Lecture 5: QED The Bohr Magneton Off-Shell Electrons Vacuum Polarization Divergences, Running Coupling & Renormalization Yukawa Scattering and the Propagator Useful Sections in Martin & Shaw: Section 5.1, Section 5.2, Section 7.1.2

The Bohr Magneton Bohr Magneton

Bremsstrahlung & Pair Production Ze Ze + Internal electron line since real electron cannot emit a real photon and still conserve energy and momentum: ~ Z3 in electron rest frame consider me = me + pe2/2me + E (conservation of energy) pe = p = E(conservation of momentum) me  me = E2/2me + E   So, for a positive energy photon to be produced, the electron has to effectively ''lose mass"  virtual electron goes ''off mass shell" Pair Production Ze ''twisted Bremsstrahlung" Matrix element will be the same; Cross-section will only differ by a kinematic factor

Vacuum Polarization e e In QED, the bare charge of Thus, we never actually ever see a ''bare" charge, only an effective charge shielded by polarized virtual electron/positron pairs. A larger charge (or, equivalently, ) will be seen in interactions involving a high momemtum transfer as they probe closer to the central charge. q  ''running coupling constant" In QED, the bare charge of an electron is actually infinite !!! Note: due to the field-energy near an infinite charge, the bare mass of the electron (E=mc2) is also infinite, but the effective mass is brought back into line by the virtual pairs again !!

 (1/q4) q3 dq Another way... = ln q ] =  dq/q Renormalization Another way... Consider the integral corresponding to the loop diagram above: At large q the product of the propagators will go as ~ 1/q4 Integration is over the 4-momenta of all internal lines where d4q = q3 dq d (just like the 2-D area integral is r dr d and the 3-D volume integral is r2 dr dcos d ) so the q integral goes like: ∞ qmin = ln q ]  (1/q4) q3 dq ∞ qmin ∞ qmin =  dq/q  logarithmically divergent !! With some fiddling, these divergences can always be shuffled around to be associated with a bare charge or mass. But we only ever measure a ''dressed" (screened) charge or mass, which is finite. Thus, we ''renormalize" by replacing the bare values of 0 and m0 by running parameters.

The Underwater Cannonball Analogy for Renormalization: Instead of changing the equations of motion, you could instead (in principle) find the ''effective" mass of the cannonball by shaking it back and forth in the water to see how much force it takes to accelerate it. This ''mass" would no longer be a true constant as it would clearly depend on how quickly you shake the ball.

Point Charges and Sweat Still unhappy? Well, if it’s any comfort, note that the electrostatic potential of a classical point charge, e2/r, is also infinite as r 0 (perhaps this all just means that there are really no true point particles... strings?? something else?? ) Does an electron feel it’s own field ??? ''Ven you svet, you smell you’re own stink, Ja? Zo..." M. Veltman

Steve’s Tips for Becoming a Particle Physicist 1) Be Lazy 2) Start Lying 3) Sweat Freely

e - e+  e- W - e

Is A Particle?

No: All Are But Shadows Of The Field

Relativistic Wave Equation The Story So Far Recap: Relativity Subjectivity of position, time, colour, shape,... Relativistic Wave Equation Spin (Dirac Equation) Antimatter Klein-Gordon Equation Asymmetry?! Yukawa Potential (Nobel Prize opportunity missed) New ''charge" + limited range ''virtual" particles exchange quanta Strong Nuclear Force Central Reality of the FIELD Feynman Diagrams (QED) Prediction of pion

6 V(r) = e-Mcr/ħ Yo = eip•r/ħ Yf = eip¢•r/ħ sheet 2 Consider the scattering of one nucleon by another via the Yukawa potential: V(r) = e-Mcr/ħ - g2 4pr In the CM, both nucleons have equal energy and equal & opposite momenta, p. Take the incoming state of the 1st nucleon (normalised per unit volume) to be: Yo = eip•r/ħ After scattering, the final state will have a new momentum vector, p¢, where |p| = |p¢ | : Yf = eip¢•r/ħ If q is the angle between p and p¢, write an expression for the momentum transfer, q , in terms of p . p p¢ q q = p¢ - p magnitude of q = 2p sin(q/2) 13

∫ ∫ ∫ ∫ [ + ] ( ) ( - ) <Yf|V(r)|Yo> = e-iq•r/ħ e- Mcr/ħ Compute the matrix element for the transition. <Yf|V(r)|Yo> = ∫ e-ip¢•r/ħ eip•r/ħ e- Mcr/ħ - g2 4pr ( ) r2 dr df d(cosq) ∫ e-iq•r/ħ e- Mcr/ħ r dr d(cosq) - g2 2 = ∫ e-iqrcosq/ħ e- Mcr/ħ r dr d(cosq) - g2 2 = ∫ e(iq+Mc)r/ħ dr -iħg2 2q = ( - ) e (iq-Mc)r/ħ e-(iq+Mc)r/ħ iħ2g2 2q = [ + ] e(iq-Mc)r/ħ iq  Mc iq-Mc r = ∞ r = 0 14

[ + ] [ + ] [ ] e-(iq+Mc)r/ħ = e (iq-Mc)r/ħ iq  Mc iq-Mc iħ2g2 2q = [ + ] e (iq-Mc)r/ħ iq  Mc iq-Mc r = ∞ r = 0 This will go to zero as r ® ∞, so we’re left with: <Yf|V(r)|Yo> = 1 - iħ2g2 2q [ + ] iq  Mc iq-Mc iq-Mc - iħ2g2 2q [ ] + iq  Mc (iq  Mc)( iq-Mc) = ħ2g2 (q2+M2c2) = 1 q2+M2 ~ (natural units)

NOTE: This is all really a semi-relativistic approximation! Really, we want to consider a 4-momentum transfer Recall that P2 = E2  p2 1 q2+M2 so, 1 q42 M2 (choosing appropriate sign convention) Known as the “propagator” associated with the field characterised by an exchange particle of mass M

dp/dE = E/p = (gm/gmv) = 1/v dp/dE = (m/mv) = 1/v Show that the relation dp/dE = 1/v , where v is the velocity, holds for both relativistic and non-relativistic limits. Classically: Relativistically: E = p2/2m E2 = p2 + m2 dE = (p/m) dp dp/dE = E/p = (gm/gmv) = 1/v dp/dE = (m/mv) = 1/v Also show that, for the present case: p2d(cosq) = - ½ dq2 q = 2p sin(q/2) cosq = cos2(q/2) - sin2(q/2) = 1 - 2sin2(q/2) q2 = 4p2 sin2(q/2) sin2(q/2) = (1 – cosq)/2 q2 = 2p2 (1 - cosq) dq2 = -2p2 dcosq p2d(cosq) = - ½ dq2 17

( ) ( ) in our case Nbeam = Ntarget = 1 Now, from the definition of cross-section in terms of a rate, the relative velocities of the nucleons in the CM, and using Fermi’s Golden Rule, derive the differential cross-section ds /dW . in our case Nbeam = Ntarget = 1 vbeam = 2v (relative velocity in CM) Normalised Volume (= 1) Rate = vbeamNbeamNtarget s Volume so, ds = dRate 1 2v given by FGR ds = 1 2v 2p ħ |<Yf|V(r)|Yo>|2 (2pħ)3 p2dp df dcosq dEtot 2 = ( ) 1 (2pħ)3 df ħ2g2 (q2+M2c2) 2v (- ½ dq2) 2p ħ = ( ) dq2df (2pħ)3 ħ2g2 (q2+M2c2) 2 p 4v2ħ 18

∫ ∫ ( ) [ ] ( ) ( ) = dq2df (2pħ)3 ħ2g2 (q2+M2c2) p 4v2ħ ds = dq2df ( ) dq2df (2pħ)3 ħ2g2 (q2+M2c2) 2 p 4v2ħ ds = dq2df 8pħ3 ħ4g4 (q2+M2c2)2 p 4v2ħ = dq2df (q2+M2c2)2 1 (4p)2 g4 2v2 Integrate this expression and take the limit as v ® c s = dq2df (q2+M2c2)2 g4 2v2(4p)2 ∫ = dq2 (q2+M2c2)2 pg4 v2(4p)2 ∫ = 1 (q2+M2c2) pg4 v2(4p)2 - [ ] q2 = 0 q2 = ∞ since integral was from cosq = -1 to 1 (where 1 = zero momentum transfer) ħ Mc = p g2 4pħc ( ) ( ) 2 2 taking v ~ c s = pg4 Mc2(4p)2 s = p × 10-30 m2 ~ 30 mb strong coupling constant (~1) “range” of Yukawa potential (~1fm)   (within a factor of 2-3)