Renormalization of Φ3 Scalar Field Theory (December 2016) 1 TODO 2 Lagrangian and Green Functions in D Dimensions

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Renormalization of φ3 scalar ﬁeld theory (December 2016)

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Examine iε in the two-point function (cf Sterman).

2 Lagrangian and Green functions in d dimensions

In these notes, we’ll use η00 = 1.

2.1 Lagrangian, counterterms, and Feynman rules Consider the scalar lagrangian with cubic interaction

2 6−d 1 1 2 2 1 2 3 L = 2 (∂φ) − 2 m φ − 6 gµ φ (2.1) where φ is the renormalized (finite) field, and m and g renormalized (finite) parameters, but not directly identifiable as physical parameters. It also depends on d to regularize the loop amplitudes, and a scale µ to make the couplings dimensionless in d dimensions. The counterterm lagrangian 2 6−d 1 1 2 2 1 2 3 Lc.t. = 2 δZ (∂φ) − 2 δm φ − 6 (δg)µ φ − (δτ)φ (2.2) contains poles in and also depends on arbitrarily chosen finite constants ci. Adding the counterterm Lagrangian to the original lagrangian, we obtain the full Lagrangian

2 6−d 1 1 2 2 2 1 2 3 L0 = L + Lc.t. = 2 (1 + δZ)(∂φ) − 2 (m + δm )φ − 6 (g + δg)µ φ − (δτ)φ (2.3)

Renormalized amplitudes are ﬁnite as → 0, but depend on m and g, as well as µ and ci. The choice of ci is determined by the renormalization scheme. If we choose an on-shell scheme, m and g will be directly related to physical properties. The ci will involve log µ, and the amplitudes will be independent of µ. Alternatively, in an MS or MS scheme, the ci only cancel poles and do not depend on µ. Physical couplings are functions of m, g, and µ. Ignoring the tadpole counterterm for now, we can deﬁne

6−d 1/2 2 −1 2 2 2 −3/2 φ0 = (1 + δZ) φ, m0 = (1 + δZ) (m + δm ), g0 = µ (1 + δZ) (g + δg) (2.4) and then rewrite 1 2 1 2 2 1 3 L0 = 2 (∂φ0) − 2 m0φ0 − 6 g0φ0 (2.5) showing that physical quantities depend only on two parameters m0 and g0. This then can be used to derive renormalization group equations.

1 The propagator and three-point vertex are given by

i 6−d , −igµ 2 (2.6) p2 − m2 + iε with counterterms 2 2 6−d −iδm + iδZ p , −iδg µ 2 (2.7)

2.2 Two-point function We now construct the two-point Green function i i i hT (φφ)i = + −iΣ − iδm2 + iδZ p2 + ··· p2 − m2 p2 − m2 p2 − m2 i Σ + δm2 − δZ p2 = 1 + + ··· p2 − m2 p2 − m2 i Σ + δm2 − δZ p2 −1 = 1 − p2 − m2 p2 − m2 i = p2 − m2 − Σ − δm2 + δZ p2 i = 2 (2.8) Γ2(p ) 2 where −iΣ is the self energy diagram, and Γ2(p ) the 1PI two-point function. Expanding 2 2 0 2 2 2 2 2 2 Σ(p ) = Σ(mp) + Σ (mp)(p − mp) + O((p − mp) ) (2.9) about the physical mass mp, we find that the denominator of the two-point Green function is 2 2 2 2 2 0 2 2 2 2 2 2 (mp − m − Σ(mp) − δm + δZ mp) + (1 − Σ (mp) + δZ)(p − mp) + O((p − mp) ) (2.10) The physical mass is determined by the solutions of 2 2 2 2 2 mp − m − Σ(mp) − δm + δZ mp = 0 (2.11) and the two-point function at the pole is given by iR hT (φφ)i = 2 2 (2.12) p − mp where the (finite) field renormalization is given by 1 R = 0 2 (2.13) 1 − Σ (mp) + δZ 2 2 Now make the one-loop approximation, setting mp = m in the terms that are already first order, to find 2 2 2 2 2 mp = m + Σ(m ) + δm − δZ m (2.14)

R = 1 + Σ0(m2) − δZ (2.15)

2 2.3 Self energy The one-loop self energy is

2 Z d 2 2 6−d 1 d ` i i −iΣ(p , m ) = −igµ 2 (2.16) 2 (2π)d `2 − m2 + iε (` + p)2 − m2 + iε evidently UV divergent for d ≥ 4 and IR divergent (for m = 0) for d ≤ 2.

ig2µ6−d Σ(p2, m2) = I 2 2 4−d 1 g2µ6−d Γ( ) Z d−4 2 2 2 2 = − d/2 dx m − x(1 − x)p − iε 2 (4π) 0 d−4 2 6−d d−4 6−d Z 1 2 2 g µ m Γ( 2 ) p = d/2 dx 1 − x(1 − x) 2 (d − 4) (4π) 0 m d−4 6−d d−4 2 6−d 2 2 Z 1 2 2 g µ (−p ) Γ( 2 ) m = d/2 dx x(1 − x) − 2 (2.17) (d − 4)(4π) 0 p with the expected UV poles in d = 4, 6, ··· .

1 d−4 d−2 R 2 2 In the massless limit we have 0 dx[x(1 − x)] = Γ( 2 ) /Γ(d − 2) so

2 6−d 2 d−4 6−d d−2 2 g µ (−p ) 2 Γ( )Γ( ) Σ(p2, 0) = 2 2 (2.18) (d − 4) (4π)d/2Γ(d − 2) with the IR pole at d = 2. We also obtain

d−6 2 6−d d−6 6−d Z 1 2 2 ∂Σ 2 2 g µ m Γ( 2 ) p 2 (p , m ) = − d/2 dx x(1 − x) 1 − x(1 − x) 2 ∂p 2(4π) 0 m d−6 6−d d−6 2 6−d 2 2 Z 1 2 2 g µ (−p ) Γ( 2 ) m = − d/2 dx x(1 − x) x(1 − x) − 2 (2.19) 2(4π) 0 p In the massless limit we have

2 6−d 2 d−6 6−d d−2 2 ∂Σ g µ (−p ) 2 Γ( )Γ( ) (p2, 0) = − 2 2 (2.20) ∂p2 2(4π)d/2Γ(d − 2)

3 2.4 Three-point function The three-point function is

6−d iΓ3(p1, p2) = − i(g + δg)µ 2 d 6−d 3 Z d ` i i i 2 + −igµ d 2 2 2 2 2 2 (2π) (` + p1) − m + iε (` + p1 + p2) − m + iε ` − m + iε 6−d 6−d 3 = − i(g + δg)µ 2 + gµ 2 I3(p1, p2, −p1 − p2) (2.21)

This has a UV divergence for d ≥ 6. From the loop integral notes, we have

−i Z 1 dβ dβ dβ δ(1 − P β ) I (p , p , p ) = Γ( 6−d ) 1 2 3 i i (2.22) 3 1 2 3 d/2 2 h i3−d/2 (4π) 0 2 2 2 2 m − β1β2p2 − β2β3(p1 + p2) − β3β1p1 so   3 6−d 6−d Z 1 P 6−d g µ Γ( 2 ) dβ1dβ2dβ3δ(1 − i βi) Γ (p , p ) = −µ 2 g + δg +  3 1 2  d/2 h i3−d/2  (4π) 0 2 2 2 2 m − β1β2p2 − β2β3(p1 + p2) − β3β1p1 (2.23) with the expected UV poles at d = 6, 8, ··· .

2 2 In the massless m = 0 limit, and with two legs onshell p1 = p2 = 0, we have

3 6−d 1 1−β2 6−d g µ d−6 Z d−6 Z d−6 2 2 2 6−d 2 2 Γ3(p1, p2) = −µ g + δg + d/2 −(p1 + p2) Γ( 2 ) dβ2 β2 dβ3 β3 (4π) 0 0 3 6−d 6−d 1 ! 6−d g µ d−6 2 Γ( ) Z d−6 d−4 2 2 2 2 2 2 = −µ g + δg + d/2 −(p1 + p2) dβ2 β2 (1 − β2) (4π) d − 4 0 6−d ! 3 2 2 6−d d−4 d−2 6−d g µ 2Γ( )Γ( )Γ( ) 2 2 2 2 = −µ g + δg + d/2 2 (2.24) (4π) −(p1 + p2) (d − 4)Γ(d − 3)

0 02 0 0 For ` = ` + p1 → 0, the denominator goes as ` (2` · p2)(−2` · p1), giving an IR divergence in d = 4. In fact, looks like a double pole, suggesting also a collinear divergence?

2.5 Four-point function The 1PI four-point function has contributions from three topologically-distinct box functions

iΓ4 = i [B(s, t) + B(t, u) + B(u, s)] (2.25)

4 where

d 6−d 4 Z d ` i i 2 iB(s, t) = −igµ d 2 2 2 2 (2π) ` − m + iε (` + p1) − m + iε i i × 2 2 2 2 (` + p1 + p2) − m + iε (` + p1 + p2 + p3) − m + iε 6−d 4 = gµ 2 I4(s, t) (2.26) where 4−d 8−d ∞ 4 P P Z δ(1 − jcjβj) jβj iΓ( 2 ) Y I4(s, t) = d/2 dβj 4−d/2 (2.27) (4π) 0 2 j=1 2 P P 2 m jβj − i

We see the expected UV divergence for d ≥ 8. Choosing ci = 1, and letting external legs be 2 2 2 2 2 on-shell (P12 = P23 = P34 = P41 = m )

8−d Z ∞ 4 δ(1 − P β ) iΓ( 2 ) Y j j I4(s, t) = dβj (2.28) (4π)d/2 2 4−d/2 0 j=1 [m (1 − β1β2 − β2β3 − β3β4 − β4β1) − β1β3t − β2β4s]

In the massless limit

8−d Z ∞ 4 P iΓ( ) Y δ(1 − jβj) I (s, t) = 2 dβ (2.29) 4 (4π)d/2 j 4−d/2 0 j=1 [−β1β3t − β2β4s]

2 If the external legs are on-shell, then in the ` → 0 limit, the denominator goes to ` (` · p1)(` · p4) and thus has an IR divergence in d = 4.

2.6 Four-point scattering amplitude First we compute the sum of all one-loop truncated diagrams i i i iM = iΓ4 + (iΓ3(s)) (iΓ3(s)) + (iΓ3(t)) (iΓ3(t)) + (iΓ3(u)) (iΓ3(u)) Γ2(s) Γ2(t) Γ2(u) 2 2 2 Γ3(s) Γ3(t) Γ3(u) = i Γ4 − − − (2.30) Γ2(s) Γ2(t) Γ2(u) The one-loop external leg corrections yield

4 ! 2 2 2 Y iR(m) Γ3(s) Γ3(t) Γ3(u) i Γ − − − (2.31) p2 − m2 4 Γ (s) Γ (t) Γ (u) i=1 i 2 2 2

5 LSZ tells us to multiply this by 4 2 2 Y pi − m p (2.32) i=1 R(m) to obtain the scattering amplitude

2 2 2 2 Γ3(s) Γ3(t) Γ3(u) iM = iR(m) Γ4 − − − (2.33) Γ2(s) Γ2(t) Γ2(u)

6 3 Renormalization in six dimensions

For convenience, deﬁne p2 F ≡ 1 − x(1 − x) (3.1) m2 3.1 Self energy in six dimensions Evaluate eq. (2.17) near six dimensions d = 6 − 2, expanding in to write

2 2 Z 1 2 2 g Γ(1 + ) 4πµ 2 1− Σ(p , m ) = 3 2 dx m F 2(4π) (1 − ) m 0 g2 1 4πµ2 Z 1 m2 = 3 − γ + 1 + log 2 dx F (1 − log F ) + O() (3.2) (4π) m 0 2 Thus using Z 1 m2 Z 1 1 m2 p2 dx F = dx m2 − x(1 − x)p2 = − (3.3) 0 2 0 2 2 12 we obtain

2 2 2 2 2 Z 1 2 2 g m p 1 4πµ m Σ(p , m ) = 3 − − γ + 1 + log 2 − dx F log F + O() (4π) 2 12 m 2 0 (3.4) In the massless limit

2 Z 1 2 Z 1 2 2 2 m p −p 2 p −p 5 2 − dx F log F → dx x(1 − x) log 2 + log(x − x ) = log 2 − p 2 0 2 0 m 12 m 36 (3.5) so g2 p2 1 8 4πµ2 Σ(p2, 0) = − − γ + + log (3.6) (4π)3 12 3 −p2 Also evaluating eq. (2.19) near d = 6 − 2

2 2 Z 1 ∂Σ 2 2 g Γ(1 + ) 4πµ − 2 (p , m ) = − 3 2 dx x(1 − x)F ∂p 2(4π) m 0 g2 1 4πµ2 Z 1 = − 3 − γ + log 2 dx x(1 − x)(1 − log F ) + O() 2(4π) m 0 g2 1 1 4πµ2 1 Z 1 = 3 − − γ + log 2 + dx x(1 − x) log F + O() (3.7) (4π) 12 m 2 0 which we could also have obtained by taking the derivative of eq. (3.4) above.

7 The massless limit can alternatively be obtained by evaluating eqs. (2.18) and (2.20) near six dimensions g2µ2(−p2)1− Γ()Γ(2 − )2 g2 p2 1 8 −p2 Σ(p2, 0) = = − − γ + − log 2(1 − ) (4π)3−Γ(4 − 2) (4π)3 12 3 4πµ2 (3.8) ∂Σ g2µ2(−p2)−Γ()Γ(2 − )2 g2 1 1 5 −p2 (p2, 0) = − = − − γ + − log (3.9) ∂p2 2(4π)3−Γ(4 − 2) (4π)3 12 3 4πµ2

3.2 Six dimensional wavefunction and mass counterterms To render the field renormalization R = 1 + Σ0(m2) − δZ finite, we choose the counterterm g2 1 δZ = − + c (3.10) 12(4π)3 φ in order to absorb the UV pole in Σ0(m2) (cf. eq. (3.7)). We evaluate √ Z 1 Z 1 2 3π 3 − 17 dx x(1 − x) log F0 = dx x(1 − x) log(1 − x + x ) = ≈ −0.0375448 (3.11) 0 0 18 where 2 F0 ≡ F |p2=m2 = 1 − x + x (3.12) to obtain " √ # ∂Σ g2 1 1 4πµ2 3π 3 − 17 (m2, m2) = − − γ + log + (3.13) ∂p2 (4π)3 12 m2 36 and thus " √ # g2 1 4πµ2 3π 3 − 17 R = 1 + c + γ − log + (3.14) (4π)3 12 φ m2 36 √ (note that 3π 3 ≈ 16.32). R is now UV finite, but still has an IR divergence when m → 0. This means we will not be able to do on-shell renormalization for the massless theory.

2 2 2 Equation (2.14) implies the mass counterterm must satisfy δm |1/ = −Σ(m )|1/ + m δZ|1/ so g2m2 1 δm2 = − + c (3.15) 2(4π)3 m Then using eq. (3.11) and √ Z 1 Z 1 π 3 dx log F = log(1 − x + x2)dx = − 2 0 3 0 0 √ Z 1 Z 1 2 2 3π 3 − 19 =⇒ dx F0 log F0 = (1 − x + x ) log(1 − x + x )dx = ≈ −0.148656 (3.16) 0 0 18

8 we evaluate " √ # g2 5 1 4πµ2 −3π 3 + 34 Σ(m2, m2) = m2 − γ + log + (3.17) (4π)3 12 m2 36

2 2 2 2 2 From this we can evaluate the physical mass mp = m + Σ(m ) + δm − δZ m namely

( " √ #) g2 1 1 5 4πµ2 −3π 3 + 34 m2 = m2 1 + c − c + −γ + log + (3.18) p (4π)3 12 φ 2 m 12 m2 36

3.3 Renormalized two-point function in six dimensions Finally, we evaluate the renormalized two-point function in six dimensions i i hT (φφ)i = 2 2 2 2 = 2 (3.19) p − m − Σ − δm + δZ p Γ2(p ) where

2 2 Z 1 2 2 2 2 2 g m 4πµ m Γ2(p ) = p − m + 3 dx F log F + cm + γ − 1 − log 2 (4π) 2 0 m 2 4πµ2 p2 − c + γ − 1 − log , φ m2 12 ∂Σ g2 1 Z 1 1 4πµ2 2 − δZ = 3 dx x(1 − x) log F + cφ + γ − log 2 (3.20) ∂p (4π) 2 0 12 m Using eq. (3.11) and √ √ 2 Z 1 ! ! m 2 π 3 2 3π 3 − 17 dx F log F0 = m − 1 − p (3.21) 2 0 6 36 we can also write these expressions as

g2 m2 Z 1 F π 3 4πµ2 m2 2 2 2 √ Γ2(p ) = p − m + 3 dx F log + cm + γ + − − log 2 (4π) 2 0 F0 3 2 m 2 √ 20 4πµ2 p2 − c + γ + π 3 − − log , φ 3 m2 12 ∂Σ g2 1 Z 1 F 1 √ 17 4πµ2 2 − δZ = 3 dx x(1 − x) log + cφ + γ + π 3 − − log 2 ∂p (4π) 2 0 F0 12 3 m (3.22)

9 In the massless (m → 0) limit m2 Z 1 F p2 Z 1 p2 x − x2 dx F log → − dx x(1 − x) log − 2 + log 2 2 0 F0 2 0 m 1 − x + x p2 p2 √ = − log − + 4 − π 3 (3.23) 12 m2 so that g2 p2 8 Γ (p2) → p2 1 − log − + c + γ − (3.24) 2 12(4π)3 4πµ2 φ 3 With on shell renormalization (3.28) this agrees with Srednicki eq. 20.12). Alternatively, using eqs. (3.8) and (3.9) in the massless limit, with δZ given by eq. (3.10) and δm2 absent g2 −p2 8 Γ (p2) = p2 1 − log + c + γ − 2 12(4π)3 4πµ2 φ 3 ∂Σ g2 −p2 5 − δZ = log + c + γ − (3.25) ∂p2 12(4π)3 4πµ2 φ 3

In the MS scheme, we have cm = −γ + log 4π and cφ = −γ + log 4π so (cf Srednicki eq. 27.4) 2 2 Z 1 2 2 2 2 2 2 g m µ m p Γ2(p ) = p − m + 3 dx F log F + −1 − log 2 − (4π) 2 0 m 2 12 ∂Σ g2 1 Z 1 1 µ2 2 − δZ = 3 dx x(1 − x) log F − log 2 (3.26) ∂p (4π) 2 0 12 m For the massless theory, MS gives g2 −p2 8 Γ (p2) = p2 1 − log − 2 12(4π)3 µ2 3 ∂Σ g2 −p2 5 − δZ = log − (3.27) ∂p2 12(4π)3 µ2 3

On-shell renormalization, only available for m 6= 0, implies R = 1 √ 4πµ2 17 − 3π 3 c + γ − log = (3.28) φ m2 3 Thus " √ # ∂Σ g2 1 Z 1 17 − 3π 3 2 − δZ = 3 dx x(1 − x) log F + ∂p (4π) 2 0 36 g2 1 Z 1 F = 3 dx x(1 − x) log (3.29) (4π) 2 0 F0

10 2 2 2 2 which vanishes at p = m . On-shell renormalization also implies mp = m so √ 1 4πµ2 1 4πµ2 3π 3 − 34 c + γ − log − c + γ − log = 12 φ m2 2 m m2 36 √ 4πµ2 51 − 6π 3 c + γ − log = (3.30) m m2 18 so that the inverse propagator with on-shell renormalization is √ √ 2 " 2 Z 1 ! ! # 2 2 2 g m 11 − 2π 3 2 3π 3 − 14 2 Γ2(p ) = p − m + 3 dx F log F + m + p (4π) 2 0 12 36 2 2 Z 1 2 2 g m F 1 2 2 = p − m + 3 dx F log + (p − m ) (3.31) (4π) 2 0 F0 12 (agreeing with Srednicki eq. 14.43) which manifestly vanishes at p2 = m2. According to Srednicki, p. 104, s Z 1 2 2 2 4 5 p p 3 −1 1 4m dx F log F = − + 2 − 2 S tanh ,S = 1 − 2 (3.32) 0 3 18 m 3m S p which I veriﬁed numerically. Mathematica gives √ 3/2 −1 y 1 √ 2 Z 5y − 24 (4 − y) tan 4−y p dx F log F = + √ y = 2 (3.33) 0 18 3 y m which agrees numerically with Srednicki. Hence " √ ! √ ! # g2 3 − 2π 3 3π 3 − 9 p2 1 Γ (p2) = p2 − m2 + m2 + p2 − S3 tanh−1 (3.34) 2 (4π)3 12 36 6 S

3.4 Three-point function in six dimensions Evaluating the three-point function in d = 6 − 2 , we get   g3µ2Γ() Z 1 dβ dβ dβ δ(1 − P β ) Γ (p , p ) = −µ g + δg + 1 2 3 i i 3 1 2  3− h i  (4π) 0 2 2 2 2 m − β1β2p2 − β2β3(p1 + p2) − β3β1p1 g3 1 = −µ g + δg + + O(0) (3.35) 2(4π)3 Choose the counterterm g3 1 δg = − + c (3.36) 2(4π)3 g

11 to cancel the UV divergence

g2 Γ (p , p ) = −µg 1 + −c + O(0) (3.37) 3 1 2 2(4π)3 g

Instead evaluate the massless three-point function (2.24) with two on-shell legs in six dimensions

g3 1 (p + p )2 Γ (p , p ) = −µ g + δg + − log − 1 2 − γ + 3 3 1 2 2(4π)3 4πµ2 g2 (p + p )2 = −µg 1 + −c − log − 1 2 − γ + 3 (3.38) 2(4π)3 g 4πµ2

MS renormalization with cg = −γ + log 4π gives

g2 (p + p )2 Γ (p , p ) = −µg 1 + − log − 1 2 + 3 (3.39) 3 1 2 2(4π)3 µ2

3.5 Four-point function in six dimensions Recall that Γ4 = B(s, t) + B(t, u) + B(u, s) (3.40) where 4 iB(s, t) = g I4(s, t) (3.41) and

Z ∞ 4 P i Y δ(1 − jβj) I (s, t) = dβ (3.42) 4 (4π)3 j [m2(1 − β β − β β − β β − β β ) − β β t − β β s] 0 j=1 1 2 2 3 3 4 4 1 1 3 2 4

In the massless limit, this is according to Srednicki 20.17 given by

4 Z ∞ 4 P 4 g Y δ(1 − jβj) g −1 hsi B(s, t) → dβ = π2 + log2 (3.43) (4π)3 j [−β β t − β β s] (4π)3 2(s + t) t 0 j=1 1 3 2 4

I checked this result numerically.

3.6 Four-point massless scattering amplitude in six dimensions The scattering amplitude is

2 2 2 2 Γ3(s) Γ3(t) Γ3(u) iM = iR(m) Γ4 − − − (3.44) Γ2(s) Γ2(t) Γ2(u)

12 where " √ # g2 1 4πµ2 3π 3 − 17 g2 1 m2 R(m) = 1+ c + γ − log + = 1+ log + const (4π)3 12 φ m2 36 (4π)3 12 µ2 (3.45) R(m) has an IR divergence as m → 0 but the expression in brackets in eq. (3.44) does not. Thus we can evaluate it in the massless limit in MS renormalization 2 g2 h s i 2 2 2 2 Γ (s) g 1 − 2(4π)3 log − µ2 − 3 g g 11 s 25 3 → = 1 − log − − (3.46) 2 h i 3 2 Γ2(s) s g s 8 s (4π) 12 µ 9 1 − 12(4π)3 log − µ2 − 3 and g4 1 π2 1 hsi B(s, t) → + log2 (3.47) (4π)3 u 2 2 t Thus the scattering amplitude is g2 g2 11 s 1 t π2 25 iM = iR(m)2 − 1 − log − + log2 + − s (4π)3 12 µ2 2 u 2 9 g2 g2 11 t 1 u π2 25 − 1 − log − + log2 + − t (4π)3 12 µ2 2 s 2 9 g2 g2 11 u 1 s π2 25 − 1 − log − + log2 + − (3.48) u (4π)3 12 µ2 2 t 2 9 This is consistent with Srednicki 20.19. Finally, with s ≈ t ≈ u, and neglecting constants, this together with eq. (3.45) gives

g2 g2 g2 g2 11 s 1 m2 iM = i − − − 1 + − log − + log (3.49) s t u (4π)3 12 µ2 6 µ2

3.7 Infrared divergences from collinear particles Srednicki very clearly explains how the emission of collinear scalar particles gives rise to IR divergences such that (eq. 26.14)

g2 δ2s |M|2 = |M|2 1 + log (3.50) obs 3(4π)3 m2 Combining this with the result above

2 2 2 2 2 2 2 2 g g g g 11 s 1 m 1 δ s |M| = − − − 1 + − log − + log + log obs s t u (4π)3 6 µ2 3 µ2 3 m2 (3.51)

13 we see that the log m2 dependences cancels giving

2 2 2 2 2 2 g g g g 3 s 1 2 |M| = − − − 1 + − log − + log δ (3.52) obs s t u (4π)3 2 µ2 3 which is ﬁnite in the m → 0 limit, but depends on the resolution of the detector. This equation can be used to derive the one-loop beta function.

14 4 Renormalization in four dimensions

4.1 Four dimensional self energy As before, deﬁne p2 F ≡ 1 − x(1 − x) (4.1) m2 Evaluate eq. (2.17) near four dimensions d = 4 − 2, expanding in to write

2 2 2 Z 1 2 2 g µ Γ(1 + ) 4πµ − Σ(p , m ) = − 2 2 dx F 2(4π) m 0 g2µ2 1 4πµ2 Z 1 = − 2 − γ + log 2 dx (1 − log F ) + O() (4.2) 2(4π) m 0 Thus 2 2 2 Z 1 2 2 g µ 1 4πµ Σ(p , m ) = − 2 − γ + log 2 − dx log F + O() (4.3) 2(4π) m 0 If p2 < 0, we can recast s −p2 m2 −p2 1 4m2 F = x(1 − x) − = S2 − y2 , y = x− 1 ,S = 1 − > 1 (4.4) m2 p2 m2 4 2 p2 then Z 1 2 Z 1/2 −p 1 1 dx log F = log 2 + dy log( 2 S + y) + log( 2 S − y) 0 m −1/2 2 y=1/2 −p 1 1 1 1 = log 2 + ( 2 S + y) log( 2 S + y) − ( 2 S − y) log( 2 S − y) − 2y m y=−1/2 2 2 y=1/2 −p S S + 2y S 2 = log 2 + log + y log − y − 2y m 2 S − 2y 4 y=−1/2 −p2 S + 1 S2 − 1 = log + S log + log − 2 m2 S − 1 4 S + 1 = S log − 2 (4.5) S − 1

Thus for p2 < 0,

q 4m2  1 s ( 2 2 2 Z 4m2 1 − p2 + 1 log(−p /m ) − 2, p → −∞ dx log F = 1 − log − 2 −→ p2 < 0 2 q 2  2 2 2 0 p 4m −p /6m , p → 0− 1 − p2 − 1 (4.6)

15 If 0 < p2 < 4m2, we can analytically continue s s 1 −p2 π p2 = = eiαT, α = ± ,T = , 0 < T < ∞ (4.7) S 4m2 − p2 2 4m2 − p2 implying that (independently of α)

Z 1 1 −iα iα 1 + S e 1 + e T 2 dx log F = S log 1 − 2 = log iα − 2 = arctan T − 2 (4.8) 0 1 − S T 1 − e T T Thus for 0 < p2 < 4m2

s r ( Z 1 4m2 − p2 p2 −p2/6m2, p2 → 0 dx log F = 2 arcsin − 2 −→ + (4.9) p2 4m2 π p 2 2 2 2 0 −2 + 2m 4m − p , p → 4m−

In the massless m → 0 limit

Z 1 Z 1 2 2 −p 2 −p dx log F → dx log 2 + log(x − x ) = log 2 − 2 (4.10) 0 0 m m so g2µ2 1 4πµ2 Σ(p2, 0) = − − γ + 2 + log + O() (4.11) 2(4π)2 −p2 Equivalently we could evaluate eq. (2.18) at d = 4 − 2

g2µ2 4πµ2 Γ(1 + )Γ(1 − )2 Σ(p2, 0) = − (4.12) 2(4π)2 −p2 Γ(2 − 2) and expand to obtain the same result. Also evaluating eq. (2.19) near d = 4 − 2

2 2 2 Z 1 ∂Σ 2 2 g µ 4πµ −1− 2 (p , m ) = − 2 2 Γ(1 + ) 2 dx x(1 − x)F ∂p 2(4π) m m 0 g2 µ2 Z 1 x(1 − x) = − 2 2 dx (4.13) 2(4π) m 0 F which is the derivative of the expression above. Taking the massless limit we get

∂Σ g2 µ2 (p2, 0) = (4.14) ∂p2 2(4π)2 p2

16 4.2 Four dimensional wavefunction and mass counterterms Since Σ0(p2) is UV ﬁnite, so is the counterterm

g2 µ2 c δZ = φ (4.15) (4π)2 m2 2

(For m = 0, we must choose cφ = 0.) Evaluating

Z x(1 − x) Z x(1 − x) 2π √ dx = dx 2 = − 1 ≈ 0.2092 (4.16) F0 1 − x + x 3 3 we obtain ∂Σ g2 µ2 1 π (m2, m2) = − √ (4.17) ∂p2 (4π)2 m2 2 3 3 so the ﬁeld strength renormalization R = 1 + Σ0(m2) − δZ is

g2 µ2 1 π c R = 1 + − √ − φ (4.18) (4π)2 m2 2 3 3 2

2 2 2 Equation (2.14) implies the mass counterterm must satisfy δm |1/ = −Σ(m )|1/ + m δZ|1/ so

g2µ2 1 δm2 = + c (4.19) 2(4π)2 m

Next we evaluate Z 1 Z 1 2 π dx log F0 = dx log(1 − x + x ) = √ − 2 ≈ −0.186201 (4.20) 0 0 3 to obtain g2µ2 1 4πµ2 π Σ(m2, m2) = − − γ + log + 2 − √ + O() (4.21) 2(4π)2 m2 3

2 2 2 2 2 Now we can evaluate the physical mass mp = m + Σ(m ) + δm − δZ m to be

2 2 2 2 2 g µ 4πµ π m = m + −cφ + cm + γ − log + √ − 2 (4.22) p 2(4π)2 m2 3

17 4.3 Renormalized two-point function in four dimensions Finally, we evaluate the renormalized two-point function i i hT (φφ)i = 2 2 2 2 = 2 (4.23) p − m − Σ − δm + δZ p Γ2(p ) where

2 2 2 2 Z 1 2 2 2 g µ 4πµ p Γ2(p ) = p − m + 2 −cm − γ + log 2 + cφ 2 − dx log F + O() (4.24) 2(4π) m m 0

Also ∂Σ g2 µ2 Z 1 x(1 − x) 2 − δZ = 2 2 −cφ − dx (4.25) ∂p 2(4π) m 0 F R 1 2 In the massless limit, with cφ = 0, we have (using 0 log(x − x )dx = −2) g2µ2 4πµ2 Γ (p2) = p2 + −c − γ + 2 + log + O() (4.26) 2 2(4π)2 m −p2

On-shell renormalization implies R = 1 so 2π cφ = 1 − √ (4.27) 3 3 Thus ∂Σ g2 µ2 2π Z 1 x(1 − x) √ 2 − δZ = 2 2 −1 + − dx (4.28) ∂p 2(4π) m 3 3 0 F which vanishes at p2 = m2.

2 2 On-shell renormalization also implies mp = m so

4πµ2 π 5π cm + γ − log = cφ − √ + 2 = 3 − √ (4.29) m2 3 3 3 so that

g2µ2 5π 2π p2 Z 1 2 2 2 √ √ Γ2(p ) = p − m + 2 − 3 + 1 − 2 − dx log F + O() (4.30) 2(4π) 3 3 3 3 m 0 which vanishes at p2 = m2.

18 4.4 Three-point function in four dimensions The three-point function in d = 4 − 2 is   3 6−d 6−d Z 1 P 6−d g µ Γ( 2 ) dβ1dβ2dβ3δ(1 − i βi) Γ (p , p ) = −µ 2 g + δg +  3 1 2  d/2 h i3−d/2  (4π) 0 2 2 2 2 m − β1β2p2 − β2β3(p1 + p2) − β3β1p1   g3µ2+2 Z 1 dβ dβ dβ δ(1 − P β ) = −µ1+ g + δg + Γ(1 + ) 1 2 3 i i   2− h i1+  (4π) 0 2 2 2 2 m − β1β2p2 − β2β3(p1 + p2) − β3β1p1 (4.31)

This is ﬁnite for m 6= 0 so we can set = 0:   g3µ2 Z 1 dβ dβ dβ δ(1 − P β ) Γ (p , p ) = −µ g + δg + 1 2 3 i i (4.32) 3 1 2  2 h i (4π) 0 2 2 2 2 m − β1β2p2 − β2β3(p1 + p2) − β3β1p1

2 2 In the massless limit, and with p1 = p2 = 0, we have ! 3 2 1+ Z 1 Z 1−β2 1+ g µ −1− −1− Γ3(p1, p2) = −µ g + δg + 2− 2 Γ(1 + ) dβ2 β2 dβ3 β3 (4π) −(p1 + p2) 0 0 3 2 1+ 2 2 ! 1+ g µ 1 γ γ π = −µ g + δg + 2− 2 2 − + − (4.33) (4π) −(p1 + p2) 2 12

This is also the result obtained from eq. (2.24).

4.5 Four-point function in four dimensions The 1PI four-point function has contributions from three topologically-distinct box functions

iΓ4 = i [B(s, t) + B(t, u) + B(u, s)] (4.34) where 1+4 iB(s, t) = gµ I4(s, t) (4.35) 2 2 2 2 2 Letting external legs be on-shell (P12 = P23 = P34 = P41 = m )

Z ∞ 4 P iΓ(2) Y δ(1 − jβj) I4(s, t) = dβj (4.36) (4π)2 2 2 0 j=1 [m (1 − β1β2 − β2β3 − β3β4 − β4β1) − β1β3t − β2β4s]

19 In the massless limit

8−d Z ∞ 4 P iΓ( ) Y δ(1 − jβj) I (s, t) = 2 dβ (4.37) 4 (4π)d/2 j 2 0 j=1 [−β1β3t − β2β4s]

2 If the external legs are on-shell, then in the ` → 0 limit, the denominator goes to ` (` · p1)(` · p4) and thus has an IR divergence in d = 4. Need to study this!