Effective Field Theory
Iain Stewart MIT
The 19’th Taiwan Spring School on Particles and Fields April, 2006 Physics compartmentalized short distance String quantum gravity Theory? electro-weak Quantum General Relativity Field Theory QCD & quarks Special nuclei Quantum Relativity Mechanics Newtonian atoms Gravity Classical Classical chemistry Mechanics Electricity & Magnetism long distance us
But, one doesn’t need nuclear physics to build a boat
Generality vs. Precision Dynamics at long distance do not depend on the details of what happens at short distance λ ∼ 1 In the quantum realm, p , wavelength and momentum are related, so
Low energy interactions do not depend on the details of high energy interactions Good: • we can focus on the relevant interactions & degrees of freedom Newton didn’t need quantum gravity calculations are simpler for projectile motion Bad: Phew! we have to work harder to probe the interesting physics at short distances Outline Top Lecture I Principles, Operators, Power Counting , Bottom
NRQED, Why the Sky is Blue, Weak Decays
Operator Counting, Matching, Using Equations of Motion • Quantum Loops, Renormalization and Decoupling
Summing Large Logarithms
α Lecture II QCD, s matching, Heavy Quark Effective Theory (An Effective Theory for Static Sources) Lecture III Soft - Collinear Effective Theory Masses particle mass 17 γ < 6 10− eV gluon ×0 (theory) 2 5 2 νe, νµ, ντ ∆msun = 8 10− eV 2 × 3 2 ∆matm = 2 10− eV e ×0.511 MeV u quark 4 MeV d quark ∼ 7 MeV µ ∼106 MeV s quark 120 MeV c quark ∼ 1.4 GeV τ ∼1.78 GeV b quark 4.5 GeV W boson ∼80.4 GeV Z boson 91.2 GeV higgs > 114 GeV t quark 174 GeV
mproton ∼ 1 GeV Lecture I EFT Concepts
1) The ingredients in any EFT: Relevant degrees of freedom, symmetries, scales & power counting
2) Renormalization: Meaning of parameters
3) Decoupling: Effects from Heavy Particles are suppressed
4) Matching: How we can encode dynamics of one theory into another
5) Running: Connecting physics at different momentum scales Example: Hydrogen non-relativistic quantum mechanics
parameters: mass me charges Qe , Qp 1 coupling α = degrees of 137 freedom: → ∞ scales: mp = 938 MeV e- me = 0.511 MeV 1 proton p meα = 3.7 keV (aBohr)− ∼ ∼ m α2 13.6 eV E = e = + corrections n − 2n2 − n2
Why not quarks? QCD? b-quark charge? e + ? weak force? m p r o t o n ? spin? Equations for Taiwan School
Iain W. Stewart1
1Center for Theoretical Physics, Massachusetts Institute of Technology, Cambridge, MA 02139
−1 meα ! (proton size) ∼ ΛQCD ∼ 200 MeV meα ! mp ∼ 1 GeV meα ! me
p2 • L p, e−, γ, b α, m L p, e−, γ α" O ( ; b) = ( ; ) + ! 2 " (1) mb e suppressed coupling changed, iα2 1 E4quatioµn2 s for Tai4wka2n School b it rsuns: Aα(0) =¯A 0 = TF (T ⊗ T 13)7 ln M − . (21g) k2 3 m2 15m12 1 ! " Iai#n W. Stewart $ p α(mW ) = 128 In Eq. (21) the matrix elemen1tCenter for Theoretical Physics, Massachusetts Institute of Technology, Cambridge, MA 02139 Insensitive to quarks2 in 2prot2 on: 2 2 q 0 p k S 3k Λ k T ∼ m αM! = 1 + −−1 ∼ −∼ − . (22) pγ e (proton smize2) 3mΛQ2CD 22m002MeV 12m2 q q Note that we disagreeshorwitht distancRef. [2e3] on thelongord distancer v0 sepin independent part of Eq. (21b). − 2 r = Λ 1 Ref. [23] has an additional non-logarithmic 1/m term. The differeQnCcDe arises because we 2 find a different value for the order k term in the non-Abelian F1 form-factor. Eqs. (21c) through (21g) agree with Ref. [23]. The diagrams in Eq. (19) and (21) have contributions from several different scales. In particular, in the language of the threshold expansion [13], the hard regime gives ln(µ/m)’s, the soft regime gives ln(µ/k)’s, and the ultrasoft regime gives ln(µ/λ)’s. There are also i ln(λ/p) and i ln(λ/k) terms from the Coulomb divergence in the potential regime. In addition to logarithms, all regimes can give constant factors. In the effective theory, terms from the hard regime are absorbed into matching coefficients such as the four-quark potential operators and the remaining terms correspond to graphs involving modes in the effective theory. These graphs are discussed in more detail below. Next consider the one-loop annihilation diagrams in QCD. Since the intermediate gluons are hard we expect these graphs to include a factor of 1/m2, thus giving hard order v0 contributions to the potential. However, the graph in Eq. (23c) also has terms enhanced by a factor of m/p, which are order 1/v contributions.
2 2 iαs C1 Cd(Nc −1) Cd A ¯A + = − 2 + 2 (1 ⊗ 1) + −2C1 (T ⊗ T ) m % Nc 8N 4Nc !" c # " # $ (N 2 −1) 1 S2 c A ¯A × ( −2)(iπ+2−2 ln 2) − CA 2 (1 ⊗ 1)+ (T ⊗ T ) ! 2Nc Nc $ iπ 1 ln 2 λ × S2 + − −ln , (23a) 12 6 6 m & ! " #$ 2 2 iαs A ¯A 1 (Nc − 1) + = − 2 CA (T ⊗ T ) + (1 ⊗ 1) m ! Nc 2Nc $ 3 µ 4 2 ln 2 iπ × S2 ln + + − , (23b) 2 m 3 3 3 ! " # $ 2 2 iαs A ¯A 1 (Nc − 1) + = − 2 (2CF − CA) (T ⊗ T ) + (1 ⊗ 1) 4m ! Nc 2Nc $ π2m 2iπm 2p µ λ × S2 + ln − 4 + 2 ln + 4 ln , (23c) p p λ m m ! " # " # " #$ 11 Equations for Taiwan School
Iain W. Stewart1
1Center for Theoretical Physics, Massachusetts Institute of Technology, Cambridge, MA 02139
m α ! −1 ∼ ∼ • Insensitivee (pr ottoo nprotsizeon) mass:ΛQCD 200 MeV Equations for Taiwan School meα ! mp ∼ 1 GeV (static proton suffices) 2 Iain W. Stewart1 me 4 hyperfine splitting ∼ µ µ α 1 e p Center for Theoretical Physics, Massachusetts Institute of Technology, Cambridge, MA 02139 mp from QCD ! −1 ∼ ∼ meα but(p universalroton size) ΛQCD 200 MeV
meα ! mp ∼ 1 GeV
Experimentmeα ! m fore m p , µ p is more accurate that QCD computations 2 L − L − " O p (p, e , γ, b; α, mb) = (p, e , γ; α ) + 2 (1) !mb "
∞ NRQED n L = L0 + # Ln (2) n#=1
2 † 0 ∇ † 0 † † L0 = ψ i∂ − + . . . ψ + Ψ i∂ Ψ + “V”(ψ ψ)(Ψ Ψ) (3) ! 2me " L ¯ − − 1 µν Why QED = ψ(iD/ m)ψ 4 F Fµν ?
Why LSM ?
SU(3) × SU(2) × U(1) L
d 1 µ 1 2 2 λ 4 τ 6 S[φ] = d x ∂µφ∂ φ − m φ − φ − φ (4) $ !2 2 4! 6! "
L Ldim-4 1 T i j " k Ldim-6 = + new LL C#ijH H #"kLL + + . . . (5) Λ % &
C = iγ2γ0
ν L (6) ' eL ( Equations for Taiwan School Equations for Taiwan School Iain W. Stewart1 Iain W. Stewart1 1Center for Theoretical Physics, Massachusetts Institute of Technology, Cambridge, MA 02139 1Center for Theoretical Physics, Massachusetts Institute of Technology, Cambridge, MA 02139
m α ! (proton size)−1 ∼ Λ ∼ 200 MeV e EquationQs CfDor Taiwan School − ! ∼ m α ! 1 ∼ ∼ meα mp 1 GeV Iain W. Stewaret1 (proton size) ΛQCD 200 MeV
1 ! Cenmtereαfor Thmeoeretical Physics, Massachusetts Instmituteeαof !Techmnolpog∼y, C1amGbreidVge, MA 02139
• −1 m α ! m + meα ! (proton sizNe)on∼-relativisticΛQCD ∼ 200 M LaeV grangian e e (no e ) 2 − − " p m α ! m ∼ L(p, e , γ, b; α, mb) = L(p, e , γ; α ) + O (1) e p 1 GeV !m2 " QED short distance theory b meα ! me is more general expand in ∞ p m , e , α LNRQED L n L 2 = 0 + # n (2) me mp p # L(p, e−, γ, b; α, m ) = L(p, e−, γ; α") + O n=1 (1) b !m2 " NRQED long distance theory whereb its easier to compute ∞ exact answer is irrelevant, work to the desired level of precision LNRQED = L + #n L (2) 0 # n n=1 Leading Order 2 † 0 ∇ † 0 † † L0 = ψ i∂ − + . . . ψ + Ψ i∂ Ψ + “V”(ψ ψ)(Ψ Ψ) (3) ! 2me " Equations for Taiwan School Equations for Taiwan School Iain W. Stewart1 Iain W. Stewart1 1Center for Theoretical Physics, Massachusetts Institute of Technology, Cambridge, MA 02139 1Center for Theoretical Physics, Massachusetts Institute of Technology, Cambridge, MA 02139 Equations for Taiwan School −1 meα ! (proton size) ∼ ΛQCD ∼ 200 MeV 1 Imainα W. Stewart −1 meα ! mp ∼ 1 GeV e ! (proton size) ∼ ΛQCD ∼ 200 MeV
1Center !for Theoretical Physics, Massachusetts Institute of Technology, Cambridge, MA 02139 meα me meα ! mp ∼ 1 GeV
• m α ! m + Non−-relativistic Lagrangian e e (no e ) ! 1 ∼ ∼ 2 meα (proton size) ΛQCD 200 MeV − − " p L(p, e , γ, b; α, m ) = L(p, e , γ; α ) + O (1) b !m2 " QED short distance theory b meα ! mp ∼ 1 GeV is more general expand in m ∞ meα ! me p e α NRQED n , , L = L0 + # Ln (2) e mp # m n=1 NRQED long distance theory where its easier to compute p2 exact answL per, ise− irrele, γ,vant,b α w, mork to theL desiredp, e− le, velγ αof" precisionO ( ; b) = ( ; ) + ! 2 " (1) mb Symmetries of QED constrain the form of NRQED: + − Charge conjugation ( e ↔ e ) ∞ Parity ( ! x → − !x ) LNRQED = L + #n L (2) Time-Reversal ( t → − t ) constrain0 # the n ’s n=1 Gauge Symmetry Spin-Statistics Theorem Equations for Taiwan School Equations for Taiwan School Equations for Taiwan School Equations for Taiwan School Iain W. Stewart1 1 Iain W. Stewart 1 Iain W. Stewart1 Iain W. Stewart 1Center for Theoretical Physics, Massachusetts Institute of Technology, Cambridge, MA 02139 1 1 1Center for TheoreticaClePnhteyrsicfso,rMTahsesoacrehtuisceatltsPIhnCysetsinitctuestr,e fMorfaTssehcaehocnrheoutlisoceagltyt,PsChIyansmisctsbi,rtiuMdtgaees,soaMfchTAuesec0th2ts1n3oI9nlosgtiytu,teCoafmTbecrhidngoelo,gyM, CAam0b2r1id3g9e, MA 02139 −1 meα ! (proton size) ∼ ΛQCD ∼ 200 MeV ! −1 ∼ ∼ ! −1 !∼ ∼ −1 ∼meα (∼proton size) ΛQCD 200 MeV meα ! mp ∼ 1 GeV meα (proton sizme)eα Λ(pQrCoDton2s0i0zeM) eV ΛQCD 200 MeV Equations for Taiwan School meα ! mp ∼ 1 GeV m α ! m m , µ meα ! mp ∼ 1 GeV e e p p meα ! mp ∼ 1 GeV Iain W. Stewart1 meα ! me m α ! m 1 e e m α ! m 2 Center for Theoretical Physics, Massachusetts Institute of Technology, Cambridge, MA 02139 L − L − e " O ep (p, e , γ, b; α, mb) = (p, e , γ; α ) + 2 (1) !mb " m α ! (proton size)−1 ∼ Λ ∼ 200 MeV 2 e QCD L − L − " O p 2 (p, e , γ, b; α, mb) = (p, e , γ; α ) + ! 2 " (1) ∞ − m α ! m ∼ 1 GeV− " p mb L(p, e , γ, b; α,emb) =p L(p, e , γ; α ) + O 2 (1) LNRQED L n L − !m2 " − " p = 0 + # n L(p, e , (γ2,)b; α, mb) =b L(p, e , γ; α ) + O (1) meα ! me ! 2 " n#=1 mb∞ LNRQED = L + #n L (2) ∞ 0 # n n=1 ∇2 LNRQED L n L p2 † 0 † 0 † † = 0 + # n L − ∞ L − " O (2) L0 = ψ i∂ − + . . . ψ + Ψ i∂ Ψ + “V”(ψ ψ)(Ψ Ψ) #(3) (p, e , γ, b; α, mb) = (p, e , γ; α ) + ! 2 " (1) n=1 NRQED n mb ! 2me " L = L0 + # Ln (2) #∇2 1 µν L † 0 −n=1 † 0 † † Why LQED = ψ¯(iD/ − m)ψ − F Fµν ? 0 = ψ i∂ +∞. . . ψ + Ψ i∂ Ψ + “V”(ψ ψ)(Ψ Ψ) (3) 4 !NRQED 2m n" 2 L = L e+ # L (2) † 0 ∇ † 0 † † 0 # n L0 = ψ i∂ − + . . . ψ + Ψ i∂ Ψ + “V”(ψ ψ)(Ψ Ψ) n=1 (3) Why LSM ? ! L" ¯ − − 1 µν 2meWhy QED = ψ(iD/ m2)ψ 4 F Fµν ? † 0 ∇ † 0 † † L0 = ψ i∂ − + . . . ψ∇+2 Ψ i∂ Ψ + “V”(ψ ψ)(Ψ Ψ) (3) SU(3) × SU(2) × U(1) L L ¯ − − 1 µν L ! † 0 " † 0 † † Why QED = ψ(iD/ m)ψ 4 F Fµν ? Why SM ? 2mL0 e= ψ i∂ − + . . . ψ + Ψ i∂ Ψ + “V”(ψ ψ)(Ψ Ψ) (3) ! 2me " 1 µν ¯ 1 µν Why L = ψ¯(iD/ − m)ψW−hy LQFED =Fψ(iD/ −? m)ψ − F Fµν ? d 1 µ 1 2 2 λ 4QEDτ 6 4 µν 4 S[φ] = d xUse:∂µφ∂ Obserφ − mvedφ d.o.f− φ. − φ (4) $ 2 2 4! 6! Why L ? ! Lorentz" Invariance SM SymmetriesWhy LSM ? Gauge Invariance SU(3) × SU(2) × U(1) L Ldim-4 1 T i j " k ×Ldim-6 × L = + new LL C#ijH H #S"kUL(L3)+USnitarityU(2+) .(H. .ermitianU(1) ) (5) Λ % &
C = iγ2γ0 Renormalizability (absorb divergences from loops in finite # of parameters) this was not in our list! ν L (6) ' eL ( Modern Definition: renormalizable order by order in power expansion ∞ n L = L0 + # Ln (7) Ken Wilson n#=1 Equations for Taiwan School
Iain W. Stewart1
1Center for Theoretical Physics, Massachusetts Institute of Technology, Cambridge, MA 02139
−1 meα ! (proton size) ∼ ΛQCD ∼ 200 MeV meα ! mp ∼ 1 GeV meα ! me
2 L − L − " O p (p, e , γ, b; α, mb) = (p, e , γ; α ) + 2 (1) !mb "
∞ NRQED n L = L0 + # Ln (2) n#=1
2 † 0 ∇ † 0 † † L0 = ψ i∂ − + . . . ψ + Ψ i∂ Ψ + “V”(ψ ψ)(Ψ Ψ) (3) ! 2me " L ¯Pow−er Counting− 1 µν in Mass Dimension Why QED = ψ(iD/ m)ψ 4 F Fµν ? Dimension of Operators Why LSM ? (Marginal, Irrelevant, Relevant Operators) SU(3) × SU(2) × U(1) L
d 1 µ − 1 2 2 − λ 4 − τ 6 Consider S[φ] = d x ∂µφ∂ φ m φ φ φ (4) $ !2 2 4! 6! " (board) L Ldim-4 1 T i j " k Ldim-6 = + new LL C#ijH H #"kLL + + . . . (5) Λ % &
C = iγ2γ0
ν L (6) ' eL ( Equations for Taiwan School Equations for Taiwan School Iain W. Stewart1 Iain W. Stewart1
1 Center for Theoretical Physics, Massachusetts Institu1tCeenotferTfeocrhTnhoeloorgeyti,caCl aPmhybsricisd,gMe,asMsaAchu0s2e1tt3s9Institute of Technology, Cambridge, MA 02139
−1 ! −1 ∼ ∼ meα ! (proton size) ∼ ΛQCD ∼ 200 MeV meα (proton size) ΛQCD 200 MeV Equations for Taiwan School ! ∼ ! ∼ meα mp 1 GeV meα mp 1 GeV Iain W. Stewart1
meα ! me 1Center for Theoretical Physics, Massachusetts Institute of Technology, Cambridge, MA 02139 meα ! me
−1 meα ! (proton size) ∼ ΛQCD ∼ 200 MeV 2 L − L − " O p meα ! mp ∼ 1 GeV (p, e , γ, b; α, mb) = (p, e , γ; α ) + 2 (1) p2 !mb " L − L ! − " O (p, e , γ, b; α, mb) =meα(p,mee , γ; α ) + 2 (1) mb ! " ∞ LNRQED L n L = 0 +p2 # n (2) L(p, e−, γ, b; α, m ) = L(p, e−, γ; α") + O (1) ∞ b n#2=1 !mb " NRQED n L = L0 + # Ln (2) 2 ∞ n#=1 NRQED ∇ n †L 0 = L0 + # Ln † 0 † † (2) L0 = ψ i∂ − + . . . ψ + Ψ i∂ Ψ + “V”(ψ ψ)(Ψ Ψ) (3) n#=1 ! 2me "
∇2 L ¯ − − 1 µν ∇2 † 0 Why QED† =0ψ(iD/ m)ψ †† 4 F0 F† µν ? † 0 † † L − L0 = ψ i∂ − + . . . ψ + Ψ i∂ Ψ + “V”(ψ ψ)(Ψ Ψ) (3) 0 = ψ i∂ + . . . ψ + Ψ i∂ Ψ + “V”(ψ !ψ)(Ψ2mΨe ) " (3) ! 2me " Why LLSM ? ¯ − − 1 µν Why QED = ψ(iD/ m)ψ 4 F Fµν ? 1 µν Why LQED = ψ¯(iD/ − m)ψ − F Fµν ? L 4 SUW(h3y) ×SMS?U(2) × U(1) L SU(3) × SU(2) × U(1) L Why LSM ? d 1 µ 1 2 2 λ 4 τ 6 S[φ] = d x ∂µφ∂ φ − m φ − φ − φ (4) d 1 µ − 1 2 2 − λ 4 − τ 6 S[φ] = d x ∂$µφ∂ φ !2 m φ φ 2 φ 4! 6 " (4) SU(3) × SU(2) × U(1) L $ !2 2 4! 6 "
1 L Ldim-4 dim 4¯T i 1 j " T i k Ldjim-6" k dim 6 = L +L new- LL C#ijH H¯#"kLL + + . . . L - (5) d 1 µ 1 2 2 λ 4 τ =6 Λ % + new LL C#&ijH H #"kLL + + . . . (5) S[φ] = d x ∂µφ∂ φ − m φ − φ − φ Λ % & (4) $ !2 C = iγ22γ0 4! 6 " eg. Standard Model C = iγ2γ0 Left-handed Lepton doublet ν L (6) ' eL ( L Ldim-4 1 T i j " k Ldim-6 = + new LL C#ijH H #"kLL + + . . . (5) Λ % &
C = iγ2γ0 Higgs SU(2) doublet
ν Exercise 1: Show that this is mostL general dim-5 operator that is (6) ' eL ( consistent with the symmetries of the Standard Model
When the Higgs gets a vev this gives a small Majorana Neutrino Mass Equations for Taiwan School
Iain W. Stewart1
1Center for Theoretical Physics, Massachusetts Institute of Technology, Cambridge, MA 02139
−1 meα ! (proton size) ∼ ΛQCD ∼ 200 MeV
meα ! mp ∼ 1 GeV
meα ! me mp, µp
p2 L − L − " O (p, e , γ, b; α, mb) = (p, e , γ; α ) + 2 (1) !mb " 2 ∞ NRQED n I. WHY THE SKY ILS BLUE= L0 + # Ln (2) n#=1
∇2 † 0 † 0 † † L0 = ψ i∂ − + . . . ψ + Ψ i∂ Ψ + “V”(ψ ψ)(Ψ Ψ) (3) 2 ! −1 2me " 2 Eγ ! ∆E ∼ meα ! a0 ∼ meα ! Matom (10) L ¯ − − 1 µν Why QED = ψ(iD/ m)ψ 4 F Fµν ? I. WHY THE SKY IS BLUE Why LSM ? 2 vµ = (1, 0, 0, 0) (11) SU(3) × SU(I2.) ×WU(H1)Y TLHE SKY IS BLUE 2 −1 Eγ ! ∆E ∼ meα ! a ∼ meα ! Matom (10) 0 2 † 0 †d 1 µ − 1 2 2 − λ 4 − τ 6 L = φv i∂S[φφv] == φdv ixv · ∂∂φµvφ∂ φ m φ φ φ (12) (4) $ !2 2 4! 6! " I. WHY THE S!KY IS B∼LUE 2 ! −1 ∼ ! 2 2 Eγ ∆E meα a0 meα Matom (10) vµ = (1, 0, 0, 0) (11) L Ldim-4 I.1 WHTYi THE SjKY" IS Bk LUELdim-6 I. WHY THE SKY IS BLUE = + new LL C#ijH H #"kLL + + . . . (5) [φv] = 3/2 Λ % & (13) 2 −1 E ! ∆E ∼Cm=αiγ2!γ0a ∼ m α ! M (10) γ e 0 e vµ a=tom(1, 0, 0, 0) (11) † † L 0 · 2 −1 2 = φv i∂ φv = φv iv ∂ φv Eγ ! ∆E ∼ meα ! a ∼ me(α1!2) Matom (10) 2 −1 0 2 E ! ∆E ∼ m α ! a int∼ m α !† M µν † λ νL σµ (10) γ e L0 = τe1 φv φvFatµoνmF + τ2 φv φvv FλµvσF (14) (6) vµ = (1, 0, 0, 0) ' eL ( (11) 2 I. WHY THE SKY IS BLUE † 0 † I. WHY THE SKY IS BLUE L = φv i∂ φv = φv iv · ∂ φv µ (12) v = (1, 0, 0, 0) I. WHY(11T)HE SKY IS BLUE [φµv] = 3/2 ∞ (13) 2 v = (1, 0, 0, 0) † λ Lµν L n L (11) † 0 † +τ3 φv φv(v i∂λ)Fµν F = 0 + # n (15) (7) 2 2 L = φv i∂ φv = φv iv · ∂ φv (12) 2 I. WHn#Y=1THE SKY IS BLUE 2 L = φ† i∂0 φ = φ† iv · ∂ φ I. WHY(1T2H) E SKY IS BLUE even number of Fµν ’s [φv] = 3/2 ! v ∼ v 2v! −1v ∼ ! (13)EI. !W∆HEY∼TmHEα2S!KYa−I1S∼BLmUαE ! M (10) int † µν I. W† HYλTHE SKYσµIS BLUEEγ ∆E meα a0 meα Matom γ e 0 e (10at)om ˜ L = τ1 φv φvFµν†F 0 + τ2 φ†v φvv FλµvσF (n) (14) no Fµν2here −1 L = φ i∂ φv = φ iv · ∂ φv Ltheory1 → L (12) (8) ! ∼ ! ∼ I. vW[!φH]Y=T3/H2vE SKY IS BLUE t(h1e3o)ry2 Eγ ∆E mµ eα µa0 meα v Matom #n (10) ∂ Fµν = 0, v ∂µφv = 0 ! [φ∼v] = 3/2 ! −1 ∼ ! ! ! ∼ ∼ 2(1!µ32)!−1 ∼−1 ∼ ! ! int † µν † Eλγ ∆E σµ meα a0 meα Matom EγEγ ∆E∆Emeαmveα =a0(1a,00,m0,e0αm)(1eα0)MatMomatom (10)(1(01)1) L 2 −1 E ! =∆Eτ1∼φmv φαvF!µνaF ∼+mτ2αφ!v φMvv FλµvσF (n)µ (10) (14) eg.† Whyλγ theµeν Sky 0is Bluee atom L v = (1, 0, 0, 0) (9) (11) int +τ3 †φv φv([φvvµ]iν∂=λ3)F/2µν†F λ σµ theory2 (1(31)5) L = τ1 φv φvFµν F + τ2 φv φvv FλµvσF [Fµν ] = 2 #n (14) (16) 2 −1 Lint † µν † λ σµ L µ †µ 0 † · Eγ ! ∆ELow∼ m energyeα ! ascatt∼eringmeα of! photMatoonsm= τ1 fφromv φvF µneutralν F +µτ2 atφvomsφvv FλµvσF (10) =v φv=v(1i(4∂=1), 0(φ1,v,00,=0, 0)φ,v0)iv ∂ φv (11)(1(11)2) even number of Fµν ’s µ 0 v = (1, 0, 0, 0) (11) µ v = (1, 0, 0, 0) v = (†1, 0, 0,λ0) µν 2 −1 (11) (11) ˜ in their ground+τ3 φ statφv(ev i∂λE)Fµ!ν F∆E ∼ m α ! a ∼ m α ! M (15) (10) no Fµν here int † µν † λ v σµ γ e † 0 0 e † atom L † λ µν L = φv i∂ φv = φv iv · ∂ φv (12) = τ1 φv+φτv3 FφµvνφFv(v +i∂λτ2)Fφµvν Fφvv FλµvσF (15)(14) µ µ † λ µν L † †0 0 † †· ∂ F = 0, v ∂ φ = 0 =Lφ=v iφ∂ iφ∂v[φ=φv]φ==v 3iφv/2i∂vφ· v∂ φv (12)(1(21)3) µν µ v even number of F ’s [τ1] = [τ2] = −3 +τ3 φv φv(v i∂λ)Fµν F (17) v(15) v v µν • µ L † 0 † · even number of F˜µν ’s Let v =L (1, †0, 00, 0) , at†oms· are static = φv i∂ φv = φv iv ∂ φv (11) (12) no Fµν here even num=beφr vofi∂Fµφν ’vs = φv iv ∂ φv (12) no F˜µν here µ † 0µ † † ˜ λ µν µ Lµ∂ F = 0, v ∂ φ = 0 +τ·3 φ[Fvnφov](Fvµ=νi2h∂eλr)eFµν F (1(51)6) Lint † [φv][φ=]µ3=ν/23/2 † λ σµ (13)(13) ∂ F = 0, v ∂=φ µφ=ν 0 i∂ φµv =v φ iv ∂ φµvν is field which destroys an atom, mass dim. [φv] = 3(/122) = τ1 φv φvFµν Fv + τ2 φv φvv Fλ(µ1v3σ)F (14) µν µ v v v µ µ 3 ∂ Fµν = 0, v ∂µφv = 0 τ1 ∼ τ2 ∼ a0 (18) even number of Fµν ’s Interactions are [cφonstrainedv] = 3/2 by gauge invariance, parity & charge( 13) no F˜ here Lint int † † µν µν † † λ λ σµ σµ µν [Fµν ] = 2 [Fµν ] = 2 (16) (16L)= τ=1 φτv φφvFφµνFF † F+ τ+2λ φτv φφvvφFvλµµνFvσFv F (14)(14) conjugation. Consider 1 v+τv3 φµvν φv(v i∂2λ)FvµνvF λµ σ (15) µ µ − [Fµν ] = 2 (16) ∂ Fµν = 0, v ∂µφv = 0 [τ1] = [τ2] = 3 (17) 2 2 evenvennunmubmerbeorf oFfµFν ’s ’s int † µν σ ∝† |A| λ ∼ τi σµ even number of Fµµνν ’s (19) L = τ1 φ φvFµν F + τ2 φ φvv FλµvσF ˜ (14) v v nnoo FF˜µµνν hheerree [τ1] = [τ2] = −3 (−17)µ µ [Fµν ] = 2 so [τ1] = [τ2] = −3 [τ1] = [τ2] = ∂3 (F1µ6ν)= 0, v ∂µφv = 0 (17) (17) 3 τ1 ∼ τ2 ∼ a0 (18) Very low energy photons[ σdo] = not−2 probe inside the atom, so expect (20) 3 [φv] = 3/2 (16) τ1 ∼ τ2 ∼ a (318) cross section0 to depend on size of atom: τ1 ∼ τ2 ∼ a0 (18) − [σ] = 2 4 6 blue light (is19 scatt) ered so σ ∝ Eγ a (21) 2 2 0 σ ∝ |A| ∼ τi stronger(19 )than red light
[σ] = −2 (20)
4 σ ∼ Eγ (21) Exercise 2: Consider QED for photon momenta much less than the mass of the electron.
By constructing an appropriate operator in a low energy EFT, → estimate the cross section for γ γ γ γ with 10 keV photons. Include factors of e in your estimate, by considering which QED graphs generate your operator. ) ) ) ) ) ) ) ) ) ) ) ) ) 2 2 0 1 2 3 4 5 6 0 1 9 8 7 2 1 1 1 1 2 1 1 1 2 1 1 1 ( ( ( ( ( ( ( ( ( ( ( ( ( φ 2 ∂ µ 3 σ m φ o F 2 t σ a g v τ µ M E λ + U F v ! 6 λ L φ ν φ v α µ 1 B ∂ v e g F · φ τ S ν m v 3 I ) † v µ i 2 i 0 + φ 3 0 ∼ − F τ , † v 2 6 0 Y a ) 4 2 2 0 1 2 φ / a λ = τ φ , ∼ K − 0 3 ∼ ∂ − 4 ] 0 λ γ = S i a = 2 2 + , | 2 ] E λ = τ = 1 τ − v ν [ ν E v A ( ] ] ! | µ φ µ ( 2 ∝ v ∼ H σ v = F 2 φ 0 [ = F φ [ ∝ σ [ φ 2 ] 1 T ∂ α ν 1 τ i µ e † v µ σ m τ v [ † v φ Y F 2 m 2 1 φ v 3 H I. WHY THE SKY IS BLUE φ τ ∼ −
= 2 † v W +
φ 2 −1
E I. WHY THE SKY IS BLUE φ Eγ ! ∆E ∼ meα ! a0 ∼ meα ! Matom (10) L µ 1 ∆ . ∂ µ
τ −1 v = (1, 0, 0, 0) 2 (11) I 2 Eγ ! ∆E ∼ meα ! a0 ∼ meα ! Matom (10) φ I. WHY THE SKY IS BLUE µ ! 2 = † 0 † µ L = φv i∂ φv = φv iv · ∂ φv (12) ∂ v = (1, 0, 0, 0) (11)
γ I. WHY THE SKY IS BLUE 2 t 2 −1 Eγ ! ∆E ∼ meα ! a0 ∼ meα ! Matom (10) 2 1 n I. WHY THE SKY IS BLUE i E [φv] = 3/2 (13) † 0 † L = φ i∂ φv = φ iv · ∂ φv (12) 2 µ−v 1 v Eγ ! ∆E ∼ meα !va0 =∼(1,m0e, 0α, 0!) Matom (10)(11) L = 2 −1 Einγt ∆E †meα a0µν meα † Matoλm σµ (10) L != τ1 φ∼v φvFµ!ν F ∼+ τ2 φ!v φvv FλµvσF (14) vµ = (1, 0, 0, 0) (11) † 0 [φv] =†3/2 (13) L = φv i∂ φv = φv iv · ∂ φv (12) L vµ = (1, 0, 0, 0) (11)
† 0 † † λ µν L = φv i∂ φv = φv iv · ∂ φv +τ3 φv φv(v i∂λ)Fµν F (12) (15) int † [φ ] =µν3/2 † λ † σ0µ † (13) L = τ φ φ F v F + τ φ φ v F = vφvFi∂ φv = φv iv ∂ φv (14) (12) 1 v v µν 2 v v Lλµ σ · even number of Fµν ’s 2 no F˜µν here [φv] = 3/2 (13) int † µν † λ σµ [φv] = 3/2 I. WHY THE SKY IS BL(1U3)E µ L =µ τ1 φv φvFµν F + τ2 φv φvv FλµvσF (14) ∂ Fµν = 0, v ∂µφv = 0 † λ µν +τ3 φv φv(v i∂λ)Fµν F (15) int † µν † λ σµ L = τ1 φv φvFµν F + τ2 φv φvv FλµvinσtF † µν † λ (14σ)µ = τ1 φv φvFµν F + τ2 φv φvv FλµvσF (14) even number of Fµν ’s L +τ φ† φ (vλi∂ )F F µν (15) ˜ 3 v v λ µν [Fµν ] = 2 ! ∼ 2 ! −1 ∼ ! (16) no Fµν here Eγ ∆E meα a0 meα Matom (10) † λ µν † λ µν µ even numµ ber of Fµν ’s +τ3 φv φv(v i∂λ)Fµν F +τ3 φv φv(v i∂λ)Fµν F (15) (15) ∂ Fµν = 0, v ∂µφv = 0 no F˜µν here even number of Fµν ’s even number of Fµν ’s s µ µ ˜ ∂ Fµν = 0, v ∂µφv = 0 no F˜µν here [τ1] = [τ2] = −3 µ (17)
0 no Fµν here ’ v = (1, 0, 0, 0) (11) µ µ µ µ [Fµν ] = 2 (16) ∂ Fµν = 0, v ∂µφv = 0 ∂ Fµν = 0, v ∂µφv = 0 ν [Fµν ] = 2 (16)
µ 3 = [Fτµν1]∼= 2τ2 ∼ a † 0 † (16) (18) [Fµν ] = 2 0 (16) L = φv i∂ φv = φv iv · ∂ φv (12) [τ1] = [τ2] = −3 (17) F v [τ1] = [τ2] = −3 (17) [τ1] = [τ2] = 3 (17) [τ1] = [τ2] = −3 −2 2 (17) f
φ σ ∝ |A| ∼ τi [φv] = 3/2 (19) (13) 3 3 o τ ∼ τ ∼ a (18) 1 2 3 0 τ1 τ2 a0 (18) µ τ1 ∼ τ2 ∼ a (18) 3 0 ∼ ∼ τ1 ∼ τ2 ∼ a0 (18)
∂ int † µν † λ σµ r [σ2 ] =2−2 L = τ1 φv φvFµν F + τ2 φv φvv FλµvσF (20) (14) σ A τi (19) 2 2 2 2 ∝ | | ∼ e µ 2σ ∝σ |∝A2 ||A∼| τi∼ τi (19) (19) σ ∝ |A| ∼ τi (19) v b [σ] = 2 4 6 † λ µν(20) σ ∝− Eγ a0 +τ3 φv φv(v i∂λ)Fµν F (21) (15) e
, [σ] = −2 (20) [σ] = −2 [σ] = −2 (20) (20) m r even number of Fµ4ν ’s6 σ Eγ a0 (21) 0 no F˜ here ∝ e Can we really use equations of motion toµ ν u 1 µ 1 2 2 4 6 3 2 4 6 4 6 simplifyσ ∝operatEγ σa0∝ors?E a L = µ∂µφ∂ φ −µ m φ − λφ + τg1(φ21)+(21τ)g2φ ∂ φ (22) γ 0 ∂1 Fµν µ= 0, 1v ∂2µφ2 v = 04 6 3 2 h 2 2 n 4 6= ∂µφ∂ φ m φ λφ + τg1φ + τg2φ ∂ φ (22) σ ∝ Eγ La0 2 − 2 − (21) = 3 2 1 2 1 eg. p τ µ!1p 1τ 12 2 1 4 6 3 2 [φv] = 3/2 (16) ν L = ∂µφ∂ φ − mµ φ − λ2φ 2+ τg14φ + τg26φ ∂ φ 3 2 (22) n !2 L!= ∂µφ2!∂ φ − m φ − λφ + τg1φ + τg2φ ∂ φ (22) ν 2 2
µ 1 1 e µ 2 2 4 6 2 3 2 2 3 2 L = ∂ φ∂ φ − m φ − λφ + τ2g φ1 +∂τµφg2=φ 1∂mφ2φ 2 4 4!λφ4 6! 6 3 2 (22) (23) µ µ 1 2 p τ ! 1 2 L∂ =φ =∂µ−φ∂mφφ−−m−φλφ− λ+φτg+ φτg1+φ τg φ ∂ φ (24) (23) ˜ ! p τ ! 1 2 2 2 −2 − − 1 2 v
F !
F 2 2 3 1 1 ! e ∂ φ = −m φ − −3λφ µ 2 2 " 4 " (623) 2 2 3 = ∂µφ∂ φ m φ λ φ + τg1φ (24) e.o.m. ∂ φ = −m φ − −4λφ L 2 ! − 2 2 − (23)
µ λ = λ + m g η (25) o 2
1 1 " 2 n ∂ µ 2 2 " 4 " 6 λ = λ + m g2τ (25) L = ∂µφ∂ φ − m φ − λ φ + τg1φ (24) 2 2 ! g1 = g1 − 4λg2 (26) g" = g 4λg (26) (board) 1 1 − 2 Exercise 3: Demonstrate the equivalence for tree level 6-pt functions at O ( τ ) , pre- and post- the use of the equations of motion. 2
I. WHY THE SKY IS BLUE
2 −1 Eγ ! ∆E ∼ meα ! a0 ∼ meα ! Matom (10)
vµ = (1, 0, 0, 0) (11)
† 0 † L = φv i∂ φv = φv iv · ∂ φv (12)
[φv] = 3/2 (13)
int † µν † λ σµ L = τ1 φv φvFµν F + τ2 φv φvv FλµvσF (14)
† λ µν +τ3 φv φv(v i∂λ)Fµν F (15)
even number of Fµν ’s 3 no F˜µν here µ µ ∂ Fµν = 0, v ∂µφv = 0 II. RENORMALIZATION
Linearity: ddp [af(p) + bf(p)] = a ddp f(p) + b ddp g(p) [Fµν ] = 2 (16) Scaling: ! ddp f(sp) = s−!d ddp f(p) ! Translation: d!dp f(p + q) = dd!p f(p) [τ ] = [τ ] = −3 (17) 1! 2 ! d−1 = dp p dΩd ! 3 Regularization and Renormalization d 2 α τ1 ∼ τ2 ∼ a0 (18) d p(p ) = 3 Regularization: How we cutoff UV infinities in loop integrals! 0 for α < 4 anIdI.α R>E4NORMALIZATION 3 2 2 Renormalization: How we pick a scheme to give definite meaning to σ ∝ |A| ∼ τ (19) i 3 parameters of a theorLineayrity: ddp [af(p) + bf(p)] = a ddp IfI(.p) +RbENdOdpRgM(p)ALIZATION 3 3 ! d −!d d ! Scaling: dIIp. f(RspE)N=OsRMAd LpIZf(ApT) ION • II. RdENORMALIZATIONd d i 1 1 Computations are IeaI. sierRE ifN OourRM regulatALIZATLorIiOn preserNearity:ve symmetriesd p [a!df(p) + bf(p)] =d!a d p f(p) + b d p g(p) 3 Translation: d p f(p + q) = d p f(p) [σ] = −2 2 − (20) (27) d ! d d d−!d d ! 16π #UV #IR d d Lineaandrity: d Scadlinpg[:af(p) + bf(p)] =! a d dp pf(fp()s+p!)b= ds p g(pd) p f(p) " # Linearity: d p [af(p) + bf(p)] = a d p f(p) + b d p g(p) d−1 d = dp p dΩdd d ! Lpreserinearvesidty :power c−!ountingd dd p by[a f!not(p )mixing+ b!f (upp) t]erms= a ofd ddiffperentf(pI )I−.+!d bREddNOp Rg(Mp!)ALIZATION Scaling: d p f(sp) = s dScpafli(np)g: Translation:d p f(sp) ==s 0d!dfpodrfp(αpf=+(pq)4) = dd!p f(p) order in the expansion ! 3 !d d!! d −!d d ! 4 6 Translation: Scalidngp :f(p + q) = d p f(p) d p f(sdp) =2 !αsd d p f(!p)d! ! σ ∝ Eγ a0 (21) Translation: d d p(p )dd−p=1f(p + q) = dd p f(p) d ! ! II. RENORMALIZATION d−1 Dimensional RLegularizationinearity: d p [a=f(pd)p+pbf(pd)Ω] =d a dd=p4f−(p2)#+ b d p g(p) = dp p dΩd !d ! ! d! ! Translation: dd−p1f(p + q) = d p f(p) ! = dp p! dΩd0 fo!r αd< 4 and α > 4−!d d ! Linearity: Scaldidnp g[a:f(p) + bf(p)] = a ddp f(p) + b dddpp fg((p)sp) = s d p f(p) d 2 α d p(p ) = ! d ! ! −!d d ! ! 1 1 Scaling: d−1 d p f(sp) = s d p f(p!)d d! µ 2 2 4 6 3 2 ! = dp p TrandsΩladtion: d 2 α d p f(p + q) = Ld =p f(p∂)µφ∂ φ − m φ − λφ + τg1φ + τg2φ ∂ φ (22) 0 for α < 4 and α > T4ranslation: d!dp fd(p +p(qp) =) d=d!p f(p) 2 2 ! ! ! ! ! ! i 1 1 2 − (27) 0 for α < 4 and α > 4 16π #UV #IR ddp(p2)α = 2 no power divergences " # ip τ1! 11 ! 2 − = 0 for α = 4 be careful ! (27) ! 16π "#UV #IR # ∂2φ = −m2φ − −4λφ3 (23)
1 1 L = ∂ φ∂µφ − m2φ2 − λ"φ4 + τg" φ6 (24) 2 µ 2 1
" 2 λ = λ + m g2η (25)
" g1 = g1 − 4λg2 (26) 3
II. RENORMALIZATION
Linearity: ddp [af(p) + bf(p)] = a ddp f(p) + b ddp g(p) 3 ! d −!d d ! Scaling: d p f(sp) = s II.d pREfN(pO)RMALIZATION Translation: d!dp f(p + q) = dd!p f(p) Linearity: ddp [af(p) + bf(p)] = a ddp f(p) + b ddp g(p) ! ! d−1 ! d −!d d ! 3 3 = dpScpaling:dΩd d p f(sp) = s d p f(p) !d d! 3 ! IIT.raRnsElaNtOioRn:MALIZATION d p f(p + q) = d 3p f(p) II. RENORMALIZATION d 2 α ! ! II. RdENpO(pRM)AL=IZATION 3 d II. RENORMdA=LIZAdTpIOpNd−1ddΩ Linearity: d p [af(p) + bf(p)] = a d p f(p) + b d p gd(p) Linearity: ddp [af(p) + bf(p)] = a ddp f(p) + b ddp g(p) ! II. RENORMALIZATION d ! d d −!d d d! ! LLinineeaarriStiytcy:a:ling: dddp p[a[fa(fp()p+) +bfb(fp)(]p=)]da=pafdd(psdpf)p(p=f)(+sp)b+ dbddpdgp(ppf)g(p(p)) ! d −!d d ! 0 for α < 4d an2dαα > 4 Scaling: d p f(sp) = s d p f(p) ! d −!d d ! d p(p ) = SSccaalilningg: : ! dddp fp(fsp()s!pd=) s=−!ds Ldidnpedafrp(itpyf):(p!)d! ddp [af(p) + bf(p)] = a ddp f(p) + b ddp g(p) Translation: d p f(p + q) = d p f(p) !d d! !d d! ! ! d −!d Trdanslati!on: d p f(p + q) = d p f(p) TTrraannsslalatitoionn: : d!ddp fp(fp(+p!+q) q=) =dd!pdScfpa(lpifn)(gp:)! d p f(sp) = s d p f(p) d−1 0 for α < 4 and α > 4 = dp p d!Ω!d ! ! !d d! ! ! d−d−1 1 Translation: d p f(p + q) = d p f(p) d−1 == ddpppp dΩdΩd = dp p dΩd ! d ! ! i 1 1 d−1 !! d 2 α = dp p dΩd ! d p(p ) = 2 − (27) dd 22αα 16π # # ddpp(p(p!)) == ! ddp(p2)α = i " 1UV 1 IR # !! 0 for α < 4 and α > 4 ddp(p2)α = − (27) 0 for α < 4 and α > 4 2 0 for α < 4 and α > 4 != 0 for α = 4 ! 16π "#UV #IR # 0 for α < 4 and α > 4 0 for α < 4 and α > 4 = 0 for α = 4 di= 41− 2#1 i 1 1 i −1 1 (27) 2 − 2 − (27) (27) 16π "#2UV d #=1IR6π#4 −#2U#V #IR i 1 1 16π "#UV #IR"# # − (27) Rescale(0) couplings" to keep them dimensionless16π2 # # i 1 1 = 0 for α = 4 g = Z µ g(µ) " UV IR # − (27) = 0 for=α0=fo4r α = 4 g(0) " 2 = 0 foeg.r α = 4g = Zg µ g(µ) 16π "#UV #IR # d = 4 − 2# d = 4 − 2# d = 4 − 2# dµ= 4 −is2 dim.reg.# parameter and acts = 0 for α = 4 g(0) = Z µ" g(µ) (0) g " like a “resolution” in MS scheme g = Z(g0)µ g(µ) " (0) " g = Zg µ g(µ) g = Zg µ g(µ) µ m " M, a ∼ 1 d = 4 − 2# µ m " M, a ∼ 1 µ m " M, a ∼ 1 µ (0) " Mass independenta regulator g = Zg µ g(µ) L = ψ¯(i/∂ − m)ψ − (ψ¯ψ)2 + . . . (28) M 2 a 2 a L = ψ¯(i/∂ − m)ψ − (ψ¯ψ) + .a. . L = ψ¯(i/∂ − m)ψ − (ψ¯ψ)2 + .(.2. 8) (28) eg. L = ψ¯(i/∂M−2m)ψ − (ψ¯ψ)2 + . . .µ m " M 2, a ∼ 1 (28) M 2 i a d4k /k + m i a d4k m δm ∼ = (29) 2 44 2 2 2 4 42 i2a d4k /k + m i a d4k m M i a$ (2π)d kk −/km+ m M i$a (2π) dδkmk−∼m m = (29) a 2 δm ∼ =gives 2 4 2 2 2 (249)2 2 L = ψ¯(i/∂ − m)ψ − (ψ¯ψ) + . . . (28) M 2 $ (2π)4 k2 − m2 M 2 $ (2π)4 k2M− m$2 (2π) k − m M $ (2π) k − m M 2 a δmcutoff ∼ Λ2 + . . . (30) M 2 a δmcutoff ∼ Λ2 + . . . (30) M 2 the cutoff indep. terms are “hidden” i a a δm ∼ 2 (29) δmdim.reg. ∼ m2 (31) M M 2 small as expected, power counting manifest 3
II. RENORMALIZATION
Linearity: ddp [af(p) + bf(p)] = a ddp f(p) + b ddp g(p) Scaling: ! ddp f(sp) = s−!d ddp f(p) ! Translation: d!dp f(p + q) = dd!p f(p) 3 ! ! II. RENORMALIZdA−T1ION = dp p dΩd 3
d d ! d Linearity: d p [af(p) + bf(p)] =II.a RdENpOfR(pM) A+LbIZAdTpIOgN(p) ddp(p2)α = Scaling: ! ddp f(sp) = s−!d ddp f(p) ! Linearity: ddp [af(p) + bf(p)] = a ddp f(p) + b ddp!g(p) Translation: d!dp f(p + q) = dd!p f(p) ! d −!d d ! 0 for α < 4 and α > 4 Scaling: ! d p f(sp) = s !d p f(p) d−1 !d d! Tra=nsladtiponp: dΩd d p f(p + q) = d p f(p) ! ! ! = dp pd−1 dΩ ddp(p2)α = d i 1 1 ! 2 − (27) 16π #UV #IR ! d 2 α " # 0 dfopr(pα )< =4 and α > 4 ! = 0 for α = 4 0 for α < 4 and α > 4 d = 4 − 2# i 1 1 − (27) i 1 2 1 16π "#U(0V) #IR # " 2 −g = Zg µ g(µ) (27) 16π "#UV #IR # = 0 for α = 4 = 0 for α = 4 µ m " M, a ∼ 1 d = 4 − 2# d = 4 − 2# (0) " a 2 gg(0) ==ZZgµ"µg(gµ()µ) L = ψ¯(i/∂ − m)ψ − (ψ¯ψ) + . . . (28) g M 2 µµ mm""MM, a, a∼∼1 1
4 4 a a2 i a d k /k + m i a d k m L = ψL¯(i=/∂ −ψ¯(mi/∂)ψ−−m)ψ(−ψ¯ψ) +(ψ.¯.ψ. )2 + . . . δm ∼ (28) (28) = (29) Decoupling Theorem M 2 M 2 M 2 $ (2π)4 k2 − m2 M 2 $ (2π)4 k2 − m2 If remaining low energy theory is renormalizable then all effects due to heavy particles appeari a as dchanges4k /k + m in thei a d4k m 4 4 a
δm ∼ 2 i a 4 d2 k 2/k =+ m 2 i a 4 2 d k2 m (29) cutoff 2
δmM∼ $ (2 π1/M) k − m M =$- (2π) k − m δm (2∼9) Λ + . . . (30)
coupling constants or are suppressed by - 2 2 4 2 2 + 2 4 2 2
M (2π) k − m+ M (2π) k − m
$ + - $ M
- requires a “physical” renormalization scheme, +
e- cutoff a 2 - + -
-
δm ∼ Λ - + . . . (30)
+ not true in MS ! ! M 2 a + (board) cutoff 2 dim.reg. a 2 δm ∼ 2 + Λ + . . . (30) M - δm ∼ m (31) e+ M 2 a δmdim.reg. ∼ m2 (31) Solution: We must implement decoupling by hand at Mµ2≈ m dim.reg. a 2 µ ≈ m δm ∼ 2 m (31) E = me M e3 1 137 βµQ≈EDm= 2 12π resolution µ = E 3 α e 136.5 βQED = 12π2 hydrogen 136.0 βQED = 0 E (MeV ) 135.5
0.1 1.0 10 100 1000 ) ) ) ) ) ) 3 8 0 2 7 9 1 2 3 3 2 2 3 ( ( ( ( ( (
3
II. RENORMALIZATION
3 Linearity: ddp [af(p) + bf(p)] = a ddp f(p) + b ddp g(p) 3 Scaling: ! II. RENdOdRpMf(AspL)I=ZAsT−!IdONddp f(p) ! II. RENORMALIZATION Translation: d!dp f(p + q) = dd!p f(p) Linearity: ddp [af(p) + bf(p)] = a ddp f(p) + b ddp g(p) 3 d ! ! d d Linearity: d−1 d p [af(p) + bf(p)] = a d p f(p) + b d p g(p) Scaling: ! = ddpdpfp(sp) =dΩs−d!d ddpIIf.(pR) E!NORMALIZATION ! d −!d d ! Sca!ldin!g: d! d p f(sp) = s d p f(p) Translation: d p fd(p + q) = d p f(p) d d Linearity: ddpp(p[a2f)(αp)=+ bf(p)] = a d!dp f(p) + b d p g(pd)!
2 Translation: d p f(p + q) = d p f(p) ! ! d ! −!d d ! = dp pd−1ScdaΩling: ! d p f(sp) = s d p f(p) m d ! ! d−!d1 d! ! Translation: =0 fodrpαp
4 3 . d 2 α )
k 0 for α < 4 and α > 4 i 1 1 . d p(p ) =! 4
π − (27) . ! 0 for α
d 16π # # 2 " UV IR # N
( 0 for α < 4 and α > 4 + d d d . O
2 Linearity: d p [af(p) + bf=(p)0] =foar αd p=f(4p) + b d p g(p) I . )
$ i 1 1 Λ d −d d
2 ! ! ! . Scaling: d p f(sp) = s d p f(p) − (27) T S ψ 2 ) # 16π #UV #IR 2
m i 1 1 ¯ !d d! " i 1 #1 − d = 4 − 2# A + p ψ Translation: d p f(p + q) = d p f(p) a 3 − (27)
R − (27) ( e 3 2 ( 2 1 I 16π # # Z UV IR 2 i 2 ! ! " # 16π #UV #IR M g = 0 for α = 4
# 2 " # I d−1 II. RE=NORdMpApLIZAdTΩIOd N (0) " a φ Λ II. RENORMALIZATION g = Z µ g(µ) a g L = 0 for α = 4 p 3 M !d = M
− = 0 for α = 4 d 2 d d d d d Λ α d A Linearity: d p [af(p) + bf(p)] = a d p f(p) + b d2p g(p) Linearity: d p [af(p) + bf(p)] = ad =d p4f(−p) +2b# d p g(p) II. RENORMALIZATION
d p(pδ ) = d a 2
∼ ! d −!d d ! d = 4 − 2# V ! d −!d d ! µ m " M, a ∼ 1
− Scaling: d p f(sp) = s d p f(p) $
M Scaling: d p f(sp) = s! d p f(p)
M d d d 1 . ! Linearity: d p [af(p) + bf(p)] = a d p f(p) + b d p g(p) U !d d! (0) " d = 4 − 2# m m !d 0 fd!or α < 4 and α > 4 g Translation: d p f(p + q) = d p f(p)g = Zg µ g(µ) " # ψ Translation: d p f(p + q) = d p f(p) (0) ! d −!d d ! b
R Scaling: d p f(sp) = s d p f(p) = e g = Zg µ g(µ) ) ∼ ! ! r ! ! !d d!
) d−1 + − " . = dp p dΩd d−1 Translation: d p f(p + q) = d p f(p) " a O = dp p dΩd (0) 2 Λ
+ ¯ ¯ p ff m ! g =! Z µ g(µ) 2 g L = ψ(i/∂ − m)ψ − (ψψ) + . . . (28) δ 2 m ! d−1 / k ! µ m " M, a ∼ 1 2 ( = dp pµ dΩmd " M, a ∼ 1 i o )
− M N d α i 1 1 π 2 k t d 2 α i d p(p ) = ! d Λ
f d p(p ) = − (27) p ) − 2 6 u d 2 α E 16π #UV #IR ( S 4 ! ! d p(p ) = " # p c µ m " M, a ∼ 1 1 m ) p 0 for α < 4 and α > 4 a ∂ / ! k 0 for α < 4 and α > 4 ( 2 f
− a
R ¯ ¯2 i 0 for α < 4 and α > 4 L = ψ(i/∂ − m)ψ − (ψψ) + . . . (28) δ d = 0 for α = 4 ¯ ¯ 4
π 2 m L = ψ(i/∂ − m)ψ − (ψψ) + . . . (28) e 4 4 f ( 2 M d p i a d k /k + m i a d k m d 2
δ M ¯ d ( ψ δm ∼ = (29) p d = 4 − 2# . i 1 1 2 4 2 2 a 2 4 2 2 ! i 1 1 − i 1 1 (27) M (2π) k − m M 2 (2π) k − m d $ ¯ ¯ $ d I − 2 (−27) (27) 2 16π # # 2 L = ψ(i/∂ − m)ψ − (ψψ) + . . . (28) 16π #UV #"IR UV IR # 16π #UV #IR 2 d I " # " # 4 4 = d (0) " i a d k /k + m i a d k m M $ ! = 0 for α = 4 g = Zg µ g(µ) 4 4
− = 0 for α = 4 = 0 for α = 4 i aδm ∼ d2 k /k +4 m2 2i=a 2 d k 4 2m 2 (29) ! L
a M (2π) k − m M (2π) k − m s δm ∼ $ = $ (29) 2 d = 4 − 2# µ m " Md,=a4 ∼− 21# 2 4 2 2 2 c4uto2ff 2a 2 a d = 4 − 2# M $ (2π) k − m M $ (2δπm) k −∼m Λ + . . . (30) = (0) " = = " 2 i (0) g = Z µ g(µ) 4 4 g = Zg µ g(µ) g M M (0) " i a d k /k + m i a d k m ] ) g = Zg µ g(µ) a ) a 2 cutoff 2
) µ m " M, a ∼ 1 ¯ ¯ δm δm ∼ Λ + . . . = (30) (29) q L = ψ(i/∂ − m)ψ − (ψψ) + . . . 2 4 2 2 (28) 2 4 2 2 p µ m " M, a ∼ 1 M 2 MM2 (2π) k − m M (2π) k − m p $ $ ∼ µ m " M, a ∼ 1 cutoff a 2 s
( a δ2 m ∼ Λ + . . . (30) + ( L = ψ¯(i/∂ − m)ψ − (ψ¯ψ) + . . . 2 (28) a ¯ a ¯ 2 M 2 dim.reg. 2 f L = ψ(i/∂ − m)ψ − (ψψ) + . . . (28) M f m 2 δm ∼ m (31) p a M b L = ψ¯(i/∂ − m)ψ − (ψ¯ψ)2 + . . . i a d4k /k(+28)m i a d4k m a 2 ( δ 2 dim.reg. 2 M p M δm ∼ = (29) i a2 d4k /k +4 m 2 i a 2 d4k 2 m δm 4 2 ∼ 2 2 m cutoff a 2 (31) f 4 4 M $ (2π) k − m M $ (2π) k − m d δm ∼ = M (29) + i a d k /k + m i a d k m δm ∼ Λ + . . . (30) δm ∼ = µ ≈Mm2 $ (2π)4 k2 − m2 M 2 $ ((229π))4 k2 − m2 a 2 d 2 4 2 2 2 4 2 2 dim.reg. 2 M p ) M $ (2π) k − m M $ (2π) k − m i a d4k /k + m i a µd4≈k mm δm ∼ m 4 (31) d 2 p δm ∼ = 3 a(29) ! 2 4 2 2 2 4 2 2 cutoff cutoff 2 a 2 M ( M $ (2π) k − m M $a (2π) k − m δm ∼ Λ + . . . (30) d e δmcutoff ∼ Λ2 + . . . 3 δm M 2 ∼ 2(3Λ0) + . . . (30) 2 e βQED = 2 M f M 2 12π µ ≈ m βQED = 12π III. WEAK HAMILTONIAN ! a a . . a . . a dim reg 2 4 cutoff 2 dim reg 2 [ δm ∼ Λ + . . . a δm ∼(30) m (31) δm ∼ m (31) δMm2dim.reg. 3∼ m2 M 2 (31) 2 e β2QED = 0 βQED = 0 dim.reg. a 2 M p M β = µ ≈b m2 cu¯d δm ∼ m (31) > QED π 2 d µ ≈ m 12 → M 3 1 e d a 2 3 dim.reg. βQE2D =Λ12π µ ≈ m α e δm µ ≈ m∼ m (31) ) 2 2 2 µ ν βQED = 12π M Λ ig k k 1 d 2 ∗ µν Λ ! ∼ βQED = 0βQED = 0 µ 3 3 = VcbVud( i) g d µ ≈ m e e 2 2 2 ( √ − − m k m Ω 2 βQED = 0 βQED = π2 2 ! " ! W " W
a 12 n Λ βQED = −
< π g 12 d 3 c b d Λ−uδΛ Λ
4 −S −S e u¯ γ P u u¯ γ P v (34) a µ L ν L ,
2 e = dφ e (32) βQED = 12π2 Λ Wilsonian vs. Continuum EFT =
" Λ × −SΛ−δΛ −SΛ
1 β = 0 $ 3 QED δΛ π −SΛ−δΛ # −SΛ $# $ E e = dφ e (32) = 4 e = dφ e (32) µ e 2
# $
− k = p p = p +β p =m0asses δ α β = 0 Λ M QED b c u QdED : Wilson effective action E 1 −SΛ−δΛ −S−ΛSΛ $δΛ
0 − ∼ ) d g 2 Λ e =−SΛ deφ e (32) n α for soft modese < <
2 $δΛ hard modes p Λ
: Λ−δΛ Λ o −S −S Z
" −SΛ p r Λ
i −S = = − e removing modes with Λ − δΛ < E < Λ Λ 4GF e ∗ c = b dφd µe u (32) y
p e Λ α ( t o = i V V u¯ γµPLu u¯ γ PLv + . . . (35)
: cb ud t −SΛ−δΛ −SΛ $δΛ m δ f d = 4 √ p − i a Λ−δΛ Λ 2 D D −S −S e = dφ e (32) r g m Λ # $# $
l Λ − δΛ < E < Λ e = dφ e (32) r d ∞ $δΛ ) E E $δ o Λ soft modes s S n 0 ! → ∞ O (µ) −SΛ − a 2 2 ≈ f d 0
= Λ − δΛ < E < Λ i e GF = √2g /(8m ) Q Q 2 W n
( Λ l −S − e e a Λ e Λ β 0 ! = g d = µ β a µ EFT ∞ n Continuum L = C(µ)O(µ) (33) r i E c ΛΛ −−δδΛΛ< 1 Iain W. Stewart1 Iain W. Stewart 1Center for Theoretical Physics, Massachusetts Institute of Technology, Cambridge, MA 02139 1Center for Theoretical Physics, Massachusetts Institute of Technology, Cambridge, MA 02139 −1 meα ! (proton size) ∼ ΛQCD ∼ 200 MeV −1 meα ! (proton size) ∼ ΛQCD ∼ 200 MeV meα ! mp ∼ 1 GeV meα ! mp ∼ 1 GeV meα ! me mp, µp meα ! me mp, µp p2 L(p, e−, γ, b; α, m ) = L(p, e−, γ; α") + O (1) b m2 2 ! b " L − L − " O p (p, e , γ, b; α, mb) = (p, e , γ; α ) + 2 (1) !mb "∞ NRQED n L = L0 + # Ln (2) n#=1 ∞ NRQED n L = L0 + # Ln (2) ∇2 † 0n#=1 † 0 † † L0 = ψ i∂ − + . . . ψ + Ψ i∂ Ψ + “V”(ψ ψ)(Ψ Ψ) (3) ! 2me " 1 µν Why LQED = ψ¯(iD/ −∇m2 )ψ − F Fµν ? † 0 4 † 0 † † L0 = ψ i∂ − + . . . ψ + Ψ i∂ Ψ + “V”(ψ ψ)(Ψ Ψ) (3) 2me Why LSM ?! " L ¯ − − 1 µν Why QED = ψ(iD/ m)ψSU(34)F× SFUµ(2ν)?× U(1) L Why LSM ? d 1 µ 1 2 2 λ 4 τ 6 S[φ] = d x ∂µφ∂ φ − m φ − φ − φ (4) SU(3) × SU(2) × U(1) L $ !2 2 4! 6! " L Ldim-4 1 T i j " k Ldim-6 d 1 =µ 1 +2 n2ew LλL C4 #ijHτ H6 #"kLL + + . . . (5) S[φ] = d x ∂µφ∂ φ − m Λφ −% φ − φ & (4) $ !2 2 4! 6! " C = iγ2γ0 1 L Ldim-4 T i j " k LνdLim-6 = + new LL C#ijH H #"kLL + + . . . (5) (6) Λ % & ' eL ( C = iγ2γ0 Top Theory 1 is understood, but it is useful∞ to L L n L theory1 have the simpler theory 2 at =low0 energies.+ # n (7) ν n#=1 L (6) Integrate out hea'viereL ( particles in 1 and match onto 2 L → L(n) theory1 theory2 (8) #n ∞ theory2 n Theory 1 & theoryL 2 =agreeL0 + in the# IR,Ln differ in UV (7) n#=1 eg. NRQED; Heavy Quark Effective Theory; Remove t,W,Z: Hweak ; Soft Collinear Effective Theory L → L(n) theory1 theory2 (8) Theory 1 is unknown#n or matching is too difficult to carry out analytically theory1 ? Bottom L(n) Construct t h e o r y 2 by writing down most (9) #n general set of interactions consistent with symmetries theory2 eg. Standard Model; Chiral Perturbation theory for low energy pion and Kaon interactions EFT Principles 1) Dynamics at low E does not depend on details of dynamics at high E 2) Build an EFT using the relevant d.o.f. and known symmetries. 3) EFT has an infinite number of operators, but only a finite number are needed for a given precision as determined by the power counting. With this precision this set closes under renormalization. 4) EFT has same infrared but different ultraviolet than the more fundamental theory. 5) Nature of high energy theory shows up as couplings and symmetries in the low energy EFT. Effective Field Theories of QCD ChPT* HTL* NRQCD* cc states pions finite T finite HDET * spectrum density unstable energetic top quark QCD particles hadrons SCET SCET jets perturbative nuclear QCD * Lattice HQET* bd states forces NNEFT* QCD