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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) 1 TODO 2 Lagrangian and Green Functions in D Dimensions Renormalization of φ3 scalar field theory (December 2016) Pdf file generated on February 17, 2018. 1 TODO Examine i" 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 coun- terterm 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 + i" 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 + i" (` + p)2 − m2 + i" 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 − i" 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 + i" (` + p1 + p2) − m + i" ` − m + i" 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 0 1 3 6−d 6−d Z 1 P 6−d g µ Γ( 2 ) dβ1dβ2dβ3δ(1 − i βi) Γ (p ; p ) = −µ 2 Bg + δg + C 3 1 2 @ d=2 h i3−d=2 A (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 + i" (` + p1) − m + i" i i × 2 2 2 2 (` + p1 + p2) − m + i" (` + p1 + p2 + p3) − m + i" 6−d 4 = gµ 2 I4(s; t) (2.26) where 4−d 8−d 1 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<jβiβjPij 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 1 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 1 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.
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