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 → ∞, ∆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 → ∞. Field strength renormalization The complete spectrum of the full interacting H and P: • Vacuum |Ω⟩, H|Ω⟩ = 0, P|Ω⟩ = 0. • Rest massive one-particle states: |λ0⟩. λ label different states, 0 means p = 0, H|λ0⟩ = m physical mass, including the correction after interaction, P|λ0⟩ = 0. • Boosted one particle states with p, includingq massless: | ⟩ | ⟩ | ⟩ | ⟩ 2 2 H λp = Ep λp , P λp = p λp , Ep(λ) = p + mλ. mλ discrete. • | ⟩ | ⟩ | ⟩ | ⟩ 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

⟨Ω|Tϕ(x)ϕ(y)|Ω⟩

Assume x0 > y0, insert Z X d3p 1 I = |Ω⟩⟨Ω| + |λ ⟩⟨λ | (2π)3 2E p p λ p(λ) Suppose ⟨Ω|ϕ(x)|Ω⟩ = 0, use iP·x −iP·x −ip·x ⟨ | | ⟩ ⟨ | | ⟩ | 0 ⟨ | | ⟩ Ω ϕ(x) λp = Ω e ϕ(0)e λp = e p =Ep Ω ϕ(0) λ0 , |λp⟩ boosted to |λ0⟩ at last step Z X 3 ⟨ | | ⟩ d p 1 ⟨ | | ⟩⟨ | | ⟩ 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 | 0 ⟨Ω|ϕ(0)|λ ⟩⟨λ |ϕ(0)|Ω⟩ 3 p =Ep 0 0 (2π) 2Ep(λ) λ Z X d4p i = e−ip·(x−y)|⟨Ω|ϕ(0)|λ ⟩|2, (2π)4 p2 − m2 + iϵ 0 λ λ For x0 < y0, similar. Finally, Källén-Lehmann spectral representation Z ∞ dM2 ⟨Ω|Tϕ(x)ϕ(y)|Ω⟩ = ρ(M2)D (x − y; M2), 2π F X0 2 2 − 2 |⟨ | | ⟩|2 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 = |⟨Ω|ϕ(0)|λ0⟩| : 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 ∞ dM2 i d4x eip·x⟨Ω|Tϕ(x)ϕ(0)|Ω = ρ(M2) 2π p2 − M2 + iϵ Z 0 iZ ∞ dM2 i = + ρ(M2) 2 − 2 2 − 2 p m + iϵ ∼4m2 2π p M + iϵ Z Z iZ ∞ dM2 i d4x eip·x⟨Ω|Tϕ(x)ϕ(0)|Ω⟩ = + ρ(M2) 2 − 2 2 − 2 p m + iϵ ∼4m2 2π p M + iϵ Compare with free field theory: Z i d4x eip·x⟨0|Tϕ(x)ϕ(0)|0⟩ = 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·x⟨Ω|Tψ(x)ψ¯(0)|Ω⟩ = 2 s + · = 2 + · p2 − m2 + iϵ p2 − m2 + iϵ p s ⟨Ω|ψ(0)|p, s⟩ = 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 ∫ 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 ∑ 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 ℓ = k − xp, Z Z 1 d4ℓ γµ(ℓ/ + xp/ + m )γ −iΣ (p) = (−ie)2 dx 0 µ 2 4 2 − 2 − − 2 − 2 2 0 (2π) [ℓ + x(1 x)p (1 x)m0 xµ + iϵ]

• Denominator ℓ2 even, numerator ℓ odd, integrate to be zero. Terms linear in ℓ 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 d4ℓ −2xp/ + 4m −iΣ (p) = −e2 dx 0 2 4 2 − 2 0 (2π) [ℓ ∆ + iϵ] − − 2 − 2 2 ∆ = x(1 x)p + (1 x)m0 + xµ For the ℓ0 integration, we do a Wick rotation: On the ℓ0 complex plane, Poles p 2 − ⇒ 0 ± | |2 − ℓ ∆ + iϵ = 0 ℓp = ( ℓ + ∆ iϵ)

Rotate the integrate path anti-clockwise.

0 ≡ 0 Change ℓ iℓE; ℓ = ℓE Minkowski ℓ2 = (ℓ0)2 − ℓ2 → − 2 − 0 2 − 2 Euclidean ℓE = (ℓE) ℓE, 2 Ω4 = 2π Z Z d4ℓ 1 i 1 1 = d4ℓ (2π)4 [ℓ2 − ∆]m (−1)m (2π)4 E [ℓ2 + ∆]m Z E Z i 1 ℓ3 i ∞ ℓ2 = dΩ dℓ E = dℓ2  E  − m 4 4 E 2 m − m 2 E 2 m ( 1) (2π) [ℓE + ∆] ( 1) (4π) 0 ℓE + ∆ Z d4ℓ 1 i(−1)m 1 1 = . (2π)4 [ℓ2 − ∆]m (4π)2 (m − 1)(m − 2) ∆m−2

This result is valid for m > 2. Similarly, for m > 3 Z d4ℓ ℓ2 i(−1)m−1 2 1 = . (2π)4 [ℓ2 − ∆]m (4π)2 (m − 1)(m − 2)(m − 3) ∆m−3

Linear ℓµ: Z d4ℓ ℓµ = 0 (2π)4 [ℓ2 − ∆]m Quadratic ℓµℓν : Z Z d4ℓ ℓµℓν d4ℓ 1 gµν ℓ2 = 4 (2π)4 [ℓ2 − ∆]m (2π)4 [ℓ2 − ∆]m m = 2, the integral over ℓ is divergent. The Wick-rotation and change ℓ = 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. We will take it to ∞ limit at last. • For small k, the integrand is almost unaffected, but it cuts off smoothly when k ≳ Λ. • Effectively introducing a fictitious heavy photon, whose contribution is subtracted from the original Feynman integral. • For the new term: → − − 2 − 2 2 → 2 → ∞ ∆ ∆Λ = x(1 x)p + (1 x)m0 + xΛ xΛ as Λ Z Z 4 4   d ℓ 1 → d ℓ 1 − 1 (2π)4 [ℓ2 − ∆]2 (2π)4 [ℓ2 − ∆]2 [ℓ2 − ∆ ]2 Z Λ ∞  2 2  i 2 ℓE − ℓE = dℓE     (4π)2 2 2 2 2 0 ℓE + ∆ ℓE + ∆     i ∆ i xΛ2 = log Λ → log (4π)2 ∆ (4π)2 ∆ Finally, (α = e2/(4π)) Z   α 1 xΛ2 Σ (p) = dx(2m − xp/) log 2 0 − − 2 − 2 2 2π 0 x(1 x)p + (1 x)m0 + xµ

• p2 Branch cut for log:

− − 2 − 2 2 x(1 x)p + (1 x)m0 + xµ = 0 1 m2 µ2 1 p ⇒x = + 0 − ± [p2 − (m + µ)2][p2 − (m − µ)2] 2 2p2 2p2 2p2 0 0

2 2 The solution is real and 0 < x < 1 for p > (m0 + µ) , threshold for creation of a two-particle state, e + γ. This is expected from the Källen-Lehmann spectral representation. • We need to find the simple pole at p2 = m2 (physical mass) for the one-particle state in the two point function. Define one-particle-irriducible (1PI) diagram: can not be split in two by removing one internal line

Define −iΣ(p): sum over all 1PI ( not including the external two line)

To the leading order Σ = Σ2. The two point function: Since p2 = p/p/, Σ = Σ(p/), commute with 1 , p/−m0

Thus there is a simple pole, shifted away from m , 0

[p/ − m0 − Σ(p/)] = 0 ⇒ m = m0 + Σ(m) , p/=m d d 2 d Close to the pole: the denominator ( dp/ p/ = 1, dp/ p = dp/ p/p/ = 2p/)   dΣ
(p/ − m) · 1 − + O (p/ − m)2 dp/ p/=m Compare with Z iZ d4x eip·x⟨Ω|Tψ(x)ψ¯(0)|Ω⟩ = 2 + ··· p/ − m + iϵ

− dΣ −1 We have Z2 = (1 dp/ ) p/=m Our calculation gives the order α correction to m and Z2,

δm =m − m0 = Σ2(p/ = m) ≈ Σ2(p/ = m0) Z   α 1 xΛ2 = m dx(2 − x) log 0 − 2 2 2 2π 0 (1 x) m0 + xµ   −−−−→Λ→∞ 3α Λ ultraviolet divergent: δm m0 log 2 4π m0 • The physical m is what the low energy experiment measures. The bare m0 is the parameter in the L which is not a direct physical observable. • The infinite δm is obsorbed into m0. • Classical rest mass of a charged point particle: divergent Z Z   Z 1 1 e 2 α dr d3r |E|2 = d3r = ∼ αΛ 2 2 4πr2 2 r2 • Here the δm ∼ −Σ2 ∼ log Λ: Chiral symmetry, suppose m0 = 0

L =iψ¯(∂/ − ieA/)ψ + ··· ¯ ¯ =iψL(∂/ − ieA/)ψL + iψR(∂/ − ieA/)ψR, 1 ∓ γ5 1 ∓ γ5 1 ± γ5 ψ = ψ, ψ¯ = ψ† γ0 = ψ†γ0 L,R 2 L,R 2 2

iαL,R Invariant under ψL,R → e ψL,R. A mass term ¯ ¯ ¯ m0ψψ = m0(ψLψR + ψRψL), not invariant. • So when m0 = 0, ψL couples to ψL, ψR couples to ψR, ¯ ⟨Ω|ψRψL|Ω⟩1PI = 0. d ⟨Ω|Tψψ¯|Ω⟩ = ⟨Ω|T(ψ ψ¯ +ψ ψ¯ )|Ω⟩ ∼ Σ ∼ p/ Σ(p)| +··· . 1PI L L R R 1PI dp/ p/=0

δm = 0. So when m0 → 0, δm → 0 required by the Chiral symmetry,

 0 δm ∝ m0 × const ∼ m0 × M ∼ m0 log(Λ/m0) • m = m0 + O(α), the expansion is expressed using the m0, differ from m by a higher order term. In fact, to compare with the experiment, the perturbation theory should be rearranged using m parameter.

  i i Σ (p/) ⟨Ω|ψ(−p)ψ¯(p)|Ω⟩ = + 2 + O(α2) p/ − m0 p/ − m0 p/ − m0   i i δm + (p/ − m)Σ′ (p/ = m) + ··· = + 2 + O(α2) p/ − m + δm p/ − m p/ − m      i i δm i δm iΣ′ (p/ = m) = − + + 2 + ··· p/ − m p/ − m p/ − m p/ − m p/ − m p/ − m i iΣ′ (p/ = m) = + 2 + (finite p dependent O(α)) + O(α2) p/ − m p/ − m iZ (up to O(α)) = 2 + ··· + O(α2) p/ − m

We will do this from the beginning systematically later. Perturbative correction to Z2 to α order, h i dΣ −1 δZ2 =Z2 − 1 = 1 − − 1 dp/ p/=m

dΣ = 2 + O(α2) dp/ p/=m Z h i α 1 xΛ2 x(1 − x)m2 = dx − x log + 2(2 − x) − 2 2 2 − 2 2 2 2π 0 (1 x) m + xµ (1 x) m + xµ

−1/2 We define renormalized field ψR = Z2 ψ: ⟨ | − ¯ | ⟩ −1⟨ | − ¯ | ⟩ Ω ψR( p)ψR(p) Ω =Z2 Ω ψ( p)ψ(p) Ω   i(1 + δZ ) =(1 − δZ + O(α2)) 2 + finite O(α) + O(α2) 2 p/ − m + iϵ i = + (finiteO(α)) + O(α2) p/ − m + iϵ

The renormalized Green’s functions are finite. ′′ Homework 1: check Σ2 (p/) is finite. Photon self-energy

• 1PI Photon self-energy: . • µ µν µν 2 µ ν 2 Ward id. q Πµν (q) = 0, Π (q) = (g q − q q )Π(q ). µ Ward id.:Operator formalism, using current conservation:∂µj = 0, and canonical anti-commutation relation ∫ ∫ µν 4 iq·x µ ν 4 iq·x µ ν qµΠ ∼ d xe qµ⟨Ω|Tj (x)j (0)|Ω⟩1PI = i d xe ∂µ⟨Ω|Tj (x)j (0)|Ω⟩1PI

0 0 ν ∼⟨Ω|δ(x )[j (x), j (0)]|Ω⟩1PI = 0 µν 2 µν µ ν 2 Π = C1(q )g + q q C2(q ) . 1 q Πµν = 0 ⇒ C (q2) = − C (q2) = −Π(q2) µ 2 q2 1 Homework 2: Complete this derivation using the canonical anti-commutation relation. • 2 1 We also expect that Π(q ) has not q2 pole. • The full propagator

µ µρ − qµqρ µ ν µ Using ∆ν = (g q2 )gρν , ∆ν ∆ρ = ∆ρ ,

• As long as Π(q2) has no 1/q2 pole, the photon is massless to all order. • At q2 = 0 pole, ⟨Ω|TA (q)A (−q)|Ω⟩ ∼ −i (g − qµqν ), µ ν q√2(1−Π(0) µν q2 1 Z3 = 1−Π(0) . Field renormalization Aµ = Z3AR,µ, the correlation function ⟨Ω|TAµ,R(q)AR,ν (−q)|Ω⟩ will be finite. Explicitly check Ward id to one-loop: Z h i d4k i(/k + m) i(/k + /q + m) iΠµν (q) = − (−ie)2 tr γµ γν 2 (2π)4 k2 − m2 (k + q)2 − m2 Z d4k kµ(k + q)ν + kν (k + q)µ − gµν (k · (k + q) − m2) = − 4e2 (2π)4 (k2 − m2)((k + q)2 − m2)

Use tr(γµ1 γµ2 γµ3 γµ4 ) = 4(gµ1µ2 gµ3µ4 − gµ1µ3 gµ2µ4 + gµ1µ4 gµ2µ3 ) Feynman parameterization : Z 1 1 1 = dx (k2 − m2)((k + q)2 − m2) [(1 − x)(k2 − m2) + x((k + q)2 − m2)]2 Z0 1 1 = dx 2 − 2 − 2 2 0 [(k + xq) + x(1 x)q m ] Change variable k = l − xq, linear lµ term integrate → 0, lµlν → gµν l2/4 Z Z 1 4 µν − 2 d l iΠ2 (q) = 4e dx 4 0 (2π) 1 µν 2 − µν 2 − − µ ν µν 2 − 2 2 g l g l 2x(1 x)q q + g (m + x(1 x)q ) (l2 − ∆)2 ∆ = −x(1 − x)q2 + m2. 0 0 Wick rotation: l = ilE. Z Z 1 4 µν − 2 d lE iΠ2 (q) = 4ie dx 4 0 (2π) − 1 µν 2 µν 2 − − µ ν µν 2 − 2 2 g lE + g lE 2x(1 x)q q + g (m + x(1 x)q ) 2 2 (lE + ∆)

• Note that the change of variable l = k + xq needs the integral to be convergent. It is nonsence to do this for a divergent integral. We need first regularized the integral. • Cut-off regularization : iΠµν (q) ∝ e2Λ2gµν , no compensating qµqν , violates the Ward Id. Break the gauge invariance. • Pauli-Villas: not break gauge invariance, but need to introduce more fermions. • We introduce another kind of regularization: Dimensional regularization, analytically continue the space-time dimension d to be non-integer such that the integration is convergent. • Compute the diagram as an analytic function of the dimensionality of spacetime, d. Dimensional regularization: • For d small, the loop-momentum integral is finite, Ward id can be proved, Lorentz symmetry is preserved. • Physical observable, after renormalization and cancellation of the divergences — is well-defined as d → 4. An example: Z Z Z ddℓ 1 dΩ ∞ ℓd−1 E = d dℓ E . d 2 2 d E 2 2 (2π) (ℓE + ∆) (2π) 0 (ℓE + ∆) R 2πd/2 area of d-dim unit sphere: dΩd = Γ(d/2) . (Peskin p249)

∫ ∞ − ∫ ∞ ( ) ∫ d 1 2 d/2−1 2−d/2 1 ℓ 1 2 (ℓE) 1 1 1− d d −1 dℓ E = d(ℓ ) = dx x 2 (1 − x) 2 E 2 2 E 2 2 0 (ℓE + ∆) 2 0 (ℓE + ∆) 2 ∆ 0 ( ) − d d 2− d 1 Γ(2 2 )Γ( 2 ) 1 2 ∆ = ; x = 2 2 Γ(2) ∆ ℓE + ∆ RWe have used the beta function: 1 α−1 − β−1 Γ(α)Γ(β) 0 dx x (1 x) = B(α, β) = Γ(α+β) Z d d   − d ϵ   ϵ d ℓ 1 1 Γ(2 − ) 1 2 2 Γ( ) 1 2 E = 2 = 2 d 2 2 d/2 2−ϵ/2 (2π) (ℓE + ∆) (4π) Γ(2) ∆ (4π) ∆ Γ(z): isolated poles at z = 0, −1, −2, ··· , − d ··· − Γ(2 2 ): poles at d = 4, 6, 8, . near d=4, ϵ = 4 d, d ϵ 2 Γ(2− ) = Γ( ) = −γ+O(ϵ) , γ = 0.5772, Euler-Mascheroni constant. 2 2 ϵ Expand w.r.t ϵ: Z    ddℓ 1 → 1 ϵ
2 ϵ E −−−→d 4 1 + log(4π) − γ + O(ϵ) 1 − log ∆ (2π)d (ℓ2 + ∆)2 (4π)2 2 ϵ 2 E   1 2 = − log ∆ − γ + log(4π) + O(ϵ) . (4π)2 ϵ Compare with Pauli-Villars Z ∞   d4ℓ 1 →∞ 1 xΛ2 −−−−→Λ log + O(Λ−1) . 4 2 2 2 0 (2π) (ℓE + ∆) (4π) ∆ 1/ϵ ∼ log Λ, log divergent. • More general

• µν In d-dim spacetime, gµν g = d, integrand

ℓµℓν 1 ℓ2gµν → . (ℓ2 + ∆)n d (ℓ2 + ∆)n

• Dirac matrices: {γµ, γν } = 2gµν , tr[1] = 4. Contraction of Dirac matrices:

µ ν µν ν µ γ γ γµ =(2g − γ γ )γµ =2γν − dγν = −(2 − ϵ)γν

Homework 3: Derive (7.85), (7.86), (7.89). Appendix, (A.46) 2 2 Continue the computation of Π2.(∆ = (m − x(1 − x)q )) ∫ ∫ 1 ddl (1 − 2 )l2 gµν − 2x(1 − x)qµqν + gµν (m2 + x(1 − x)q2) iΠµν (q) = −4ie2 dx E d E d 2 2 0 (2π) (lE + ∆) Z d − 2 2 µν − d lE (1 d )lEg 1 d d 1 1− d µν = (1 − )Γ(1 − )( ) 2 g d 2 2 d/2 (2π) (lE + ∆) (4π) 2 2 ∆ 1 d 1 2− d µν = Γ(2 − )( ) 2 (−∆g ) (4π)d/2 2 ∆ Z 1 1 Γ(2 − d ) iΠµν (q) = − 4ie2 dx 2 2 d/2 2− d 0 (4π) ∆ 2 × [−∆gµν + gµν (m2 + x(1 − x)q2) − 2x(1 − x)qµqν ] 2 µν µ ν 2 =(q g − q q ) · iΠ2(q ) Z −8e2 1 Γ(2 − d ) Π (q2) = dx x(1 − x) 2 2 (4π)d/2 ∆2−d/2 Z 0 − 2 1   → 8e − 2 − − 2 dx x(1 x) log ∆ γ + log(4π) (4π) 0 ϵ − d − d − d Quadratic div. cancelled: (1 2 )Γ(1 2 ) = Γ(2 2 ). Finally, we have the O(α) correction in Z3   2α 2 Z = 1 + Π (0) = 1 − − log m2 − γ + log(4π) 3 2 π ϵ

• We have replaced m0 to m. • In fact here e should be e0 the bare charge up to higher α orders. Superficial degree of divergence(SDOD) A typical diagram,

• propagators: negative power of internal momentum k, R • Loop number of d4k, increase the k power; • Some kinds of vertex also introduce loop momentum in the numerator. We can naively count the power of k in the integral — superficial degree of divergence,

D ≡ (powers of k in the numerator) − (powers of k in the denominator)

We expect: Diagram ∝ ΛD, Λ cutoff. D ≥ 0 divergence, D = 0 log divergence, D < 0 No divergence. Z 4 ··· 4 M ∼ d k1 d kL i 2 2 2 ((ki − pi) − m ) ··· (kL) We expand M w.r.t p, or take derivative w.r.t p Z d d4k ··· d4k M ∼ 1 L (k − p ) µ − 2 − 2 2 ··· 2 i i µ dpi ((ki pi) m ) (kL) Reduce the SDOD by one. Superficial: not always correct. • The divergence of the subdiagram may be worse than the total diagram. We constraint ourselves to 1PI diagrams. • Some symmetry may reduce the degree of the divergence. • Trivial diagram with no propagators and no loops has D = 0 , no divergence. • We will consider the 1PI diagrams: all other diagrams are composed with 1PI diagrams. Superfacial degree of devergence in QED QED: for a 1PI diagram • Ne: external electron leg; • Nγ : external photons; • Pe: internal electron propagators; • Pγ : internal photon propagators; • V: vertices; • L: loops. Superficial degree of divergence:

D ≡ 4L − Pe − 2Pγ Superficial degree of divergence: QED

• From momentum conservation, each vertex has a δ function, decrease the integral of internal momentum, the overall δ function does not constraint the internal momentum,

L = Pe + Pγ − V + 1,

• 1 V = 2Pγ + Nγ = 2 (2Pe + Ne) 3 D = 4(P + P − V + 1) − P − 2P = 4 − N − N . e γ e γ γ 2 e • D depends only on the number of the external legs of each type. • Divergent amplitudes D ≥ 0: There are only finite number of diagrams with a small number of external legs. 3 D = 4 − N − N . γ 2 e Seven types of divergent diagrams: • No external legs,Nγ = Ne = 0: D = 4, Vacuum buble diagram, ⟨Ω|Ω⟩, no external legs. An unobservable shift of the vacuum energy, no contribution to the physical S matrix. • Nγ = 1, Ne = 0: , Z 4 ¯ ν ⟨Ω|Aµ(x)|Ω⟩ = −ie d y⟨Ω|ψ(y)γ ψ(y)|Ω⟩1PI⟨Ω|TAν (y)Aµ(x)|Ω⟩ ,

Ignore the photon leg, Z 4 −iq·x µ ¯ µ = −ie d x e ⟨Ω|Tjµ(x)|Ω⟩1PI , j = ψ(x)γ ψ(x)

iP·x −iP·x From Lorentz inv, using Aµ(x) = e Aµ(0)e , † −1 ν −1 ν −1 U(Λ)Aµ(x)U (Λ) = (Λ ) µAν (Λ x) = Λµ Aν (Λ x),

iP·x −iP·x ⟨Ω|Aµ(x)|Ω⟩ =⟨Ω|e Aµ(0)e |Ω⟩ = ⟨Ω|Aµ(0)|Ω⟩= cµ † ν =⟨Ω|U(Λ)Aµ(0)U (Λ)|Ω⟩ = Λµ cν = 0 Another reason: C−1jµC = −jµ ⇒ C−1AµC = −Aµ, C|Ω⟩ = |Ω⟩.

−1 ⟨Ω|Aµ|Ω⟩ = ⟨Ω|C AµC|Ω⟩ = −⟨Ω|Aµ|Ω⟩= 0 In general, for odd number of jµ(x) in the correlator, µ ν µ ν ⟨Ω|Tj (x1) ··· j (xn)|Ω⟩1PI = 0 and ⟨Ω|TA (x1) ··· A (xn)|Ω⟩ = 0. • 3 external γ legs, Nγ = 3: D = 1

⟨Ω|TA (x)A (y)A (z)|Ω⟩ Z µ ν ρ

4 4 4 µ1 µ2 µ3 = d x1d x2d x3⟨Ω|Tj (x1)j (x2)j (x3)|Ω⟩1PI ×⟨ | | ⟩⟨ | | ⟩⟨ | | ⟩ Ω Aµ1 (x1)Aµ(x) Ω Ω Aµ2 (x2)Aν (y) Ω Ω Aµ3 (x3)Aρ(z) Ω = 0. µ1 µ2 µ3 ⟨Ω|Tj (x1)j (x2)j (x3)|Ω⟩1PI = 0

• For one-loop diagram, we can explict calculate: Homework 4: Peskin (10.1) • 2 external e legs, D = 1 self-energy diagram:

2 = −iΣ(p/) = A0 + A1p/ + A2p + ··· ,

2 1 dn | p/p/ = p . An = n! dp/n (Σ) p/=0, Ai independent of p.. • − 1 In iΣ, p in propagators, /k+p/−m

d 1 1 ( ) = dp/ /k + p/ − m (/k + p/ − m)2

Reduce the degree of divergence by one. • So only A0 and A1 are divergent: A0 ∼ δm, D = 1; A1 ∼ δZ2, D = 0.

• iαL,R Chiral symmetry, m → 0, L invariant under ψL,R → e ψL,R: ¯ ⟨Ω|ψ(0)ψ(0)|Ω⟩ ∼ 4A0 not invariant under chiral transformation. A0 should be 0 when m = 0. So A0 ∼ m, −iΣ ∼ a0m log Λ + a1p/ log Λ + (finite terms) • 2 external electrons + one external γ, Nγ = 1, Ne = 2, D = 0 Log divergence: expand w.r.t external momentum, only the constant term is infinite

∼ −ieγµ log Λ + (finite terms). • Nγ = 2, D R= 2, Photon Self-energy diagram: Log divergence. µν 4 iq·x µ ν iΠ (q) = d xe ⟨Ω|Tj (x)j (0)|Ω⟩1PI µν From Ward id.: qµΠ = 0

µν µν µν 2 µ ν 2 qµΠ = 0 ⇒ Π ∼ (g q − q q )Π(q )

• There is no constant term and linear terms w.r.t q in iΠµν . • Π has no 1/q2 terms — 1PI diagrams.— photon no mass correction. • 4 external γ legs: D = 0 Ward id: for each leg there should be a factor like gµν kσ − gµσkν , D = 0 − 4 = −4, finite.

In summary, ”primitively” divergent amplitudes in QED: Three amplitdes, µ Σ, Π, Γ , four divergent constants A0, A1, Π, Γ. For general d dimensions QED, D = dL − Pe − 2Pγ , Using the relations: L = Pe + Pγ − V + 1 and V = 2Pγ + Nγ = Pe + Ne/2 d − 4 d − 2 d − 1 D = d + ( )V − ( )N − ( )N 2 2 γ 2 e

• d < 4 : More vertices, lower degree of divergence, total number of divergent Feynman diagrams is finite. ”SUPER-RENORMALIZABLE” • d = 4 : Only a finite number of amplitudes are superficially divergent, and at each loop-order these amplitudes are divergent. ”RENORMALIZABLE” • d > 4: More vertices , higher degree of divergence, Every amplitude: higher order diagrams — higher number of vertices — higher divergence — number of divergent diagrams is infinite. All amplitude will be divergent at a sufficient high order. ”NON-RENORMALIZABLE” Superficial Degree of divergence: Naively correct for most cases. Theory with more symmetries, the SDOD may not be correct, and we need to study it in detail. A most general situation: • bosons field B and fermions, F. • Interaction: g∂δBbFf, g: coupling constant. • A diagram: nB internal Bosons, NB external bosons, nF, internal fermions, NF, external Fermions, V vertices,l loops.) •

2nB + NB = Vb, 2nF + NF = Vf l = nB + nf − (V − 1), Suppose all the ∂ in the vertices contributes to the D

D = dl + Vδ − 2nB − nF

Eliminate l, nB and nF, d − 2 d − 1 D =Vr − N − N + d, 2 B 2 F d − 2 d − 1 r =δ + b + f − d = −dim[g] 2 2 • Generalize to more general case:different interaction i, X d − 2 d − 1 D = r V − N − N + d, i i 2 B 2 F i d − 2 d − 1 r =δ + b + f − d = −dim[g ] i i 2 i 2 i i X d − 2 d − 1 D = r V − N − N + d, i i 2 B 2 F i d − 2 d − 1 r =δ + b + f − d = −dim[g ] i i 2 i 2 i i Another way to derive: − − • NB NF L − d 2 − d 1 Suppose a vertex: ηB F in , dim[η] = d 2 NB 2 NF. • The amplitude M with NB, NF: dim[M] = dim[η] = d − d−2 N − d−1 N . Q 2 B 2 FP • M ∼ Vi D M − SDOD: i i gi Λ , D = dim[ ] i Vidim[gi] We also have: • There is one ri > 0 or dim[gi] < 0 : Non-renormalizable. • ri < 0 or dim[gi] > 0: superrenomalizable • ri = 0 or dim[gi] = 0: Renormalizable. This is a most general result. Eg. 1 1 λ L = (∂ ϕ)2 − m2ϕ2 − ϕn 2 µ 2 n! N: external lines; P: propagators; L: Loops, L = P − V + 1, Each vertex has n propagators/legs: nV = N + 2P. h   i   d − 2 d − 2 D = dL − 2P = d + n − d V − N 2 2   • − d−2 dim[λ] = n 2 + d • d = 4: n = 4, [λ] = 0, renormalizable; n > 4, nonrenormalizable • d = 2, n arbitrary: [λ] = 2, superrenormalizable. • d = 3:n = 4, [λ] = 1, super-renormalizable; n = 6, [λ] = 0, renormalizable; n > 6, [λ] < 0, nonrenormalizable. Electron vertex and charge renormalization Let’s look at four point fermion correlation function to one-loop:

and also three diagrams with loop on the right fermion line. • We have considered the fermion and photon selfenergy to one-loop, ′ now we will consider the 1PI vertex correction, −ie0Γµ(p , p), with p′, p on-shell.

• Γµ: an expression of p, p′, γµ. Since it is a vector,

Γµ = γµA + (p′ + p)µB + (p′ − p)µC.

A,B, C are Lorentz scalar, could involve p/, p/′. Γµ = γµA + (p′ + p)µB + (p′ − p)µC.

• We consider the on-shell combination u¯(p′)Γµ(p′, p)u(p), p/u(p) = mu(p), u¯(p′)p/′ = u¯(p′)m, and p2 = p′2 = m2. A, B, C are functions of q2 = −2p′ · p + 2m2. • ′ µ ′ ′ µ From Ward id., qµu¯(p )Γ (p , p)u(p) = 0, using qµu¯(p )γ u(p) = 0

′ µ ′ 2 qµu¯(p )Γ (p , p)u(p) = q C = 0 ⇒ C = 0 • Using the Gordon id. h i p′µ + pµ iσµν q u¯(p′)γµu(p) = u¯(p′) + ν u(p) 2m 2m ′µ µ µ µν (p + p ) → 2mγ − iσ qν . iσµν q Γµ(p′, p) ∼ γµF (q2) + ν F (q2) 1 2m 2

F1 and F2 are called form factors. • To lowest order F1 = 1, F2 = 0. Higher orders: F1 term is Log divergent and F2 finite. • The physical charge can be measured in the static Coulomb interaction of two charges as the momentum transfer q → 0: e2 ∼ e2 V(r) = 4πr , in momentum space V(q) |q|2 , which is related to → − ′ ′ − e + e e + e diagram at a photon pole (p1 p1 = p2 p2 = q), ie2 u¯(p′ )γµu(p )( )u¯(p′ )γ u(p ) 1 1 q2 2 µ 2 In the nonrelativistic limit and p′ → p, the largest component come from u¯(p′)γ0u(p) → 2mξ′†ξ, ξ is the spin wave function, for the same electron instate and outstate, ξ′†ξ = 1.

ie2 ie2 −iV(q)4m2 ∼ u¯(p′ )γ0u(p )( )u¯(p′ )γ u(p ) ∼ 4m2 1 1 −|q|2 2 0 2 −|q|2 • ⟨ | − ′ − ′ ¯ ¯ | ⟩ Ω ψ( p2)ψ( p1)ψ(p2)ψ(p1) Ω , at photon pole: ′ ′ ¯ ¯ ⟨Ω|ψ(−p2)ψ(−p1)ψ(p2)ψ(p1)|Ω⟩ ′ µ ′ ∼ SF(p2)(−ie0Γ (q))SF(p2)(iΠµν )SF(p1)(−ie0Γµ(q))SF(p1) Z i Z ig S (p) −−−−→on shell 2 , iΠ (q2) −−−−→on shell − 3 µν + ... F p/ − m µν q2 • The physical amplitude for 2 → 2 is related with the correlation function by LSZ: in the q → 0 limit ( ) ( ) − ′ − ′ −iM ∼ lim − iZ 1/2u¯(p )(p/ − m) − iZ 1/2u¯(p )(p/ − m) ′ 2 2 2 2 1 1 p ,p 1,2 1,2 β2 β1 on shell ′ ′ × ⟨Ω|ψ (−p )ψ (−p )ψ¯ (p )ψ¯ (p )|Ω⟩ ( β2 2 β1 1 α)1 2( α2 1 ) × − −1/2 − − −1/2 − iZ2 (p/2 m)u(p2) iZ2 (p/1 m)u(p1) α2 α1 ( ) ( ) ′ iZ ∼Z1/2u¯(p ) − ie Γµ(q → 0) Z1/2u(p ) − 3 2 2 0 2 2 q2 ( ) × 1/2 ′ − → 1/2 Z2 u¯(p1) ie0Γµ(q 0) Z2 u(p1) 2 ′ ie ′ →u¯(p )γµu(p )( )u¯(p )γ u(p ) 1 1 q2 2 µ 2 µ → −1 µ We expect Γ (q 0) = Z1 γ , thus − µ → 1/2 − µ 1/2 −1 ie0Γ (q 0)Z2Z3 = ( ie)γ , i.e. e0Z2Z3 Z1 = e • The form factors: iσµν q −ie Γµ(p′, p)Z Z1/2 = −ie(γµF (q2) + ν F (q2)) 0 2 3 1 2m 2

2 F1(q → 0) = 1. Vertex function:evaluation

Γµ = γµ + δΓµ

ν µ µ γ γ γν = −2γ , ν µ ρ δ νµ µ ν ρ δ ρ δ µ µ νρ ρ ν δ γ γ γ γ γν = (2g − γ γ )γ γ γν = 2γ γ γ − γ (2g − γ γ )γ γν =2γργδγµ − 2γµγδγρ − 2γµγργδ =4γµgδρ − 2γδγργµ − 4γµgδρ = −2γδγργµ Feynman parametrization: Z 1 1 2 = dx dy dzδ(x + y + z − 1) − 2 ′2 − 2 2 − 2 3 ((k p) + iϵ)(k m + iϵ)(k m + iϵ) 0 D D = x(k2 − m2) + y(k′2 − m2) + z(k − p)2 + iϵ = k2 + 2k · (yq − zp) + yq2 + zp2 − (x + y)m2 + iϵ = (k + (yq − zp))2 + 2yzp · q + y(1 − y)q2 + z(1 − z)p2 − (x + y)m2 + iϵ use 2q · p = (q + p)2 − q2 − p2 = −q2; p2 = p′2 = (q + p)2 = m2 = (k + (yq − zp))2 + yxq2 − (x + y)2m2 + iϵ = ℓ2 − ∆ + iϵ , ∆ = −yxq2 + (x + y)2m2 , ℓµ = (k + yq − zp)µ

If we give photon a small mass µ, ∆ = −yxq2 + (x + y)2m2 + zµ2. µ → µ ν → 1 µν 2 ′ Change to ℓ, Numerator, using ℓ 0, ℓ ℓ 4 g ℓ , k = k + q, (we will use Pauli-Villars)   Numerator =u¯(p′) /γk µ/k′ + m2γµ − 2m(k + k′)µ u(p)  ′ 1 →u¯(p ) − γµℓ2 + (−/q + zp/)γµ((1 − y)/q + zp/) 2  + m2γµ − 2m((1 − 2y)qµ + 2zpµ) u(p) (1) The final form should be γµA + (p′ + p)µB + qµC, use {γµ, γν } = 2gµν , ′ ′ ′ p/γµ = 2pµ − γµp/, pu/ (p) = mu(p), u¯(p )p/ = u¯(p )m, and u¯/qu(p) = 0, and x + y + z = 1,  1
Numerator = u¯(p′) γµ − ℓ2 + (1 − x)(1 − y)q2 + (1 − 2z − z2)m2 2  + (p′ + p)µmz(z − 1) + qµm(z − 2)(x − y) u(p) (2)

From x ↔ y symmetic in the denominator, the last term with (x − y) is integrated to be zero, as expected by Ward id. ′µ µ µ µν Using the Gordon id.(p + p ) → 2mγ − iσ qν .  1
Numerator = u¯(p′) γµ − ℓ2 + (1 − x)(1 − y)q2 + (1 − 4z + z2)m2 2  µν − iσ qν mz(z − 1) u(p) (3)

The ℓ2 term integration is divergent.

Homework 5: derive (1), (2) and (3). Pauli-Villars regularization: Replace in the photon propagator, subtracting a fictitious heavy photon

1 → 1 − 1 (k − p)2 + iϵ (k − p)2 + iϵ (k − p)2 − Λ2 + iϵ 2 2 2 2 ∆ → new term ∆Λ = −xyq + (1 − z) m + zΛ 1 → 1 − 1 3 3 3 D D DΛ Do the Wick rotation and integrate: for m > 2 Z d4ℓ 1 i(−1)m 1 1 = . (2π)4 [ℓ2 − ∆]m (4π)2 (m − 1)(m − 2) ∆m−2 for m > 3 Z d4ℓ ℓ2 i(−1)m−1 2 1 = . (2π)4 [ℓ2 − ∆]m (4π)2 (m − 1)(m − 2)(m − 3) ∆m−3 Z 4  2 2    d ℓ ℓ − ℓ i ∆Λ 4 2 3 2 3 = 2 log . (2π) [ℓ − ∆] [ℓ − ∆Λ] (4π) ∆ The convergent terms got correction of 1/Λ2 → 0, in the Λ → ∞. The form factors: iσµν q −ie Γµ(p′, p)Z Z1/2 = −ie(γµF (q2) + ν F (q2)) 0 2 3 1 2m 2 µ −1 µ − µ Recall the definition Z1, Γ (0) = Z1 γ = (1 δZ1)γ , and the relation between the bare charge and physical charge: − µ → 1/2 − µ ie0Γ (q 0)Z2Z3 = ( ie)γ , 1 i.e. e = e Z Z1/2Z−1 = e (1 + δZ − δZ + δZ ). 0 2 3 1 0 2 1 2 3 Recover a small mass µ of photon ∆ = −yxq2 + (x + y)2m2 + zµ2. ∫ [ ] α 1 zΛ2 (1 − 4z + z2)m2 −δZ = dxdydzδ(x + y + z − 1) log + 1 − 2 2 2 − 2 2 2 2π 0 (1 z) m + zµ (1 z) m + zµ ∫ [ ] α 1 zΛ2 (1 − 4z + z2)m2 = dz(1 − z) log + − 2 2 2 − 2 2 2 2π 0 (1 z) m + zµ (1 z) m + zµ ∫ [ ] α 1 xΛ2 x(1 − x)m2 δZ = dx − x log + 2(2 − x) 2 − 2 2 2 − 2 2 2 2π 0 (1 x) m + xµ (1 x) m + xµ

It can be shown that −δZ1 + δZ2 = 0. In fact, from Ward Id. we can 1/2 show Z1 = Z2, and thus e = e0Z3 . Homework 6: Show that −δZ1 + δZ2 = 0. hint: see Peskin page 222. µ µ µν 2 The form factors: Γ = γ (1 + δΓ) + σ qν (··· ), upto eO(α ) iσµν q −ieZ Γµ(p′, p) = −ie(Γµ(p′, p) + δZ γµ) = −ie(γµF (q2) + ν F (q2)) 1 1 1 2m 2 2 F1(q ) = 1 + δΓ + δZ1,

• We have replaced all the bare parameters into physical ones. • 2 F2(q ) is automatically finite. • 2 If µ → 0, there are infrared divergences in the last two terms of F1. z → 1, x → 0, y → 0. Z1 = Z2

Yn Xn µ −kµ⟨TJ (k) ϕ(ki)⟩ = ⟨Tϕ(k1) ... i∆ϕ(ki + k) . . . ϕ(kn)⟩ i=1 i=1 QED: consider U(1) global symmetry ψ → eiαψ, Jµ = −ψγ¯ µψ, correlation funtion ⟨Jµ(k)ψ(−(p + k))ψ¯(p)⟩, Ward Id:   µ ¯ ¯ ¯ −kµ⟨J (k)ψ(−(p + k))ψ(p)⟩ =i ⟨ψ(−p)ψ(p)⟩ − ⟨ψ(−(p + k))ψ(p + k)⟩

µ S(p + k)[−ikµΓ (p + k, p)]S(p) = S(p) − S(p + k), where i S(p) = ⟨ψ(−p)ψ¯(p)⟩ = . p/ − m − Σ(p) µ −1 −1 −ikµΓ (p + k, p) = S (p + k) − S (p) . − µ → 1 µ µ → ∼ Z2 2 2 Γ (p + k, p) Z1 γ as k 0, S(p) i p/−m , at around p = m , k → 0 − −1 − −1 iZ1 /k = iZ2 /k

Z1 = Z2 Perturbative proof of the Ward id: (Peskin 7.4) µ If an amplitude is expressed as M = ϵµ(k)M (k), then

µ kµM (k) = 0 Ward-Takahashi identity. We first consider not on-shell correlation function and at last useLSZ. • A diagram, remove an external photon γ(k), get a simpler diagram for a simpler amplitude M0. • Reinsert the photon somewhere in the simpler diagram, we obtain a contribution to M(k). • Sum over all possible diagrams for M0 and then summing over all possible insertions in each of these diagrams, we obtain M(k). • The Ward id is true individually for each diagram contribution to M0. Two kinds of possibilities of insertions: • Attach to an elecctron line running out of the diagram to two external points. • Attach to an internal electron loop. Attach to a electron line runs between external points.

• Before insertion: Momenta in the propagators: ′ p, p1 = p + q1, p2 = p1 + q2, ··· , p = pn−1 + qn. • Insert after the ith vertex:

µ ϵµ(k) → kµ : − iekµγ = −ie[(p/ + /k − m) − (p/ − m)], i  i  i i i i (−ie/k) = −ie − p + /k − m p − m p − m p + /k − m /i /i /i /i • After insertion i:

• Insertion after the (i − 1)th vertex:

• The first term cancels with the second term of previous expression. • Simlilar cancellation for ajacent insertions. • After summing of the insertion points along this line, only unpaired term at the ends remain. q = p′ + k. • From correlation function to S-matrix, take the pole of i i , or /q−m p/−m lim /q→m (−i(/q − m))(−i(p/ − m)) × · · · , thus the right hand side → 0. /p→m  X  µ i i lim (−i(/q − m)) k ( ··· (−ieγµ) ··· ) (−i(p/ − m)) /q→m /q − m p/ − m /p→m insertions  i i = lim (−i(/q − m))e ( ··· ) /q→m q − /k − m p − m p→m / / /  i i − ( ··· ) (−i(p/ − m)) /q − m p/ + /k − m Insertion on the internal fermion loop:

µ M0: → kµM ,

( ) i i i i (−ie/k) = −ie − p + /k − m p − m p − m p + /k − m /i /i /i /i • Insert between 1 and 2

• Sumover all n insertion:

• Shifting the integration variable from p1 → p1 + k in the seocnd term, the two term cancel. • The shift of the variable needs the integraion be finite first. We need first do regularization and then shift. If the regularization conflict with the shift, like brute force cut-off regularization, the Ward id may be violated. Assemble the pieces: Suppose the correlation function M has 2n externel electron lines, n ingoing and n outgoing. M0 one less external γ(k). • To form kµM from M0: sum over all diagrams from M0, and sum over all insertions for each diagram of M0. • For each diagram of M0, insertions on a fermion loop sum up to give zero. • For each diagram, sum over all insertions on through-going fermion lines give

This is the Ward-Takahashi identity for correlation function in QED. • Go to S-matrix, we need to extract poles for external legs of the left hand side, but the right hand side does not have all the poles, thus gives zero. µ kµM (k; , p1, ··· , pn; q1, ··· , qn) = 0 • We only need the fermions to be on-shell. Photons do not need to be on-shell. Renormalization

The techniques are common to all loop calculations: 1 Draw the diagram(s) and write down the amplitude. 2 Introduce Feynman parameters to combine the denominators of the propagators. 3 Complete the square in the new denominator by shifting to a new loop momentum variable ℓ. 4 Write the numerator in terms of ℓ. Drop odd powers of ℓ, and µ ν → 1 2 rewrite even powers using identities like ℓ ℓ 4 ℓ . 5 Perform the momentum integral by means of a Wick rotation and four-dimensional spherical coordinates. If the integral is divergent, we need first do regularization. Renormalization Review: • We use the bare parameters m0, e0 and bare fields to do the perturbation calculation of correlation function. • We meet UV divergence in the loop integral. • Regularize the integral using a cut-off or dimensional regularization. • Then absorb the infinity in the bare parameters m0, e0 and get the relations between physical parameters m and e with the bare parameters. • Rewrite the S-matrix using these physical parameters m and e, then as the Λ → ∞, the physical observable should be finite. Renormalization:

• The divergence comes from the UV integration. The theory should not be used to arbitrarily high energy scale. There should be a cut off of the theory. • Nature creates the world which have a great property: Low energy physics is not sensitive to the high energy theory details. Newton mechanics , low energy world, — relativity, high energy, high speed world. Condense matter physics, quantum mechanics — High energy particle physics, relativistic QED, — Electroweak theory • We do not need to be precise about high energy physics, the high energy effects can be obsorbed into the parameters of the low energy theory. Renormalization procedure • Regularization : Make the UV divergent finite, —Energy Cut off Λ, Dimensional regularization, space-time dimension <4. • Obsorb the ”would be” divergence into the parameters which to be fixed by some special experiments at a typical energy scale. • Then there is no divergence in the theory, and we can predict other experiments using these parameters. Renormalized Perturbation theory

• Originally, we used the bare parameters as expansion parameters, and express the physical observables using the renormalized parameters. • We could use the renormalized parameters as the expansion parameters to do the perturbation theory from the beginning. — Renormalized perturbation theory. Renormalized Perturbation Theory: ϕ4 Use the ϕ4 theory as an example, Bare Lagrangian with bare parameters Z 1 1 λ L = d4x (∂ ϕ )2 − m2ϕ2 − 0 ϕ4 . 2 µ 0 2 0 0 4! 0 Superficial DoD D = 4 − N, N: external legs.

Ignore the unobservable vacuum diagram, there are 3 infinite constants — absorbed into 3 unobservable bare parameters: bare mass m0, bare coupling constant λ0, bare field strength ϕ0. Renormalized Perturbation Theory: ϕ4 • 1/2 Bare field ϕ0 = Z ϕ, ϕ renormalized fields, ⟨ | | ⟩ iZ ··· iZ ··· Ω ϕ0ϕ0 Ω = 2− 2 + = 2− 2− 2 + , p m p (m0 δm ) ⟨ | | ⟩ i ··· Ω ϕϕ Ω = p2−m2 + • LSZ reduction formula: Using bare field Z Yn Ym 4 ipi·xi 4 −ikj·yj d xie d yje ⟨Ω|T{ϕ0(x1) ··· ϕ0(xn)ϕ0(y1) ··· ϕ0(ym)|Ω⟩ 1 j=1 √ √  n  m  Y Zi Y Zi −−−−−→ ⟨p1 ··· pn|S|k1 ··· km⟩ 0 2 2 2 2 p →Ep p − m + iϵ k − m + iϵ 0 i i=1 i j=1 j k0→E 0 ki Using the renormalized physical field Z Yn Ym 4 ipi·xi 4 −ikj·yj d xie d yje ⟨Ω|T{ϕ(x1) ··· ϕ(xn)ϕ(y1) ··· ϕ(ym)|Ω⟩ 1 j=1  n  m  Y i Y i −−−−−→ ⟨p1 ··· pn|S|k1 ··· km⟩ 0 2 2 2 2 p →Ep p − m + iϵ k − m + iϵ 0 i i=1 i j=1 j k0→E 0 ki • Physical coupling: four point amplitude when scattering with zero momentum, at s = 4m2, t = u = 0,

Y4 2 − 2 √ pi m + iϵ λ0 4 −iλ =iM = lim √ ⟨Ω|ϕ0(p1) ··· ϕ0(p4)|Ω⟩ = ( Z) s→4m2, Z Zλ t=u→0 i=1 Y4 2 − 2 ⟨ | ··· | ⟩ = lim (pi m + iϵ) Ω ϕ(p1) ϕ(p4) Ω s→4m2, t=u→0 i=1

• − 2 − 2 2 − Define δZ = Z 1, δm = m0Z m = m (Zm 1) . 2 δλ = λ0Z − λ = (Zλ − 1)λ • In terms of renormalized physical quantities: Z 4 1 2 1 2 2 λ 4 L = d x Z(∂µϕ) − m Zmϕ − Zλϕ . Z 2 2 4! 1 1 λ 1 1 1 = d4x (∂ ϕ)2 − m2ϕ2 − ϕ4 + δ (∂ ϕ)2 − δ ϕ2 − δ ϕ4 2 µ 2 4! 2 Z µ 2 m 4! λ

The δZ,m,λ terms in the last line are called counterterms. Zλ, Z, Zm are called renormalization constants. δZ, δm, δλ are called counterterm coefficients. Z 1 1 λ 1 1 1 L = d4x (∂ ϕ)2 − m2ϕ2− ϕ4 + δ (∂ ϕ)2 − δ ϕ2 − δ ϕ4 2 µ 2 4! 2 Z µ 2 m 4! λ

• Take the counterterms as perturbation, and do the perturbation calculation, using the counterterms to cancel the divergence from loop diagrams. • Feynman rules: (Homework 7: derive the right two )

• Determine counterterms: Renormalization conditions

The first condition contains two conditions: pole position, mass renormalization, and the residue, field renormalization. Compute the amplitude: • Compute the desired amplitude as the sum of all possible diagrams created from the propagator and vertices above. • The loop integrals in the diagrams will often diverge, so one must introduce a regulator. • The result of this computation will be a function of the three unknown parameters δZ, δm, and δλ. • Adjust (or “renormalize”) these three parameters as necessary to maintain the renormalization conditions. • After this adjustment, the expression for the amplitude should be finite and independent of the regulator. This procedure, using Feynman rules with counterterms, is known as renormalized perturbation theory. The previous method using bare quantities is called bare perturbation theory. One-loop structure of ϕ4 theory 4-point amplitude:

2 iM(s, t, u) = −iλ + (−iλ) [iV(s) + iV(t) + iV(u)] − iδλ. (4)

From the renormalization condtion: iM(s = 4m2, t = u = 0) = −iλ, we then have 2 2 δλ = −λ [V(4m ) + 2V(0)] . Use dimensional regularization

where ϵ = 4 − d. We then have the δλ

the amplitude is then finite: Two point function and Renormalization constants δZ and δm

Renormalization condition: On-shell renormalization condition

2 2 d 2 2 M (p )| 2 2 = 0 , M (p )| 2 2 = 0 p =m dp2 p =m Expand M2 at p2 = m2 to one-loop

Γ(1− d ) • − λ 2 Renormalization constants: δZ = 0 , δm = 2(4π)d/2 (m2)1−d/2 • 4 To one-loop, δZ = 0, this is special for ϕ theory. The first nonzero 2 contribution to δZ comes from two-loop, ∼ O(λ ). Field strength renormalization in Yukawa theory.

The last line Summary of Renormalized perturbation theory 1 Absorb the field-strength renormalizations into the Lagrangian by rescaling the fields. 2 Split each term of the Lagrangian into two pieces, absorbing the infinite and unobservable shifts into counterterms. 3 Specify the renormalization conditions, which define the physical masses and coupling constants and keep the field-strength renormalizations equal to 1. 4 Compute amplitudes with the new Feynman rules, adjusting the counter-terms as necessary to maintain the renormalization conditions. Renormalization of QED

• The bare Lagrangian → renormalized fields and parameters:

L − 1 i 2 ¯ − ¯ − ¯ − 1 µ 2 = (F0,µν ) + (iψ0∂/ψ0 m0ψ0ψ0) e0ψ0A/0,µψ0 (∂µA0 ) 4 2ξ0 1 1 = − Z (Fi )2 + (iZ ψ∂/ψ¯ − mZ ψψ¯ ) − eZ ψ¯A/ψ − (∂ Aµ)2 3 4 µν 2 m 1 2ξ µ √ √ Z1 e µ µ Zm e0 =e 1/2 = 1/2 , A0 = Z3A , ψ0 = Z2ψ, m0 = m = m + δm Z2 Z2Z3 Z3 • Expand Z1,2,3 perturbatively w.r.t e or α, (Notice: δm ≠ δm)

Z1,2,3 =1 + δ1,2,3, Z2m0 = mZm = m + δm , 1 1 L = − (Fi )2 + iψ∂/ψ¯ − eψ¯A/ ψ − mψψ¯ − (∂ Aµ)2 4 µν µ 2ξ µ 1 − δ (Fi )2 + iδ ψ∂/ψ¯ − eδ ψ¯A/ψ − δ ψψ¯ 3 4 µν 2 1 m • Then we do the perturbation using the renormalized parameters. Non-renormalization of ξ

bare photon propagator, R ⟨ ⟩ (2) − d4k −ik·(x−y) (2) i TA0,µ(x)A0,ν (y) = iG0,µν (x y) = i (2π)4 e G0,µν (k). • Previous discussion of Ward id., we have   k k −iξ k k G(2) (k) = g − µ ν A(k2) + 0 µ ν 0,µν µν k2 k2 k2

The ξ0 term is the same as free bare propagator, no loop correction. • ⟨ ⟩ −1⟨ ⟩ For Renormalized fields, TAµ(x)Aν (y) = Z3 TA0,µ(x)A0,ν (y) , we just choose ξ0Z3 = ξ,   k k −iξ k k G(2)(k) = g − µ ν A(k2)Z−1 + µ ν µν µν k2 3 k2 k2 ξ term does not need counter term. Renormalized Lagrangian:

Z1,2,3 =1 + δ1,2,3, Z2m0 = mZm = m + δm , 1 1 L = − (Fi )2 + iψ∂/ψ¯ − eψ¯A/ ψ − mψψ¯ − (∂ Aµ)2 4 µν µ 2ξ µ 1 − δ (Fi )2 + iδ ψ∂/ψ¯ − eδ ψ¯A/ψ − δ ψψ¯ 3 4 µν 2 1 m Feynman gauge

Homework 8: derive the Feynman rule for the counterterms Renormalization constants determined by the renormalization condition. Recall definition of 1PI and amputated amplitudes:

Now, Σ, Π, Γµ also include counterterm contributions.

i p/→m i ⟨Ω|ψ(−p)ψ¯(p)|Ω⟩ = −−−→ p/ − m + Σ(p) p/ − m µν µ ν 2 µν µ ν 2 −i(g − q q /q ) q2→0 −i(g − q q /q ) ⟨Ω|Aµ(−q)Aν (q)|Ω⟩ = −−−−→ q2(1 − Π(q)) q2 Renormalization condition: d Σ(p/ = m) = 0; Σ(p/)| = 0; (5) dp/ p/=m Π(q2 = 0) = 0; −ieΓµ(p′ − p = 0) = −ieγµ (6) Four renormalization constants — Four conditions. Electron self-energy:

−iΣ(p/)|one−loop = −iΣ2(p/) + i(δ2p/ − δm) Dimensional regularization:

d | Renormalization condition:Σ(p/ = m) = 0; dp/ Σ(p/) p/=m = 0, Homework (Homework 9: Derive the results on the previous page from Feynman diagram, using the dimensional regulariztion.) Photon self-energy: to one-loop µν 2 µν µ ν 2 2 µν µ ν 2 Π (q) = i(q g − q q )Π(q ) = i(q g − q q )(Π2(q ) − δ3) Renormalization condition:Π(q2 = 0) = 0

Z   2α m2 Πˆ (q2) ≡ Π (q2) − Π (0) = − dx x(1 − x) log 2 2 2 π m2 − x(1 − x)q2

Homework 10: derive the above three equations from Feynman diagram using dimensional regularization. Interpretatoin of Πˆ 2 2 Analytic struction of Πˆ 2(q ). • q2 < 0, Πˆ real, photon propagator in t or u channel. • When m2 − x(1 − x)q2 < 0, log provides an imaginary part,Π(ˆ q2) has a cut begin at q2 = 4m2, threshold of creating a real e+e− pair. Consistent with Källen-Lehman-representation. • For any fixed q2, the imaginaryp part comes from the integration 1 ± 1 − 2 2 between x = 2 2 β, β = 1 4m /q , Modification of the potential In the non-relativistic limit, V(r) for unlike charges Z d3q −e2 V(x) = eiq·x 3 2 2 (2π) |q| [1 − Πˆ 2(−|q| )]

For q2 ≪ m2, α 4α2 V(x) = − − δ(3)(x) r 15m2 Stronger at small distance. Hydrogen atom: Energy level shift Z   4α2 4α2 ∆E = d3x|ψ(x)|2 · − δ(3)(x) = − 15m2 15m2|ψ(0)|2

For the 2S state, ∆E = −1.123 × 10−7eV. A (small) part of Lamb shift. More precisely, Z d3q −e2 V(x) = eiq·x [1 + Πˆ (−|q|2)] (2π)3 |q|2 2 Z 2 ∞ iQr ie Qe ˆ − 2 ≡ | | = lim 2 dQ 2 2 [1 + Π2( Q )] , (Q q ) µ→0 (2π) r −∞ Q + µ α = − + δV(r) r Pole Q = iµ gives the Coulomb potential. There is also the cut beginning at Q = 2im. The real part of Πˆ 2 does not contribute, When r ≫ 1/m, q ≈ 2m dominant. Approximate the integrand in this region, and t = q − 2m

  α α e−2mr V(r) = − 1 + √ + ··· . r 4 π (mr)3/2

• The range of the correction term ∼ 1/m. δ function was a good approximation. The radiative correction to V(r) is called the Uehling potential. • At r ≳ 1/m, virtual e+e− makes the vacuum a dielectric medium. The apparent charge is less than the true charge. At smaller distances we begin to penetrate the polarization cloud and see the bare charge. This phenomenon is known as vacuum polariza- ation. The opposite limit: small distance or −q2 ≫ m2. Z   2α m2 Πˆ (q2) = − dx x(1 − x) log 2 π m2 − x(1 − x)q2 Z h   i 2α −q2 m2 ≈ dx x(1 − x) log + log(x(1 − x)) + O( ) π m2 q2 h   i α −q2 5 m2 = log − + O( ) 3π m2 3 q2 effective coupling:

2 α α   αeff(q ) = = , A = exp(5/3) 1 − Π(ˆ −q2) − α −q2 1 3π log Am2

Effective charge: larger at small distance. Penetrating the screening cloud of virtual e+e−. Set r = 1/q, αeff(r),distance dependent coupling constant. Unitarity and Optical theorem S-Matrix Unitarity: SS† = 1, S = 1 + iT matrix . 1 = (1 − iT†)(1 + iT) = 1 + iT − iT† + T†T ⇒ −i(T − T†) = T†T

Consider 2 → 2 process:(|k1k2⟩ → |p1p2⟩)  n Z  X Y d3q 1 ⟨p p |T†T|k k ⟩ = i ⟨p p |T†|{q }⟩⟨{q }|T|k k ⟩ 1 2 1 2 (2π)3 2E 1 2 i i 1 2 n i=1 i From the definition X
4 (4) ⟨{qi}|T|k1k2⟩ = (2π) δ k1 + k2 − qi M(k1, k2 → {qi}) Z X  Yn 3  − M → − M∗ → d qi 1 i[ (k1, k2 p1p2) (p1p2 k1, k2)] = 3 (2π) 2Ei X
n i=1 4 (4) ∗ × (2π) δ k1 + k2 − qi M (p1, p2 → {qi})M(k1, k2 → {qi}) i
4 (4) times a overall (2π) δ k1 + k2 − p1 − p2 In general, in abbreviation form : P R − M → − M∗ → M∗ → M → i[ (a b) (b a)] = f dΠf (b f) (a f) Unitarity and Optical theorem P R − M → − M∗ → M∗ → M → i[ (a b) (b a)] = f dΠf (b f) (a f) a, b can be more general states. If a = b, forward scattering X Z ∗ 2ImM(a → a) = dΠfM (a → f)M(a → f) f

For two particle states, scattering cross section: X Z 1 ∗ σtot = dΠfM (a → f)M(a → f) 2E12E2|v1 − v2| f

In c.m.s., using E1 + E2 = Ecm, |v1| = pcm/E1, |v2| = pcm/E2,

ImM(k1, k2 → k1, k2) = 2Ecmpcmσtot Optical theorem in Feynman Diagrams • For s < s0, s0 threshold for internal particles , amplitude M(s) real, satisfies M(s) = M∗(s∗) this can be analytically continued to the complex plane. • 4 For s > s0, M(s) has imaginary part, for example, one-loop ϕ amplitude in ϕ4 theory, ∫ 1 1 M(s) ∼ −λ2[V(s)+V(u)+V(t)], V(s) ∼ − dx(· · ·−log[m2−x(1−x)p2]) 32 0 For s = p2 > 4m2, log term is complex. • There is a cut above s0. ReM(s + iϵ) = ReM(s − iϵ), ImM(s + iϵ) = −ImM(s − iϵ) DiscM(s) = 2iImM(s + iϵ) • Cutkosky rule for Disc. of Feynman diagram: (i) Cut the internal propagators of the diagram and put them on shell simultaneously 1 → − 2 − 2 (ii) For each cut, p2−m2+iϵ 2πiδ(p m ), then perform the loop integral (iii) sum all possible cuts. ,

k = k1 + k2. 0 In the c.o.m frame, k = (k , 0). Integrate over q0, four poles, 1 1 q0 = k0 ± (E − iϵ), q0 = − k0 ± (E − iϵ) 2 q 2 q

Close the integration contour downward, only consider the 0 − 1 0 − q = 2 k + (Eq iϵ) residue which contributes to the discontinuity, picking up the residue is equivalent to 1
→ − 2πiδ (k/2 + q)2 − m2 (k/2 + q)2 − m2 + iϵ −2πi ∼ δ((k0/2 + q0) − E ) 2(k0/2 + q0) q Z   λ2 d3q 1 1 1 1 iδM = − 2πi − 2 (2π)4 2E (k0 − E )2 − E2 2E (k0 + E )2 − E2 Z q q q q q q 2 3   − λ d q 1 1 − 1 1 = 2πi 4 0 0 0 0 2 (2π) 2Eq k (k − 2Eq) 2Eq k (k + 2Eq)

• 0 1 0 − The second term from the residue of q = 2 k + (Eq iϵ), which would not contribute to the discontinuity. • When k0 > 2m, there is a pole on the integral contour, k0 → k0 ± iϵ, imaginary part different,

1 1 ∓ 0 − 0 = P 0 iπδ(k 2Eq) k − 2Eq ± iϵ k − 2Eq

• 0 Discontinuity: imaginary part −2πiδ(k − 2Eq), equivalent to 1
→ −2πiδ (k/2 − q)2 − m2 (k/2 − q)2 − m2 + iϵ The original integral: relabel the internal momentum as p1, p2, Z Z d4q d4p d4p = 1 2 (2π)4δ(4)(p + p − k) (2π)4 (2π)4 (2π)4 1 2 • The discontinuity: replace the propagators by on shell condition 1 → −2πiδ(p2 − m2) 2 − 2 i pi m + iϵ • Thus, we have Disc =2iImM(k) Z 3 3 i d p1 1 d p2 1 |M |2 4 (4) − = 3 3 (k) (2π) δ (p1 + p2 k) 2 (2π) 2E1 (2π) 2E2 Unitarity relation. • The argument can be generalize to any diagram. Vertex function µ µ µ µ µ Γ = γ + δΓ , δΓ (0) = γ (δF1(0) + δ1) iσµν q −ieΓµ(p′, p) = − ie(γµF (q2) + ν F (q2)) 1 2m 2 iσµν q = − ie(γµ(1 + δF (q2) + δ ) + ν F (q2)) 1 1 2m 2 ′ p ,p on-shell µ −−−−−−−→ − ieγ (1 + δF1(0) + δ1) q→0,

Homework 11: derive the above two equitions from Feynman diagram, check in dim reg, δ1 = δ2. Universality of charge renormalization for different fermions: p p ⇒ −1 Z1 = Z2 e = e0Z1 Z2 Z3 = e0 Z3

• e only depends on Z3 photon field renormalization, does not depend on fermions. • ′ If we have a second fermions, µ, do the same renormalization, Z2, ′ Z1 are different from electron.√ √ The theory still has one Z3. ′ ′ ⇒ ′−1 ′ Z1 = Z2 e = e0Z1 Z2 Z3 = e0 Z3. • This is the result of the Ward id. from U(1) gauge symmetry. Form Factor F2— anomalous magnetic moment

iσµν q −ieΓµ(p′, p) = − ie(γµF (q2) + ν F (q2)) 1 2m 2

F2 is finite and is a prediction of QED. Consider an electron interacts with a fixed (t-independent) classical E-M cl potential: Aµ Z 3 cl µ µ ¯ µ ∆H = d x e Aµ j , j (x) = ψ(x)γ ψ(x)

Leading order, in momentum space:

′ M 0 − − ′ µ ˜ cl ′ − i δ(2π)δ(p p) = ieu¯(p )γ u(p)Aµ (p p) ′ 0 − cl δ(p p) means the t independent of Aµ . Considering the quantum correction from vertex: ′ M 0 − − ′ µ ′ cl ′ − i δ(2π)δ(p p) = ieu¯(p )Γ (p , p)u(p)Aµ (p p)

F1 and F2 contain the influence of the E-M field on the electron. To leading order F1 = 1, F2 = 0. Coulomb interaction

cl ˜ cl 0 ˜ In static E-field, Aµ (x) = (ϕ(x), 0, 0, 0), Aµ (q) = ((2π)δ(q )ϕ(q), 0, 0, 0)

iM = −ieu¯(p′)Γ0(p′, p)u(p)ϕ˜(q)

• For spatial slowly varying field, ϕ(q) concentrated around q = 0. • Then, in the nonrelativistic limit, u¯(p′)γ0u(p) = u†(p′)u(p) ≈ 2mξ′†ξ , • The amplitude takes the form ˜ ′† iM = −ieF1(0)ϕ(q) · 2mξ ξ . • This is the Born approx V(x) = eF1(0)ϕ(x), F1(0) is the charge in the unit of e, for electron F1(0) = 1. In fact we use this to define e. • At leading order F1(0)=1, and radiative correction should satisfy 2 δF1(q )|q2→0 + δ1 = 0. cl cl Magnetic moment: Set Aµ (x) = (0, A (x)). h   i iσiν q iM = +ie u¯(p′) γiF + ν F u(p) A˜ i (q) . 1 2m 2 cl  √    p · σξ √ (1 − p · σ/2m)ξ u(p) = √ ≈ m p · σξ¯ (1 + p · σ/2m)ξ • F1 term:   p′ · σ p · σ u¯(p′)γiu(p) =2mξ′† σi + σi ξ 2m 2m   (p′ + p)i iϵijkqjσk =2mξ′† − ξ 2m 2m The first term: nonrelativistic qm, (∇ + eA)2 = ∇2 + e∇ · A + eA · ∇ + e2A2. Thus we keep only the second term. • F2 term: keep only leading term,     iσiν q −iϵijkqjσk u¯(p′) ν u(p) = ξ′† ξ 2m 2m Vertex expanded to linear in qj: h   i iσiν q iM = + ie u¯(p′) γiF + ν F u(p) A˜ i (q) 1 2m 2 cl  −  ′† i ijk j k ˜ i =i(2m)eξ ϵ q σ [F1(0) + F2(0)] ξAcl(q) 2m  −1 = − i(2m)eξ′† σk[F (0) + F (0)] ξB˜ k(q) , B˜ k(q) = −iϵijkqiA˜ j (q) 2m 1 2 cl

• This is the Born approximation of an electron scattering from a potential. The magnetic interaction is e σ V(x) = −⟨µ⟩· B(x), ⟨µ⟩ = [F (0) + F (0)]ξ′† ξ m 1 2 2   • e magnetic moment of electron in standard form µ = g 2m S. g is called Landé g-factor,

g = 2[F1(0) + F2(0)] = 2 + 2F2(0) • The leading 2 is the standard prediction of the Dirac Eq. • QED predict g = 2 + O(α). The correction is called anomalous magnetic moment. • This discussion can also used to other charged particles. Thus we get a correction to the g-factor of the electron: g − 2 α a ≡ = ≈ 0.0011614. e 2 2π

Experiments give ae = 0.0011597. Homework: Homework 12: Peskin page 344, problem 10.2 Renormalized Green’s function & Bare Green’s function • 1/2 Bare field ΦB(x) and renormalized field ΦR(x): ΦB(x) = Z ΦR(x). Bare Green’s function: (n) ≡ ⟨ | | ⟩ GB (x1,..., xn) Ω ΦB(x1) ... ΦB(xn) Ω 1/2 n = (Z ) ⟨Ω|ΦR(x1) ... ΦR(xn)|Ω⟩ n/2 (n) = Z GR (x1,..., xn) If Φ’s are different, differnt Z’s are used. • Generating functional for Bare . Z Z 4 4 SB = d xL[ΦB] = d xLR[ΦR] + Lct = SR + Sct (7)

R R D { 4 } [ ΦB] expR i(S[ΦB] + d xJB(x)ΦB(x)) ZB[JB] = (8) [DΦB] exp{iS[ΦB]}

n (n) 1 δ ZB[JB] G (x1,..., xn) = B in δJ (x ) ... J (x ) B 1 B n JB=0 Renormalized Green’s function & Bare Green’s function • 1/2 RRenormalizedR Green’s funct: define JR = Z JB, JBΦB = JRΦR R R D { 4 } [ ΦR] expR i(SR[ΦR] + Sct + d xJR(x)ΦR(x)) ZR[JR] = ZB[JB] = [DΦR] exp{iS[ΦR] + Sct}

n (n) 1 δ ZR[JR] GR (x1,..., xn) = n i δJR(x1) ... JR(xn) JR=0 n 1 δ ZB[JB] = in Zn/2δJ (x ) ... J (x ) B 1 B n JB=0 −n/2 (n) = Z GB (x1,..., xn) Similar for Connected Green’s function. For on-shell renormalization scheme: in momentum space, near physical pole i iZ G(2)(p2) → , G(2)(p2) → R 2 − 2 B 2 − 2 p mph + iϵ p mph + iϵ Renormalized & Bare Amputated Green’s function • Amputated Green’s funct: ∫ (n) 4 4 (2) (2) (n) GB,conn(x1,..., xn) = d y1 ... d ynGB (x1, y1) ... GB (xn, yn)GB,Amp(y1,..., yn) ∫ (n) 4 4 (2) (2) (n) GR,conn(x1,..., xn) = d y1 ... d ynGR (x1, y1) ... GR (xn, yn)GR,Amp(y1,..., yn)

(n) n/2 (n) (2) (2) using GB (x1,..., xn) = Z GR (x1,..., xn), GB = ZGR : (n) −n/2 (n) GB,Amp(y1,..., yn) = Z GR,Amp(y1,..., yn) • 1/2 Bare & Renormalized 1PI: similar as before JR = Z JB

δWB[JB] 1/2 δWR[JR] 1/2 φB,cl = = Z = Z φR,cl(x) δJB Z δJR 4 ΓB[φB,cl] ≡ WB[JB[φB,cl]] − d xJB[φB,cl]φB,cl(x) Z 4 = WR[JR[φR,cl]] − d xJR[φR,cl]φR,cl(x) ≡ ΓR[φR,cl]

n (n) δ ΓR[φR,cl] n/2 (n) ΓR (x1,..., xn) = = Z ΓB (x1,..., xn) δφR,cl(x1) . . . δφR,cl(xn) Renormalization beyond leading order Multiloop amplitude: could have divergent subdiagram • A otherwise finite diagram with divergent subdiagram:

• Nested or overlapping divergence: two divergent diagrams share a propagator Infinities come from: • k2 very large: x,y,z very close, and w can be farther away

Insert into the self-energy diagram

• 2 2 The log Λ comes from both k1 and k2 very large region. • 2 2 The log p log Λ comes from k2 very large but k1 small region . • 2 2 There would also be another log p log Λ comes from k1 very large but k2 small region . Π(p2) · log Λ2 ∼ log p2 log Λ2 • Previous (p2 polynomial) × (log Λ)n terms: Local divergences, Fourier transform back to δ or derivative of δ functions. • (Non-polynomial of p2)×(log Λ)n terms: Non-local divergernces, a local divergence surrounded by an ordinary non-divergent process. Cancellation of these overlapping divergences: two types of counterterms • Insertions of one-loop counterterm vertices into the one-loop selfenergy diagrams: O(α)×O(α) = O(α2)

Cancel: regions k1 small k2 large, k2 small k1 large • Only local divergences are left, cancel by an additional O(α2) local counterterm δ3

• We can repeat this process to higher loop. BPHZ theorem: Bogoliubov and Parasiuk, completed by Hepp, and refined by Zimmermann, for a general renormalizable quantum field theory, to any order in perturbation theory, all divergences are removed by the counterterm vertices corresponding to superficially divergent amplitudes. In other words, any superficially renormalizable quantum field theory is infact rendered finite when one performs renormalized perturbation theory with the complete set of counterterms. A two-loop example: ϕ4

Crossing symmetry: only six (propagator corrections cancels)

• The last term is the s-channel piece of the counterterm. • The last term is a constant to cancel the momentum independent divergences. • The p dependent divergences are cancelled in the sum of the other diagrams: we will show this. Recall one-loop ϕ4

Counterterm

Separate to two terms: We will show: All the momentum dependent divergences are cancelled in each group. The sum gives: • V(p2) − V(4m2) is finite

 2 • 2 2 ∝ 2 − d → 2 The term left is (V(4m )) Γ (2 2 ) ϵ : a constant, double pole, can be cancelled by δ3. • 2 ≫ 2 2 − 2 2 ∼ 2 p2 p m : (V(p ) V(4m )) log m2 • Higher order: Feynman parameterization the first two factor, and insert V(q2) Z i Γ(2 − d ) 1 1 V(q2) = − 2 dx d/2 2 2 2−d/2 2 (4π) 0 [m − x(1 − x)q ]

Use identity: Z 1 1 wα−1(1 − w)β−1 Γ(α + β) = dw . α β − α+β A B 0 [wA + (1 w)B] Γ(α)Γ(β) Complete the square: The denominator, −[(1 − w) + wx(1 − x)]ℓ2 − P2 + m2; ℓ: a shifted momentum; P2: a 2 2 complicte function of p, p3,x, y, w. At w → 0, P (w) = y(1 − y)p + O(w) Change variable and after integration:

R − 1 1−d/2 One Γ(4 d) pole, another from w integration 0 dww f(w). Separate the pole at w = 0 Z Z Z 1 1 1 dw w1−d/2f(w) = dw w1−d/2f(0) + dw w1−d/2[f(w) − f(0)] 0 0 0 • The second piece is

At d → 4, the residue of 1/ϵ is a constant, which can be absorbed into O(λ3) vertex counterterm. • The first piece:

R   ∼ − iλ3 2 1 − 2 − − 2 Nonlocal divergence: 2(4π)4 ϵ 0 dy log[m y(1 y)p ] The “t + u” counterterm diagram:

R   ∼ iλ3 2 1 − 2 − − 2 Nonlocal divergence: 2(4π)4 ϵ 0 dy log[m y(1 y)p ]

• The nonlocal divergences in Group II cancel. • There are double poles not cancel: the vertex counterterm contains double pole. • The finite terms contain double logrithms, λ3 log2 p2 as p → ∞.