2012 Matthew Schwartz

II-6:

1 Introduction Now we are ready to do calculations in QED. We have found that the Lagrangian for QED is 1 = F2 + i ψ¯ D ψ m ψ¯ ψ (1) L − 4 µν − with D µ ψ = ∂µψ + ieA µ ψ. We have also introduced quantized Dirac fields d3 p 1 ψ( x) = ( as us e − ipx + bs † vs eipx) (2) (2 π) 3 p p p p s 2 ωp X Z d3 p p1 ψ¯ ( x) = ( bs v¯ s e − ipx + as † u¯ s eipx ) (3) (2 π) 3 p p p p s 2 ωp X Z The spinor creation and annihilation operatorsp for spinors must anticommute by the spin-statis- tics theorem: s † s ′ † s s ′ s † s ′ † s s ′ a p , a q = a p, a q = b p , bq = b p, b q = 0 (4)

n o′  ′ n o  s s † s s † 3 3 a , a = b , b = δss ′ (2 π) δ ( p q) (5) p q p q − n o n o A basis of spinors for each momentum pµ can be written as

√p σ ξs √p σ ηs us ( p) = · , vs ( p) = · (6) √p σ¯ ξs √p σ¯ ηs · ! − · ! T T with ξ1 = η1 = (1 , 0) and ξ2 = η2 = (0, 1) . These spinors satisfy

2 us ( p) u¯s ( p) = p + m s =1 X2 (7) vs ( p) v¯s ( p) = p m − s =1 X We also calculated the Feynman for a Dirac spinor: d4 p i( p + m) 0 T ψ(0) ψ¯ ( x) 0 = eipx (8) h | { }| i (2 π) 4 p2 m2 + iε Z − In this lecture we will derive the Feynman rules for QED and then perform some important cal- culations.

2 QED Feynman rules The Feynman rules for QED can be read right off of the Lagrangian just as in scalar QED. The only subtlety is possible extra minus signs coming from anti-commuting spinors within the time- ordering. First we write down the Feynman rules, then derive the supplementary minus sign rules. A propagator is represented with a squiggly line

i pµ pν = − gµν (1 ξ) (9) p2 + iε − − p2  

¡ 1 2 Section 2

Unless we are explicitly checking gauge invariance, we will usually work in Feynman gauge, ξ = 1 , where the propagator is

igµν = − ( Feynman gauge) p2 + iε A spinor propagator is a solid line with an arrow i( p + m)

¡ = p2 m2 + iε − The arrow points to the right forparticles and to the left for anti-particles. For internal lines, the

arrow points with momentum¡ flow. External photon lines get polarization vectors

b =ǫ µ( p) ( incoming) (10) ⊗ ⋆

b = ǫ ( p) ( outgoing) (11) ⊗ µ Here the blob means the rest of the diagram. External fermion lines get spinors, with u spinors for and v spinors for . =us ( p) ⊗ =u¯ s ( p) ⊗¡ = v¯ s ( p) ¡⊗ =vs ( p) ¡⊗ External spinors are on-shell (they are forced to be on-shell by LSZ). So for external spinors, we can use the equations of motion ¡ ( p m) us ( p) = u¯ s ( p)( p m) = 0 (12) − − ( p + m) vs ( p) = v¯ s ( p)( p + m) = 0 (13) which will simplify a number of calculations. Expanding the Lagrangian

1 2 µ µ = F + ψ¯ ( iγ ∂µ m) ψ e ψ¯ γ ψ A µ (14) L − 4 µν − − µ we see the interaction is int = e ψ¯ γ ψ A µ. Since there is no factor of momentum, the Feynman rule is the sameL for any− combination of incoming or outgoing fields (unlike in scalar QED) e+ e− e− + e = = = e− = ieγµ (15) e− + + −

e e ¡

¡

¡ ¡

µ The µ on the γ will get contracted with the µ of the photon which will either be in the gµν of the photon propagator (if the photon is internal) or the ǫ µ of a polarization vector (if the photon is external). µ µ The γ = γαβ as a matrix will always get sandwiched between spinors, as in µ µ u¯ γ u = u¯α γαβ uβ (16) for e − e − , or v¯ γµ u for e+ e − annihilation, etc. The barred spinor always goes on the µ left, since the interaction is ψ¯ A µ γ ψ. If there is an internal fermion line between the ends, the fermion propagator goes between the end spinors:

γν γµ

p p p i( p2 + m) 1 2 3 = ie 2 u p γµ γν u p ǫ2 q ǫ1 q ( ) ¯( 3) 2 2 ( 1 ) µ( 2 ) ν( 1 ) (17) − p2 m + iε ¡ − ν µ ǫ1 ǫ2 QED Feynman rules 3

µ µ µ µ µ µ where the photon momenta are q1 = p2 p1 and q2 = p3 p2 . In this example, the 3 γ-matrices get multiplied and then sandwiched between− the spinors.− To see explicitly what is a matrix and what is a vector, we can add in the spinor indices

µ i( p2 + m) ν µ i( p2 + m) βγ ν u¯( p3 ) γ γ u ( p1 ) = u¯α ( p3 ) γ γ uδ( p1 ) (18) p2 m2 + iε αβ p2 m2 + iε γδ 2 − 2 − If we tied the ends of the diagram above together we would get a loop

p1 γν γµ µ ǫν ǫ (19) 1 p p 2 → → p2

For fermion loops we use the same convention as for scalar loops that the loop momentum goes in the direction of the particle-flow arrow. In the loop, since any possible intermediate states are allowed, we must integrate over the momenta of the virtual spinors as well as sum over their possible spins. The uδ u¯α¡then in Eq. (18) gets replaced by a propagator which sums over all possible spins. This is done automatically since the numerator of the propagator is ( p2 + m) δα = s s s uδ u¯α . We also must integrate over all possible momenta constrained by momentum conser- vation at each vertex. So the loop in Eq. (19) evaluates to P d4 p d4 p i = ( i e) 2 1 2 (2 π) 4 δ4( p + p M − − (2 π) 4 (2 π) 4 2 − Z i( p1 + m) βγ i( p2 + m) δα p ) ǫ2 ⋆( p) ǫ1 ( p) γµ γν 1 µ ν αβ p2 m2 + iε γδ p2 m2 + iε  1 − 2 −  The extra minus sign is due to spin-statistics as will be explained shortly. Contracting all the µ µ µ µ µ spinor indices and replacing p1 by p + k and p2 by k

p + k d4 k i( p + k + m) i( k + m) ǫν µ 2 2 ⋆ 1 µ ν i = 1 ǫ2 = e ǫ ǫ Tr γ γ (20) M p p µ ν (2 π) 4 ( p + k) 2 m2 + iε k2 m2 + iε k Z " − − # where the¡ trace is a trace of spinor indices. Computing Feynman diagrams in QED will often involve taking the trace of products of γ-matrices. A useful general rule is that the spinor matrices always get multiplied together in the direc- tion opposite to the particle-flow arrow, which allows us to read off Eqs. (17) and (20) easily from the corresponding diagrams.

2.1 Signs

Recall that spinors anti-commute within a time-ordered product:

T ψ( x) ψ( y) = T ψ( y) ψ( x) (21) { } − { } Minus signs coming from such anti-commutations appear in the Feynman rules. It is easiest to see when they should appear by example. Consider Moller Scattering ( e − e − e − e − ) at tree-level. There are two Feynman diagrams, → for the t-channel (in Feynman gauge)

p1 p3 µ igµν ν i t = = ( ie) u¯( p3) γ u( p1 ) − ( ie) u¯( p4) γ u( p2 ) (22) M ± − ( p p ) 2 − 1 − 3

p2 p4 ¡ 4 Section 2 and u-channel

p1 p 3 µ igµν ν i u = = ( ie) u¯( p3 ) γ u( p2 ) − 2 ( ie) u¯( p4) γ u( p1 ) (23) M p4 ± − ( p p ) − 1 − 4

p2 ¡

The question is what sign should each diagram have? To find out, recall that these Feynman diagrams represent S-matrix elements. By the LSZ reduction theorem, they represent contributions to the Fourier transform of the Green’s function G ( x , x , x , x ) = Ω T ψ( x ) ψ¯ ( x ) ψ( x ) ψ¯ ( x ) Ω (24) 4 1 2 3 4 h | { 1 3 2 4 }| i with external removed and external spinors added. The first non-zero contribution to this Green’s function in perturbation theory comes at order e2 in an expansion of free fields

G = ( ie) 2 d4 x d4 y (25) 4 − Z Z 0 T ψ( x ) ψ¯ ( x ) ψ( x ) ψ¯ ( x ) ψ¯ ( x) A( x) ψ( x) ψ¯ ( y) A( y) ψ( y) 0 (26) × 1 3 2 4 D n   o E where the big () indicate that the spinors inside are contracted. More explicitly, we can write G ( x , x , x , x ) = ( ie) 2 γµ γν d4 x d4 y 4 1 2 3 4 − β1 β2 β3 β4 Z Z ¯ ¯ ¯ µ ¯ ν 0 T ψα 1 ( x ) ψ ( x ) ψα 2 ( x ) ψ ( x ) ψ ( x) A ( x) ψβ2 ( x) ψ ( y) A ( y) ψβ4 ( y) 0 (27) × 1 α 3 3 2 α 4 4 β1 β3  In this form, we can anti-commute the spinors within the time-ordering before performing any contractions. To get Feynman diagrams out of this Green’s function, we have to perform contractions, which means creating fields from the vacuum and then annihilating them. To be absolutely cer- tain about the sign coming from the contraction, it is easiest to anti-commute the fields so that the fields which annihilate spinors are right next to the fields which create them. For the t- channel diagram, the top line is created by ψ¯ ( x3) annihilated by ψ( x) , created by ψ¯ ( x) and annihilated by ψ( x1 ) , and similarly for the bottom line. So we need

G ( x , x , x , x ) = ( ie) 2 γµ γν d4 x d4 y 4 1 2 3 4 − β1 β2 β3 β4 Z Z µ ν ¯ ¯ ¯ ¯ 0 T A ( x) A ( y) ψα 1 ( x ) ψ ( x) ψβ2 ( x) ψ ( x ) ψα 2 ( x ) ψ ( y) ψβ4 ( y) ψ ( x ) 0 (28) ×h | { 1 β1 α 3 3 2 β3 α 4 4 }| i Contractions of these spinors in this order gives the t-channel diagram in Eq. (22). For the u-channel, ordering the fields so that the contractions are in order gives

G ( x , x , x , x ) = ( ie) 2 γµ γν d4 x d4 y 4 1 2 3 4 − − β1 β2 β3 β4 Z Z µ ν ¯ ¯ ¯ ¯ 0 T A ( x) A ( y) ψα 1 ( x ) ψ ( x) ψβ2 ( x) ψ ( x ) ψα 2 ( x ) ψ ( y) ψβ4 ( y) ψ ( x ) 0 (29) ×h | { 1 β1 α 4 4 2 β3 α 3 3 }| i So that the u-channel diagram in Eq. (23) has a sign out front. The result is that the matrix element for Moller Scattering has the form −

u p γµ u p u p γµ u p u p γµ u p u p γµ u p 2 ¯( 3) ( 1 ) ¯( 4) ( 2 ) ¯( 4) ( 1 ) ¯( 3 ) ( 2 ) = t + u = e M M M ( p p ) 2 − ( p p ) 2 ( 1 − 3  1 − 4  ) A shortcut to remembering the relative minus sign is simply to note that G ( x , x , x , x ) = 4 1 2 3 4 − G4( x1 , x2 , x4, x3) . A minus sign from interchanging the identical fermions at x3 and x4 is exactly what you would expect from Fermi-Dirac statistics. The overall sign of the sum of the matrix elements is an unphysical phase, but the relative sign of the t- and u-channels is important for 2 2 the cross-term in the = u + t and has observable effects. |M| |M M | γ-matrix identities 5

One can do the same exercise for loops. For example, a loop like this

x1 x y x2 ¡ comes from a term in the perturbative expansion of the time-ordered product for 2 photon fields of the form (leaving all spinor indices implicit)

2 µ ν ¯ α ¯ β G = ( ie) ( γαβ ) 0 T A ( x ) A ( x ) ψ ( x) A ( x) ψ( x) ψ ( y) A ( y) ψ( y) 0 (30) 2 − h | { 1 2 }| i To get the spinors into the order where they are created and then immediately destroyed, we need to anti-commute ψ( y) from the right to the left. That is, we use

ψ¯ ( x) ψβ ( x) ψ¯ ( y) ψδ( y) = ψδ( y) ψ¯ ( x) ψβ ( x) ψ¯ ( y) (31) α γ − α γ Thus the Feynman rule for this fermion loop should be supplemented with an additional sign. − As an exercise, you should check that adding more to a fermionic loop does not change this overall minus sign. In summary, the Feynman rules for fermions must be supplemented by a factor of 1 for interchange of external identical fermions. Diagrams like s and t channel •exchanges, − which would be present even for non-identical particles, do not get an extra minus sign. The 1 supplements diagrams which exist from new channels, coming from interchanging two− external identical fermions, such as the u-channel diagram in Eq. (23). 1 for each fermion loop. • −

3 γ-matrix identities

Before beginning the QED calculations, let’s derive some useful identities about γ matrices. We will often need to take traces of products of γ matrices. These can be often be simplified using

the cyclic property of the trace Tr[ AB C] = Tr[ B CA] (32)

We will often also use γ5 iγ0 γ1 γ2 γ3 which satisfies ≡ 2 γ = 1 , γ γµ = γµ γ (33) 5 5 − 5 To keep the spinor indices straight, we sometimes write 1 for the identity on spinor indices. So, γµ ,γν = 2 gµν 1 (34) { } and Tr[ gµν 1 ] = gµν Tr[ 1 ] = 4 gµν (35) The gµν are just numbers for each µ and ν and pull out of the trace. Then Tr[ γµ] = Tr[ γ γ γµ] = Tr[ γ γµ γ ] = Tr[ γ γ γµ] = Tr[ γµ] (36) 5 5 5 5 − 5 5 − where we have cycled the γ5 matrix in the second step. Thus Tr[ γµ] = 0 (37) Similarly Tr[ γµ γν ] = Tr[ γν γµ + 2 gµν 1 ] = Tr[ γµ γν] + 8 gµν (38) − − which leads to Tr[ γµ γν] = 4 gµν (39) In a similar way, you can show Tr[ γα γβ γµ] = 0 (40) 6 Section 4 and more generally that the trace of an odd number of γ-matrices is zero. For 4 γ-matrices, the result is (Problem ??) Tr[ γα γµ γβ γν]=4( gαµ gβν gαβ gµν + gαν gµβ ) (41) − you will use this last one a lot! You can remember the signs because adjacent indices give + and the other one gives . A contraction of− vector indices is also a type of trace. For example,

gµν gµν = 4 (42) However, to avoid confusion, we will keep vector index contractions explicit and use Tr[ ] only for traces over spinor indices.

4 e + e − → µ+ µ −

The , µ− , is a particle which is identical to the electron as far as QED is concerned, except heavier. Studying processes with therefore provides simple tests of QED. Indeed, the simplest tree level scattering process in QED is e+ e − µ+ µ− , which we calculate at tree-level here and at 1-loop in Lecture III-6. The leading-order contr→ ibution is

k µ k ν p1 p3 i g ξ µν (1 ) k 2 i = = ( ie) v¯ ( p ) γµ u ( p ) − − − ( ie) u¯ ( p ) γν v ( p ) α 2 αβ β 1 h 2 i δ 3 δι ι 4

M p2 p4 − k − ¡

µ µ µ µ µ where k = p1 + p2 = p3 + p4 . Each of these spinors has a spin, thus we should properly write s 1 uα ( p1 ) and so on. It is conventional to leave these spin labels implicit. Since the spinors are on- shell, we can use the equations of motion p1 u( p1 ) = mu( p1 ) and v¯( p2 ) p2 = mv¯( p2 ) . Thus, − µ µ v¯α ( p ) γ uβ ( p ) k = v¯( p ) p u( p ) + v¯( p ) p u( p ) = m v¯( p ) u( p ) m v¯( p ) u( p ) = 0 (43) 2 αβ 1 2 1 1 2 2 1 2 1 − 2 1 implying that the k µ kν term does not contribute, as expected by gauge invariance. So, 2 e µ = v¯( p ) γ u( p ) u¯( p ) γµ v( p ) (44) M s 2 1 3 4 2 where s = ( p1 + p2 ) as usual. To calculate 2 we need the conjugate amplitude. To get this, we first recall that |M| † † γµ γ0 = γ0 γµ and γ0 = γ0 (45) So, ¯ µ † † µ † † µ † † † µ ¯ µ ( ψ1 γ ψ2 ) = ( ψ1 γ0 γ ψ2 ) = ψ2 γ γ0 ψ1 = ψ2 γ0 γ ψ1 = ψ2 γ ψ1 (46) This nice transformation property is another reason why using ψ¯ instead of ψ † is useful. Then,

2 † e µ = v¯( p ) γ u( p ) u¯( p ) γµ v( p ) (47) M s 4 3 1 2 and therefore 4 2 e µ ν = [ v¯( p ) γ u( p ) ][ u¯( p ) γµ v( p )][ v¯( p ) γ u( p )][ u¯( p ) γν v( p ) ] (48) |M| s 2 2 1 3 4 4 3 1 2 The grouping is meant to emphasize that each term in brackets is just a number for each µ and each set of spins. Thus 2 a product of these numbers. For example, we could also have written |M| 4 2 e µ ν = [ v¯( p ) γ u( p ) ][ u¯( p ) γ v( p ) ][ u¯( p ) γµ v( p )][ v¯( p ) γν u( p )] (49) |M| s 2 2 1 1 2 3 4 4 3 which shows that 2 is the contraction of two tensors, one depending only on the initial state, and the other depending|M| only on the final state. e+ e− → µ+ µ− 7

4.1 Unpolarized scattering The easiest thing to calculate from this is the cross section for scattering assuming spin is not measured. The spin sum can be performed with a Dirac trace. To see this, we will sum over the µ+ spins using s s s s v ( p ) v¯ ( p ) = v¯ ( p ) v ( p ) = ( p m µ 1 ) αβ (50) α 4 β 4 β 4 α 4 4 − s s X X and over the µ− spins using s s s s 1 uα ( p3) u¯ β ( p3) = u¯ β ( p3) uα ( p3) = ( p3 + m µ ) αβ (51) s s X X We have written each sum two ways to emphasize that these are sums over vectors of complex numbers corresponding to external spinors, not over fermion fields. Thus, no minus sign is s s s s induced from reversing the order of the sum: uα u¯ β = u¯ β uα . Using these relations ′ ′ s ′ µ s s ν s ′ s µ ν s [ u¯ ( p ) γ v ( p )][ v¯ ( p ) γ u ( p )] = [ u¯ ( p ) γ ( p m µ 1 ) δι γ u ( p )] 3 4 4 3 β 3 βδ 4 − ια α 3 s ′ s s ′ X X X 1 µ 1 ν =( p3 + m µ ) αβ γβδ( p4 m µ ) δι για µ − ν =Tr[( p + m µ) γ ( p m µ) γ ] 3 4 − which is a simple expression we can evaluate using γ-matrix identities. Let us also assume that we do not know the polarization of the initial states. Then, if we do the measurement many times, we will get the average over each polarization. This leads to con- 1 1 e+ tractions and traces of the initial state, with a factor of 4 ( 2 each for the incoming and incoming e − ) to average over our ignorance. Thus we need

4 1 2 e ν µ µ ν = Tr[( p + me ) γ ( p me ) γ ] Tr[( p + m µ) γ ( p m µ) γ ] (52) 4 |M| 4 s 2 1 2 − 3 4 − spins X These traces simplify using trace identities: α β ρ σ ρ α σ β 2 α β Tr[( p + m µ) γ ( p m µ) γ ] = p p Tr[ γ γ γ γ ] m Tr[ γ γ ] 3 4 − 3 4 − µ =4( pα pβ + pα pβ pρ pρ gαβ ) 4 m2 gαβ 3 4 4 3 − 3 4 − µ So, 1 4 e4 2 = pα pβ + pα pβ (( pρ pρ) + m2 ) gαβ pα pβ + pα pβ (( pσ pσ) + m2 ) gαβ 4 |M| s 2 1 2 2 1 − 1 2 e 3 4 4 3 − 3 4 µ spins X 8 e4   = ( p p + p p + m2 p + m2 p + 2 m2 m2 ) s 2 13 24 14 23 µ 12 e 34 e µ with pij pi pj. We can simplify this further with Mandelstam variables ≡ · 2 2 2 2 s =( p1 + p2 ) = ( p3 + p4) = 2 me + 2 p12 = 2 m µ + 2 p34 t =( p p ) 2 = ( p p ) 2 = m2 + m2 2 p = m2 + m2 2 p 1 − 3 2 − 4 e µ − 13 e µ − 24 u =( p p ) 2 = ( p p ) 2 = m2 + m2 2 p = m2 + m2 2 p 1 − 4 2 − 3 e µ − 14 e µ − 23 After some algebra, the result is

4 1 2 2 e 2 2 2 2 2 2 2 = t + u + 4 s( me + m µ) 2( me + m µ) (53) 4 |M| s 2 − spins X h i

4.2 Differential cross section For 2 2 scattering of particles of different mass, the cross section in the center-of-mass frame is given→ by Eq. (???) of Lecture I-5:

dσ 1 Qp = f 2 (54) 2 2 | | dΩ 64 π E Qp |M|   CM CM | i | 8 Section 5

In the center–of-mass frame, there are only two variables on which the cross section depends:

ECM and scattering angle between the incoming electron and the outgoing muon. We take

Q Q p = ( E, kQ ) , p = ( E, k ) , k = E2 m2 (55) 1 2 − | | − e

p 2 2

Q Q p = ( E,pQ ) p = ( E, p ) , p = E m (56) 3 4 − | | − µ Then, q

2 2 2 s = ( p1 + p2 ) = 4 E = ECM

2 2 2 2 Q t = ( p p ) = m + m 2 E + 2 k Qp

1 − 3 e µ − · Q

Q 2 2 2 2 Q u = ( k + Qp ) = m + m 2 E 2 k p − e µ − − · And so 4 dσ e Qp t2 u2 s m2 m2 m4 m2 m2 m4 = 2 2 2 | | + + 4 ( e + µ) 2( e + 2 e µ + µ) (57) d Ω 32 π ECM s kQ − | | h i 2 α Qp

4 Q 2 2 2 2 = 6 | | E + ( k Qp ) + E ( me + m µ) (58) 16 E kQ ·

| |  Q

The only angular dependence comes from the k Qp term

·

Q Q Q k Qp = k p cos θ (59) · | k | So, 2 dσ α Qp

4 Q 2 2 2 2 2 2 = 6 | | E + k Qp cos θ + E ( me + m µ) (60) dΩ 16 E kQ | | | | 2 | |  α e where = 4 π and

Q 2 2 2 2 k = E m , Qp = E m (61) | | − e | | − µ p q Which is the general result for the e+ e − µ+ µ− rate in the center of mass frame. → Taking me = 0 for simplicity gives kQ = E and this reduces to | | dσ α2 m2 m2 m2 = 1 µ 1 + µ + 1 µ cos2 θ (62) dΩ 4 E2 − E2 E2 − E2 CM   !

If in addition we take m µ = 0, which is the ultra high energy limit, we find 2 dσ α 2 = 2 (1 + cos θ) (63) dΩ 4 ECM which is the same thing we had from the naive sum over spin states back in Lecture I-5. Recall that scattering with spins transverse plane gave 1 and scattering with spins in the plane M ∝ gave cos2 θ, so this agrees with our previous analysis. You can check explicitly by choosing explicitM spinors∝ that our intuition with spin scattering agrees with QED even for the polarized 4 π α 2 cross-section. Integrating the differential cross section over θ gives σ0 = 2 for the total cross 3 EC M section at tree-level. The 1-loop correction to the total cross section will be calculated in Lec- ture III-6.

5 Rutherford scattering e − p+ → e − p+

Now let’s go back to the problem we considered long ago, scattering of an electron by a Coulomb potential. Recall the classical Rutherford scattering formula

dσ m2 e4 = e (64) dΩ p4 4 θ 4 sin 2

Rutherford scattering e − p+ → e − p+ 9 Q where p = Qp = p is the magnitude of the incoming electron momentum, which is the same as | i | | f | the magnitude of the outgoing electron momentum for elastic scattering. Rutherford calculated this using classical mechanics to describe how an electron would get deflected in a central poten- tial, as from an atomic nucleus. We recalled in Lecture I-5 that Rutherford’s formula is repro- duced in quantum mechanics through the Born approximation, which relates the cross section to 2 the Fourier transform of the Coulomb potential V( r) = e : 4 π r

2 2 2 2 2 2 2 2 4

dσ m m Q e m e m e e ˜ 2 e 3 − i k · xQ e e = 2 V ( k) = 2 d xe = 2 = θ (65)

Q 2 dΩ born 4 π 4 π 4 π Qx 4 π k 64 π4 sin4    Z | |  ! 2

θ

Q Q Q where k = pQi pf is the momentum transfer satisfying k = 2 p sin . − 2 We also reproduced these results from field theory, taking the non-relativistic limit before doing the calculation. We found that the amplitude is given by a t-channel diagram.

dσ e4 (2 m ) 2 (2 m ) 2 = e p (66) d Ω 64 π2 E2 t2   QFT cm 2 2 where the 2 me and 2 m p factors come from the non-relativistic normalization of the electron and 2 2 2 θ proton states. Since t = ( p p ) = 2 p (1 cos θ) = 4 p sin and E = m p in the center-of- 3 − 1 − − − 2 cm mass frame, we reproduce the Rutherford formula. We will now do the calculation in QED. This will allow us to reproduce the above equation, but it will also give us the relativistic corrections. In this whole section, we neglect any internal structure of the proton, treating it, like the muon, as a pointlike particle. A discussion of what actually happens at extremely high energy, E m p, is given in Lecture IV-7. CM ≫

5.1 QED amplitude As far as QED is concerned, a proton and a muon are the same thing, up to the sign of the charge, which gets squared anyway, and the mass. So let’s start with e − µ− e − µ− . The amplitude is given by a t-channel diagram →

p 1 p3 i g ξ k µ k ν µν (1 ) k 2 i = = ( ie) u¯( p ) γµ u( p ) − − − ( ie) u¯( p ) γν u( p ) M − 3 1 h ( p p ) 2 i − 4 2 1 − 3

p2 p4 ¡

µ µ µ + − + − µ ν with k = p1 p3 . As in e e µ µ the k k term drops out for on-shell spinors, as expected by gauge− invariance. So this→ matrix element simplifies to

2 e µ = u¯( p ) γ u( p ) u¯( p ) γµ u( p ) (67) M t 3 1 4 2 with t = ( p p ) 2 . Summing over final states and averaging over initial states, 1 − 3 4 1 2 e µ ν = Tr[( p + me ) γν ( p + me) γ ] Tr[( p + m µ) γµ( p + m µ) γ ] (68) 4 |M| 4 t2 1 3 4 2 spins X This is remarkably similar to what we had for e+ e − µ+ µ− → 4 1 2 e µ ν = Tr[( p + me) γν( p me ) γ ] Tr[( p + m µ) γµ( p m µ) γ ] (69) 4 |M| 4 s 2 1 2 − 3 4 − spins X In fact, the two are identical if we take the e+ e − µ+ µ− formula and replace → ( p ,p ,p ,p ) ( p , p ,p , p ) (70) 1 2 3 4 → 1 − 3 4 − 2 10 Section 5

These changes send s t, or more generally → s = ( p + p ) 2 ( p p ) 2 = t (71) 1 2 → 1 − 3 t = ( p p ) 2 ( p p ) 2 = u (72) 1 − 3 → 1 − 4 u = ( p p ) 2 ( p + p ) 2 = s (73) 1 − 4 → 1 2 These replacements are not physical, since p2 p3 produces a momentum with negative energy, which cannot be an external particle. It’s→ just − a trick, called a symmetry which lets us recycle tedious algebraic manipulations. You can prove crossing symmetries in gen- eral, even for polarized cross sections with particular spins, and there exist general crossing rules. However, rather than derive and apply these rules, it is often easier simply to write down the amplitude that you want and inspect it to find the right transformation. With the crossing symmetry we can just skip to the final answer. For e+ e − µ+ µ− we had → 1 2 e4 2 = [ t2 + u2 + 4 s( m2 + m2 ) 2( m2 + m2 ) 2 ] (74) 4 |M| s 2 e µ − e µ spins X Therefore, for e − p+ e − p+ we get → 1 2 e4 2 = [ u2 + s 2 + 4 t( m2 + m2 ) 2( m2 + m2 ) 2 ] (75) 4 |M| t2 e p − e p spins X

5.2 Corrections to Rutherford’s formula

Now let’s take the limit m p me to get the relativistic corrections to Rutherford’s formula. In this limit we can treat the proton≫ mass as effectively infinite, but we have to treat the electron mass as finite to go from the non-relativistic to the relativistic limit. As the proton mass goes to infinity, the momenta are

µ µ µ µ Q p1 = ( E,pQ i) , p2 = ( m p, 0) , p3 = ( E,pf ) , p4 = ( m p, 0) (76) The scattering angle is defined by

2 2 2 Q Qp p = p cos θ = v E cos θ (77)

i · f Q where p = Qpi = pf and | | | | 2 p me v = = 1 2 (78) E r − E is the electron’s relativistic velocity. Then, to leading order in me/m p, p = E2 (1 v2 cos θ) (79) 13 − p12 = p23 = p34 = p14 = Em p (80)

2 p24 = m p (81) and

2 2 2 Q t = ( p p ) = ( Qp p ) = 2 p (1 cos θ) (82) 1 − 3 − i − f − − So that 1 8 e4 2 = [ p p + p p m2 p m2 p + 2 m2 m2 ] 4 |M| t2 14 23 12 34 − p 13 − e 24 e p spins X 8 e4 = [ E2 m2 + E2 m2 v2 cos θ + m2 m2 ] 4 v4 E4(1 cos θ) 2 p p e p − (83) 2 e4 m2 = p [2 v2 (1 cos θ)] v4 E2 (1 cos θ) 2 − − − e4 m2 θ = p 1 v2 sin2 v4 E2 sin4 θ − 2 2   11

2 Note that each term in the top line of this equation scales like m p, as does the final answer, so dropping subleading terms in Eq. (76) is justified. Since the center-of-mass frame is essentially dσ 1 2 the lab frame, the differential cross section is given by d Ω = 2 2 : 64 π Ec m |M| 4 dσ e 2 2 θ = 1 v sin , E m p (84) dΩ 64 π2 v2 p2 sin4 θ − 2 ≪ 2  

This is known as the Mott formula. Note that the limit m p exists: there is no depen- → ∞ p dence on the proton mass. For slow velocities we can use v 1 and p E me so v . m e Thus, ≪ ≪ ∼ ∼ dσ e4 m2 = e (85) d Ω π2 p4 4 θ 64 sin 2 2 which is the Rutherford formula. In particular, note that the normalization factors me worked out correctly. In the very high energy limit, E me , one can no longer assume that the final state proton is also at rest. However, one can now≫ neglect the electron mass, so that v = 1 . Then the momenta are

µ µ µ ′ µ µ µ µ Q p = ( E,pQ ) , p = ( m p, 0) , p = ( E ,p ) , p = p + p p (86) 1 i 2 3 f 4 1 2 − 3

′ Q with Qp = E and p = E . For me = 0 in the proton rest frame, following the same steps as | i | | f | above, we find the cross section:

e4 ′ ′ dσ E 2 θ E E 2 θ , me E (87) = θ cos + − sin dΩ 64 π2 E2 sin4 E 2 m p 2 ≪ 2   ′ ′ where E is the final state electron’s energy. As m p , E E and this reduces to the v 1 limit of the Mott formula, Eq. (84). → ∞ → → These formulas characterize the scattering of pointlike particles off of other pointlike parti- cles. Note that the final forms in which we have written these cross sections depend only on properties of the initial and final electrons. Thus they are suitable to experimental situations in which electrons are scattered off of Hydrogen gas and the final state proton momenta are not measured. Such experiments were carried out in the 1950s, notably at Stanford, and deviations of the measured cross section from the form of Eq. (87) led to the conclusion that the proton must have substructure. More shockingly, at very high energy, this pointlike scattering form was once again observed, indicating the presence of pointlike constituents within the proton, now known as quarks. We will discuss these important e − p+ scattering experiments and their theo- retical interpretation in great detail in Lecture IV-7.

6 Compton scattering The next process worth studying is the QED prediction for Compton scattering γe − γe − . By simple relativistic kinematics, Compton was able to predict the shift in wavelength of→ the scat- tered light as a function of angle 1 ∆λ = (1 cos θ) (88) m − but he could not predict the intensity of radiation at each angle. In the classical limit, for scattering soft radiation against electrons, J. J. Thompson had derived the formula dσ π α2 = π r2 (1 + cos2 θ) = (1 + cos2 θ) (89) dcos θ e m2 r r α r e is the classical electron radius e = m , defined so that if the electron were a disk of radius , the cross section would be π r2 . The 1 comes from radiation polarized in the plane of scattering and the cos2 θ from polarization out of the plane, just as we saw for e+ e − µ+ µ− in Lecture I- 5. From QED we should be able to reproduce this formula, plus the relativistic→ corrections. 12 Section 6

There are two diagrams:

p1 p3 i( p1 + p2 + m) i = = ( ie) 2 ǫ µ ǫ⋆ν u¯( p ) γν γµ u( p ) s in out 3 p p 2 m2 2 M p4 − ( 1 + 2 )

p2 − ¡

p3 p1 2 µ ⋆ν µ i( p2 p4 + m) ν i t = = ( ie) ǫ ǫ u¯( p ) γ − γ u( p ) M − in out 3 ( p p ) 2 m2 2 2 − 4 −

p2 p4 ¡

So the sum is ν µ µ ν γ ( p1 + p2 + m) γ γ ( p2 p4 + m) γ = e2 ǫ µ ǫ⋆ν u¯( p ) + − u( p ) (90) M − in out 3 s m2 t m2 2  − −  We would next like to calculate the unpolarized cross-section.

6.1 Photon polarization sums To square this and sum over on-shell physical polarizations, it is helpful to employ a trick for the photon polarization sum. There is no way to write the sum over transverse modes in a k µ k ν 2 Lorentz invariant way, since the only available tensors are gµν and 2 , but on-shell k = 0 so µ ν k k k k µ k 2 is undefined. In fact, the physical polarizations can be defined as not only orthogonal to but orthogonal to any other light-like reference vector r µ as long as r µ is not proportional to k µ. µ µ µ For example, if k = ( E, 0, 0, E) then the canonical polarizations ǫ1 = (0, 1 , 0, 0) and ǫ2 = (0, µ 0, 1 , 0) are orthogonal to k¯ = ( E, 0, 0, E) . More generally, if k µ = E, kQ then choosing the µ − reference vector as r µ = k¯ where µ  k¯ = ( E, kQ ) (91) − will uniquely determine the two transverse polarizations. Other choices of reference vector r µ lead to transverse polarizations which are related to the canonical transverse polarizations by little group transformations (Lorentz transformations which hold k µ fixed). For example, with µ µ µ µ µ k = ( E, 0, 0, E) choosing r = (1 , 0, 1 , 0) leads to ǫˆ 1 = (0, 1 , 0, 0) and ǫˆ 2 = (1 , 0, 1 , 1) . Since ǫˆ 2 = ǫ µ 1 k µ 2 + E there will be no difference in matrix elements calculated using these different polariza- tion sets by the Ward identity. In fact, invariance under change of reference vector provides an important constraint on the form that matrix elements can have. This constraint will be effi- ciently exploited in the calculation of amplitudes using spinor helicities in Lecture IV-3. With the choice r µ = k¯ µ, you should verify that (see Problem ??)

2 i⋆ i 1 µ ν µ ν ǫ ǫ = gµν + ( k k¯ + k¯ k ) (92) µ ν − 2 E2 i =1 X Now, suppose we have an amplitude involving a photon. Writing = ǫ µ Mµ we find M 2 ⋆i ⋆ i ⋆ 1 ⋆ ⋆ = ǫ M Mν ǫ = M Mµ + ( kµ M Mν k¯ν + k¯µ M Mν kν) (93) |M| µ µ ν − µ 2 E2 µ µ pols i X µ By the Ward identity, k Mµ = 0, and therefore, we can simply replace

i⋆ i ǫ ǫ gµν (94) µ ν → − polarizations i X in any physical matrix element. Note that this replacement only works for the sum of all rele- vant diagrams – individual diagrams are not gauge invariant, as you can explore in Problem 1. Compton scattering 13

6.2 Matrix element Returning to the Compton scattering process, we are now ready to evaluate 2 summed over spins and polarizations. 2 includes terms from the t-channel and s-channel|M| diagrams squared |M| ⋆ ⋆ as well as their cross terms ( t s + s t). To see what’s involved, let’s just evaluate one piece in the high energy limitM whereM we canM setM m = 0. In this limit

2 e in out ⋆ µ ν t = ǫ ǫ u¯( p ) γ ( p p ) γ u( p ) (95) M − t µ ν 3 2 − 4 2 and so, using Eqs. (7) and (94),

4 2 e µ ν t = Tr[ p γ ( p p ) γ p γν( p p ) γµ] (96) |M | t2 3 2 − 4 2 2 − 4 spins/ pols X ν µ µ µ µ µ Now use γ pγν = 2p and q = p p = p p to get − 2 − 4 3 − 1 4 4 2 e e 2 t = 4 Tr[ p qp q] = 16 (2( p q)( p q) p q ) (97) |M | t2 3 2 t2 3 · 2 · − 23 spins/ pols X 2 2 Using p3 = p2 = 0, we can simplify this to

4 4 2 e e 2 4 s 4 p12 t = 16 2 (2 p13 p24 + 2 p23 p13 ) = 8 2 ( t + ut) = 8 e = 8 e (98) |M | t t − t p24 spins/ pols X t t 1 1 Note that one of the factors of canceled, so the divergence at = 0 is not t 2 but simply t . Including all the terms ν µ µ ν γ ( p1 + p2 + m) γ γ ( p2 p4 + m) γ = e2 ǫin ǫout ⋆ u¯( p ) + − u( p ) (99) M µ ν 3 s m2 t m2 2  − −  Then summing/averaging over spins and polarizations

ν µ µ ν 1 γ ( p1 + p2 + m) γ γ ( p2 p4 + m) γ 2 = e4 Tr ( p + m) + − (100) 4 |M| 3 s m2 t m2 pols X   − −  µ ν ν µ γ ( p1 + p2 + m) γ γ ( p2 p4 + m) γ ( p + m) + − (101) × 2 s m2 t m2  − −  This is a bit of a mess, but after some algebra, the result is rather simple

1 p p 1 1 1 1 2 2 = 2 e4 24 + 12 + 2 m2 + m4 (102) 4 |M| p12 p24 p12 − p24 p12 − p24 pols X      

6.3 Klein-Nishina formula Let’s start with the low energy limit, ω m, where it makes sense to work in the lab frame. Then ≪

p1 = ( ω, 0, 0, ω) p2 = ( m, 0, 0, 0) (103) p = ( ω ′, ω ′ sin θ, 0, ω ′ cos θ) p = p + p p = ( E ′,p′) (104) 4 3 1 2 − 4 2 2 Note that the on-shell condition p3 = m implies 0 = p p p = ωm ωω ′(1 cos θ) mω ′ (105) 12 − 14 − 24 − − − so ω ω ′ = (106) 1 + ω (1 cos θ) m − 14 Section 6 which is the formula for the shifted frequency as a function of angle. There is no QED in this relation – it is just momentum conservation and the same as Compton’s formula for the wave- length shift 1 1 1 ∆λ = = (1 cos θ) (107) ω ′ − ω m − but it is still a very important relation! Then since p = ωm and p = ω ′ m, we get a simple formula for 2 12 24 |M| 1 ω ′ ω 2 = 2 e4 + 2(1 cos θ)+(1 cos θ) 2 (108) 4 |M| ω ω ′ − − − pols X   ω ′ ω =2 e4 + sin2 θ (109) ω ω ′ −   Now we need to deal with the phase space. In the lab frame, we have to go back to our gen- eral formula 1 2 1 2

dσ = dΠLIPS = d ΠLIPS (110) Q (2 E1 )(2 E2 ) Qv1 v2 |M| 4 ωm |M| and | − | d3 p 1 d3 p 1 dΠ = 3 4 [(2 π) 4 δ4( pµ + pµ pµ pµ)] (111) LIPS (2 π) 3 2 E ′ (2 π) 3 2 ω ′ 1 2 − 3 − 4 Z Z Z 3 The δ-function fixes the 3-momenta when we integrate over d p4, leaving the energy constraint 1 1 d Π = ω ′2 d Ωdω ′ δ E (112) LIPS 4(2 π) 2 ω ′ E ′ Z Z X  1 ω ′ = d cos θdω ′ δ E (113) 8 π E ′ Z X  Now we want to integrate over ω ′ to enforce the energy constraint E ′ + ω ′ = m + ω. But we have to be a little careful because E ′ and ω ′ are already constrained by the electron’s on-shell condi- tion E ′2 =m2 + p′2 = m2 + ( ω ′ sin θ) 2 + ( ω ′ cos θ ω) 2 − =m2 + ω ′2 + ω2 2 ωω ′ cos θ − So, dE ′ E ′ = ω ′ ω cos θ (114) dω ′ − Thus 1 ω ′ dΠ = d cos θdω ′ δ( ω ′ + E ′( ω ′) m ω) LIPS 8 π E ′ − − Z 1 Z ω ′ dE ′ = d cos θ (1 + ) − 1 8 π E ′ dω ′ Z 1 ω ′ ω ′ ω cos θ − 1 = d cos θ 1 + − 8 π E ′ E ′ Z   1 ( ω ′) 2 = d cos θ 8 π ωm Z where ω ′ now refers to Eq. (106) not the integration variable. And so dσ 1 1 ( ω ′) 2 ω ′ ω = 2 e4 + sin2 θ (115) d cos θ 4 ωm 8 π ωm ω ω ′ − or more simply   dσ π α2 ω ′ 2 ω ′ ω = + sin2 θ (116) d cos θ m2 ω ω ω ′ −     This is the Klein-Nishina formula. This was first calculated by Klein and Nishina in 1929 and was one of the first tests of QED. Compton scattering 15

Substituting in for ω ′ using Eq. (106), dσ π α2 2 ω ω2 = 1 + cos2 θ (1 + cos2 θ)(1 cos θ) + ( ) (117) dcos θ m2 − m − O m2   Note that in the limit m → ∞ dσ π α2 = [1 + cos2 θ] (118) dcos θ m2 This is the Thompson scattering cross section for classical electromagnetic radiation by a free electron. We have calculated the full relativistic corrections.

6.4 High-energy behavior Next, consider the opposite limit, ω m. In this limit, we will be able to understand some of the physics of Compton scattering, in≫ particular, the spin and polarization dependence and the origin of an apparent singularity for exactly backwards scattering, θ = π. At high energy, the center-of-mass frame makes the most sense. Then p = ( ω, 0, 0, ω) , p = ( E, 0, 0, ω) (119) 1 2 − p = ( E, ω sin θ, 0, ω cos θ) , p = ( ω, ω sin θ, 0, ω cos θ) (120) 3 − − 4 so that

p12 = ω( E + ω) (121)

p24 = ω( E + ω cos θ) (122) and 1 p p E + ω cos θ E + ω 2 2 e4 24 + 12 = 2 e4 + (123) 4 |M| ≈ p12 p24 E + ω E + ω cos θ pols X     2 2 2 m For ω m, E = √m + ω ω 1 + 2 and ≫ ≈ 2 ω  

1 2 4 1 + cos θ 1 4 e + 2 (124) 4 |M| ≈  4 m  pols 2 + 1 + cos θ X 2 ω   We have only kept the factors of m required to cutoff the singularity at cos θ = 1 . The cross section for ω m is − ≫ 2 dσ 2 π 1 2 π α 1 + cos θ 1 + 2 (125) d cos θ ≈ 64 π2 (2 ω) 2 4 |M| ≈ 2 ω2 4 m  pols ! 2 + 1 + cos θ X 2 ω   Near θ = π as ω m we see that the cross section becomes very large (but still finite). In this region of phase space,≫ the photon and electron bounce off each other and go back the way they came. Or in more Lorentz invariant language, the direction of the outgoing photon momentum is the same as the direction of incoming electron momentum. Let us now try to understand the origin of the θ = π singularity. Since the matrix element can be written in the massless limit as p p t s 2 e4 24 + 12 2 e4 + (126) p12 p24 ≈ − s t for ω m and   h i ≫

2 t 2 p24 = 2 ω (1 + cos θ) (127) ≈ − − we see that the origin of the pole at θ = π is due to the t-channel exchange. Looking back, the t- channel matrix element is 2 e in out ⋆ µ ν t = ǫ ǫ u¯( p ) γ ( p p ) γ u( p ) (128) M − t µ ν 3 2 − 4 2 16 Section 6

1 1 1 Since this scales like we might expect the cross section to diverge as 2 2 . In fact t t ∼ (1 + cos θ) this would happen in a scalar field theory, such as one with interaction g φ3 for which g2 g4 , 2 (129) M ∼ t |M| ∼ t2 which has a strong t2 pole. In QED, we calculated the t-channel diagram in the massless limit and found

1 2 4 s 4 1 t = 2 e = 4 e (130) 4 |M | − t 1 + cos θ 1 X This gives the entire t pole, so we do not have to worry about interference for the purpose of understanding the singularity. Where did the other factor of t come from to cancel the pole? For θ = π φ with φ 0, the momenta become − ∼ p = ( ω, 0, 0, ω) , p = ( ω, 0, 0, ω) (131) 1 2 − p3 = ( ω, ωφ, 0, ω) p4 = ( ω,ωφ, 0, ω) (132) − − t = ω2 φ2 (133) − 1 1 1 1 So a t pole goes like φ 2 , but t2 goes like φ 4 . But notice the momentum factor in the matrix ele- ment also vanishes as p p : 2 → 4 p p = ωφk, k µ = (0, 1 , 0, 0) (134) 2 − 4 − So, 2 2 e ⋆ e ⋆ t = u¯( p3) ǫ ( p2 p4) ǫ u( p2 ) = u¯( p3) ǫ kǫ u( p2 ) (135) M ω2 φ2 in − out − ωφ in out Thus one factor of φ is canceling. This factor came from the spinors, and is an effect of angular momentum conservation. ωφ If we include the electron mass, as in Eq. (125), we would have found 2 instead of ω 2 φ 2 + m 1 e ωφ which is finite even for exactly backwards scattering. So there is not really a divergence. Still, the cross section becomes very large for nearly backwards scattering. More discussion of these types of infrared divergences is given in Lecture III-6. Let us further explore the singular t 0 region by looking at the helicity structure. To sepa- → rate the helicities of the spinor product in t first observe that it has 3 γ-matrices, so it pre- serves helicity. We can see this by insertingM factors of

1 + γ5 1 γ5 1 = + − (136) 2 2 and using

γµ(1 + γ )=(1 γ ) γµ (137) 5 − 5 (1 + γ ) ψL = 0 ψ¯ (1 γ ) = 0 (138) 5 L − 5 (1 γ ) ψR = 0 ψ¯ (1 + γ ) = 0 (139) − 5 R 5 For future reference ¯ ¯ ¯ ψ ψ = ψL ψR + ψR ψL (140) ¯ µ ¯ µ ¯ µ ψ γ ψ = ψL γ ψL + ψR γ ψR (141) ¯ µ ν ¯ µ ν ¯ µ ν ψ γ γ ψ = ψL γ γ ψR + ψR γ γ ψL (142) ¯ µ α β ¯ µ α β ¯ µ α β ψ γ γ γ ψ = ψL γ γ γ ψL + ψR γ γ γ ψR (143) and so on. The general rule is that an odd number of γ-matrices couples RR and LL while an even number couples LR and RL. In particular, our interaction with 3 γ-matrices couples RR and LL. Thus, 2 e ⋆ ⋆ t = [ u¯L ( p ) ǫ kǫ uL ( p ) + u¯R( p ) ǫ kǫ uR( p )] (144) M − ωφ 3 in out 2 3 in out 2 Historical comment 17 which is helicity conserving. This is consistent with general properties of QED. For massless electrons, as we saw in Lecture II-4, there are no interactions between right- and left-handed states in QED and so u( p2 ) and u¯( p3 ) should either be both right-handed or both left-handed. Now recall our explicit electron polarizations in the massless limit

0 0 0 1 uR =√ 2 E  uL =√ 2 E  (145) 1 0  0   0          For the photons, we need to use the helicity eigenstates1

µ 1 µ 1 ǫR = (0, 1 , i, 0) , ǫL = (0, 1 , i, 0) (146) √2 √2 − Note that 0 2 0 0 − R 0 0 L 2 0 √2 ǫ = γ1 iγ2 =  , √2 ǫ = γ1 + iγ2 = −  (147) − − 0 2 − 0 0  0 0   2 0      So     ⋆ ⋆ ǫL uR = ǫRuL = u¯RǫL = u¯L ǫR = 0 (148) Thus everything is right-handed or everything is left-handed. This has an important physical implication. Consider shooting a laser beam at an high energy beam of electrons. Lasers are polarized, so let us convert the polarization of our laser to left-handed (left circularly polarized light). Such a beam will dominantly back-scatter only left- handed electrons. This is a useful way to polarize your electron beam. It also directly connects helicity for spinors to helicity for spin 1 particles.

7 Historical comment

Considering only 2 2 scattering involving electrons, positrons, muons, anti-muons and photons there are quite a number→ of historically important processes in QED. Some examples are γe − γe − : Compton Scattering. • Observed→ in 1923 by American physicist Arthur Holly Compton. The differential scat- tering formula is called the Klein-Nishina formula (Oskar Klein and Yoshio Nishina, 1929). This was one of the first results obtained from QED, and was crucial in convincing people of the correctness of Dirac’s equation. Before this, all that was known was the classical Thomson scattering formula, which was already in disagreement with experiment by the early 1920s. The Klein-Nishina formula agreed perfectly with available experi- ments in the late 1920s. However, at higher energies, above 2 MeV or so, it looked wrong. It was not until many years later that the discrepancy was due to the production of e − e+ pairs with the annihilating into some other electron, and to . e − e − e − e − : Moller Scattering. • → First calculated in the ultrarelativistic regime by Danish physicist Christian Moller. In the non-relativistic regime it is called Coulomb scattering or Rutherford scattering. Moller calculated the cross section based on some guesses and consistency requirements, not using QED. The cross section was calculated in QED only a few years later by Bethe and Fermi. Moller scattering was not measured until 1950 by Canadian physicist Lorne Albert Page. This is partly because people did not consider it interesting until renormal- ization was understood and radiative corrections could be measured.

1. To see that the convention for “left” and “right” is being used consistently, it is easy to check that

Q Q · Q S · Qp S p 1 ǫ = − ǫ and u = − u using pµ = ( E, 0, 0, p ) and S = S or S = V in Eqs. 116?? and 120?? E L L E L 2 L z z 12 z 12 respectively of Lecture II-3. 18 Section 7

e+ e − e+ e − : Bhabha scattering. • First→ calculated by Indian physicist Homi Jehengir Bhabha in 1936. The positron was not discovered until 1932, so it would be a while before the differential cross section that Bhabha predicted could be measured in the lab. However, the total cross section for e+ e − e+ e − was important for physics from the 1930s onward. → γγ γγ: Halpern Scattering. • →In 1933, German physicist Otto Halpern realized that QED predicted light could scatter off of light. There is no tree-level contribution to this process in QED, with the first contribution coming from a box diagram at 1-loop. Heisenberg and his students Hans Euler and Bernhard Kockel were able to show that this box diagram was finite. They expressed the result in terms of an effective Lagrangian now known as the Euler- Heisenberg Lagrangian (see Lecture IV-9). Light-by-light scattering was not observed until 1997. In going beyond the box diagram, Euler and Heisenberg encountered diver- gences in the loop graphs, concluding that “QED must be considered provisional” [Schweber p. 119] A more thorough discussion of some of these processes can be found in Schweber or Pais. By the late 1930s, QED was still only a tree-level field theory. Moreover, there were a number of problems with it: the self-energy of the electron and the corrections to the energy levels of the Hydrogen atom are two examples. None of these things had been measured well enough yet to see a difference, but the theorists were still very aware that QED was apparently only valid at leading order in αe. Bohr and Dirac were ready to give up on QED by the mid 1930s. In justifying his dismissal, Dirac famously said “because of its extreme complexity, most physicists will be glad to see the end of it” [Dirac, 1937]. Bohr was ready to give up on conservation of energy. Even in 1939, he expected the resolution would be a “radical departure” even more radical than with quantum mechanics. People expected QED to break down at around the classical electron radius, E 100 MeV. There is a nice discussion of this fascinating historical period in Pais’s book and also∼ Schweber’s book, from where these quotes were taken. It wasn’t really until 1947 at the famous Shelter Island conference that data finally showed that there were finite effects subleading in αe, which gave theorists something precise to calcu- late. The next year, Schwinger came out with his celebrated calculation for radiative corrections to the electron magnetic moment: g 2 = α e . That and Lamb’s measurement of splitting − π between the 2 S1/2 and 2 P1/2 levels of the Hydrogen atom – the hyperfine structure – and Bethe’s calculation of it using QED made people take QED seriously again. It wasn’t until the 1970s, with the proof of the renormalizability of non-Abelian gauge theories, and experimental confirmation of the theory of strong interactions, quantum chromodynamics, that people were finally convinced that quantum field theory was a consistent description of nature.