Introduction to the SM (5)

Transcription

1 Y. Grossman The SM (5) TES-HEP, July 12, 2015 p. 1 Introduction to the SM (5) Yuval Grossman Cornell

2 Y. Grossman The SM (5) TES-HEP, July 12, 2015 p. 2 Yesterday... Yesterday: Symmetries Today SSB the SM

3 SSB Y. Grossman The SM (5) TES-HEP, July 12, 2015 p. 3

4 Breaking a symmetry Y. Grossman The SM (5) TES-HEP, July 12, 2015 p. 4

5 Y. Grossman The SM (5) TES-HEP, July 12, 2015 p. 5 SSB By choosing a ground state we break the symmetry We choose to expend around a point that does not respect the symmetry PT only works when we expand around a minimum What is the different between a broken symmetry and no symmetry? SSB implies relations between parameters

6 Y. Grossman The SM (5) TES-HEP, July 12, 2015 p. 6 SSB Symmetry is x x and we keep up to x 4 f(x) = a 2 x 4 2b 2 x 2 x min = ±b/a We choose to expand around +b/a and use u x b/a f(x) = 4b 2 u 2 + 4bau 3 + a 2 u 4 No u u symmetry The x x symmetry is hidden A general function has 3 parameters c 2 u 2 + c 3 u 3 + c 4 u 4 SSB implies a relation between them c 2 3 = 4c 2 c 4

7 Y. Grossman The SM (5) TES-HEP, July 12, 2015 p. 7 SSB in QFT When we expand the field around a minimum that is not invariant under a symmetry φ v + h It breaks the symmetries that φ is not a singlet under Masses to other fields via Yukawa interactions φx 2 (v + h)x 2 = vx Gauge fields of the broken symmetries also get mass D µ φ 2 = µ φ + iqa µ φ 2 A 2 φ 2 v 2 A 2

8 The SM Y. Grossman The SM (5) TES-HEP, July 12, 2015 p. 8

9 The SM Input: Symmetries and fields Symmetry: 4d Poincare and SU(3) C SU(2) L U(1) Y Fields: 3 copies of QUDLE fermions Q L (3, 2) 1/6 U R (3, 1) 2/3 D R (3, 1) 1/3 L L (1, 2) 1/2 E R (1, 1) 1 One scalar φ(1, 2) +1/2 Y. Grossman The SM (5) TES-HEP, July 12, 2015 p. 9

10 Y. Grossman The SM (5) TES-HEP, July 12, 2015 p. 10 Then Nature is described by Output: the most general L up to dim 4 L = L kin + L Higgs + L Y ukawa This model has a U(1) B U(1) e U(1) µ U(1) τ accidental symmetry Initial set of measuremnts to find the parameters SSB: SU(2) L U(1) Y U(1) EM Fermion masses, gauge couplings and mixing angles The SM pass (almost) all of it tests

11 The gauge interactions Y. Grossman The SM (5) TES-HEP, July 12, 2015 p. 11

12 Y. Grossman The SM (5) TES-HEP, July 12, 2015 p. 12 The gauge part SU(3) C SU(2) L U(1) Y Three parts, each look so different... SU(3) C U(1) EM QED - photon interaction: Perturbation theory QCD - gluon interaction: Confinement and asymptotic freedom Electro-weak: SSB and massive gauge bosons

13 L kin and SU(2) U(1) Four gauge bosons DOFs The covariant derivative is W µ a (1, 3) 0 B µ (1, 1) 0 D µ = µ + igw µ a T a + ig Y B µ Two parameters g and g Y is the U(1) charge of the field D µ work on T a is the SU(2) representation T a = 0 for singlets. T a = σ a /2 for doublets Write D µ for L(1, 2) 1/2 and E(1, 1) 1 Y. Grossman The SM (5) TES-HEP, July 12, 2015 p. 13

14 Y. Grossman The SM (5) TES-HEP, July 12, 2015 p. 14 Explicit examples D µ = µ + igw µ a T a + ig Y B µ Write D µ for L(1, 2) 1/2 and E(1, 1) 1 ( µ + i2 gw µa σ a i2 ) g B µ D µ L = L D µ E = ( µ ig B µ ) E HW: Using φ(1, 2) 1/2 write D µ φ

15 Y. Grossman The SM (5) TES-HEP, July 12, 2015 p. 15 QED Where is QED in all of this? Q = T 3 + Y We can write explicitly for L(1, 2) 1/2 and φ(1, 2) 1/2 L L = ν L e L φ = φ+ φ 0 This is arbitrary. It becomes usefull once we have SSB

16 SSB in the SM L Higgs = λφ 4 µ 2 φ 2 = λ(φ 2 v 2 ) 2 We measure the fact that µ 2 > 0 by having SSB The minimum is at φ = v φ has 4 DOFs. We can choose φ 1 = φ 2 = φ 4 = 0 φ 3 = v It leads to: SU(2) L U(1) Y U(1) EM We call the remaining symmetry EM Could we choose the vev in the neutral direction? We left with one real scalal field: the Higgs boson Y. Grossman The SM (5) TES-HEP, July 12, 2015 p. 16

17 Spectrum Y. Grossman The SM (5) TES-HEP, July 12, 2015 p. 17

18 Y. Grossman The SM (5) TES-HEP, July 12, 2015 p. 18 Gauge boson masses W 1, W 2, W 3, B Gauge bosons masses from D µ φ 2 (HW: do it) Diagonalzing the mass matrix the masses are M 2 W + = M 2 W = 1 4 g2 v 2 M 2 Z = 1 4 (g2 +g 2 )v 2 M 2 A = 0 The mass eigenstates W ± = 1 2 (W 1 iw 2 ) tan θ W g g Z = cos θ W W 3 sin θ W B A = sin θ W W 3 + cos θ W B We have a θ W rotation from (W 3, B) to (Z, A)

19 Y. Grossman The SM (5) TES-HEP, July 12, 2015 p. 19 L Yuk and fermion masses There is no way to write a mass term (why?) The Yukawa part of the leptons L Y uk = y ij L Li E Rj φ m ij L Li E Rj m ij = vy ij i, j = 1, 2, 3 are flavor indices y is a general complex 3 3 matrix and we can choose a basis where m is diagonal and real Neurinos are massless m ij = y v = diag(m e, m µ, m τ )

20 Interactions Y. Grossman The SM (5) TES-HEP, July 12, 2015 p. 20

21 Y. Grossman The SM (5) TES-HEP, July 12, 2015 p. 21 Charged current interactions g 2 ν el W µ γ µ e L + h.c. Only left-handed fields take part in charged-current interactions. Therefore the W interaction violate parity The Wlν interaction is universal Can be used to measure g µ ν µ A g 2 /m 2 W G F W ν e e

22 Y. Grossman The SM (5) TES-HEP, July 12, 2015 p. 22 Neutral currents L int = e sin θ cos θ (T 3 sin 2 θ W Q) ψγ µ ψ Z µ + e Q ψγ µ ψ A µ, We define Q = T 3 + Y Photon coupling is parity invariant e = g sin θ Z couples to both LH and RH fermions but in a parity violating way The coupling to the Z is larger. So why we call it weak interaction? Once we know e and g we know θ

23 Y. Grossman The SM (5) TES-HEP, July 12, 2015 p. 23 The ρ = 1 relation We get the following testable relation ρ MW 2 MZ 2 = 1 tan θ W g cos2 θ W g The above is a signal of SSB

24 Y. Grossman The SM (5) TES-HEP, July 12, 2015 p. 24 Experimental tests High energy: Open your pdg and check W and Z decays to leptons. What do you expect to see? Z decays to lepton actually measures sin 2 θ W 0.23 HW: Calculate Γ(Z ν ν)/γ(z e + e ), get sin 2 θ W from the data and check the ρ = 1 prediction Also low energy data tests

25 Y. Grossman The SM (5) TES-HEP, July 12, 2015 p. 25 Experimental tests of ρ = 1 From the ρ = 1 relation m 2 W m 2 Z = cos 2 θ W ( ) sin 2 θ W Z decays to leptons Γ(Z ll) (T 3 Q sin 2 θ W ) 2 L,R Γ(Z l + l ) (1/2 sin 2 θ W ) 2 + (sin 2 θ W ) 2 1/8 Γ(Z ν ν) (1/2) 2 1/4 r Γ l /Γ inv 1/6 PDG: Γ l = 3.37% and Γ inv = 20.00% r 1/6