π). In the high-energy limit, Ecm is the only dimensionful quantity 2in this process, so dimensional analysis dictates that σT Ecm. Since, we knew intuitively fromthe beginning that σT α2 , we only had to work to get the factor of (4π/3) in total scatteringcross-section. Here, scattering cross-section is zero for Ecm 2mµ . But, the threshold energy forwhich scattering cross-section is finite will be the rest mass energy of the incoming particle.2.5Polarized electron and positron annihilationIn our previous discussion, we have used unpolarized incident and scattered particles to find theaverage scattering cross-section. Now, we will take definite polarization of incident and scatteredparticles using helicity operator to calculate scattering cross-section. But, we can check averagescattering cross-section by taking an average of all the possible scattering cross-section corresponding polarization of incident and scattered particles. Here, our Feynman diagram will be same for9

e e µ µ and lowest order in α. This calculation of polarized cross-section will help us tounderstand how the angular dependence appears in unpolarized cross-section. Here, we used helicity projection operator to project out the desired left and right-handed spinor for incoming andoutgoing particles. Throughout this section, we work in high energy limit. Without putting helicityoperator in scattering amplitude gives the same result as got for the unpolarized case.2.6HelicityThe helicity gives the projection of the directions of spin and the particle’s momentum. If the 3momentum p and spin both point in the same direction, the helicity has its maximum value (positivevalue), while if they point in opposite directions, the helicity has its maximum negative value. If pand spin are at right angles, the helicity is zero. Here, we are taking γ 5 matric as!0 1γ5 01Now, we will take only one set of polarizations at a time. To do this, our projections operators ontoright-and left-handed spinors, respectivelyI γ5 20 00 1!I γ5 21 00 0!Now, we can make replacement in amplitude for right-handed spinor asv(p′ )γ µ u(p) v(p′ )γ µ I γ52 u(p) Scattering amplitude for e R eL µR µL after replacementie2iM 2q I γ5I γ5′′ µv(p )γu(k)γνu(p)v(k )22(2.21)Here, we have taken right handed electron and the simplification of initial current density in Feynmandiagram is given by′v(p )γµ I γ52 †′0† µu(p) v (p )γ γ I γ52 u(p)Since, we know that γ µ , γ 0 and γ 5 are hermitian matrix†′v (p ) I γ52 µ 0γ γ †u(p)We know that right handed electron corresponds to a left-handed positron. Hence, the amplitudevanishes unless the electron and positron have their opposite helicity or equivalently unless their10

spinor have the same helicity.Now, the sum over the electron and positron spins in the modulus square amplitude.X′v(p )γµspin I γ52 u(p)2 X ′v(p )γµspin I γ52 u(p)u(p)γν I γ52 v(p )′Since we know that the spins sums of ferimons and antifermionsXus (p)us (p) /p msXs′′′v s (p)v s (p) /p mIn high energy limit, we can take (m 0)Xus (p)us (p) /psX′′v s (p)v s (p) p/s′By using above spins sum of ferimons and antifermions, we can write modulus square half amplitudeas XI γ5I γ5I γ5′ µν2′ µu(p) p/ γ v(p )γ/pγ222spinNow, using trace technology T r p/′ γ µ I γ52 pγ ν/ I γ52 Tr"γ σ p′σ γ µ γ ρ pρ γ νγ σ p′σ γ µ γ ρ pρ γ νγ σ p′σ γ µ γ ρ pρ γ ν γ 5 444γ σ p′σ γ µ γ 5 γ ρ pρ γ ν γ 5 4# i1 h σ µ ρ ν ′T r γ γ γ γ pσ pρ γ σ γ µ γ ρ γ ν p′σ pρ γ σ γ µ γ ρ γ ν γ 5 p′σ pρ4 i1 h σ µ ρ ν ′T r 2γ γ γ γ pσ pρ 2γ σ γ µ γ ρ γ ν p′σ pρ γ 54(2.22) 1 σ µ ρ ν ′ 1 σ µ ρ ν ′T r γ γ γ γ pσ pρ T r γ γ γ γ pσ pρ γ 522(2.23)Now, using the gamma matrices identitiesT r[γ µ γ ν γ ρ γ σ ] 4 [g µν g ρσ g µρ g νσ g µσ g νρ ]T r[γ µ γ ν γ ρ γ σ γ 5 ] 4iǫµνρσ11

Above identities makes the trace in Minkowski matrices, since p′σ , pρ are vectors so we can takethese out from trace. I γ5I γ5ν′ µ p/γT r p/ γ22 1 T r (γ σ γ µ γ ρ γ ν ) T r γ σ γ µ γ ρ γ ν γ 5 p′σ pρ (2.24)2hi2 g σµ g ρν g σρ g µν g σν g µρ iǫµνρσ p′σ pρ (2.25)Final expression for modulus squared half amplitudeXspin v(p′ )γ µ I γ52 u(p) 2 2 [p′µ pν g µν (p′ · p) p′ν pµ iǫσµρν p′σ pρ ](2.26)Similarly, for muon modulus squared half amplitude given byXspin u(k)γµ I γ52 v(k ′ ) 2 2 kµ kν′ gµν (k · k ′ ) k ν kµ′ iǫσµρν k σ k ′ρ(2.27) Further, we can write M 2 for e R eL µR µL interactionXspin M 2 4e4 ′µ νp p g µν (p′ · p) p′ν pµ iǫαµβν p′α pβ4q kµ kν′ gµν (k · k ′ ) kν kµ′ iǫσµρν k σ k ′ρ 4e4 2(p · k)(p′ · k ′ ) 2(p · k ′ )(p′ · k) ǫαµβν ǫσµρν p′α pβ k ρ k ′σq416e4(p · k ′ )(p′ · k)q4(2.28)(2.29)(2.30) Further, we can write differential cross-section for e R eL µR µL interaction in center mass frame as dσdΩ cm1 k 1 X M 22 16π 2 E2Ecmcm 4 spinsIn center of mass frame momentum product is given byq2 (p p′ )2 4E 2 2E 2qEcmE 2 m2µ , E k 2p · k p′ · k ′ E 2 E k cos θp · p′ p · k′ p′ · k E 2 E k cos θdσdΩ dσdΩ cmcm1 k 1 16e4(p · k ′ )(p′ · k)2 16π 2 E42Ecmcm 4 q 2 k 1 e41E 2 E k cos θ2 π2 E42Ecm4qcm12(2.31)

In the high energy limit (E mµ ) our final expression for differential cross-section is dσ α22 e R e L µ µ (1 cos θ)R L2dΩ4Ecm(2.32)We can also get the other non vanishing helicity amplitude intuitively without repeating the whole process. For example this reaction e R eL µL µR , the replacement in the modulus square amplitude 55of µ L µR will be γ to γ on the left-hand side. Thus ǫρµσν replaced by ǫρµσν on the right-handside.We can easily see that, α2dσ e R e L µ µ(1 cos θ)2 L R2dΩ4Ecm(2.33)Similarly, for other helicity amplitude dσ α22 e L e R µ (1 cos θ)R µL 2dΩ4Ecm α2dσ e L e R µ µ(1 cos θ)2 L R2dΩ4Ecm(2.34)(2.35)The other twelve helicity cross-sections are zero because we know that right handed electron corresponds to a left-handed positron and the amplitude vanishes unless the electron and positronhave their opposite helicity. Hence, this gives four non zero helicity amplitude. Adding all sixteencontributions, and dividing by four will give us average unpolarized cross-section.2.7Crossing symmetryCrossing symmetry defined as, the s-matrix element for any process involving a particle with momentum p in the initial state is equal to the s-matrix element for an identical process but particlereplaced by anti-particle with momentum k (k p) in the final state. It is one of the mostimportant elements of calculations which makes use of the analytical properties of the scatteringamplitudes. It relates various amplitudes, for example, helicity amplitudes, in one channel to thosein other channels, in which all incoming and outgoing particles have been interchanged.13

Chapter 3Compton scatteringIn Compton scattering, a photon collides with an electron, loses some of its energy and deflectedfrom its original direction of travel. In this scattering assuming that electron to be initially free.It is an inelastic scattering of a photon with an electrically charged particle. We will calculate theunpolarized cross-section for this section, to lowest order in α.3.1Feynman diagramThere are two diagrams that contribute to Compton scattering at tree-levelp′p′րk′p k kրր′kkրppFigure 3.1: Feynman diagram for Compton scatteringIn the Feynman diagram, p and k are the 4-momentum of the electron and photon before thecollision, and p′ , k ′ their 4-momentum after the collision. Since the fermion portion of the diagrams′is identical. Using εν (k) and ε µ (k ) to denote the polarization vector of the initial and final photon,we have the expression of Scattering amplitude for a given interaction.14

iMu(p′ )( ieγ µ )ε µ (k ′ ) i(p/ k/ m)( ieγ ν )εν (k)u(p)(p k)2 m2i(p/ k/′ m)( ieγ µ )εµ (k ′ )u(p)(3.1)(p k)2 m2"#γ µ (pγ ν (p/ k/ m)γ ν/ k/′ m)γ µ′2 ′u(p) (3.2) ie εν (k)εµ (k )u(p ) (p k)2 m2(p k ′ )2 m2 u(p′ )( ieγ ν )εν (k) As we found M 2 expression for e e µ µ interaction in chapter 2. Similarly, we will do forcompton scattering but before this we need a simplified solution for iM . Since, p2 m2 andk 2 0, the denominators of the propagators are(p k)2 m2 (p k ) m ′ 222p · k 2p · k(3.3)′(3.4)To simplify the numerators in iM using Dirac gamma matrix algebra.ννννν(p/ m)γ u(p) (2p γ p/ γ m)u(p) 2p u(p)(3.5)Using above simplification of numerator in iM , we obtained iM ie2εν (k)ε µ (k ′ )u(p′ ) γ ν k/′ γ µ 2γ ν pµγ µ k/γ ν 2γ µ pν u(p)2p · k 2p · k ′(3.6)Here, we are considering the scattering of an unpolarized photon by an unpolarized electron, withoutregard to their polarizations after the scattering. So, we will find the polarization sum to get theaverage cross-section.Photon Polarization SumsThe amplitude expression retains freely specified spin and polarization states for the electrons andphotons. Experimental Compton scattering involves unpolarized photons colliding with electrons,and so we must average over these states. We are considering an arbitrary QED process comprisingan external photon with momentum k to get the expression for photon polarization sum. Since thescattering amplitude always contains ε µ (k), so we can extract this factor and defined M (k) to bethe rest of the scattering amplitude M .Xǫ ε µ (k)M (k) 2 Xǫ15ε µ εν M µ (k)M ν (k)(3.7)

For simplicity, we took k vector in the 3-direction: k µ (k, 0, 0, k). Then the corresponding twotransverse polarization vector areεµ1 (0, 1, 0, 0)(3.8)εµ2 (0, 0, 1, 0)(3.9)Now, we will sum our scattering amplitude M (k) over two transverse polarization vector. Then wehaveX ε µ (k)M (k) 2 M 1 (k) 2 M 2 (k) 2(3.10)ǫClassically, we know that the current density j µ is conserved µ j µ 0. If the property still holds inthe quantum theory, we can dot kµ with M µ (k) to obtainkµ M µ (k) 0(3.11)The amplitude M vanishes when the polarization vector εµ (k) is replaced by kµ . This relation isknown as the Ward identity.Now, we can see for k µ (k, 0, 0, k) the ward identity takes the formkM 0 (k) kM 3 (k) M0X M3ε µ εν M µ (k)M ν (k) M 1 2 M 2 2ε µ εν M µ (k)M ν (k) gµν M µ (k)M ν (k)ǫX0ǫSo, photon polarization sum is given byXǫ3.2ε µ εν gµνThe Klein-Nishima FormulaWe want to average the modulus square amplitude over the initial electron and photon polarizations,and sum over the final electron and photon polarizations.1X M 24 spin "e4gµρ gνσ · T r (p/′ m) ·4γ ν k/′ γ µ 2γ ν pµγ µ k/ 2γ µ pν 2p · k2p · k ′! #γ σ k/γ ρ 2γ ρ pσγ ρ k/′ γ σ 2γ σ pρ (/p m) 2p · k2p · k ′To simplify above expression we used the trace technology and Mandelstam variables: st u (p k)2 2p · k m2 2p′ · k ′ m2 ;(3.12)(p p) 2p · p 2m 2k · k ;(3.13)′2′2′(k ′ p)2 2k ′ · p m2 2k · p′ m216(3.14)

The momentum conservation at vertex implies s u t 2m2 . Now, we can put Mandelstamvariable to our scattering amplitude in lab frame. 2 3224 22 22 41X24 (s m )t (3s 2m s 3m )t 4s(s m ) t 2(s m ) M 2e4 spin(s m2 )2 (t s m2 )2Now, rewriting s, u and t in terms of p · k, p · k ′ and k · k ′ , we finally obtain" 2 #′11X1p·k1124 p·k42 M 2e m 2m 4 spinp·kp · k′p · k p · k′p · k p · k′To get the expression for a differential cross-section, we must decide a frame of reference where ourcalculation becomes simpler. The easiest choice is lab frame because we can find the dynamics ofparticles after collision separately. But in the lab frame, the electron is initially at rest. But afterthe collision, The energy of the electron is typically ten orders of magnitude larger than that of thephoton.′′′k′ (ω , ω sin θ, 0, ω cos θ) րθ k (ω, ω ẑ)p (m, 0)p′ (E ′ , p′ )Figure 3.2: Compton scattering in Lab frameSince, we know that Compton’s formula for the shift in photon wavelength.ω′ ωω1 (1 cosθ)mNow, the phase space integral in Lab frame isZZdΠ dΠ Z18πd3 k ′ 1 d3 p′ 1(2π)4 δ (4) (k ′ p′ k p)(2π)3 2ω ′ (2π)3 2E ′Z(d cos θ)(ω ′ )2ωm(3.15)(3.16)The differential cross-section is given bydσ 14Pspin M 24EA EB vA vB Πf d3 pf 1(2π)3 2Ef17 .(2π)4 δ (4) (pA pB Σpf )(3.17)

Now, we will plug everything into our differential cross-section formula which we have written above.Since, in lab frame vA vB 1, then we find 1 1 1 (ω ′ )2 1 Xdσ M 2 .dcosθ2ω 2m 8π ωm4 spinNow, our general cross-section formula isdσπα2 2dcosθm ω′ω 2 ω′ω ′ sin2 θωω Above expression correspond Klein-Nishina formula for differential cross-sectionIn the limit (this low energy limit) ω 0 and ω ′ /ω 1, then the cross-section becomes πα2dσ 2 1 cos2 θdcosθmThis is the familiar Thomson cross section for scattering of classical electromagnetic radiation by afree electron where the energy of the photon is less than the rest mass energy of electron.3.3High Energy BehaviorIn the centre of mass frame the 4-momenta of the particles may be writtenk (ω, 0, 0, ω)p k′(E, 0, 0, ω) (ω, ω sin θ, 0, ω cos θ)′ (E, ω sin θ, 0, ω cos θ)pTo analyze the high energy behavior of the Compton scattering cross-section, it is easiest to workin the center-of-mass frame.րk (ω, ω ẑ)θ p (E, ω ẑ)p′Figure 3.3: Compton scattering in center-of-mass frame18

We can see that for θ π, the term (p · k/p · k ′ ) becomes very large, while the other terms areall smaller. Thus for E m and θ π, we have p·kE ω1X4 M 2 2e4 · 2e·4 spinp · k′E ωcosθThe differential cross-section in center of mass frame is given by1 11dσ ··dcosθ2 2E 2ω ω8π(E ω) 2e4 (E ω)(E ω cos θ) dσ2πα2 2dcosθ2m s(1 cos θ)Where s 2ω(E ω) m2In the high energy limit s m2 , so we can drop the electron mass term if we supply an equivalentcuttoff near θ π. In this way, we can approximate the total Compton scattering cross-section. Wefind that the total cross-section behaves at high energy asσtotal s 2πα2logsm2In the high energy limit (s m2 ) the differential cross-section has singularity at (θ π) whichmeans we have singularity for backward photon.3.4ResultWe have two tree level Feynman diagram for Compton scattering which gives one amplitude for eachdiagram. We have shown that in the lab frame if the photon has more energy than the rest massenergy of the electron, Then the calculation for differential cross-section produces Klein-Nishinaformula. But if photon has less energy than the re

atomic physics, and astrophysics. Also, quantum field theory has led to new bridges between physics and mathematics. Since this thesis is about scattering cross-section in quantum electrodynamics, so it is instructive to start with an introduction of quantum electrodynamics (QED). It is a field theory of interaction between light and matter.