Infrared Divergences

2012 Matthew Schwartz III-6: Infrared divergences

1 Introduction

We have shown that the 1, 2 and 3 point functions in QED are UV ﬁnite at one loop. We were able to introduce 4 counterterms ( δm , δ1 , δ2 , δ3) which canceled all the inﬁnities. Now let us move on to four point functions, such as

Ω T ψ( x ) ψ¯ ( x ) ψ( x ) ψ¯ ( x ) Ω (1) h | { 1 2 3 4 }| i This could represent, for example, Møller scattering ( e − e − e − e − ) or Bhabha scattering → ( e+ e − e+ e − ). We will take it to be e+ e − µ+ µ− for simplicity, since at tree-level this process only has→ an s-channel diagram. Looking→ at these 4-point functions at one-loop will help us understand how to combine previous loop calculations and counterterms into new observables, and will also illustrate a new feature: cancellation of infrared divergences. Although important results and calculational techniques are introduced in this lecture, it can be skipped without much loss of continuity with the rest of the text.

Recall that in the on-shell subtraction scheme we found δ1 and δ2 depended on a fictitious photon mass m γ. This mass was introduced to make the loops finite and is an example of an infrared regulator. As we will see, the dependence on IR regulators, like m γ, drops out not in differences between the Green’s functions at different scales, as with UV regulators, but in the sum of different types of Green’s functions contributing to the same observable at the same scale. The general principle by which infrared divergences cancel is the same as the principle by which ultraviolet divergences cancel: only physical, observable quantities are guaranteed to be finite. For ultraviolet divergences, it turns out that a simple proxy for the set of observables is

the set of Green’s functions of renormalized fields φ1 ( x1 ) φ2 ( x2 ) . These Green’s functions are not observable, and often not gauge invariant, buth are still UV finite.i For infrared divergences, Green’s functions are not good enough. In fact, S-matrix elements or even differences of S- matrix elements at different scales are not good enough. As we will see, infrared divergences only generally cancel after cross sections for processes involving different initial or final states are combined. In this lecture, we will perform one of the most important calculations in QED. We will show that although the cross section for the 2 2 process e+ e − µ+ µ− is infrared divergent at → → order e4 , as is the cross section for the related 2 3 process e+ e − µ+ µ− γ, their sum is R → → infrared finite. More precisely, we will find from calculating

2 2Re + + (2)

 ×   

¡ ¡ ¡   that     3e2 σ( e+ e − µ+ µ− (+ γ)) σ( e+ e − µ+ µ− ) + σ( e+ e − µ+ µ− γ) = σ 1 + R (3) → ≡ → → 0 16π2 4 e R + − + − where σ = 2 is the tree-level cross section for e e µ µ at E = Q. Other diagrams 0 12 πQ → CM contributing to the total e+ e − µ+ µ− cross section at 1-loop are also discussed, but this combi- nation is the most important.→ While this QED cross section is very diﬃcult to measure, its analog in QCD: e+ e − q¯q(+ g) , to be discussed in Lecture IV-2, is an important precision calculation which has been→ well conﬁrmed by data and provides strong constraints on beyond-the- standard model physics.

1 2 Section 2

We will see how having to sum over ﬁnal states (and sometimes initial states) with diﬀerent particle multiplicities is related to a muon not being physically separable from its surrounding cloud of soft photons. Trying to make this photon cloud more precise leads naturally to the notion of jets. Similarly, trying to understand the initial state radiation contribution leads naturally to the notion of parton distribution functions. The total cross section calculation is so important that we will calculate it two ways, with a Pauli-Villars UV regulator and a photon mass IR regulator, and with dimensional regularization for both the UV and the IR, showing that the total cross section is regulator independent.

2 e + e − → µ+ µ − (+ γ) At leading order, the cross section for e+ e − µ+ µ− involves a single Feynman diagram →

p1 p3 2 eR µ µ i 0 = = i 2 v¯( p2 ) γ u( p1 ) u¯( p3) γ v( p4) (4)

M p2 p4 Q ¡

2 2 2 where Q = ( p1 + p2 ) = ECM = s is the square of the center-of-mass energy. We already studied this process at tree-level in Lecture II-6 and found that, in the high energy limit Q me , m µ, the diﬀerential cross section is ≫ dσ e4 = R (1 + cos2 θ) (5) dΩ 64π2 Q 2 The total tree-level cross section is then a simple integral 2 π 1 dσ e4 σ ( Q 2 ) = dφ d cosθ = R (6) 0 dΩ 12πQ 2 Z0 Z− 1 What we would like to calculate is the next-to-leading order correction to σ0, which begins at ( α3) . O For an S-matrix calculation only amputated graphs are necessary (see Lecture III-4). In this case, there are ﬁve relevant 1-loop graphs in QED

+ + + + (7)

¡ ¡ ¡

The next-to-leading order ( α3 ) result is the interference between these graphs (of order α2 ) and the original graph (of orderO α). In addition to loop corrections to the 4-point function, we will also need to calculate real emission graphs to cancel the infrared divergences. Real emission graphs correspond to processes which are the same order in perturbation theory as the loops but involve more final state particles. We will do the loops first, then the real emission graphs, and then show that we can + − take m γ 0 after all the contributions are combined into the full cross section σ = σ( e e → tot → µ+ µ− ( + γ)) . An important simplifying observation is that since, as far as QED is concerned, the electron and muon charges Q e and Q µ can be anything, the infrared divergence must cancel order by 2 2 order in Q e and Q µ separately. The tree-level cross section scales like σ0 Q e Q µ. The loops in 3 3 2 2 2 2 2 ∼ Eq. (7) scale like Q e Q µ, Q eQ µ, Q e Q µ, Q e Q µ and Q e Q µQ X respectively, where Q X is the charge of the particles going around the vacuum polarization loop, which can be anything. In 3 particular, we will focus on the cancellation of divergences proportional to σ0 Q eQ µ. This cancellation gives the critical demonstration of infrared finiteness, and is phenomenologically relevant. Other loop contributions will be discussed afterwards. e+ e− → µ+ µ− (+ γ) 3

2.1 Vertex Correction The vertex correction is

p1 2 eR µ µ i Γ = p + =i 2 v¯( p2 ) γ u( p1 ) u¯( p3)Γ2 ( p) v( p4) (8)

M p2 Q ¡

p4 ¡

µ µ µ 2 2 where p = p1 + p2 is the photon momentum entering the vertex with p = Q . In this equation, µ 2 Γ2 ( p) refers to the ( e ) contribution to 1PI vertex function, for which we do not introduce any O µ new subscripts for readability. Conveniently, we have already computed Γ2 ( p) for a general oﬀ- shell photon in Lectures III-3 and III-5. Thus we can just copy over those results. Recall the general vertex function Γ µ( p) was parameterized in terms of two form factors

iσ µν Γ µ( p) = F ( p2 ) γµ + p F ( p2 ) (9) 1 2 m ν 2

Here m can represent either the electron or muon mass. We also do not write mR, since mass renormalization will not be relevant to the calculations in this lecture. We found that the second 2 form factor at order eRwas

2 1 2 2 eR z(1 z) m 4 F2 ( p ) = dxdydzδ( x + y + z 1) − + ( eR) (10) 4π2 − (1 z) 2 m2 xyp2 O Z0 − − m 2 In the high-energy limit, 2 0, this form factor vanishes, F ( p ) 0. This makes sense, since p → 2 → F2 couples right- and left-handed spinors which are uncoupled in massless QED. The ﬁrst form factor was both ultraviolet and infrared divergent. Regulating the ultraviolet 2 divergence in F1 ( p ) with Pauli-Villars and the infrared divergence with a photon mass, we found that F ( p2 )=1+ f( p2 ) + δ + ( e4 ) (11) 1 1 O R where

e2 1 zΛ2 p2 (1 x)(1 y) + m2 (1 4z + z 2 ) f( p2 ) = R dxdydzδ( x + y + z 1) ln + − − − 8π2 − ∆ ∆ 0 with Z ∆ = (1 z) 2 m2 xyp2 + zm2 (12) − − γ For e+ e − µ+ µ− we need p2 = Q 2 and we can take m = 0 for the high-energy limit ( Q m), → ≫ The counterterm is set by F1 (0) = 1 which normalizes the electric charge to what is measured at large distances. In the previous lecture we calculated δ1 for ﬁnite m. Now with m = 0 we ﬁnd

2 2 eR 1 Λ δ1 = f(0) = 2 ln 2 (13) − − 8π 2 m γ !

Evaluating f( Q 2 ) is more challenging. It has the form

e2 1 1 − x (1 x y)Λ2 Q 2 (1 x)(1 y) f( Q 2 ) = R dx dy ln − − + − − (14) 8π2 xyQ 2 + (1 x y) m2 xyQ 2 + (1 x y) m2 Z0 Z0 " − − − γ − − − γ # The ﬁrst term is IR ﬁnite and gives

1 1 − x (1 x y)Λ2 3 1 Λ2 dx dy ln − − = + ln + ( m γ) (15) xyQ 2 + (1 x y) m2 4 2 Q 2 O Z0 Z0 " − − − γ # − 4 Section 2

2 Note that the ln Λ has the right coeﬃcient to be canceled by δ1 . More generally, the divergences in the the vertex correction and δ1 will always cancel for arbitrarily complicated processes involving a photon-fermion vertex. This is simply because the divergent part of the counterterm µ was determined by calculating the 1PI contributions to Ω T ψA ψ¯ Ω . In the divergent h i region of loop momentum, the external scales are irrelevant. Thus the divergences for the 3- point function are the same whether or not it is embedded in a larger diagram, and therefore they will always be canceled by δ1 . The second term in Eq. (14) is IR divergent but UV ﬁnite. Moreover, for real Q 2 there is a pole in the integration region. Fortunately, there is a small imaginary part in the denominator (due to the iε prescription) which makes the integral converge. Since x and y are positive we can perform the integral by taking Q 2 Q 2 + iε, which gives → 1 1 − x 2 2 2 2 Q (1 x)(1 y) 1 2 m γ m γ π 5 dx dy − − = ln 2ln + ( m γ) xyQ 2 + (1 x y) m2 − 2 Q 2 iε − Q 2 iε − 3 − 2 O Z0 Z0 − − − γ − − − − So that, 2 2 2 2 2 eR 2 m γ m γ 2π 7 f( Q ) + δ = ln 3ln + ( m γ) (16) 1 16π2 − Q 2 iε − Q 2 iε − 3 − 2 O ( − − − − ) Then we use lim ln ( Q 2 iε) = lnQ2 iπ (17) ε → 0 − − − to write 2 2 2 2 2 eR 2 m γ m γ π 7 f( Q ) + δ1 = 2 ln 2 (3+2πi) ln 2 + 3πi + ( m γ) (18) 16π ( − Q − Q 3 − 2 − O )

2 2 2 π 2 2 m γ Note that the has combined with the π coming from the expansion of ln 2 to give π 2 − 3 − − Q the term. 3 To evaluate the cross section at next to leading order, we need the ﬁrst subleading term in + 2 . The ( e6 ) term in this comes from |M Γ M 0| O R 4 † † eR µ µ + = Tr[ p γ p γµ] Tr[ p Γ p γµ] + c.c. (19) M ΓM 0 M 0 M Γ Q 4 2 1 3 2 4 In the high-energy limit in which we are interested, the σ µν term in Γ µ gives an odd-number of γ-matrices in the second trace, forcing the contribution of the F2 form factor to vanish. This is consistent with F itself vanishing for p2 m2 . So we have simply 2 ≫ 1 1 † + † = 2Re[ f( Q 2 ) + δ ] 2 (20) 4 M ΓM 0 M 0 M Γ 1 4 |M 0| spins spins X X 6 with f( Q) just a number. Thus the total loop (Virtual) correction at order eR is given by

2 2 2 2 2 eR 2 m γ m γ 7 π σV = 2( f( Q ) + δ1 ) σ0 = 2 σ0 ln 2 3ln 2 + (21) 8π ( − Q − Q − 2 3 )

2 2 m γ Sudakov An important qualitative feature of this result is the ln Q 2 term. This is known as a double logarithm, and is characteristic of infrared divergences. Sudakov logarithms play an important role in many areas of physics, such as the physics of jets and of parton-distribution functions , to be discussed briefly in Sections 3 and 4.2 and in more detail in Part IV. 2 The fact that σV ( Q ) is divergent cannot be remedied by comparing the cross section at different scales. Indeed, the difference between cross sections at different scales is

2 2 2 2 2 2 eR 2 m γ Q 2 2 m γ σV ( Q 1 ) σV ( Q 2 ) = 2 σ0 ln 2 3ln 2 + ln 2 (22) − 8π ( − Q 1 − Q 1 Q 2 ) e+ e− → µ+ µ− (+ γ) 5

In this difference, all the subtraction-scheme dependent constants drop out, as does the Sudakov 2 2 double log, ln m γ. However, infrared divergent single logarithms remain. This is just because differences of logarithms are the logarithm of a ratio but differences of double logarithms are not a double logarithm of a ratio. As we will see, the resolution is that a cross section like this is not in fact an observable: only when we include contributions proceses with different final states can we find an observable which is independent of m γ.

2.2 Real emission graphs Next we calculate the cross section for e+ e − µ+ µ− γ. To fourth order in the muon charge, the only diagrams have the photon coming oﬀ of→ a muon

p3 p1 p3 p1

pγ i = p + pγ M p p2 p4 p2

¡ ¡

4 2 6 The cross section for this process starts oﬀ at order Q µQ e eR, so it is the same order at tree-level as the interference between e+ e − µ+ µ− at tree-level and at 1-loop. → The diagrams give in the limit Q m in Feynman gauge, ≫ e2 i = i R v¯( p ) γµu( p ) u¯( p ) S µα v( p ) ǫ⋆ (23) M Q 2 2 1 3 4 α with

µα α i µ µ i α S = ieR γ γ γ γ (24) − " p3 + pγ − p4 + pγ # where ǫα the ﬁnal-state photon polarization. The unpolarized cross section is therefore given by

4 1 2 eR µν σR = dΠ = L Xµν (25) 2 Q 2 LIPS |M| 2 Q 6 Z with the initial spin-averaged electron tensor given by 1 1 1 L µν = v¯( p ) γµu( p ) u¯( p ) γνv( p ) = Tr[ p γµp γν] = pµpν + pνpµ Q 2 gµν (26) 4 2 1 1 2 4 2 1 1 2 1 2 − 2 e ± Xspins ⋆ and, using ǫα ǫ = gαβ , pols β −

P µν µα βν ⋆ X = d ΠLIPS u¯( p3) S v( p4) v¯( p4) S u( p3) ǫα ǫ β Z µ ± spins ǫXpols. h i = d Π Tr[ p S µα p Sαν] − LIPS 3 4 Z where in this case 3 d pj 1 4 4 dΠLIPS = 3 (2π) δ ( p p3 p4 pγ) (27) (2π) 2Ep − − − j =3, 4,γ j Y µ µ µ with p = p1 + p2 . µ µ Now note that p L µν = p Xµν = 0. This would be true even if we did not sum over spins, by the Ward identity for the intermediate photon. In particular, since X µν is a Lorentz covariant function only of pµ (the other momenta are integrated over), it must have the form

X µν = ( pµpν p2 gµν) X( p2 ) (28) − 6 Section 2

Then using Eq. (26) we ﬁnd

2 µν µ ν ν µ 1 2 µν µ ν 2 µν 2 4 2 Q µν L Xµν = p p + p p p g ( p p p g ) X( p ) = Q X( Q ) = g Xµν (29) 1 2 1 2 − 2 − − 3 2 2 µ µ 2 1 µν where p = Q = 2 p p and X( Q ) = 2 g Xµν have been used. Thus, 1 2 − 3 Q 4 eR µν 2π µν σR = g Xµν = σ0 − g Xµν (30) − 6 Q 4 Q 2 4 e R + − + − where σ = 2 is the tree-level cross section for e e µ µ from before. 0 12 πQ → We have conveniently included dΠLIPS in Xµν so that its deﬁnition would be equivalent to the cross section for γ⋆ µ+ µ− γ where γ⋆ is a photon of mass Q. That is, → 2 ⋆ + − eR µν Γ( γ µ µ γ) = gµνX (31) → − 2 Q

One can interpret the gµν in this last formula as a sum over polarizations of the oﬀ-shell photon, which can mean− either a transverse polarization sum or a sum over all polarizations – µ since p Xµν = 0 the unphysical polarizations do not contribute. The result is that the unpolarized cross section factors into e+ e − γ⋆, which gives just a normalization since there is no phase space, and γ⋆ µ+ µ− γ. More precisely→ →

4π ⋆ + − σR = σ Γ( γ µ µ γ) (32) 0 e2 Q → R This is a useful general result: since we sum over spins, all spin-correlations between the initial and ﬁnal state average out and the cross section can be calculated by considering two sub-processes, the creation and subsequent decay of an intermediate state. This is actually a special case of the narrow-width approximation, to be discussed in Lecture III-10. We have reduced the problem to the calculation of σ( γ⋆ µ+ µ− γ) in the γ⋆ rest frame. To calculate this cross section, it is helpful to use Mandelstam invariants:→

2 2 s ( p + p ) Q (1 x γ) (33) ≡ 3 4 ≡ − 2 2 t ( p + pγ) Q (1 x ) (34) ≡ 3 ≡ − 1 2 2 u ( p + pγ) Q (1 x ) (35) ≡ 4 ≡ − 2 2 2 2 with 0 6 s 6 Q and βQ 6 t, u 6 Q or equivalently 0 6 x γ 6 1 and 0 6 x1 , x2 6 1 β. As we will − γ see, the cross section is infrared divergent if the ﬁnal state photon with momentum pµ is mass- 2 2 2 2 less. We will therefore allow for pγ = m γ 0. In this case, s + t + u = Q + m γ or equivalently

x + x + x γ = 2 β (36) 1 2 − 2 β m γ s t u m2 γ⋆ where Q 2 . (In general, + + = i and here only and the real outgoing photon have non-zero≡ masses.) P The xi variables are easier to use in this calculation than s, t and u. They can be thought of as the energy of the outgoing states in the γ⋆ rest frame. For example,

2 ( p p4) 2 p4 p E4 x1 = 1 − = · = 2 (37) − Q 2 Q 2 Q

µ E3 2 E γ where p = ( Q, 0, 0, 0) in the rest frame has been used. Similarly, x = 2 and x γ = β. 2 Q Q − Since there are only two independent Lorentz invariant kinematical quantities for 4-body scattering, we can take these to be x1 and x2 . In terms of x1 and x2 , the phase space reduces to

β 2 1 − β 1 − Q 1 − x 1 d Π = dx dx (38) LIPS 128π3 1 2 Z Z0 Z1 − x 1 − β e+ e− → µ+ µ− (+ γ) 7

You can check this in Problem 1 (we derive a similar formula in d-dimensions with β = 0 in Sec- tion A.3 below; this factor of 128π3 is the same one appearing in the formula for 3-body decay rates, as you found in Problem ??). The limits of integration in Eq. (38) are the boundary of the surface bounded by the constraints on xi listed above. After some straightforward Dirac algebra, we ﬁnd

2 2 2 µα αµ 8eR 2 2 (1 x1 ) + (1 x2 ) 2 Tr[ p3S p4S ] = x1 + x2 + β 2( x1 + x2 ) − − + 2 β (39) (1 x1 )(1 x2 ) − (1 x1 )(1 x2 ) 2 − − − − β m γ with = Q 2 as before. Before evaluating the cross section by integrating this expression, let us explore where the infrared divergence is coming from. If we set m γ = 0 then the cross section would be e2 Γ( γ⋆ µ+ µ− γ) = R dΠ Tr[ q S µα q Sαµ] → 2 Q LIPS 1 2 4Z 1 1 2 2 QeR x1 + x2 = 3 dx1 dx2 32π (1 x1 )(1 x2 ) Z0 Z1 − x 1 − − which is divergent from the integration region near x = 1 or x = 1 . Suppose x 1 meaning the 1 2 2 ∼ µ− has energy E Q and its momentum is therefore pµ Q , 0, 0, Q . Thus by momentum 3 ∼ 2 3 ∼ 2 2 conservation, the sum of the µ+ and photon momenta must be pµ + pµ = Q , 0, 0, Q which is

4 γ 2 − 2 Q lightlike. This implies 0 = p pγ = E Eγ(1 cosθ) where θ is is the angle between Qp and p . 4 · 4 − 4 γ Therefore E4 0 or Eγ 0, which is known as a soft singularity, or cosθ 1 implying the photon and µ+∼ are in the∼ same direction, which is the region where there is a ∼collinear singularity. In general, infrared divergences come from regions of phase space where massless particles are either soft or collinear to other particles. Anticipating the infrared divergence, we have regulated it with a photon mass. Then the cross section is convergent. The only terms which contribute as β 0 are → β 1 − β 1 − 1 − 2 2 2 x 1 x1 + x2 2 π dx1 dx2 = ln β + 3lnβ + 6 + ( β) (40) 0 1 − x 1 − β (1 x1 )(1 x2 ) − 3 O and Z Z − − β 1 − β 1 − 2 2 1 − x 1 (1 x1 ) + (1 x2 ) β dx1 dx2 − 2 − 2 = 1 + ( β) (41) − 0 1 − x 1 − β (1 x1 ) (1 x2 ) − O Therefore Z Z − − 4 2 2 2 ⋆ + − QeR 2 m γ m γ π Γ( γ µ µ γ) = 3 ln 2 + 3ln 2 + 5 (42) → 32π ( Q Q − 3 ) and, from Eq.(32) 2 2 2 2 eR 2 m γ m γ π σR = 2 σ0 ln 2 + 3ln 2 + 5 (43) 8π ( Q Q − 3 ) Recalling 2 2 2 2 eR 2 m γ m γ 7 π σV = 2 σ0 ln 2 3ln 2 + (44) 8π ( − Q − Q − 2 3 ) We see that all infrared divergent terms precisely cancel, and we are left with

3e2 σ + σ = R σ (45) R V 16π2 0 So we see that if we include the virtual contribution and the real emission, the IR divergences cancel. The result is 3e2 σ = σ 1 + R (46) tot 0 16π2 Now we need to interpret the result. 8 Section 3

3 Jets + − + − 6 We have found that the sum of the e e µ µ cross section σV at order eR, from the p3 →

p1 + − + − 6 graphs p + and the e e µ µ γ cross section, σR also at order eR, from

p2 → ¡

p4 ¡

p3 p3

p5 the graphs + p5 was infrared and ultraviolet ﬁnite. Photons emitted oﬀ p4

¡ p4 ¡ of final state particles, such as the muons in this case, are known as final state radiation. The explanation of why one has to include final state radiation to get a finite cross section is that it is impossible to tell whether the final state in a scattering process is just a muon or a muon plus an arbitrary number of soft or collinear photons. Trying to make this more precise leads naturally to the notion of jets. For simplicity, we calculated only the total cross section for e+ e − annihilation into states containing a muon and antimuon pair, inclusive over an additional photon. One could also calculate something less inclusive. For example, experimentally, a muon might be identified as a track in a cloud chamber or an energy deposition in a calorimeter. So one could calculate the cross section for the production of a track or energy deposition. This cross section gets contributions from different processes. Even with an amazing detector, there will be some lower limit Eres on the energy of photons that can be resolved. Even for energetic photons, if the photon is going in exactly the same direction as the muon there would be no way to resolve it and the muon separately. That is, there will be some lower limit θres on the angle which can be measured between either muon and the photon. With these experimental parameters,

σtot = σ2 → 2 + σ2 → 3 (47) where + − + − + − + − σ2 → 2 = σ( e e µ µ ) + σ( e e µ µ γ) (48) → → E γ

+ − is the rate for producing for producing something that looks just like a µ µ pair and

+ − + − σ2 → 3 = σ( e e µ µ γ) (49) → E γ >Er e s and θ γµ >θr e s

is the rate for producing a muon pair in association with an observable photon. The cross section for muons plus a hard photon is now infrared finite due to the energy cutoff, even for E Q and θ 1 . Unfortunately, the phase space integral within these cuts, res ≪ res ≪ even with m γ = 0, is complicated enough to be unilluminating. The result, which we quote from [Ellis, Sterling & Webber] is that the rate for producing all but a fraction Q of the total Er e s energy in a pair of cones of half-angle θres is 2 eR 1 Q 3 Eres σ → = σ ln ln 1 + 3 2 3 0 8π2 θ 2E − − 4 Q res res π2 7 E 3 E 2 E + res + res + θ ln res (50) 12 − 16 − Q 2 Q O res Q To calculate σ2 → 2 one cannot take m γ = 0 since the two contributions are separately infrared divergent. Conveniently, since we have already calculated σtot = σ2 → 2 + σ2 → 3 , we can just read off that 2 eR 1 Q 3 Eres

σ → = σ σ → = σ 1 ln ln 1 + 3 + 2 2 tot − 2 3 0 − 8π2 θ 2E − − 4 Q res res This result was first calculated by Sterman and Weinberg in 1977. They interpreted σ2 → 2 as the rate for jet production, where a jet is defined as a two-body final state by the parameters θres and Eres. More precisely, these paramaters define a Sterman-Weinberg jet. Jets 9

Sterman-Weinberg jets are not the most useful jet definitions in practice. There are many other ways to define a jet. Any definition is acceptable as long as it allows a separation into finite cross sections for σ2 → 2 (the two-jet rate), σ2 → 3 (the three jet rate), and σ2 → n (the n-jet rate), which starts at higher order in perturbation theory. A jet definition simpler than Sterman-Weinberg simply puts a lower bound on the invariant mass of the photon-muon pair, 2 2 ( pγ + pµ ± ) > MJ . This single parameter limits both the collinear and soft singularities. An invariant mass cutoff is sometimes known as a JADE jet after the JADE (Japan, Deutschland, England) experiment which ran at DESY in Hamburg from 1979-1986. 2 2 2 2 Restricting ( pγ + pµ ± ) >MJ implies t>MJ and u>MJ in the notation of Eqs. (33)-(35), or 2 MJ equivalently, x < 1 βJ and x < 1 βJ where βJ = 2 . Then the cross section is 1 − 2 − Q e2 1 − βJ 1 − βJ x2 + x2 σ = R σ dx d x 1 2 2 → 3 8π2 0 1 2 (1 x )(1 x ) Z0 Z1 − x 1 − 1 − 2 2 2 2 2 eR 2 MJ MJ π 5 = σ 2ln + 3ln + + ( βJ ) 8π2 0 Q 2 Q 2 − 3 2 O where the MJ Q limit has been taken in the second line. One does not have to take this limit, ≪ however, the limit shows, as with Eq. (50), a general result: In physical cross sections, an experimental resolution parameter acts as an IR regulator • In other words, we did not need to introduce m γ. In practice, it is much easier to calculate the total cross section using m γ than by using a more physical regulator associated with the details of an experiment. An important qualitative feature of results like the two- or three-jet rates is that for very 2 2 M Q e R 2 MJ > small resolution parameters, J , it can happen that 8π 2 ln Q 2 1 . In this limit, the per- ≪ 2 2 2 e4 e R 2 MJ turbation expansion breaks down, since an order R correction of the form 8π 2 ln Q 2 would be of the same order. Thus, to be able to compare to experiment, one should not take MJ too small. As a concrete example, the experiment BABAR at SLAC measured the decay of B mesons to kaons and photons ( B Kγ) . They were sensitive only to photons harder than → Eres = 1.6 GeV. In other words, they could not distinguish a kaon in the final state from a kaon plus a photon softer than this energy. To compare to theory, they needed a calculation of the rate for B Kγ with the γ energy integrated up to Eres. The rate has a term of the form Er e s → ln2 1 in it, which has a quantitatively important effect. Since the logarithm is large, higher m B ∼ orders in perturbation theory are also important. The summation of these Sudakov double logarithms to all orders in perturbation theory was an important impetus for the development of new powerful theoretical tools, in particular, Soft-Collinear Effective Theory (see Lecture IV-10) in the 2000s. While these muon-photon packets are hard to see in QED, they are easy to see in QCD. In QCD, the muon is replaced by a quark and the photon replaced by a gluon. The quark itself, and the additional soft gluons turn into separate observable particles, such as pions and kaons. Thus a quark in QCD turns into a jet of hadrons . These jets are a very real and characteristic phenomenon of all high-energy collisions. We have explained their existence by studying the infrared singularity structure of Feynman diagrams in quantum field theory. In modern collider physics, it is common to look not at the rate for jet production for a fixed resolution parameter, but instead to look at the distribution of jets themselves. To do this, one needs to define a jet through a jet algorithm. For example, one might cluster together any observed particles closer than some θres. The result would be a set of jets of angular size θres. dσ Then one can look at the distribution of properties of those jets, such as where mJ is the dm J jet mass, defined as the invariant mass of the sum of the 4-momenta of all the particles in the jet. It turns out that such distributions have a peak at some finite value of mJ . However, at any order in perturbation theory, one would just find results like dσ = 1 ln m J which grow arbi- dm J m J Q trarily large at small mass. Calculating the mass distribution of jets therefore requires tools beyond perturbation theory, some of which are discussed in Lecture IV-10. 10 Section 4

4 Other loops

Now let us return to the other loops in Eq. (7). The box and crossed box diagrams

+ (51)

¡ ¡ are UV ﬁnite. To see this, note that the loop integrals for either graph will be of the form

d4k 1 1 1 1 d4k (52) (2π) 4 k2 k2 k k ∼ k6 Z Z where k pi has been taken to isolate the ultraviolet divergent region. These graphs are there- ≫ fore ultraviolet ﬁnite. So no renormalization is necessary. The interference of these graphs with 3 3 the tree-level graph contributes at order Q e and Q µ in the electron and muon charges, which is the same order as

(53)

¡ ¡ and similar cross terms. Besides the UV ﬁniteness of the loops, there is nothing qualitatively new in these graphs. You can explore them in problem 5.

4.1 Vacuum polarization correction Next, we consider the vacuum polarization graph and its counterterm:

i β = + (54)

¡ ¡

The interference between the tree-level amplitude for e+ e − µ+ µ− and these graphs gives a 6 → contribution to the cross section at order eR. This contribution is proportional to the square of the charges of whatever particle is going around the loop. For a loop involving a generic charge, there are no corresponding real emission graphs of the same order in that charge: thus any infrared divergences must cancel among between these graphs alone. We evaluated these graphs in the previous lecture for an oﬀ-shell photon. Copying over those results, the sum of the loop and its counterterms in this case give an interference contribution 2Re( β ) which leads to a correction to the cross section of the form M 0M 2 ∆σβ = 2Re[Π( Q )] σ (55) − 0 with e2 1 m2 Π( Q 2 ) = R Q 2 dxx(1 x) ln j (56) 2π2 j − m2 Q 2 x(1 x) j 0 j ! X Z − −

For this physical application we have to sum over all particles j with masses m j and charges Q j which can go around the loop. This sum therefore includes electrons, muons, quarks, and every- thing else with electric charge in the standard model. Other loops 11

A more suggestive way to write the vacuum polarization contribution is through an effective charge. Recall that it was these same vacuum polarization graphs which contributed to the run- ning of the Coulomb potential. In the Coulomb potential, the virtual photon is spacelike, with p2 > 0. In Lecture III-2, we found that for p2 m2 , the effective charge at 1-loop was − − ≫ 2 2 2 2 2 eR p eeff( p ) = eR 1 + ln − (57) − 12π2 m2 2 with the convention that eR eeff( m ) . ≡ − Now look at the correction to the cross section, with just one virtual fermion for simplicity and Q 2 m2 . Then we can use ≫ e2 m2 m Π( Q 2 ) = R ln + regular as 0 (58) 12π2 Q 2 Q → 4 − σ Q 2 e R Now recalling 0( ) = 12 πQ 2 we find 4 2 2 2 eR eR Q 4 σ( Q ) = 1 + 2Re ln − + ( eR) 12πQ 2 12π2 m2 O 4 2 2 1 2 eR Q 6 = eR + ln − + ( eR) 12πQ 2 12π2 m2 O 1 = e ( Q 2 ) 4 + ( e6 ( Q 2 )) 12πQ 2 | eff − | O eff − Including the final-state radiation and virtual correction from the muon vertex, we also have e4 ( Q 2 ) 3e2 ( Q 2 ) σ = eff 1 + eff + ( e6 ) (59) 12πQ 2 16π2 O eff And thus The entire effect of the vacuum polarization graph is encapsulated in the scale-dependent • effective charge This is true quite generally, and explains why an effective charge is such a useful concept. You may have noticed that in the limit m 0, the effective charge in Eq. (57) appears to be infrared divergent. However, since → 4 2 2 2 2 2 eR p1 eeff( p1 ) eeff( p2 ) = ln − (60) − − − 12π2 p2 − 2 as long as the effective charge measured at some scale is finite, the charge at any other scale will be finite. In particular, we can measure the charge before neglecting the electron mass, then run the charge up to high energy. Or more simply, measure the electric charge through the e+ e − + − → µ µ cross section at some scale p1 and predict the effect at p2 (if we do this, however, the finite effect from the vertex correction and final state radiation contribution cannot be measured). Although we only showed the agreement for a single virtual fermion, since the same vacuum polarization graphs correct Coulomb’s law as the e+ e − µ+ µ− cross section, the agreement will hold with arbitrary charged particles. If there are many→ particles, it is unlikely that Q will be much much larger than all their massess. Of course, if Q m j for some mass, that particle has little effect (the logarithm in Eq. (56) goes to zero). But≪ we may measure the cross section at various Q above and below some particle thresholds. In this case, the effective charge changes, sometimes even discontinuously. Physical observables (such as cross sections) are not discontin- uous, since finite corrections to the cross section exactly cancel the discontinuities of the effective charge. 1 This will be discussed more in the context of quantum chromodynamics in Lecture IV-2.

1. The effective charge is regulator and subtraction scheme dependent. In the on-shell scheme, the effective charge is very difficult to calculate through particle thresholds. It is therefore more common to use dimensional regularization with minimal subtraction to define the effective charge. In particular, in QCD, where the thresholds are very important for the effective strong coupling constant α s , MS is almost exclusively used, and there the effective charge is known to 4-loop order. 12 Section 4

By the way, the way we have defined the effective charge, through the Coulomb potential µ 2 where p is spacelike, eeff( p ) is naturally evaluated at a positive argument. Here we see that −+ − + − 2 to use the same charge for e e µ µ it must be evaluated at a negative argument, eeff( Q ) with Q 2 > 0. In fact, it is natural→ for a process with a time-like intermediate state to have− a Q 2 factor like ln 2 with a non-zero imaginary part. This imaginary part is actually required − m by unitarity, as will be discussed in Lecture III-10. It also has a measurable effect, through terms like the π2 which contributed to the real part of the virtual amplitude in going from Eq. (16) to Eq. (18). This π2 does contribute a non-zero amount to the cross section. In fact, since π2 is not a small number, π2 corrections can sometimes provide the dominant subleading contribution to a cross section. For example, they can be shown to account for a large part of the approximate doubling of the pp e+ e − cross section at next-to-leading order [L. Magnea and G. Sterman, Phys. Rev. D 42, 4222→ (1990)].

4.2 Initial state radiation Finally, we need to discuss the contributions to the e+ e − µ+ µ− ( γ) cross section to third order in the electron charge and ﬁrst order in the muon charge.→ In other words, the following diagrams

+ + + (61)

¡ ¡

In the same way that final state radiation was necessary to cancel the infrared singularity of the vertex correction involving the photon, the sum of these diagrams will be finite. The radiation coming off the electrons in this process is known as initial state radiation. These real emission graphs are closely related to the real emission graphs with the photon coming off the muons, and their integrals over phase space have infrared divergences. However the infrared divergent region is a little different and the physical interpretation of the divergences is very different. Let us suppose that the sum of the diagrams in Eq. (61) gives a finite total cross section for e+ e − µ+ µ− (+ γ) we call σ . Then we should be able to calculate a more exclusive 2-jet cross → tot section, as in the previous section, for producing less than Eres of energy outside of cones of half- angle θres around the muons. In this case, however, there is no collinear singularity with the photons going collinear to the muons. Instead, the infrared divergences come from the intermediate electron propagator going on shell. This propagator has a factor of 1 1 1 2 = − = − (62) ( pe pγ) 2 pe pγ 2Eω(1 cosθeγ) − · − where θeγ is the angle between the outgoing photon and the incoming electron, E is the electron energy and ω is the outgoing photon energy. Thus the singularity comes from the region with θeγ 0 or ω 0, but not where the photon goes collinear to the muon. So if we try to calculate → → + − the σ → = σ σ → where σ → is the rate for producing µ µ and a photon with ω > E , 2 2 tot − 2 3 2 3 res we would find an unregulated collinear singularity associated with θeγ 0 and both σ → and → 2 3 σ2 → 2 are therefore infinite! As you might guess, we’re missing something. First of all, the collinear singularity does not actually produce an infinite cross section since it is cut off by the electron mass (the electron mass does not regulate the ω 0 soft divergence, just the collinear divergence). We actually → already calculated a similar cross section with finite m for Compton scattering e − γ e − γ. Indeed, it is easy to see that the collinear singularities associated with an intermediate→ electron going on-shell are the same in the two processes. For Compton scattering, we found in Lecture II-6 that the differential cross section for ω m was ≫ 4 dσ eR 1 + cosθ 1 2 + 2 (63) d cosθ ≈ 32πω  4 m θ  2 ω 2 + 1 + cos   Other loops 13 with θ the angle between the outgoing photon and incoming electron. Integrating over θ gives

e4 1 4ω2 σ = R + ln 1 + (64) 32πω2 2 m2 ω + − + − This is finite, although extremely large as . The cross section σ → for e e µ µ γ m → ∞ 2 3 → would have a similar factor. ω E What are we to make of this large ln m factor? For final state radiation, as long as res and θres were not very small, the cross section for σ2 → 3 was not too large. More importantly, it was independent of the electron mass. In fact, it is intuitively obvious that the electron mass should be irrelevant to the cross section at high energy. So why is it appearing here? The resolution of this dilemma is easiest to understand by thinking about scattering protons instead of electrons (this part may not make sense until you have made it through Lecture IV- 7). A proton is superficially made up of two up quarks and one down quark, but really it is a complicated bound state of those quarks interacting through the exchange of gluons, which are massless spin 1 particles like the photon. When one collides protons at high energy, there is an interaction among one quark in one proton and one quark in another (or more generally, between gluons, quarks or antiquarks). But only a small fraction of the energy of the proton is usually involved in the scattering with the rest just passing through. One way to understand − 1 − 1 this is that the proton has a size of order rp = m p (1 GeV) . Thus at energies Q GeV, − 1 ∼ ≫ only a small dot of size Q rp inside the big proton can be probed. In practice, it’s impos- ≪ sible to calculate the probability that a certain quark will be involved in a short-distance collision, but we can parameterize these probabilities with non-perturbative objects called parton distribution functions (PDFs), fi( x, Q) where x is the fraction of the proton’s energy that the quark involved in the collision has at some short distance scale Q. The PDFs will be formally defined and discussed in Lecture IV-7. Now we can understand better the collinear divergence associated with initial state radiation. At ultra-high energies, when electrons and positrons collide, it is impossible for all of the energy of the electron to go into the hard collision. Instead, only some fraction x of the electron’s energy will participate, with the rest of the energy continuing along the electron’s direction in the form of radiation (photons). One can define a function fi( x, Q) for the electron, where i = − e or γ (or, technically speaking, anything else). In QED, these functions fi( x, Q) are calculable. They are called Weizsäcker-Williams distribution functions (WWDFs). For example, the probability of finding a photon inside an electron with energy ω = zQ in a collision at energy Q is 2 2 2 eR 1+(1 z) Q fγ( z) = − ln (65) 8π2 z m2

You can derive this in Problem 6. If m Q or if z 1 then fγ( z) is enormously large. In par- ≪ 8π ≪ ticular, this logarithm becomes bigger than e 2 and perturbation theory breaks down. Thus, even though we can in principle calculate the WWDFs, doing so is impractical, and in fact, unnecessary. More simply, if m = 0, as we have taken in most of this lecture, the WWDFs depend on the infrared regulator and therefore become scheme dependent. The collinear part of the infrared regulator dependence is exactly canceled by the collinear singularity from the virtual graph in Eq. (61). The remaining soft singularity of the virtual diagram is canceled by real emission diagrams involving soft radiation into the ﬁnal state, as would be included in a calculation of the total cross section, or the rate for 2 jets around the muons. Another way to think about WWDFs is that they include the eﬀects of graphs like

+ (66)

¡ ¡ 14 Appendix A in which the photon is in the initial state. In these the collinear singularity is naturally cut off by the electron mass, or the infrared regulator if the electron is massless. Either way, the incoming radiation represents the electron containing a photon, which is parametrized with the WWDFs. How do we deal with the initial state collinear singularity in practice? It turns out that for real experiments, the details of the WWDFs and how the initial state infrared divergences cancel are almost never important. For example, consider the LEP collider at CERN which ran during the 1990s. For much of its life, this machine collided electrons and positrons at a center-of-mass energy near the Z boson mass: Q 91 GeV. At this energy the Z boson is produced resonantly, ≈ almost always involving all of the energy of the electrons, with no phase space left for initial state radiation. Actually, since the electrons and positrons have variable energy in a typical beam, real or virtual soft photons were often emitted from the initial state to bring the Z to the resonance peak, a process called radiative return. The result was that you could just measure the decay of the Z and ignore the initial state completely. Thus, you only need the final state loops. The decay width is calculable, finite, and does not depend on whether it was e+ e − or 2 + − 3e something else that produced the Z. In fact, Z µ µ (+ γ) gets precisely the 2 correction 16 π + − + −→ in QED we calculated for the σtot( e e µ µ (+ γ)) rate we calculated in Eq. (46). In fact, → because the Z decays not just to muons, but also to quarks which have charges 2 or 1 , this 2 ± 3 ± 3 3e Q 2 correction becomes 16 π 2 i and is therefore a way to test the standard model. In particular, the branching ratio for Z bb¯ has proven a particularly powerful way to look for physics beyond → the standard model, since it happens to be sensitive not just to loops involving electrons, but also to loops involving hypothetical particles (such as charged Higgses). By the way, there is actually an interesting difference with regards to the necessary inclusion of initial state radiation in the QED and QCD (a non-Abelian gauge theory to be introduced in Lecture IV-1). In QED, there’s an important theorem due to Bloch and Nordsieck (1931) which says

Infrared singularities will always cancel when summing over ﬁnal state radiation in QED • with a massive electron as long as there is a ﬁnite energy resolution.

In QCD, this is not true. It was found by Doria et al [Nucl.Phys.B 168 (1980) 93] that at 2- loops, infrared singularities in QCD with massive quarks will not cancel summing over 2 n → processes only; one also needs to sum over 3 n processes. The uncanceled singularity, however, → vanishes as a power of the quark mass and therefore disappears if m q = 0. Thus, in the high- energy limit of QCD, where the mass can be neglected, one can get an infrared ﬁnite answer summing only cross sections with 2 particles in the initial state. (This result has nothing to do with QCD being asymptotically free, and would hold even if there were enough ﬂavors so that QCD were infrared free, like QED.) The more general theorem, due to Kinoshita, Lee and Nauenberg (KLN) is that

Infrared divergences will always cancel when summing over ﬁnal and initial states. • The key to this theorem is the association of infrared divergences with states degenerate in energy. The Block-Nordsieck theorem is a special case of the KLN theorem for QED.

Appendix A Dimensional regularization

The calculation of the total cross section for e+ e − µ+ µ− (+ γ) at next-to-leading order can → also be done in dimensional regularization. Repeating the calculation this way helps illustrate regulator independence of physical quantities and will give us some practice with dimensional regularization. Dimensional regularization 15

A.1 e +e − → µ + µ − The first step is to calculate the tree-level cross section in d = 4 ε dimensions. This is of course non-singular as ε 0, however, we will need the ( ε) parts of− the cross section for the virtual → O correction. We work in the limit E = Q me , m µ so that we can treat the fermions as mass- CM ≫ less. We first write an expression for a general e+ e − γ⋆ X process, then specialize to e+ e − µ+ µ− . → → → To calculate the cross section for e+ e − γ⋆ X, we use the observation from Section 2.2 that the cross section factorizes into e+ e − →γ⋆ and→ γ⋆ X. In d-dimensions we can still write → → 4 1 2 eR µν σR = dΠ = L Xµν (67) 2 Q 2 |M| LIPS 2 Q 6 Z with the electron tensor exactly as in Eq. (26): 1 Q 2 L µν = Tr[ p γµp γν] = pµpν + pνpµ gµν (68) 4 2 1 1 2 1 2 − 2 ⋆ The other tensor Xµν is the matrix element squared for a generic γ X final state averaged over γ⋆ spins integrated over the associated Lorentz-invariant phase space.→ This definition makes the total decay rate have the form 2 ⋆ eR µν Γ( γ X) = g Xµν (69) → − 2 Q ⋆ with the gµν coming from a polarization sum over the γ , assuming the Ward identity holds. Indeed, the− Ward identity does hold in d-dimensions, since dimensional regularization preserves gauge invariance, and so we can still write X µν = ( pµpν p2 gµν) X( p2 ) (70) − 2 1 µν However in in d-dimensions, X( Q ) = 2 g Xµν and − ( d − 1) Q 4 µν ( d 2) Q 2 1 d 2 2 µν L Xµν = − X( Q ) = − Q g Xµν (71) 2 − 2 d 1 And therefore − 4 + − eR 2(4 − d) d 2 µν σ( e e X) = µ − g Xµν (72) → − 4Q 4 d 1 4 − σ Q 2 e R Using 0( ) = 12 πQ 2 as before we can write this alternatively as

+ − 2(4 − d) 3π d 2 µν 2(4 − d) 6π d 2 ⋆ σ( e e X) = σ0 µ − ( g Xµν) = σ0 µ − Γ( γ X) (73) → Q 2 d 1 − Qe2 d 1 → − R − which reduces to Eq. (32) in 4 dimensions. ⋆ + − For the tree-level process, we need γ µ µ for which Xµν is just like L µν but with the phase space tacked on. Then, →

µν 2 µν µ ν ν µ 2 gµνX = gµν(2 Q g 4p p 4p p ) dΠ = 2( d 2) Q dΠ (74) − − 3 4 − 3 4 LIPS − LIPS Z Z Here dΠLIPS is the d 2 phase space. Since there is no angular dependence in the spin-summed γ⋆ µ+ µ− this phase− space is straighforward to evaluate → d − 1 d − 1 d d p3 d p4 1 d dΠLIPS = (2π) d − 1 d − 1 δ ( p3 + p4 p) (75) (2π) (2π) (2E3)(2E4) − Z Z p Q p x 2 E We ﬁrst rescale the momenta by i = 2 ˆ i to make them dimensionless. We also use i = Q i as the energy components of the rescaled momenta. Then, evaluating the p4 integral over the δ- function we get d − 2 d − 1 2 − d Q 1 d pˆ 3 d ΠLIPS =(2π) 2 δ( x3 + x4 2) 2 Q x3 x4 − Z Z 16 Appendix A

µ where x4 is an implicit function of pˆ 3 determined by spatial momentum conservation and the ˆ mass-shell conditions. Explicitly x4 = Qp3 = x3. So, d − 2 d − 2 Q 1 d x3 dΠ = δ(2x 2) dΩd − LIPS 4π Q 2 x 3 − 1 Z Z 3 Z Q d − 2 1 = Ω 4π 2 Q 2 d − 1 4 − d − d 4π 2 2 = 2 Q √πΓ d − 1 2 Combining this with Eqs. (73) and (74) 4 − d 2 d + − + − 2(4 − d) 4π 2 3√π( d 2) σ0 ( e e µ µ ) = σ0 µ 2 − (76) → Q d d + 1 2 Γ 2 which reduces to σ0 in d = 4.

A.2 Loops Next, we will compute the loop amplitude in pure dimensional regularization. The easiest way µ to do the calculation is by evaluating the form factor, which corrects the ieRγ vertex. Then we can use the result for the phase-space integral in d-dimensions we have− already calculated. To make sure we get all the factors of d correct, we will compute the loop from scratch. The loop gives

k + q2 4 − d µ ie µ 2 u q v q k R ¯( 2 )Γ2 ( 1 ) = p k q − − 1

q1 ¡

ν µ ν 4 − d 3 d u¯( q ) γ ( k + q ) γ ( k q ) γ v( q ) 2 d k 2 2 1 1 = eRµ − − (2π) d [( k + q ) 2 + iε][( k q ) 2 + iε][ k2 + iε] Z 2 − 1 2 We can simplify this using q v( q ) = u¯( q ) q = q2 = q2 = 0 and q q = Q . Using Feynman 1 1 2 2 1 2 1 · 2 2 parameters for the three denominator factors we get

1 1 − x d µ µ 2 4 − d d k u¯( q2 ) N ( k, q1 , q2 ) v( q1 ) u¯( q2 )Γ2 v( q1 ) = ieRµ dx dy d 2 2 3 (77) − (2π) (( k + xq2 yq1 ) + Q xy + iε) Z0 Z0 Z − with N µ =2[( d 2) k2 + 4k q 4k q 2 Q 2 ] γµ 4[( d 2) k µ + 2 q µ 2 q ] k (78) − · 2 − · 1 − − − 2 − 1 Shifting k µ k µ xq µ + yq µ and dropping terms linear in k turns the numerator into → − 2 1 N µ =2[( d 2) k2 + Q 2 ( (2 d) xy + 2 x +2 y 2)] γµ 4( d 2) k µk (79) − − − − − 2 2 k µkν k gµν B k µk γα gανk µkν k γµ Using d as discussed in Appendix , we can then replace = d giving → → ( d − 2) 2 ddk k2 + Q 2 ( (2 d) xy + 2 x +2 y 2) Γ µ = 2iγµe2 µ4 − d dxdy d − − (80) 2 − R (2π) d ( k2 + Q 2 xy + iε) 3 Z This has two terms: the k2 term is UV divergent, and the Q 2 term is IR divergent. The k2 term can be evaluated with d < 4 using ddk k2 d/4 1 d = i Γ 2 (81) d 2 3 d/2 d (2π) ( k ∆ + iε) (4π) 2 − 2 − 2 Z − ∆ Dimensional regularization 17 with ∆ = Q 2 xy to get − 2 2 d − 4 − d d 1 1 − x d ( 2) 2 4 − d Γ Γ d k k i 4π 2 2 2 dx dy d = (82) (2π) d ( k2 ( Q 2 xy) + iε) 3 16π2 Q 2 Γ( d 1) Z0 Z0 Z − − − − 4 − d i 4π 2 1 γ 1 = E + + ( ε ) (83) 16π2 Q 2 ε − 2 2 O UV − UV So the UV divergent part only has a single 1 pole, coming from the Γ 4 − d term. There is ε U V 2 no difference between εUV and ε. We write εUV only to remind us of the origin of the singularity and that it is finite for εUV > 0. In the Q 2 term in Eq. (80) integral is convergent in d = 4, but then the integrals over Feynman parameters would be divergent. Thus we must perform the k integral in d > 4 dimensions. In this case, we can use ddk 1 i 1 d d 2 3 = − d/2 d Γ 3 (84) (2π) ( k ∆ + iε) 2(4π) 3 − 2 − 2 Z − ∆ and then perform the x and y integrals to get 1 1 − x ddk Q 2 ( (2 d) xy + 2 x +2 y 2) dx dy − − (85) (2π) d ( k2 + Q 2 xy + iε) 3 Z0 Z0 4 − d d − 4 d 4 − d Γ Γ Γ 2 i 4π 2 2 2 2 d 8d+24 = − (86) 16π2 Q 2 Γ( d 2) 4( d 2) − − − 4 − d 2 2 i 4π 2 4 4 + 2 γE 54 + 24γE 6 γE + π = + − + − − + ( εIR) (87) 16π2 Q 2 − ε2 ε 12 O − IR IR 1 This term has a 2 pole, which is characteristic of infrared soft-collinear divergences. ε R Remember, εIR is the same as εUV but we must assume εIR < 0 ( d > 4) for this integral to be finite. Finally, we need the counterterm in the on-shell scheme (we need to use the on-shell scheme if we are to identify eR with the charge measured at Q = 0). The graph gives

4 − d 2 µ p = iδ eRµ u¯( q ) γ v( q ) (88) − 1 2 1

q1 ¡

We already computed this counteterm with a Pauli-Villars regulator and photon mass in Eq. 2 2 δ e R Λ (13), ﬁnding 1 = π 2 ln 2 , which is UV and IR divergent. The calculation in pure dimen- − 16 m γ sional regularization involves evaluating the loop at Q = 0. Taking Q 0 in Eq. (80) gives → 2 d 2 4 − d ( d 2) d k 1 δ1 = ieRµ − (89) d (2π) d k4 Z This integral is scaleless and formally vanishes in dimensional regularization. That is

δ1 = 0 (90) which is all we need to calculate the cross section. Nevertheless, as discussed in Appendix B, it can be revealing to formally separate the UV divergent region (which converges for d < 4) from the IR divergent region (which converges for d > 4) in a scaleless integral. In this case, we ﬁnd

2 2 4 − d ( d 2) i 1 1 2 4 − d 1 1 1 δ1 = ieRµ − = eRµ (91) d 8π2 ε − ε − 8π2 ε − ε UV IR UV IR 18 Appendix A

The εUV part of this cancels the divergent part of the integral for Q > 0 in Eq. (83) with the prefector from Eq. (80), as it must. Indeed, including the counterterm then makes all of the divergences formally IR divergences. Combining the UV and IR divergent pieces, the result

µ µ 2 Γ2 = γ f( Q ) (92) where 4 − d d 4 − 1 − d 2 d Γ Γ 2 2 2 2 µ 2 2 2 d 7 d+16 f( Q ) = eR4( 16π) − (93) Q 2 d − 1 d2 6 d+8 − Γ 2 −

4 − 2 − γE 2 d 2 e 4πe µ 2 1 3 π = R + + 1 + ( ε) (94) − 2π2 Q 2 ε2 4ε − 48 O −

4 − 3 iπ 2 − γE 2 d 2 e 4πe µ 2 1 + 7π 3πi = R + 4 2 + 1 + + ( ε) (95) − 2π2 Q 2 ε2 ε − 48 8 O ! where these are now all ε = εIR. The virtual contribution to the cross section is then

4 − 2 − γE 2 d 2 e 4πe µ 2 1 3 7π σd = 2σdRe [ f( Q 2 )] = σd R + + 1 (96) V 0 − 0 π2 Q 2 ε2 4ε − 48 e2 4πe − γE µ2 4 − d 1 13 5π2 29 = σ R + + + ( ε) (97) − 0 π2 Q 2 ε2 12ε − 24 18 O where Eq. (76) has been used.

A.3 Real emission contribution Next, we compute the real emission contribution. We can use the d-dimensional factorized form from Section A.1. In this case, we need γ⋆ µ+ µ− γ which comes from these diagrams: →

q2 p p i R = γ + γ (98) M p p

q1 q1

¡ ¡

The associated tensor is X µν = µ4 − d dΠ Tr[ q S µα q Sαν] (99) − LIPS 1 2 Z with S µα the same as in Eq. (24) and this dΠ being d 1 dimensional. As is the case with LIPS − the m γ regulator, it is easiest to express the result of this trace in terms of the xi variables. Here, x1 , x2 and x γ are deﬁned as they were in Eqs. (33) to (35) with β = 0, which is equivalent to 2 qi p xi = · (100) Q 2

p Q, , , x E In the center-of-mass frame, = ( 0 0) and so i = 2 Q with i the energy of the particle. These satisfy x1 + x2 + x γ = 2. We then ﬁnd the relevant spin-summed matrix element squared is

x2 + x2 + d − 4 x2 µν 4 − d µα αµ 2 4 − d 1 2 2 γ g Xµν = µ dΠ Tr[ p S p S ] = 4e ( d 2) µ dΠ (101) − LIPS 3 4 R − LIPS (1 x )(1 x ) Z Z − 1 − 2 This correctly reduces to Eq. (39) with β = 0 when d = 4. Dimensional regularization 19

Next we need to express the phase space in terms of x1 and x2 . We start with

d − 1 d − 1 d − 1 3 − 2 d d q1 d q2 d pγ d dΠLIPS = (2π) δ ( q1 + q2 + pγ p) (102) 2E1 2E2 2Eγ − Z Z Z Z Q Ei E γ q q x Qq x Let us ﬁrst rescale the momenta by i = 2 ˆ i and use i = 2 Q = ˆ i and γ = 2 Q . This gives 2 d − 3 Q 1 d − 2 d − 2 1 dΠLIPS = 3 x1 dx1 dΩd − 1 x2 dx2 dΩd − 1 δ( x1 + x2 + x γ 2) (103) 4π Q x x x γ −

Z Z Z 1 2 Q Now we have to be careful since x γ is an implicit function of the 3-momenta Qq1 and q2 :

2Eγ 2 2 2 2 2 2 2 Q x γ = = ( Qq + q ) = E + E 2E E cosθ = x + x 2x x cosθ (104) Q Q 1 2 Q 1 2 − 1 2 1 2 − 1 2

p p p Q where Qq q = E E cosθ. Since there is θ-dependence in the integrand, we cannot simply per- 1 · 2 − 1 2 form the δ-function integral. Instead we expand using the explicit form for dΩd from Appendix B: d − 4 d − 3 2 2 dΩ d − = dΩd − sin θdθ = dΩd − (1 z ) dz (105) 1 2 2 − where z cosθ is deﬁned for the last step. So ≡ 2 d − 3 1 d − 4 Q Ωd − 2 Ωd − 1 d − 3 d − 3 2 2 1 dΠLIPS = 3 dx1 x1 dx2 x2 dz(1 z ) δ( x1 + x2 + x γ 2) 4π Q − x γ − Z Z Z Z− 1 Now note that from Eq. (104), x2 + x2 x2 z = 1 2 − γ (106) 2x1 x2 also using x1 + x2 + x γ = 2,

2 (1 x1 )(1 x2 )(1 x γ) 1 z = 4 − −2 2 − (107) − x1 x2 Thus

2 2 Q d − 4 Q 4 − d 4π 1 2 d ΠLIPS = 3 dx1 dx2 dx γ δ( x1 + x2 + x γ 2) 128π Γ( d 2) (1 x )(1 x )(1 x γ) − Z − Z − 1 − 2 − 2 d − 4 2 1 1 4 − d Q Q 1 2 = 3 dx1 dx2 (108) 4π 128π Γ( d 2) (1 x1 )(1 x2 )(1 x γ) − Z0 Z1 − x 1 − − − with x γ = 2 x1 x2 . This is our ﬁnal result for the 3-body phase space. Now it is− just− a matter of integrating Eq. (101) with (108). The result is

2 2 d − 4 2 d − 4 2 d 1 1 4( d 2) x1 + x2 + x γ Γ Γ − 2 2 2 2 dx1 dx2 = 4( d 3)( d 4d + 8) − d − d − d 3d − 6 0 1 − x 1 3 2 3 2 2 2 − − (1 x ) (1 x ) (1 x γ) Γ Z Z − 1 − 2 − 2 64 16 = + 8π2 + 52 + ( ε) (109) ε2 ε − O Combining this with Eqs. (73), (101) and (108) and factoring out the tree-level cross section gives d − 4 2 d 2 d − 4 2 Γ Γ d 2 Q 3 ( d 3)( d 2)( d 4d + 8) 2 2 σR = σ0 eR 2 2 − − − (110) 4πµ 32π d 1 Γ 3d − 6 Γ( d 2) − 2 − e2 4πe − γE µ2 4 − d 1 13 5π2 259 =σ R + + + ( ε) (111) 0 π2 Q 2 ε2 12ε − 24 144 O Finally, adding in Eq. (97):

e2 4πe − γE µ2 4 − d 1 13 5π2 29 σd = σ R + + ( ε) (112) V 0 π2 Q 2 − ε2 − 12ε 24 − 18 O 20 Appendix A gives a total cross section of 3e2 σd + σd = σ R + ( ε) (113) R V 0 16π2 O which is ﬁnite as ε 0 and exactly the result we found with Pauli-Villars and a photon mass, Eq. (45). →