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 cal- culation 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 natu- rally 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 pro- cesses 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 cancel- lation 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 p3 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 .