Introduction to the Standard Model of Particle Physics

Introduction to the Standard Model of Particle Physics February 3, 008

Contents Pre-requisits 3. Special Relativity........................... 3. Transition rates............................ 5 Quantum Field Theory 7. The free scalar eld......................... 7. The interacting scalar eld..................... 4.3 Spin Fields............................. 0.4 The interacting fermionic eld.................... 8 3 Quantenelectrodynamics QED 37 3. The electromagnetic eld...................... 37 3. Lagrangian of QED.......................... 44 3.3 Magnetic moment of electron.................... 46 3.4 Renormalisation........................... 49 4 Quantum Chromodynamics QCD 5 4. The QCD-Lagrangian........................ 5 4. Running coupling........................... 56 4.3 Connement.............................. 56 4.4 Phase diagram of QCD....................... 58 5 Electroweak Theory Quantum Flavourdynamics, QFD 59 5. Lagrange density of electroweak theory............... 60 5. The Higgs sector........................... 64 5.3 Spontaneous Symmetry Breaking.................. 68 5.4 The mass matrix and the Cabibbo angles............. 7 5.5 CP-Violation in the Standard model................ 78 6 Beyond the Standard Model SM 79 6. A hint of Supersymmetry...................... 80 A Auxiliary calculation to Fermi's trick 8

B Supplement 83 B. Landé-factor............................. 83

Chapter Pre-requisits. Special Relativity The Minkowski space is a four dimensional space the following metric: g µν =. One point in the Minkowski space is a contravariant vector: x µ c t = x µ = 0,,, 3 and c = = x µ = t x The scalar product of space-time dierences. x µ = g µν x ν is a covariant vector. x y = x y µ g µν x y ν. = x y µ x y µ = x y 0 x y Symmetry transformation? What leaves x y µ x y µ invariant? Poincaré transformations Λ, a: x µ x µ = Λ µ ν x ν + a µ.3 Composition: Λ, a Λ, a = Λ Λ, Λ a + a.4 3

Invariance: In components: or Λ T g Λ = g.5 Λ ρ µ g ρσ Λ σ ν = g µν Λ σ µ Λ σ ν = g µ ν = δ µ ν Lorentz group Λ, 0: Component of unity orthochron: Parity P: x x det Λ = ± Λ µ σ = Λ σ µ.6 det Λ =, Λ 0 0 > 0.7 Λ P =.8 Time reversal T: x 0 x 0 Λ T =.9 Generators of Lorentz group: Boost along x-axis: γ Λ µ ν = γ v c = e ωk µ ν γ v c γ, γ = v c.0 rapidity ω v ω = arctanh c. 4

and generator K Lorentz algebra Boosts K i, rotations J i : K =. Transition rates 0 0 0 0 0 0 0 0 0 0 0 0 0 0.. [J i, J j ] = i ε ijk J k [K i, K j ] = i ε ijk J k.3 [J i, K j ] = i ε ijk K k Perturbation theory for computing decay rates and cross sections H 0 φ n = E n φ n.4 Perturbation H : V φ mφ n d 3 x = δ mn. H 0 + H ψ = i ψ t What is the transition rate from φ i at T to φ f at T?.5 φx = n c n t φ n x e ient From equation.5: dc f t dt c n T = δ ni.6 φ t = ih 0 + H φ = i n c n t d 3 r φ f H φ n } {{ } f H n = i f H i e ie f E it + OH e ie f E nt c i t = + OH t c f t = i dt f H i e ie f E it.7 T 5

Transition amplitude A fi : A fi = c f T A fi = i T dt f H i e ie f E it.8 or T A fi T = i φ f xh φ i x d 4 x.9 φ n x = φ n xe ient Transition probability: H time-independent lim A fi = f H i T T T dt e ie f E it T T dt e ie f E it Fermi s trick = T π f H i δe f E i.0 Transition rate Γ: A fi Γi f = lim = π f H i δe f E i. T T Integrating over nal states ρ: phase space density: Γ [i f] = de f ρe f π f H i δe f E i. Γ [i f] = π f H i ρe i.3 In general: Γ [i f] = π T fi ρe i T fi = f H i + f H n n H i. 6

Chapter Quantum Field Theory. The free scalar eld Spin 0, neutral particles, e.g. Π 0, described by a real scalar eld ϕ: Property under Lorentz transformations: ϕ x = ϕx. ϕ x = ϕx scalar. The equation of motion, free, up to second order in derivatives unique if local is called Klein-Gordon equation: µ µ + m ϕx = 0.3 µ µ = t = t = = p µ. : d Alembert operator, m : mass of the scalar particle The Klein-Gordon equation can be derived out of Boosts from the rest frame equation of motion: E m ϕx = 0 unique.4 The most fruitful approach to Elementary Particle Physics is via the action principle. Lagrange density of a free scalar eld: Lx = [ µϕx µ ϕx m ϕ x]..5 7

Action S: S[ϕ] = d 4 x Lx = d 4 x µ ϕx µ ϕx m ϕ x.6 Action principle: or and δs[ϕ] = 0 δs δϕx = 0 δϕy δϕx = δ4 x y δ µ ϕy δϕx results in the following equation which equals.3. = y µ δ 4 x y.7 L ϕ L µ µ ϕ = 0.8 Classical solutions of the Klein-Gordon equation, take e.g. ϕ = ϕx : Solutions are plane waves: + m ϕx = 0. ϕx = e ±ikx.9 kx = k µ x µ k = m, k 0 = ±ω = ± k + m. General solution: linear superposition of plane waves. d 3 k ϕx = e ikx π 3 α ω k + e ikx α k.0 }{{} d 4 k π 4 δk m ω k = k. 8

QFT: ϕx φx operator The expectation value φx is a classical eld. φ obeys canonical commutation relations: [φ x, t, φ y, t] = i δ 3 x y.. φ y, t is the canonical conjugated momentum. Let a be the lattice parameter of a crystal. }{{} Classical mechanics a a 0 eld theory QM [x, p] = i a 0 QFT [φ, Π 0 ] = i φx still obeys the Klein-Gordon equation + m φ = 0. φx = d 3 k [ π 3 e ikx a ω k + e ikx a ] k. Inserting. into. results in [ a k, a ] k = π 3 ω δ 3 k k.3 [ a k, a ] [ k = 0 = a k, a ] k Fock space 0 : normalised vacuum state: 0 0 = a k 0 = 0. 0 is the lowest energy state! a annihilates the vacuum. Heisenberg picture: t 0 = 0.4 All states are generated by applying a, a on 0. a, a are annihilation and creation operators, respectively. One particle states: k = a k 0..5 The states k are orthogonal: k k = 0 a k a k 0 = 0 [a k, a k] 0 = π 3 ω δ 3 k k.6 9

General one-particle state: f = d 3 k π 3 ω f k a k 0.7 Example: two state system spin With aa + a a = a 0 = a 0 = 0 a = 0 Realisation: 0 0 a = 0 0 0 = a = 0 0 0 =. 0 Realised in spin system. Pauli matrices: σ 0 = 0, σ = 0 i i 0, σ 3 = 0 0. Note that: [σ i, σ j ] = i ε ijk σ k {σ i, σ j } = δ jk σ ± = σ ± iσ 0 σ + =, σ 0 0 = 0 0 0..8 N-Particle states particles: a k a k 0 3 particles: a k 3 a k a k 0..9 Have Bose symmetry, as a k a k = a k a k.0 0

Energy-momentum is additive. Take some state β, then a k β is a state one additional particle momentum k. Annihilation: a k β is a state, where a particle momentum k is removed from the state β. Example a general particle state f see equation.7: a k f > = a d 3 k k π 3 f k a k 0 ω d 3 k = π 3 f k [a k, a k ] ω }{{} Interpretation of φx:.4π 3 ωδ 3 k k = ωf k 0. states dened particle number n have vanishing expectation values of φ, as φ creates and annihilates a particle, see equation.. This follows from 0 a 0 = 0 a 0 = 0 k a k = k a k = 0. 0 0 φx 0 = 0.... coherent states: φ behaves like a classical wave. α = { N exp d 3 } k π 3 ω α ka k 0 }{{} α 0 { N = exp and α k coecient function. d 3 } k π 3 ω α k = α 0 α 0.3 α α = via 0 α k α k m α k n... α k 0 δ [ nm ωπ 3 ] n α φx α = d 3 k { π 3 e ikx α ω k + e ikx α } k.4

Using [ d 3 n! ak k ] n π 3 ω α k a k 0 = = [ d n! n α k 3 k π 3 ω α k a k = α k n! [ d 3 k π 3 ω α k a k ] n 0 see equation.6 ] n 0.5 and similarily [ d 3 k ] n [ n! π 3 ω α k a k a d 3 k ] n k = n! π 3 ω α k a k α k.6 Symmetries: By partial integration of equation.5 one gets: S[φ] = d 4 x Lx = d 4 x φx[ µ µ m ]φx..7. Invariance of S[φ] under orthochronous Poincaré transformations and x µ = Λ µ νx ν + a µ see equation.3 φ x = φx.8 µ µ = µ µ Unitary Representation: UΛ, a Λ T gλ = g + m φ x = + m φx = 0 φx = φ x = U Λ, a φx UΛ, a UΛ, a φx U Λ, a = φx = φλx + a.9 On Fockspace: UΛ, a 0 = 0 UΛ, a a k U Λ, a = e ik a a k k µ = Λ µ νk ν

. Invariance of S[φ] under Parity transformations x µ = Λ µ P ν xν.30 Λ P =. Unitary Representation intrinsic parity η P = ±. On Fockspace: UP φx U P = η P φx UP φ x, t U P = η P φ x, t.3 UP 0 = 0 UP a k U P = η P a k.3 Parity reverses 3-momentum of particle: Scalar elds: η P = + Pseudo scalar elds: η P = e.g. Π 0 Parity: x x p p What about Parity transformations of pseudovectors like e.g. the angular momentum L: L = x p? So what about e.g. x L or p L? L x p pseudo vector x L x L p L p L pseudoscalars 3. Invariance of S[φ] under time reversal x µ = Λ µ T ν xν.33 Λ T =. Anti-unitary transformation V 3

a b c V c a + c b = c V a + c V b.34 V V = V V =.35 a V b = b V a.36 We have [ V ΛT φ x, t V Λ T ] = φ x t.37 On Fockspace: V Λ T 0 = 0 V Λ T a k V Λ T = a k.38 Note that a = V Λ T a b = V Λ T b.39 and a b = b a = a b.40 4. Charge conjugation: complex., 3., 4. CPT-invariance is required for any local, relativistic QFT. But CP, P, T violation is permitted and realised.. The interacting scalar eld In this chapter some basic concepts on scattering/perturbation theory are introduced. Interaction of a real scalar eld a static potential V x, e.g. a localised potential produced by a nucleus. Langrange density H = H 0 + H : Lx = L 0 x + L x.4 = ϕx µ µ m ϕx }{{} V xϕ x }{{} L 0x L x L x = V xϕ x L 0 x is the Lagrange density of a free scalar eld. interaction density. QM revisited: interaction picture L x is the Lagrange i t t = H t t.4 4

H is the interaction Hamiltonian see.5. t < T = i adiabatic t > T = f the solution t 0 = T where Ut, t 0 describes a unitary time evolution: t Ut, t 0 = + i dt H t t } 0 {{} = T exp{ i so that the time is ordered. We have rst order term, see page 5 t Iterate.4 in its innitesimal form: This denes the S-Matrix: t = Ut, t 0 t 0.43 + i t t 0 dt t t 0 dt H t H t +... t 0 dt H t }.44 i t Ut, t 0 = H tut, t 0.45 t + t = t i t H t t S = = i t H t t lim t 0 t + Ut, t 0..46 Back to eld theory: H t = d 3 x Lx, t.47 = d 3 x V x :φ x, tφ x, t: where φ x, t is a operator, which includes annihilation and creation of particles and : : denotes normal ordering: :a k a k : = + a k a k..48 5

Example: transition amplitude for transition We have Consider weak interactions: Then f S i = δ fi i from i = k = a k 0 at t 0 to f = k = a k 0..49 A fi = f S i = k S k = k i dt H t +... k R = k + i d 4 x L x +... k.50 V x 0. d 4 x V x k φx 0 0 φx k..5 The factor in.5 stands for the two permutations of a and a, included in φ x, which contribute. They are: a a and aa, because there is neither an overlap between three particles and one particle nor between one and 0, the annihilated vacuum state 0. Furthermore δ fi = f i = k k = 0 a k a k 0 = 0 [a k, a k] 0 = π 3 ω δ 3 k k.5 The last equation follows from equation.4 on page 9. Interpretation:. δ fi : no interaction k = k.. state k scatters once at V x into state k. d 3 k { 0 φx k = 0 π 3 ω e ik x a k + e ik x a } k a k 0 0 a = a 0 = 0 6

follows and similarily d 3 k 0 φx k = 0 = π 3 d 3 k π 3 ω e ik x ω e ik 0 [a k, a k] 0 x ω π 3 δ 3 k k = e ikx.53 k φx 0 = e ik x..54 Interpretation: Amplitudes for annihilating/ creating particles momentum k/ k at spacetime point x. We infer: A fi = f S i = δ fi i d 4 x V xe ik x e ikx [ ] = δ fi + i dt d 3 x V xe i k k x e ik 0 k 0 x 0 = δ fi i δk 0 k 0 Ṽ q.55 Ṽ q = d 3 x V xe i q x Interpretation revisited: q = k k. 3-momentum transfer.56. State k scatters at V x 'strength' Ṽ q into state k where q = k k.. Energy is conserved as k 0 = k 0. Final remark: Relation between the scattering amplitudes in momentum space and the form/ range of potential in space-time: Example: V x = V 0 π 3/ Ṽ q = V 0 exp { l 3 exp { } l q } x l.57.58 The potential V x and its Fourier transformation V q are plotted in gures. and. for one V 0 -l-combination. 7

Figure.: Potential V x for V 0 =000, l=0.5 Figure.: Fourier transformed potential V q for V 0 =000, l=0.5 8

Remark on self-interaction and Feynman rules: Figure.3: Feynman diagram for self-interaction. V x :φ x, tφ x, t: λ :φ x, tφ x, tφ x, tφ x, t:.59 4! 4, 3 4! : :, 4, 3 4! a a a a, 4 3 λ aa 4 Remark on complex elds: φ = φ + i φ.60 L 0 = φ x µ µ m φx.6 L = V xφφ.6 L = L 0 + L In general: L = L [φφ ] It follows that L is invariant under global U-transformation of φ: φx e iα φx.63 9

µ α = 0. φ x φ x e iα L[φ] L[φ e iα ] = L[φ].64 Noether theorem: µ j µ = 0 equation of motion.65 j µ = L δφ µ φ.66 Q = d 3 x j 0 Q = 0 Noether theorem for internal Symmetry: L φ L µ = 0 equation of motion µ φ δl = L φ δφ + L µ φ µδφ L = µ δφ + L µ φ µ φ µδφ L = µ δφ = 0 µ φ }{{} j µ µ j µ = 0 and Q = d 3 x j 0 = + d 3 x i j i = 0 No boundary terms..3 Spin Fields Motivation: Algebra of Lorentz group, see.4 on page 5. Boosts K i, Rotations J i N i = J i + i K i N i = J i i K i.67 0

SU-algebra [ ] N i, N j = i ε ijk N k.68 We have -dim. representations of the Lorentz group, the spin representations. Example: { } i Λ L = exp σ iω i iv i { } i Λ R = exp σ iω i + iv i left-handed right-handed.69 ω : rotation, v : boost, σ i : Pauli matrices The left-handed spin representation Λ L can be mapped to the right-handed spin representation Λ R by parity transformation. Dirac equation: γ µ are 4 4 matrices Standard representation: ψx = ψ x. ψ 4 x iγ µ µ mψx = 0.70 {γ µ, γ ν } = γ µ γ ν + γ ν γ µ = g µν Cliord algebra.7 γ 0 = γ i = 0 0 0 σ i σ i 0 Pauli matrices σ i see.8 on page 0. Remarks:. ψx consists of a two-component left-handed and a two-component right-handed spinor. Chiral representation: γ 0 = γ i = 0 0 0 σ i σ i 0.7.73

. γ µ transforms as a vector under Lorentztransformations. Equation of motion.70 from Lagrange density: the Dirac conjugate ψ = ψ γ 0. L D = ψ i γ µ µ m ψx.74 L D ψ L D µ = L D = 0.75 µ ψ ψ For the result of equation.75 see equation.70. Also L D L ψ D µ µψ = 0 m ψ i µ ψ γ µ = 0 Classical solution: i γ µ µ m i γ ν ν m ψx =.76 {γµ, γ ν } µ ν +m ψx }{{}}{{} g µν ν µ = [g µν ν µ + m ] ψx = [ + m ] ψx.77 We have ψx e ±ipx plane wave.78 i γ µ µ m e ±ipx = p m e ±ipx.79 p := γ µ p µ = γ 0 p 0 γ p γ p γ 3 p 3. A solution to the Dirac equation reads, s = ± ψx u s p e ipx ψx v s p e ipx.80 p m u s p = 0 = p + m v s p..8 Equation.8 is satised u s = p 0 + m v s = p 0 + m χ s σ p p 0 +m χ s σ p p 0 +m ε χ s ε χ s.8

χ = 0, χ = 0 0, ε = 0 p 0 = + p + m, σ : Pauli matrices ε describes the metric in spin space. Additional identity: u s p ū s p = p/ + m s=± v s p v s p = p/ m.83 s=± As for the scalar eld the general solution is given by the Fourier integral: d 3 p { ψx = e ipx π 3 v s p βs p + e ipx u s p α s p }.84 p 0 s=± β s and α s are independent of each other. One gets the Hamiltonian density H D via a Legendre transformation of L D : H D = Π ψ L D It follows Π = L D ψ, Π = L D ψ = 0 Π = ψ i γ 0 = i ψ..85 H D = ψ i γ 0 ψ ψ i γ µ µ m ψ = ψ i γ + m ψ.86 Hamiltonian: H D = = d 3 x ψ i γ + m ψ d 3 x ψ i γ 0 γ + γ 0 m ψ.87 =. Inserting.84 into.87 leads to d 3 p H D = π 3 α p s p α s p p 0 β s p βs p p 0.88 0 s=± 3

from H D = γ 0 i γ + m u s p = p 0 u s p γ 0 i γ + m v s p = p 0 v s p.89 d 3 p π 3 s=± α s p α s p β s p β s p.90 Negative energy states lead to unbounded Hamiltonian, no classical interpretation! Quantisation: ψx = d 3 p π 3 p 0 s=± anti-commutation relations { e ipx v s p b s p + e ipx u s p a s p }.9 {a r p, a s p } = δ rs π 3 p 0 δ 3 p p {b r p, b s p } = δ rs π 3 p 0 δ 3 p p.9 and Remarks: {a, a } = {b, b } = {a, b } = {a, b } = 0.93. The anti-commutation relations ACR are a manifestation of the Spin-statics theorem: Spin n+ particles have fermi-statistics ACR, Pauli principle, spin n particles have Bose-statistics.. Electric charge Noether: J µ = e ψ γ µ ψ Q = d 3 x J 0 = e d 3 x ψ ψ = d 3 p e π 3 p 0 a s p a s p + b s p b s p.94 s=± Please notice the positive sign in equation.94, that turns into a minus sign again! Fockspace: Construction as for scalar eld, but ACR anti-symmetric states 4

0 : normalised vacuum state, 0 0 = a s p 0 = 0 b s p 0 = 0 One-particle states: e p, s = a s p 0 e + p, s = b s p 0 e p, s / e + p, s describe electron/ positron momentum p and spin s = ± s z in rest-frame. Remark: Prediction of e +, e identical mass is triumpf of the Dirac theory. Orthogonality: e p, s e p, s = 0 a s p a s p 0 = 0 {a s p, a s p} 0 = π 3 p 0 δ s s δ 3 p p Two-particle states: e p, s e p, s = a s p a s p 0 Pauli principle e p, s e p, s = a s p a s p 0 N-particle states: = a s p a s p 0 = e p, s e p, s.95 a s p... a s n p n b r q... b r m q m 0 Finally, ψx = d 3 k π 3 k 0 r=± { e ikx v r k b r k + e ikx u r k a r } k 0 ψx e p, s = u s p e ipx e + p, s ψx 0 = vs p e ipx ψ: Annihilation of an electron/ creation of a positron at x. 0 ψx e + p, s = v s p e ipx e p, s ψx 0 = ūs p e ipx 5

ψ: Annihilation of a positron/ creation of an electron at x. Symmetries: S D [ψ, ψ] = d 4 x ψ i γ µ µ m ψ.96. Invariance of S D [ψ, ψ] under orthochronous Poincaré transformations: where S satises and U is unitary. Dirac adjoint spinor: x µ = Λ µ ν x ν + a µ see.3.97 UΛ, a ψx U Λ, a = S Λ ψλx + a S Λ γ µ SΛ = Λ µ ρ γ ρ.98 UΛ, a ψx U Λ, a = ψλx + a SΛ.99 The invariance of S is to show: d 4 x ψ i γ µ µ m ψx d 4 x ψλx + a S i γ µ µ m ψλx + a = d 4 x ψx S i γ µ ν Λ µ ν m S ψx = d 4 x ψx S i Λ ν µ γ µ ν m S ψx = d 4 x ψx S i S γ ν S ν m S ψx = d 4 x ψ i γ µ µ m ψx.00 General bilinears: a ψ ψ scalar: m ψ ψ pseudo scalar later b ψ γ µ ψ vector pseudo vector later c ψ σ µν ψ tensor, σ µν = i [γµ, γ ν ]. Invariance of S D [ψ, ψ] under Parity Λ P = 3 see equation.8.0 6

Unitary representation: UP ψ x, t U P = γ 0 ψ x, t UP e p, s = e p, s e +, e are parity 'eigen states'. Relative intrinsic parity can be measured: UP e + p, s = e + p, s.0 3. Invariance of S D [ψ, ψ] under time reversal Λ T = see equation.9.03 3 Anti-unitary transformation V : V T ψ x, t V T = ST ψt x, t.04 and ST = i γ γ 5.05 γ 5 = i 4! ε µνρσ γ µ γ ν γ ρ γ σ γ 5 = i γ 0 γ γ γ 3 ε 03 = {γ 5, γ µ } = 0 We have V T e p, s = s e p, s V T e + p, s = s e + p, s.06 4. Charge conjugation C and C : e + e UC ψx U C = SC ψ T x.07 SC = i γ γ 0 0 ε = ε 0 0 ε = 0 UC e p, s = e + p, s.08 UC e + p, s = e p, s.09 7

Bilinears: scalar: ψ ψx generator pseudo-scalar: i ψ γ 5 ψ generator vector: ψ γ µ ψ 4 generators pseudo-vector: ψ γ5 γ µ ψ 4 generators 3 tensor: ψ σ µν ψ 6 generators 6 generators of the Lorentzgroup Remark: σ µν = i [γµ, γ ν ].0 γ µ γ ν γ ρ ε µνρσ γ σ γ σ γ 5 γ σ.4 The interacting fermionic eld a rst glimps of QED Classical: Langrangian density Lx = L D x + L x. = ψxi γ µ µ mψx + e A µ x }{{} ψx γ µ ψx }{{} L D x L x L x = e A µ x ψx γ µ ψx. Since ψxγ µ ψx transforms as a vector under Lorentz transformations, A µ x has to transform as a vector: A µ is a vector eld. Remark: Lx is invariant under Quantisation in interaction picture A µ x Λ µ ν A ν x.3 ψx e ie αx ψ ψx ie αx ψe A µ x A µ x + µ αx j µ = e ψ γ µ ψ.4 i t t = H t t.5 8

H t = d 3 x L opx.6 = e d 3 x A µ x : ψxγ µ ψx: A µ can be either a background eld classical or a quantum eld A µop : it is bosonic as a vector spin and commutes ψ, ψ, also: :A µ : = A µ creation/ annihilation operators a µ, a µ. Subtleties concerning the quantisation of A µ later, physical state γ a 0 Equation.5 is solved by t = Ut, t 0 t 0.7 { t } { t } Ut, t 0 = T exp i H t dt = T exp ie d 4 x A µ : ψγ µ ψ : t 0 t 0.8 L couples e ± to the electromagnetic eld A µ, the photon. Similarily we couple µ ± and τ ± to A µ : ψ e ψx = ψ µ ψ τ.9 and L D = ψ e iγ µ µ m e ψ e + ψ µ iγ µ µ m µ ψ µ + ψ τ iγ µ µ m τ ψ τ.0 Computation of transition amplitude Initial state at t 0 : m e = 0, 5 MeV m µ = 05, 7 MeV m τ = 784 MeV L x = e [ ψe A ψ e + ψ µ A ψ µ + ψ τ A ψ τ ]. t 0 = i = e p... e p n e + q... e + q m γk... γk l. and µ s, τ s. Final state at t + : t = f = e p... e p ne + q... e + q mγk... γk l.3 9

e.g.: µ p... µ p nµ + q... µ + q mγk... γk l This is related to the S matrix element. Here, we are interested in In general: e p +... + γk l e p +... + γk l e + e µ + µ. e + k + e k µ p + µ + p S fi = f t = = lim t 0 t + = f T exp Expanding T exp{i d 4 x L } leads to First order: t = Ut, t 0 i f Ut, t 0 i.4 { ie L x = ea µ : ψγ µ ψ :. } d 4 xa µ : ψγ µ ψ : i S fi = f i +ie f T d 4 xa µ : }{{} ψγ µ ψ : i + δ fi + ie f d 4 x d 4 x T L x L x i +! +... + + ien f d 4 x... d 4 x n T L x... L x n i + n! +....5 d 4 x f A µ x : ψxγ µ ψx: i.6 d 4 x f... a µ +... a µ... b sb r +... a sb r +... b s a r +... a sa r i We have the following processes: Time ordered diagrams. Scattering of e + emission/absorption of γ.. Creation of e + e pairs emission/absorption of γ. 30

3. Annihilation of e + e pairs emission/absorption of γ. 4. Scattering of e emission/absorption of γ. Higher order processes are composed out of rst order processes, e.g. second order or i γ, f γ Problem: Convergence of expansion in i e + e f µ + µ α = e 4π = 37. Series is an asymptotic series: does not converge. All orders are innite renormalisation. Programme: Write down all diagrams for a given order in α for matrix element f S i. Sort out combinatorics normal ordering, compute the remaining integrals. Feynman rules/ Loop integrals Reminder: dierential cross section page 6 dσ = dγ[i f] Φ = π4 V 4 δ 4 p A + p B p C p D A fi dρ f Φ for two particle scattering. Φ: particle ux, normalisation of the states i, f and A fi = M fi = f H i..7 Example: V d 3 p C V d 3 p D dρ f : π 3 p 0 C π 3 p 0 D e + e µ µ + Cross section dσ: dσ = V d 3 p 3 V d 3 p 4 T π 3 p 0 3 π 3 p 0 4 F µ + p 4 µ p 3 S e + p e p }{{}}{{} spins S-Matrix element phasespace density F is the incident particle ux..8 3

Dierential cross-section dσ per unit volume V for e e + µ µ + : dσ = T F d 3 p 3 π 3 p 0 3 d 3 p 4 π 3 p 0 4 µ + p 4 µ p 3 S e + p e p.9 spins incident particle ux F, and unit volume V =. Consider the term between the absolut value bars in.9 rst: S-Matrix: S = T e ie d 4 x A νx : ψγ ν ψx: ψ γ ν ψx = ψ e γ ν ψ e x + ψ µ γ ν ψ µ x. Expansion of S-matrix element for e e + µ µ + :.30 µ + p 4 µ p 3 S e + p e p = = ie µ+ p 4 µ p 3 T d 4 x d 4 x A ν x A µ x Consider the states in.3: Fermionic eld operator d 3 p ψx = π 3 p 0 A µ commutes ψ, ψ: : ψ γ ν ψx: : ψ γ µ ψx : e + p e p + Oe 4.3 e + p e p = b e p a e p 0 s=±/ { e ipx v s p b p + e ipx u s p a p }.3 µ + p 4 µ p 3 S e + p e p = = ie d 4 x d 4 x 0 T A ν x A µ x 0 µ + p 4 µ p 3 T : ψ γ ν ψx: : ψ γ µ ψx : e + p e p + + Oe 4 = ie d 4 x d 4 x 0 T A ν x A µ x 0 µ + p 4 µ p 3 : ψ γ ν ψx: : ψ γ µ ψx : e + p e p + + Oe 4 = ie d 4 x d 4 x 0 T A ν x A µ x 0 µ + p 4 µ p 3 : ψ µ γ ν ψ µ x: : ψ e γ µ ψ e x : e + p e p + + Oe 4.33 3

Counting annihilation/creation operators: a, b : µ + p 4 µ p 3 : ψ µ γ ν ψ µ x: : ψ e γ µ ψ e x : e + p e p = = µ + p 4 µ p 3 : ψ µ γ ν ψ µ x: 0 0 : ψ e γ µ ψ e x : e + p e p.34 Further reduction of the last part of.34: 0 : ψ e γ µ ψ e x : e + p e p = 0 : ψ e γ µ ψ e x : b e p a e p 0.35 For the fermionic eld operator ψx see.3. Further reduction of the right side of.35 leads to 0 : ψ e γ µ ψ e x : b e p a e p 0 = 0 ψ e x b e p 0 γ µ 0 ψ e x a e p 0.36 Expectation value 0 ψ e 0 ψ e x a e p 0 = d 3 p π 3 p 0 e ipx s=±/ the commutator trick: 0 ψ e x a d 3 p e p 0 = π 3 p 0 e ipx s=±/ u s p 0 a p a e p 0.37 u s p 0 {a e p, a e p } 0..38 For more information about the result of the anti-commutation relation above see.9 on page 4. So one gets for 0 ψ e : {a s p, a r p } = π 3 p 0 δ rs δ p p 0 ψ e x a e p 0 = e ipx u e p.39 Analogous one can calculate the expectation value 0 ψ e +. In summary the expectation values 0 ψ e, 0 ψ e + : 0 ψ e x a e p 0 = e ipx u e p.40 0 ψ e x b e p 0 = e ipx v e p.4 It follows a further simplication of.35: 0 : ψ e x γ µ ψ e x : b e p a e p 0 = v e p γ µ u e p e ip+px Similarily for the muon:.4.43 0 b µ p 4 a µ p 3 : ψ µ x γ ν ψ µ x: 0 = ū µ p 3 γ ν v µ p 4 e ip3+p4x.44 33

Plug the results in.3: µ + p 4 µ p 3 S e + p e p ie d 4 x d 4 x 0 T A ν x A µ x 0 e ip3+p4x e ip+px ū µ p 3 γ ν v µ p 4 v e p γ µ u e p.45 Now consider the photon propagator: 0 T A ν x A µ x 0 = ig µν lim ɛ 0 + d 4 k e ikx x π 4 k + iɛ Momentum conservation d 4 x d 4 x d 4 k e ik p px e ik p3 p4x π 4 k + iɛ the square of the total energy s = p + p leads to µ + p 4 µ p 3 S e + p e p = s π4 δp + p p 3 p 4.46 ig µν s ie π 4 δp + p p 3 p 4 ū µ p 3 γ ν v µ p 4 v e p γ µ u e p i s ie π 4 δp + p p 3 p 4 ū µ p 3 γ ν v µ p 4 v e p γ ν u e p.47 Now the term between the absolut value bars in.9 is calculated. The next step is the averaging over the spins in the initial and the nal state see.9: ū µ,s p 3 γ ν v µ,s p 4 v e,r p γ ν u e,r p = 4 s,s,r,r = ū µ,s p 3 γ ν v µ,s p 4 v µ,s p 4 γ ρ u µ,s p 3 s,s v e,r p γ ν u e,r p ū e,r p γ ρ v e,r p.48 r,r [ ] v r p γ ν u r q = u r q γ ν γ 0 v r p = ū r q γ ν v r p. Consider the rst sum in.48 rst: ū µ,s p 3 γ ν v µ,s p 4 v µ,s p 4 γ ρ u µ,s p 3 = Trp/ 3 + m µ γ ν p/ 4 m µ γ ρ see.83 s,s 34

Similarily one gets for the second sum in.48: v e,s p γ ν u e,s p ū e,s p γ ρ v e,s p = Trp/ m e γ ν p/ + m e γ ρ s,s In summary one gets as an intermediate result for the average over the spins in the initial and the nal state see.48: ū µ,s p 3 γ ν v µ,s p 4 v e,r p γ ν u e,r p = 4 s,s,r,r = [p/ 4 Tr 3 + m µ γ ν p/ 4 m µ γ ρ] ] Tr [p/ m e γ ν p/ + m e γ ρ.49 In high energy limit s m µ, m e one can drop m e, m µ in the traces of.49. So.49 turns into ū µ,s p 3 γ ν v µ,s p 4 v e,r p γ ν u e,r p = [ 4 4 Tr p/ 3 γ ν p/ 4 γ ρ] ] Tr [p/ γ ν p/ γ ρ s,s,r,r With the traces Tr γ α γ ν γ β γ ρ = 4 g αν g βρ + g ρα g νβ g αβ g νρ.50 one gets for.48: [ ] ū µ,s p 3 γ ν v µ,s p 4 v e,r p γ ν u e,r p = 4 p p 4 p p 3 +p p 4 p p 3 4 s,s,r,r High energy limit revisited p p 3 = p p 4 = s 4 cos ϑ, p p 4 = p p 3 = s + cos ϑ 4 scattering angle one gets the nal result for.48: 4 cos ϑ = p p 3 p p 3.5 s,s,r,r ū µ,s p 3 γ ν v µ,s p 4 v e,r p γ ν u e,r p = s + cos ϑ.5 Back to the dierential cross-section dσ per unit volume see.9. When one plugs in all results calculated above one gets for the dierential cross-section dσ per unit volume dσ = T F d 3 p 3 π 3 p 0 3 d 3 p 4 π 3 p 0 [π 4 δp + p p 3 p 4 ] e4 s 4 s + cos ϑ.53 35

With Fermi's trick and α = e 4π [π 4 δp + p p 3 p 4 ] = π 4 δp + p p 3 p 4 V T = V T π 4 δp + p p 3 p 4 d 4 x e ixp+p p3 p4 one gets for the dierential cross-section dσ per unit volume dσ = α F d 3 p 3 p 0 3 d 3 p 4 p 0 δp + p p 3 p 4 + cos ϑ.54 4 In high energy limit and the CMS-system, i.e. p + p = 0, one gets for the dierential cross-section p 0 + p 0 s dσ dω 3 dσ dω 3 = α F With the ux F 0 d p 3 p 3 the dierential cross-section d 3 p 4 p 4 δ s p 3 p 4 δ p 3 + p 4 + cos ϑ. F = p 0 p 0 p p 0 dσ dω 3 = v A E A E B.55 dσ = α dω 3 4 + cos ϑ.56 dσ and the total cross-section σ = dω 3 dω 3 σ total e e + µ µ + = 4πα 3.57 for e e + µ µ + are calculated. 36

Chapter 3 Quantenelectrodynamics QED 3. The electromagnetic eld Maxwell's equations: µ F µν x = j ν x 3. ε µνρσ ν F ρσ x = 0 3. eld strength F µν and 4-vector potential A µ x. 3.3 trivially satises 3.: F µν x = µ A ν x ν A µ x 3.3 ε µνρσ ν ρ A σ = 0 j ν x is the 4-vector current density: j ν x = ρx jx 3.4 and F µν x = 0 E E E 3 E 0 B 3 B E B 3 0 B E 3 B B 0 3.5 We use Heaviside units rational, that is removing factors of 4π from the 37

equation. Maxwell's equations see 3. reads 3.4 and 3.5: Relation to cgs units: E = ρ B = E t + j 3.6 B = 0 E = B t α = e H 4π c 37 e H = 4π e cgs 3.7 e cgs = e H 4π = e SI 4πε 0 3.8 E H = E cgs 4π B H = B cgs 4π The most important values in the cgs-system: Remarks: e cgs = 4.8 0 0 g cm 3 /s, c = 3 0 0 cm/s cgs =.05 0 7 erg s, erg = g cm /s = 0 7 J. Inhomogeneous Maxwell equation 3.: conserved current! 3.9 ν µ F µν = ν j ν = 0 3.0. A µ carries a redundancy, the gauge degrees of freedom: F µν is invariant under A µ x A µ x + µ αx F µν F µν + [ µ, ν ] α = F µν 3. This redundancy can be removed by imposing a constraint on A µ gauge xing condition: Lorentz gauge Landau: µ A µ x = 0 µ F µν = A ν = j ν 3. consistent 3.0. For j ν = 0, each component A ν satises the Klein-Gordon equation. 38

Lagrangian density: Lx = 4 F µνx F µν x A µ x j µ x 3.3 Quantisation of free eld: A µ x = d 3 k [ π 3 e ikx a k µ k + e ikx a µ ] k 0 3.4 k 0 = k, and the commutators [a ν k, a µ k] = g µν π 3 k 0 δ k k Remarks: [a ν k, a µ k ] = 0. We have the Klein-Gordon equation: but = [a ν k, a µ k ] 3.5 A µ x = 0 3.6 µ A µ x ik µ a µ... 0. k µ [a ν k, a µ k] = k ν π 3 k 0 δ k k µ F µν [A] 0 3.7 Can we do better than 3.4? No, it was not possible to construct A op µ A opµ = 0 + covariant.. Fockspace: vacuum 0 0 0 = one particle states norm a µ k 0 = 0 3.8 a µ k 0 0 a ν k a µ k 0 = 0 [a ν k, a µ k] 0 3.9 = g µν π 3 k 0 δ k k µ = ν = i : positive norm states µ = ν = 0 : negative norm states No physical Hilbertspace no probability interpretation. 39

Remedy: Fockspace contains physical subspace F phys µ F µν [A µ ] physical states! = 0 or Evidently 0 F phys. Construction of F phys : k µ a µ k physical states = 0 3.0 µ A µ physical states = 0 α 0 k = k kµ a µ k = a 0 k ˆk a k ˆk = k k. α k = ê a k α k = ê a k α 3 k = a 0 k + ˆk a k 3. ê i ˆk = 0 ê i ê j = δ ij Commutators: Physical states: [α 0, α 0 ] = [α 3, α 3 ] = 0 [α 0, α i ] = [α 3, α i ] = 0 [α 0, α ] = π3 k 0 δ [α i, α i ] = π3 k 0 δ 3. α 0 k physical state = 0 3.3 One particle states: α 0 k 0, α k 0, α k 0 but zero-norm states 0 α 0 k α 0 k 0 = 0 3.4 Physical Hilbertspace H: for = 0 40

H = F phys. We have two one particle states in H: k, ε = α k 0 3.5 Polarisation ê, ê, g and 0 ε i = êi i =, 3.6 General: k, ε ε 0 = 0 and ε k = 0, ε C 3 E- and B-eld operators: E i x = F 0i = 0 A i i A 0 Ex = = Bx = d 3 k { π 3 ik 0 a k ˆk a 0 e ikx a ˆk a 0 e ikx} 0 d 3 k π 3 k 0 ik 0{ ê α k + ê α k e ikx ê α k + ê α k e ikx} d 3 k {ˆk π 3 ik 0 α0 k k e ikx ˆk α 0 k e ikx} 0 d 3 k π 3 i k { kx ê i α i k e ikx ê i α i k e ikx} 3.7 0 Hamiltonian depends on α i, α i. From follows the canonical momentum Hamiltonian density Lx = 4 F µν F µν = E B 3.8 Π i = L 0 A i = F 0i = E i 3.9 H = Π 0 A + 4 F µν F µν = E E B + E A 0 = E + B + EA 0 3.30 4

The last equation follows from E = 0 for ρ = 0. H = d 3 x H = d 3 x E + B 3.3 Use eld operators H d 3 k d 3 k [ π 3 π 3 k 0 k 0 ik0 α i k α i k + α i k α i ] k π 3 δk k = d 3 k [ π 3 k 0 k0 α i k α i k + α i k α i ] k 3.3 Normal ordering causes α i k α i k + α i k α i k to become α i k α i k. Inserting the result of normal ordering in equation 3.3 leads to: but... H d 3 k π 3 k 0 k0 α i k α i k 3.33 Interlude: The Casimir eect Experiment: Lamoreaux et al 997 Solution for E, B: plane waves Boundary conditions: E ε e ±ikx, B ˆk E 3.34 ˆn E x=0, L = 0 ˆn B x=0, L = 0 3.35 the polarisation ε and E ε sink x x e ikyy+kzz k0 t 3.36 k 0 = k k x = nπ, n =,,... L Casimir energy: 0 H 0 L = d 3 k π 3 0 α i k α i k 0 = d k L π k nπ + δ0 3.37 L n= k = 0, k y, k z. 4

Equation is divergent for large momenta UV. Assume for the moment that the is cut o regularly for high energies/momenta: 0 H 0 L = V d k }{{} L π Area n= where r Λ x Λ 0, r Λ x Λ. R Λ n = E L = E L E : E L = Euler-McLaurin: 0 dn R Λ n = Bernoulli numbers Since E L = 0 H 0 L = 0 k nπ + rλ k nπ L + L V π L 3.38 R Λ n 3.39 n= dk k k nπ + rλ k nπ L + L [ V R Λ n π L n= n= 0 dn R Λ n ] + R Λ0 3.40 [ R Λ n + ] n! B n R n Λ 0 + R Λ0 3.4 B = 6, B 4 = 30,... 3.4 E = V [ π L R Λ 0 + ] 70 R3 Λ 0 +... 3.43 R Λ n = = 0 nπ L dk k k nπ + L }{{} k dk k r Λ k r Λ k 3.44 R Λ n = π L R Λ 0 = 0 nπ L rλ nπ L R 3 Λ 0 = π L 3 3.45 43

R i>3 Λ 0 depends on r Λ : R i>3 Λ 0 ΛL i 3. Finally: E = π V/L 70 L 3 for Λ or Force/Area: F = d ε dl = π 40 ε = E V/L = π 70 L 3 3.46 0 7 c.3 L4 L[m] 4 pa m 4 Summary: Force/Area: nπ L E = π V/L 70 L 3 R Λ n = π L = π 70 V/L L 3 L rλ nπ L i 3 R i>3 Λ 0 ΛL + OR4 Λ 0 for Λ 3.47 E d V/L F = dl = π 40 0 7 c.3 L4 L[m] 4 pa m 4 Idea: plates, out plates For high frequencies the dierence between 'plates' and 'no plates' becomes smaller. E = 0 H 0 L 0 H 0 L= is nite. Divergent parts cancel. Computation is performed regularisation nπ Λ : r Λ k + L 3. Lagrangian of QED In equation 3.3 on page 39 the Langrangian density of the electromagnetic eld coupled to an external current j µ x was presented. In QED, j µ describes the coupling to the electron-, muon-, tau-current j µ x = e ψ γ µ ψx 3.48 44

where ψ is given by ψ = Together the Dirac term we get ψ e ψ µ ψ τ 3.49 L QED = 4 F µνx F µν x + ψ i γ µ D µ m ψ 3.50 where D µ is the covariant derivative: D µ ψx = µ iea µ x ψx 3.5 and m = m e 0 0 0 m µ 0 0 0 m τ Gauge invariance: gx = e i αx U and ψx gx ψx = e iαx ψx = ψ g ψx ψx g x = ψx e ieαx = ψ g 3.5 A µ x i e g D µ g = A µ x + µ αx = A g D µ x g D µ g x 4 F µν F µν 4 F µν F µν 3.53 ψ id/ m ψ ψ g igd/g m gψ = ψ id/ m ψ 3.54 D/ = γ µ D µ gg =. Remark: µ µ iea µ is called 'minimal coupling'. A µ is a connection Zusammenhang. Consequences of gauge invariance: Classical action of QED: S QED [A, ψ, ψ] = d 4 x L QED x L QED x from equation 3.50. Gauge invariance: S[A g, ψ g, ψ g ] = S[A, ψ, ψ] 45

Innitesimal = S[A + µ α, + iαe ψ, ψ iαe] S[A, ψ, ψ] = 0 d 4 δ δ x µ αx + ieαx ψx δa µ x δψx ψx δ δ ψx S e.g.: ψ = ψ = 0: d 4 δ x µ αx S[A, 0, 0] = 0 δa µ x d 4 δs[a, 0, 0] x αx µ = 0 δa µ x αx arbitrary: also µ µ δs[a] δa µ x = 0 δ S δa µ xδa ν y A=0 = 0 µ ρ ρ g µν µ ν δx y = 0 Completion of Feynman rules: Physical p states: k, ε i = α i k 0 = ê i a k 0 i =, 3.55 0 the 4-vector ε i = êi General states: ε µ ε µ = k, ε = ε µ α µ k 0 = ε a k 0 3.56 ε 0 = 0, ε µ k µ = 0, ε k = 0. Norm k, ε ε, k = ε ε π 3 k 0 δ k k 3.57 initial state ε µ nal state ε µ 3.3 Magnetic moment of electron Consider D µ = µ iea µ L D = ψ i γ µ D µ m ψ H D = ψ i γ 0 γ i D i + γ 0 m ψ }{{} 3.58 H 46

Evaluate non-relativistic limit of H : First H = γ 0 i γ D + m γ 0 i γ D + m = γ D + m i γ D + m = γ i γ j D i D j + m = D + [γ i, γ j ] D i D j + m = D i e 4 [γ i, γ j ] F ij + m 0 σ and Σ k k = σ k 0 [γ i, γ j ] = iε ijk Σ k H = D e ε ijk F ij + m = D + e Σ B + m. 3.59 Then Remember Σ = S H = m + p e A m + e m Σ B + Om 3.60 i t ψ = p e A m + m + e m S B 3.6 Spin coupling: e g e S m B = µ e B gyromagnetic factor g e =. Experiment: g e =.00393... e m g e =.76085977044 0 s T Anomalous magnetic moment: j op x = e : ψx γ µ ψx: µ op = d 3 x x j op x, t 3.6 e k, s = a s k 0. Expectation value: Non-interacting: e k, r µ e k, s = e d 3 x x e k, r : ψ γ ψ : When one proceeds as scattering, one gets g e =. e k, s 3.63 47

Magnetic moment of an electron in general: µt = d 3 x x jx 3.64 Electron: t = 0. e p, r µ e p, s = e = e = e = e d 3 x x e p, r : ψ x, 0 γ ψ x, 0: e p, s d 3 x x 0 a r p : ψ x, 0 γ ψ x, 0: a s p 0 d 3 x x 0 a r p ψ x, 0 0 0 ψ x, 0a s p 0 d 3 x e ip p x x ū r p γ u s p Anomalous magnetic moment Classically the interaction is given by =... 3.65 e ψ γ µ A µ ψ Looking for quantum corrections to the Vertex: Lowest order in α and A: e Γ µ q, q + p A µ p Γ µ i m σ µν p ν α π σ µν = i [γµ, γ ν ]. 3.66 For σ µν see page 8. Γ µ A µ x ie m σ α µν π ν A µ x = ie α = 4m π σ µν F νµ x α Σ π B e m One gets the last equation, because E = 0. µ µ + e m α Σ π = e m + α Σ π }{{} S 3.67 3.68 48

3.4 Renormalisation α higher order term α vacuum polarisation Computation from Feynman rules: Π µν p = ie d 4 q π 4 Tr i i q/ m γµ p/ + q/ m γν 3.69 [v ie Two Problems:. integral is divergent d 4 q q. poles: q = 0, q + p = 0. Remedy: d 4 q π 4 Tr q/ ] q γµ q/ + p/γν p + q. Regularisation: Π Π Λ p, m Casimir = p p µ 3.70. Wick rotation: t M it E, p 0 M ip0 M, p µp µ p E µ p E µ. There is a one-to-one relation between Euklidean Correlation functions G E and Minkowski Correlation functions G M. How to implement : Photon propagator: A µ ν ν g µρ µ ρ A ρ Z A A ren. µ ν ν g µν µ ρ A ren. ρ 3.7 Demand nite. Z A = + Z A α + Oα lim Z A + Π Λp, m Λ Structure of Π Λ Π Λ p, m = α renormalisation scale µ. Hence Z A ] [f 0 ln p p Λ + f + O Λ +... Z A = α f 0 ln Λ µ + α finite + Oα + Π Λp, m = + α [ ln 49 p µ ] + finite + Oα

In summary we demand Computation: µ d dµ Observables = 0 renormalised Π: Π Λ p, m Π Λ 0, m. Π µν p = p g µν p µ p ν Πp 3.7 UV-asymptotics: Πp = α π i m dx x x ln m x xp 0 = α π i m /p dx x x ln m /p x x 0 Πp α ] [ln p m 3π m 5 +O }{{} 3 p finite=c 3.73 α eff p = α α p 3π ln m c lim eff p 0 = 0 lim α eff p m c e 3π/α = here c = e 5 3, αeff c m = α -loop β-function RG-equation: µ d dµ Z A Z A = 3π α = 6π α 50

Chapter 4 Quantum Chromodynamics QCD The theory of strong interactions provides the nuclear forces that keep nuclear cores together. Peculiar properties: α s = g s 4π. asymptotic freedom: α s p 0. Nobel Prize 004 for Gross, Wilczek and Politzer. connement: V q q r r for large distances. Millenium Prize riddle Jae, Witten. 3. selnteraction of gauge elds Gauge elds are colour charged. Evidence for ± 3 e, ± 3 e charged hadronic constituents spin : quarks q J. Joyce, and gluonic jets Nobel Prize 969 for Gell-Mann, Zweig. 4. The QCD-Lagrangian Hadronic current: j µ x = e q Q q qx γ µ qx 4. Q u,c,t = 3, Q d,s,b = 3. Hadronic states are invariant under SU3 transformations in colour space: qx U qx µ U = 0 q α x U αβ q β x α, β =,, 3 5

SU3 is non-abelian. Innitesimal U SU3: U U = UU = 3, det U =. U = 3 + i δϕ a λ a, a =,..., 8 λ a : generators of SU3 Lie-Algebra of SU3 λ = 0, λ = λ Compare to the generators of SU σ a : Quarks: [σ a, σ b ] = i ε abc σ c qx = q x q x q 3 x Dirac spinors q i, where i, i =,, 3, indicates the colour. 4. In table 4. the avours of quarks, their current masses and their charges are listed. Generation rst second third Charge Mass [ev].5-4 50-350 70 Quark u c t 3 Quark d s b 3 Mass [ev] 4-8 80-30 4.-4.4 Table 4.: Quarks and some of their properties. Lagrangian: Bound states:. Mesons: q q. Baryons: qqq e.g.: L 0 quark = q qx i / m q qx 4.3 π + u d + u d + u 3 d3 p ε αβγ u α u β d γ ε αβγ u α u β d γ det U 5

π + and p are gauge invariant. where f abc are structure constants. [λ a, λ b ] = i f abc λ c Tr λ = 0 Tr λ a λ b = δ ab {λ a, λ b } = 4 3 δab + d abc λ c d abc = Tr λ a {λ b λ c } The Lagrangian L 0 quark x is invariant under q Uq q q U. L 0 quark x q q U i / m Uq 4.4 With U U = 3 : q q U i / m Uq = L 0 quark x 4.5 Gauge principle as in QED: We demand but qx Ux qx qx qx U x L quark x L quark x L 0 quark x L0 quark x + q q U /U q /δ αβ γ µ D αβ µ 4.6 see page 45. SU3: D αβ µ = µ δ αβ i g s A αβ µ, [t a, t b ] = i f abc t c t c = λc 53

f abc is anti-symmetric: and λ = f 3 = f 47 = f 56 = f 46 = f 57 = f 345 = f 367 = 3 f 458 = f 678 = 0 0 0 0 0 0 0 λ 4 = λ 7 = 0 0 0 0 0 0 0 0 0 0 0 0 i 0 i 0, λ =, λ 5 = 0 i 0 i 0 0 0 0 0, λ 8 = 0 0 i 0 0 0 i 0 0 3, λ 3 = 0 0 0 0 0 0, λ 6 = Quark colour charge: t a t a = C F C F = 4 3. Gluon colour charge: f abc f abd = C A δ cd C A = 3. 0 0 0 0 0 0 0 0 0 0 0 0 0 0,, tr adj. t c t d = C A δ cd tr fud. t c t d = δcd λ Transformation of A αβ µ = A a a αβ µ : }{{} t a A µ x Ux A µ x U x i g s Ux µ U x = i g s Ux D µ U x D µ x Ux D µ U x 4.7 As in QED we dene the eldstrength F µν : i g s F µν = [D µ, D ν ] = i g s µ A ν ν A µ + ig s [A µ, A ν ] 4.8 and F µν U F µν U [A µ, A ν ] = i f abc A b A c λ a 54

or Fµν a = µ A a ν ν A a µ g s f abc A b A c. Pure gauge theory: Yang-Mills L Y M x = Tr F µν F µν = 4 F µν a F aµν 4.9 Full Lagrangian: Lx = 4 F a µν F aµν + q q id/ m q q 4.0 gauge symmetry U SU3: Lx Lx q U q q q U A µ U A µ U i g s U µ U Tr F µν F µν Tr U F µν U U F µν U = Tr F µν F µν Parameters: g s or α s = g s 4π where α s is the strong ne structure constant. m u,d,s,c,b,t are the quark masses. Gauge xing Lorentz Lx Feynman rules for QCD Gauge xing: ζ µ A µ p/+m Quark propagator: i p m +iε δαβ Gluon propagator: i [ ] Ghost propagator:i p +iε k +iε g µν Lx + ζ [ ] ζ kµkν k +iε Tr µ A µ + c D µ c } {{ } ghosts Quark-gluon vertex: i g s t a αβ γ µ 3-gluon vertex: g s f abc [g µν p q ρ g νρ q r µ + g ρµ r p ν ] 4-gluon vertex: ig s[f ead f ebc g µν g σρ g µρ g νσ + +f eac f ebd g µν g σρ g µσ g νρ + +f eab f ecd g µσ g νρ g µρ g νσ ] 4. 55

4. Running coupling As in QED we compute the running coupling. Reminder QED page 50 Euklidean computed from β-function: p m Cpt. αp αµ = 4. αµ 3π ln p µ β QED α p p αp = α 3π + Oα3 > 0 β QED = β 0 α β α 3 +... By comparison of coecients one gets: β 0 = 3π 4.3 QCD: β QCD p p α s p = π 33 N f α s + Oα 3 s 4.4 β 0 = π 33 N f 4.5 α s p = = α s µ + α s µ β 0 ln β 0 ln p Λ QCD p µ 4.6 { µ = Λ QCD exp β } 0 α s µ asymptotic freedom: α s p β 0 ln p 0 µ 4.3 Connement no coloured asymptotic states. Example: q q-pair Λ QCD 7 +5 3 MeV 56

. Denition of q q-state: qy qx not gauge invariant but qy P exp +i g s. gauge trafo: x y dz µ t a A a µz qx }{{} x i g s dz µ t a A a z µ z=y A µ x Ux A µ x U x i g s Ux µ U x see page 54. + i g s dz µ t a A a µz = Uz + i g s dz µ t a A a µz U z + µ U z dz µ }{{} P exp +i g s x y lim x y dz µ A µ z qyp exp +i g s Uy P exp +i x in quenched QCD no dynamical quarks Three quarks: Connement Remarks: strong coupling is not enough!! V r = V 0 + κ r e r + f r y x dz µ A µ z qx 0 y g s dz µ A µ z U x mass gap in Yang-Mills pure glue [Millenium Prize Jae, Witten] perturbation theory fails non-perturbation methods lattice: space-time grid 6 4 lattices operator product expansions/sum rules... renormalisation group methods solve theory via relations between correlation functions. 3. Area law of Wilson loop WC x,y = Tr P exp +i g s C x,y dz µ A µ z U z+dz 57

QED: { exp i e } d 4 x j µ x A µ x wordline of an electron j µ = C x,y dz µ δx z WC x,y exp { F qx q y } 0 e σa 4. dynamical quarks 4.4 Phase diagram of QCD Order parameter: chiral condensate: qq = σ Polyakov loop L e Fq Remark on phase transitions: 58

Chapter 5 Electroweak Theory Quantum Flavourdynamics, QFD In 930 Pauli suggested the existence of the neutrino ν. It was discovered from 953 to 959 by Reines. In the years 933 and 934 Fermi worked out a theory of the β-decay. β-decay: n p + e + ν e Fermi interaction: H = G d 3 x [px γ µ nx][ex γ µ νx] + h.c. 5. G is the Fermi constant: G =. 0 5 GeV. Important: parity violation in β-decay! H = G [ β px γ µ g ][ ] A γ 5 nx ex γ µ γ 5 νx + h.c. g V G β =.47 0 5 GeV g A g V =.55 Weak interaction distinguishes between left- and right-handed particles. Universality of weak interaction. γ 5 and handedness. Fermions revisited: compare to page. 59

Up = p 0 + m χ = 0 χ s σ p p 0+m χ s, χ = 0 and standard representation γ 0 0 =, γ i 0 σ i = 0 σ i 0 0, γ 5 = 0 For m = 0 we equate p = p 0 = p. With ˆp = p p we get for Up: Up = χ p s σ ˆp χ s Spin-orientation/Helicity: σ ˆp χ ± = ±χ ±. Dene: U ± p = p m = 0: γ 5 U ± p = ±U ± p. We dene left- and right-handed spinors: χ± ±χ ± ψ L/R = γ 5 ψ = ψ ± for m = 0 γ 5 ψ L/R = ψ L/R chirality 5. Lagrange density of electroweak theory Fermi interaction via gauge theory: Consider µ e + ν e + ν µ Gauge principle for the time being we consider massless fermions: ψνe ψνµ ψντ Leptons: Ψ e =, Ψ ψ µ =, Ψ e ψ τ = µ ψ τ Quarks: Ψ q = ψu, Ψ ψ q = d Free Lagrangian for the electron: ψc, Ψ ψ q = s L 0 x = Ψ el i γ µ µ Ψ el + ψ er i γ µ µ ψ er x ψt ψ b 60

The second summand is necessary because of QED. L 0 is invariant under global SU rotations of Ψ L : Ψ L U Ψ L U = e iω SU ω = ω a σa Singlet ψ R : ψ R ψ R. σ a are the generators of SU Lie-algebra [ ] σ a, σb abc σc = i ε, ε3 = and Pauli-matrices σ i, i =,, 3 see 0. Local symmetry gauging via minimal coupling L 0 Lx = Ψ el i γ µ D µ Ψ el + ψ er i γ µ µ ψ er and gauge transformations D µ = µ + i g W µ W µ = Wµ a σ a W µ x Ux W µ x U x i g Ux µ U x = i g Ux D µ U x { Ψx exp i ωx γ } 5 Ψx Ux ΨL = ψ R ΨL Ψx =. ψ R Within this notation L reads L = Ψ i γ µ D µ Ψ D µ = µ + i g W µ W µ = Wµ a T a σ a T a = 0 0 0 6

Coupling via Wµ a σ a = Wµ 3 Wµ i Wµ Wµ + i Wµ Wµ 3 where Wµ 3 is neutral and Wµ, Wµ are charged. With W ± = W ± i W we have the interaction, e.g.: g Ψe,L γ µ W µ ψ νe,l = g ψe γ µ W µ γ 5 ψ νe } {{} ψ νe,l = g + γ 5 ψe γ µ Wµ ψ νe = g ψ γ 5 e γ 0 γ µ Wµ ψ ν 5. Diagrammatically neutral gauge boson W 3 µ: is not the photon: no L-R-symmetry additional U gauge boson Z 0 triumph of theory is not Z 0 : no coupling to right-handed fermions Existence of neutral currents, e.g. ν µ + e ν µ + e or ν µ + n ν µ + n. Consider W 3 µ = cos θ W Z 0 µ + sin θ W A µ where A µ represents the photon and θ W is the Weinberg angle weak mixing angle. sin θ W = 0.376, result at SLAC: sin θ W = 0.3098 ± 0.0008 Orthogonal combination: U gauge transformation B µ = sin θ W Z 0 µ + cos θ W A µ Ψ L e i Y L ω ψ R e i Y R ω } ΨL ψ R e iy ω ΨL ψ R 6

Hypercharge Y = Y L 0 0 0 Y L 0 0 0 Y R. 5.3 B µ commutes W µ! SU U The hypercharge Y equals the dierence between the electric charge Q in e and the third component of the isospin I 3 : For right-handed fermions: Y = Q. For left-handed fermions, e.g.: Particle Y Q I 3 ν L 0 e L - u L 6 3 d L 6 3 Interaction term: Ψ = ΨL ψ R Y = Q I 3. Ψ i γ µ D µ Ψ D µ = µ + i g W µ + i g B µ Y where W µ is explained on page 6 and for the Hypercharge Y see page 63. Full Lagrangian: W a µν = µ W a ν ν W a µ g ε abc W b µ W c ν B µν = µ B ν ν B µ L EW = 4 Wa µν W aµν 4 B µν B µν + ψ i γ µ D µ ψ 5.4 General gauge transformation: Neutral gauge bosons: Ψ e i g ωa I a +i g ω Y Ψ 5.5 Ψ γ µ [ g cos θ W Z 0 µ + sin θ W A µ T 3 + g sin θ W Z 0 µ + cos θ W A µ Y ] Ψ 63

Coupling to Photon: g sin θ W T 3 + g cos θ W Y = e Q = e T 3 + e Y g sin θ W = e g cos θ W = e Current: e A 0 µ j em. µ j em µ = Ψ γ µ Ψ = ψ R γ µ ψ R + Ψ L γ µ Ψ L Coupling to Z 0 : g cos θ W T 3 g sin θ W Y = = e cos θ W T 3 sin θ W Y sin θ W cos θ W e sin θ W Γ 3 sin θ W Q 5.6 Current: Z 0 µ j µ nc e sin θ W j µ nc = Ψ L γ µ T 3 sin θ W Q }{{} g L = g V g R = ψ γ µ Γ 3 L γ5 Q sin θ W ψ Ψ L + ψ R γ µ sin θ W Q ψ R }{{} g R = g V +g R Electron: = ψ νel γ µ ψ νel ψ el γ µ ψ el sin θ W j µ em Problems:. Masses for W ±, Z 0 : explicit mass-terms break gauge invariance!. Masses for matter elds: couple left- to right-handed elds break weak gauge invariance! 3. Neutrino mixing Resolution to. and.: Higgs-mechanism: masses via spontaneous symmetry breaking. 5. The Higgs sector. W ±, Z 0 are massive, e.g. m Z = 9.88GeV, m W = m Z cos θ W.. Matter-elds are massive: a m W Tr W b m ψ ψ ψ = mψ ψr ψ L + m ψ ψl ψ R 64

a and b are not gauge invariant: { ψ exp i γ 5 } ω ψ 5.7 m W Tr W i g m W Tr D W { m ψ ψ ψ mψ ψ exp i + γ } { 5 ω exp i γ } 5 ω ψ Explanation to the last transformation: { ψ ψ exp i γ } 5 γ 0 = ψ γ 0 exp { i + γ 5 = m ψ ψ e i γ 5 ω ψ = m ψ ψr e iω ψ L + ψ L e iω ψ R } { ψ exp i + γ } 5 5.8 Innitesimal: ψ ψ i ψ γ 5 ψ Assume we have scalar eld φ φ = v 0. "Massterms": doublet φ: φx = L Y x = h ψ ψ R φ Ψ L + Ψ L φ ψ R Yukawa term 5.9 φ x φ x φ x, φx Ux φ x, U SU φ x Ψ L x φ x U x Ux Ψ L x = φ x ψ L x 5.0 Yukawa term L Y x U L Y x gauge invariant under SU! Hypercharge: φx e i Y H ω φx Y H = Y L Y R =. L Y x ei Y ω L Y x Electric charge: ψ R φ Ψ L = ψ R e i Y R ω φ e i Y L Y R ω e i Y L ω Ψ L = ψ R φ Ψ L 5. 65

φ x : Y H + I 3 = φ x : Y H + I 3 = 0 φ = 0 v In summary Lx is gauge invariant, φ couples to W µ and to B µ! 0 Mass term for fermion: φ 0 = v h e ψer φ 0 Ψ L + h.c. = h e v ψer ψ L + h.c. 5. m e = h e v Kinetic term: µ φ µ φ D µ φ D µ φ Higgs Lagrangian: for electron D µ = µ + i g W µ + i g B µ Y H W µ = Wµ a σ a L H x = D µ φ D µ φ h e ψ er φ Ψ el + h.c. V φ φ 5.3 V is gauge invariant, as φ φ φ φ U φ U U φ = φ φ ei Y H ω φ e i Y H ω e i Y H ω φ 5.4 0 Mass of Z 0, W ± : Take φ 0 = v D µ φ 0 = iv g W + µ g B µ g W µ 3 5.5 g B µ g W µ g 3 = sin θ W B µ cos θ W W µ 3 cos θ W = g Z 0 µ cos θ W 66

D µ φ 0 Dµ φ 0 = v g This provides mass terms for Z 0, W ± : 8 Wµ W + µ + cos Z µ Z 0 µ 5.6 θ W m Z = v g cos θ W m W = v g and Full Lagrange density: sin θ W = m W m Z Lx = L EW x + L H x see page 63 and page 65 = 4 Wa µν W aµν 4 B µν B µν + Ψ i γ µ D µ Ψ h ψ ψ R φ Ψ L + Ψ L φ ψ R + D µ φ D µ φ V φ φ Z 0 µ = cos θ W W 3 µ sin θ W B µ A µ = sin θ W W 3 µ + cos θ W B µ and currents e sin θ W Z 0 µ j µ nc, e A µ j µ em j em µ = ψ R γ µ ψ R + Ψ L γ µ Ψ L j nc µ = ψ γ µ Γ 3 L γ5 Q sin θ W ψ Parameters: Measurements:. Fine structure constant g, sin θ W, ν, h ψ α = e 4π = g sin θ W 4π. Fermi coupling constant G F = g 8m W = 37.0359... see QED section = v =.6639 0 5 GeV 67

3. The Z 0 -boson mass m Z = g v cos θ W = 9.88 GeV 4. Fermion masses m f = h f v Mass hierarchy is not understood. 5.3 Spontaneous Symmetry Breaking. Simple example: O-model, φ complex eld invariance global Hamiltonian density Minimum: φ 0 = 0, Minimum: φ 0 = L = µ φ µ φ µ φ φ λφ φ 5.7 φ e iω φ, µ ω = 0 φ φ e iω 5.8 H = µ φ µ φ + V φ φ 5.9 µ φ µ φ = t φ t φ + φ φ V φ φ = µ φ φ + λ φ φ µ λ eiθ Mass: m = V φ φ φ0 = µ Masses: θ = 0, φ = φ + iφ, V φ φ0 = µ, V φ φ0 = 0. massive boson radial mode massless boson Goldstone boson Spontaneous Symmetry breaking: theory rests in given minimum. Remark: In QM the ground state is symmetric! In QFT for d > spontaneous symmetry breaking SSB is possible, for d <= no SSB for a cont. symmetry can occur Mermin-Wagner-Coleman, but discrete SSB. For d < no SSB can occur: QM: d = 0. Lagrangian: φx = v + σx }{{} radialmode + i πx }{{} Goldstone 5.0 68

v = µ λ. L = [ ] µ σ µ σ + µ π µ π [ µ v + σ + π ] [v 4 λ + σ + π ] = [ µ σ µ σ m σ σ ] + µπ µ π + interaction-terms + const. 5. m σ = µ. U gauge theory: Abelian Higgs model Lx = 4 B µν B µν + D µ φ D µ φ µ φ φ λφ φ, µ < 0 B µν = µ B ν ν B µ D µ = µ + i g Y B µ Again we have φ 0 = v Mass-term for B µ : R as minimum: D µ φ 0 D µ φ 0 = g Y v B µ B µ 5. mass m B = g Y v. dofs: φ, φ, B ± µ σ, π, B ± µ, B L µ + + [+ω] + + + ± in B µ ± in order to distinguish between the two helicity states and L in Bµ L stands for longitudinal. Perform gauge trafo on φ e iω, ω = arctan π v+σ. φ e iω φ = cos ω + i sin ω v + σ + iπ = v + σ + π = v + σ 5.3 σ = v + σ + π v = σ + π v +... Unitary gauge. The gauge eld has 'eaten up' the Goldstone Boson Higgs-Kibble dinner. 69

3. Electroweak theory: SU U SSB U em 4 gen. gen. 3 Goldstone bosons are eaten up. φ 0 = 0 v where φ is neutral. U em is unbroken Symmetry. For the mass terms of W ± and Z 0 see page 67. We have used that φ = e i ωa T a +Y H ω φ 0 to gauge away the Goldstones. We chose SU gauge transformation such that 0 U φx = ρ + v e iωt 3 +Y H φ 0 = φ 0 5.4 This is left invariant under the U-transformations It follows that e i ω T 3 +Y H 5.5 a b D µ φ D µ φ = µρ µ ρ + m W + ρ W + v µ W µ + + m Z + ρ Z 0 v µ Z 0µ 5.6 For m W and m Z see page 67. V φ φ = V v + ρ 4 v + ρ = µ v + ρ + λ 4 = m ρ ρ + ρ v + ρ 4 v 5.7 the Higgs mass: m ρ = λ v 5.8 where λ is a parameter. 70

c Leptons: electron In general Ψ e i γ µ D µ Ψ e = Ψ e i γ µ µ Ψ e e A 0 µ j em µ e Zµ 0 j nc µ + see page 64 sin θ W + j µ cc = Ψ e γ µ T + i T Ψ e e W + sin θ µ j cc µ + Wµ j cc µ+ W = Ψ e γ µ T + i T γ 5 h ψ ψ R φ Ψ L Ψ L φ ψ R = m e ψe ψ e + ρ v 0 ψ = Ψ e Ψ µ Ψ τ Ψ q allows for mass matrix. Mixing! Kobayashi-Maskawa matrix v, v, v 3, δ. Feynman rules: Propagators: i g : µν q +i ε Feynman gauge : : qµ qν i g µν m W q m W +i ε qµ qν i g µν m Z q m Z i +i ε : q m ρ +i ε 5.9 Ψ e 5.30 Selected vertices: : ie { } k k µ3 g µµ + k k 3 µ g µµ 3 + k 3 k µ g µ3µ : ie cos θ { } W sin θ k W k µ3 g µµ + k k 3 µ g µµ 3 + k 3 k µ g µ3µ : ie { } g µµ 3 g µµ 4 + g µµ 4 g µµ 3 g µµ g µ3µ 4 : ie cos θ { } W sin θ gµµ W 3 g µµ 4 + g µµ 4 g µµ 3 g µµ g µ3µ 4 : ie cos θ W { } sin gµµ θ W { 3 g µµ 4 + g µµ 4 g µµ 3 g µµ g µ3µ 4 } : ie sin gµµ θ W 3 g µµ 4 + g µµ 4 g µµ 3 g µµ g µ3µ 4 5.3 7