3/16/07 Lecture 20 outline Bare and renormalized fields, and counterterms. ψ = Z1/2ψ , Aµ = Z1/2Aµ , • B 2 R B 3 R e Z Z1/2 = e Z . = + . B 2 3 R 1 LB LR Lc.t. 1 = F F µν + ψ¯ (i∂/ e A/ m )ψ , LR −4 Rµν R R − R R − R R

1 = δ (F )2 + ψ¯ (iδ ∂/ δ e A/ δ )ψ . Lct −4 3 Rµν R 2 − 1 R R − m R Where δ = Z 1, δ = Z 1, δ = Z 1, and δ = Z m m. We have 1 1 − 2 2 − 3 3 − m 2 0 −

1/2 eBZ2Z3 = eRZ1,

−1 µ 2 −1 µ where the Z1 will cancel the Z1 in Γ (q =0)= Z1 γ . Gauge invariance requires Z = Z , since then δ = δ and the counterterm pieces • 1 2 1 2 have the same gauge invariance. Sure enough, direct calculation shows Z1 = Z2 (to

all orders in perturbation, theory, and exactly)! So eR = √Z3e0 = ephys. Shows that renormalized charge is same for all species (e.g. electron and muon and anti-proton all have exactly the same effective charge). Note that ∆(eA) = 1, gives ∆(e)=2 1 D = /2, so • − 2

 −1 αB = µ Z3 αR.

4 Take d/d ln µ and use dαB/d ln µ = 0 to get (just as we did for λφ )

− − d − 0 = αZ 1 + β(α, )Z 1 + β(α, )α Z 1. 3 3 dα 3

where β(α, ) = dα/d ln µ. To have a smooth  0 limit, we need β(α, ) = α + β(α), → − and then da β(α) = α2 1 . dλ − Now let us compute a . Z 1 = 1 Π(0), so a is minus the coefficient of the 1/ • 1 3 − 1 pole in Π(0). Let us compute it. Loop correction to photon propagator, from virtual electron/positron loop: •

iΠµν (p) = i(p2g pµpν)Π(p2). 2 µν − 1 Z = (1 Π(0))−1. To 1-loop, we take δ = Π(0). To one-loop, find (See Peskin p. 251 3 − 3 for details)

1 1 µν − 2 µν µ ν 2 D/2 1 2 2 D Π2 (p)=(p g p p )( 8e /(4π) ) dxx(1 x)Γ(2 2 D)/∆ , − − Z0 − −

where ∆ = m2 x(1 x)p2. We only care about the −1 term. Result to one loop: e − −

α − 2 Π(0) =  1 + finite. − π 3 in MS, choose δ to cancel the 1/ term only, so δ = α −1 2 + higher loop. Finally, we 3 3 − π 3 −1 −k 2 have Z3 = 1 + k ak , with a1 =2α /3π to one loop. This then givesP

dα da 2α2 β(α) = = α2 1 = + higher loops. d ln µ dλ 3π

The beta function is positive. Qualitatively similar to λφ4: the theory is not asymp- • totically free. Integrate 1-loop beta function:

− 2 µ α 1(µ) = ln( ). −3π Λ

Makes sense only for µ < Λ, i.e. in the IR. Λ is a UV cutoff. Get α as µ Λ; this →∞ → is the Landau pole. Looks bad, but we’ll see the the energy scale where it blows up is so fantastically large that we don’t need to worry (something new should fix it in the UV, e.g. grand unification can do the job). It does not run to zero in the IR, because there are no massless charged particles. It runs toward zero until it gets to the energy scale of the 3 lightest charged particle, me =0.5MeV , and then it stops running. So 137 = 3π ln(Λ/me).

Gives Λ = me exp(137π), which too huge to worry about the apparent Landau pole there.

(Other charged particles will bring the scale of Λ down to Λ = me exp(137π/Nf ) where

Nf is the effective number of charged particles, but it’s still huge.) Picture of vacuum polarization, of electron-positron pairs. QED vs QED. In QED, we have gauge invariance ψ eief(x)ψ, local U(1) transfor- • → mations. Generalize to local SU(N ) gauge transformations: ψ exp(igT af (x))ψ, where c → a T a are traceless, Hermitian N N matrices (a =1 ...N 2 1), and ψ is a N column vec- c × c c − c tor. Gauge conserved color charge. Need covariant derivatives, ∂ D = ∂ + igAaT a, µ → µ µ µ a b c i.e. introduce gauge fields, “gluons”. The Ta matrices do not commute, [T , T ] = ifabcT :

2 a the group is “non-Abelian.” The effect of this is that the Aµ kinetic terms are more compli- cated. The physics of this is that the gluons carry color charge (unlike the photon, which carries no electric charge). Get 3 and 4 gluon interaction diagrams. Added contributions to 1-loop correction to gluon propagator. Get finally

α2 β(α) = ( 11N +2N ) . 6π − c f

The flavors contribute positively, as in QED. But the colors contribute negatively: they

anti-screen charges! So the beta function can be negative, if 11Nc > 2Nf . This is asymp- totic freedom. Integrating the 1-loop result gives

− (11N 2N ) µ α(µ) 1 = c − f ln( ). 6π Λ

To have α > 0, we need µ > Λ (opposite from QED). Note α(µ ) 0, weak in → ∞ → UV = asymptotic freedom. Explains successes of parton model (quarks) for high energy

scattering. For QCD, Nc = 3, and Nf = 6. For energies below the top and bottom mass, use N eff = 4. Observe e.g. α(100GeV ) 0.1, so Λ 200MeV . f ∼ ∼ On the other hand, α for µ Λ: forces are strong in IR, below scale Λ. Can → ∞ → explain confinement of quarks (there is a million dollar prize, waiting to be collected, if you prove it in detail)!

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