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Renormalization, One-Loop QED Calculations

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Renormalization, One-Loop QED Calculations Ultraviolet divergence and Renormalization In QED, when we calculation higher order corrections, we meet loop diagrams, e.g. Z d4k i −igµν ∼ ; (2π)4 (=k − p=) − m k2 linear divergent. • Problem: k integrated to infinity, U-V (ultraviolet) divergence. However, QED can not be correct to the infinite energy, and there must be a fundamental energy cut off of QED. me ∼ 0:5MeV, Cutoff Λ ∼ mµ ∼ 105MeV. This means the integration depends on the fundamental scale ∼ ln Λ. • ∆p ! 1, ∆x ! 0, point interaction. Large distance physics can not see the details of the small distance physics. For e.g., Fluid mechanics does not need to know the fundamental physics of molecules. The information of the more fundamental molecular physics is absorbed into the macro constants such as viscosity. • The low energy physics is not sensitive to the high energy physics. We can absorb the high energy contribution to low energy constants describing the low energy physics, such as the coupling constants and masses and rescale factors of fields. • The mass, coupling in the original Lagrangian are not physically observed ones in the low energy physics, and we call them bare mass and bare coupling. • We will cancel the infinities in the integral by the bare parameters, such that the physical observables are correctly calculated. This is the basic idea of Renormalization in QFT. Regularization: We need to first make the infinite integral finite using some finite parameter such as an energy-momentum cut-off Λ. Taking some limit of the parameter, we recover the infinite integral, e.g. Λ ! 1. Field strength renormalization The complete spectrum of the full interacting H and P: • Vacuum jΩi, HjΩi = 0, PjΩi = 0. • Rest massive one-particle states: jλ0i. λ label different states, 0 means p = 0, Hjλ0i = m physical mass, including the correction after interaction, Pjλ0i = 0. • Boosted one particle states with p, includingq massless: j i j i j i j i 2 2 H λp = Ep λp ; P λp = p λp , Ep(λ) = p + mλ. mλ discrete. • j i j i j i j i Multiparticle states: H λp = E λp ; pqλp = p λp . E and p are 2 2 total energy & momentum, Ep(λ) = p + mλ, mλ center of mass energy, continuously changing. Consider the analytic structure of the two-point correlation function, in some field theory, L[ϕ], ϕ(x) is the field in the lagrangian hΩjTϕ(x)ϕ(y)jΩi Assume x0 > y0, insert Z X d3p 1 I = jΩihΩj + jλ ihλ j (2π)3 2E p p λ p(λ) Suppose hΩjϕ(x)jΩi = 0, use iP·x −iP·x −ip·x h j j i h j j i j 0 h j j i Ω ϕ(x) λp = Ω e ϕ(0)e λp = e p =Ep Ω ϕ(0) λ0 , jλpi boosted to jλ0i at last step Z X 3 h j j i d p 1 h j j ih j j i Ω ϕ(x)ϕ(y) Ω = 3 Ω ϕ(x) λp λp ϕ(y) Ω (2π) 2Ep(λ) Z λ X 3 d p 1 −ip·(x−y) = e j 0 hΩjϕ(0)jλ ihλ jϕ(0)jΩi 3 p =Ep 0 0 (2π) 2Ep(λ) λ Z X d4p i = e−ip·(x−y)jhΩjϕ(0)jλ ij2; (2π)4 p2 − m2 + iϵ 0 λ λ For x0 < y0, similar. Finally, Källén-Lehmann spectral representation Z 1 dM2 hΩjTϕ(x)ϕ(y)jΩi = ρ(M2)D (x − y; M2); 2π F X0 2 2 − 2 jh j j ij2 ρ(M ) = (2π)δ(M mλ) Ω ϕ(0) λ0 λ ρ(M2): Spectral density function ρ(M2) = 2πδ(M2−m2)Z+(nothing else until bound states or M2 & (2m)2) One particle state: • 2 Z = jhΩjϕ(0)jλ0ij : Probability for ϕ to create or annihilate a one-particle eigenstate of H from the vacuum. field-strength renormalization constant (or wave function renormalization constant). Interaction correction to the wavefunction. • m: physical mass, the exact energy eigenvalue at rest, physical observable. Different from m0 bare mass in the lagrangian. Transform to momentum space: Z Z 1 dM2 i d4x eip·xhΩjTϕ(x)ϕ(0)jΩ = ρ(M2) 2π p2 − M2 + iϵ Z 0 iZ 1 dM2 i = + ρ(M2) 2 − 2 2 − 2 p m + iϵ ∼4m2 2π p M + iϵ Z Z iZ 1 dM2 i d4x eip·xhΩjTϕ(x)ϕ(0)jΩi = + ρ(M2) 2 − 2 2 − 2 p m + iϵ ∼4m2 2π p M + iϵ Compare with free field theory: Z i d4x eip·xh0jTϕ(x)ϕ(0)j0i = p2 − m2 + iϵ • Free theory: Z = 1, no interaction correction for the field strength. • Free theory: No continuum intermediate multiparticle states. Generalize to Dirac Fermions: Non-trivial transformation under boost Z P iZ us(p)u¯s(p) iZ (p= + m) d4x eip·xhΩjT (x) ¯(0)jΩi = 2 s + · = 2 + · p2 − m2 + iϵ p2 − m2 + iϵ p s hΩj (0)jp; si = Z2u (p) Electron Self-energy Go to momentum space, • The first term: Just the free propagator, m0, bare mass in the Lagrangian. • The second term: electron selfenergy Σ2 to one-loop Z d4k i(=k + m ) −i −iΣ (p) = (−ie)2 γµ 0 γ 2 4 2 − 2 µ − 2 − 2 (2π) k m0 + iϵ (p k) µ + iϵ To avoid the infrared divergence we use a small µ for photon mass. Feynman parameterization Z d4k i(=k + m ) −i −iΣ (p) = (−ie)2 γµ 0 γ 2 4 2 − 2 µ − 2 − 2 (2π) k m0 + iϵ (p k) µ + iϵ The two denominators: combine into a single quadratic polynomial in momentum. Z Z 1 1 1 1 1 = dx = dxdyδ(x + y − 1) − 2 2 AB 0 [xA + (1 x)B] 0 [xA + yB] 1 (k2 − m2 + iϵ)((p − k)2 − µ2 + iϵ) Z 0 1 1 = dx dy δ(x + y − 1) [x(k2 − m2 + iϵ) + y((p − k)2 − µ2 + iϵ)]2 Z0 Z 0 1 1 = dx dy δ(x + y − 1) [k2 − 2k · py + p2y − m2x − µ2y + iϵ]2 Z0 0 1 1 = dx [k2 − 2(1 − x)k · p + (1 − x)p2 − xm2 − (1 − x)µ2 + iϵ]2 Z0 0 1 1 = dx 2 − · 2 − − 2 − 2 2 0 [k 2x k p + xp (1 x)m0 xµ + iϵ] x, y are called Feynman parameters: Z Z 1 1 1 1 1 = dx = dxdyδ(x + y − 1) − 2 2 AB 0 [xA + (1 x)B] 0 [xA + yB] Differentiate with respect to B Z 1 n−1 1 − ny n = dxdyδ(x + y 1) n+1 AB 0 [xA + yB] A more general formula can be proved by induction Z 1 1 X (n − 1)! = dx ··· dx δ( x − 1) ··· 1 n i ··· n A1A2 An 0 [x1A1 + x2A2 + xnAn] Repeated differentiation gives Z Q − P 1 1 X xmi 1 Γ( m ) = dx ··· dx δ x − 1 i P Q i m1 m2 mn 1 n i P m A A ··· An i Γ(m ) 1 2 0 xiAi i This formula is even true when mi are not integers. Z Z 1 d4k γµ(=k + m )γ −iΣ (p) = (−ie)2 dx 0 µ 2 4 2 − · 2 − − 2 − 2 2 0 (2π) [k 2x k p + xp (1 x)m0 xµ + iϵ] The denominator 2 − · 2 − − 2 − 2 k 2x k p + xp (1 x)m0 xµ + iϵ − 2 − 2 − − 2 − 2 =(k xp) + x(1 x)p (1 x)m0 xµ Change variable ` = k − xp, Z Z 1 d4` γµ(`= + xp= + m )γ −iΣ (p) = (−ie)2 dx 0 µ 2 4 2 − 2 − − 2 − 2 2 0 (2π) [` + x(1 x)p (1 x)m0 xµ + iϵ] • Denominator `2 even, numerator ` odd, integrate to be zero. Terms linear in ` in the numerator can be discarded. • µ µ α αµ α µ α Use γ γµ = 4, γ γ γµ = 2g γµ − γ γ γµ = −2γ , the numerator µ γ (xp= + m0)γµ = −2xp= + 4m0 Z Z 1 d4` −2xp= + 4m −iΣ (p) = −e2 dx 0 2 4 2 − 2 0 (2π) [` ∆ + iϵ] − − 2 − 2 2 ∆ = x(1 x)p + (1 x)m0 + xµ For the `0 integration, we do a Wick rotation: On the `0 complex plane, Poles p 2 − ) 0 ± j j2 − ` ∆ + iϵ = 0 `p = ( ` + ∆ iϵ) Rotate the integrate path anti-clockwise. 0 ≡ 0 Change ` i`E; ` = `E Minkowski `2 = (`0)2 − `2 ! − 2 − 0 2 − 2 Euclidean `E = (`E) `E, 2 Ω4 = 2π Z Z d4` 1 i 1 1 = d4` (2π)4 [`2 − ∆]m (−1)m (2π)4 E [`2 + ∆]m Z E Z i 1 `3 i 1 `2 = dΩ d` E = d`2 E − m 4 4 E 2 m − m 2 E 2 m ( 1) (2π) [`E + ∆] ( 1) (4π) 0 `E + ∆ Z d4` 1 i(−1)m 1 1 = : (2π)4 [`2 − ∆]m (4π)2 (m − 1)(m − 2) ∆m−2 This result is valid for m > 2. Similarly, for m > 3 Z d4` `2 i(−1)m−1 2 1 = : (2π)4 [`2 − ∆]m (4π)2 (m − 1)(m − 2)(m − 3) ∆m−3 Linear `µ: Z d4` `µ = 0 (2π)4 [`2 − ∆]m Quadratic `µ`ν : Z Z d4` `µ`ν d4` 1 gµν `2 = 4 (2π)4 [`2 − ∆]m (2π)4 [`2 − ∆]m m = 2, the integral over ` is divergent. The Wick-rotation and change ` = k − xp may not be valid. We first need to make the integration finite. Pauli-Villars regularization procedure: to make it finite 1 ! 1 − 1 (p − k)2 − µ2 + iϵ (p − k)2 − µ2 + iϵ (p − k)2 − Λ2 + iϵ Λ: a very large mass.
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