Paul Langacker. The Standard Model and Beyond, Second Edition Answers to Selected Problems

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1 Paul Langacker The Standard Model and Beyond, Second Edition Answers to Selected Problems

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3 Contents Chapter Review of Perturbative Field Theory 1.1 PROBLEMS 1 Chapter 3 Lie Groups, Lie Algebras, and Symmetries PROBLEMS 5 Chapter 4 Gauge Theories PROBLEMS 9 Chapter 5 The Strong Interactions and QCD PROBLEMS 11 Chapter 6 Collider Physics PROBLEMS 15 Chapter 7 The Weak Interactions PROBLEMS 17 Chapter 8 The Standard Electroweak Theory PROBLEMS 19 Chapter 9 Neutrino Mass and Mixing PROBLEMS 5 Chapter 10 Beyond the Standard Model PROBLEMS 7 i

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5 C H A P T E R Review of Perturbative Field Theory.1 PROBLEMS Problem.: 0.04 dσ d cos θ (fm ) cos θ σ (fm ) s (GeV) Left: dσ/d cos θ in units of fm as a function of cos θ for s = 5m, 6m, and 7m (from upper to lower). Right: σ(s) in units of fm as a function of s Problem.5: [ ] M fi = ( iκ) i i s m + 3 u m 3 dσ d cos θ = M fi 3πs, u = p (1 + cos θ), p = s 4m 1 / Problem.6: M fi = 4iσ 4 Problem.7: dγ d cos θ = αm A 8 (1 4m M A ) 3/ cos θ, Γ = αm A 1 (1 4m M A ) 3/ 1

6 The Standard Model and Beyond Problem.8: (a) ( i M fi = ( ig) s µ + i ) t µ, s 4m t = p (1 cos θ), p =, E = (b,c) dσ d cos θ = M fi +1 3πs, σ = 1 dσ d cos θ. d cos θ s, dσ d cos θ (fm ) π + π - π + π - x=4.0 x=4. x=4.4 σ (fm ) π + π - π + π cos θ x Left: dσ/d cos θ in units of fm for x = 4, 4., and 4.4. Right: σ in units of fm vs x. ( ) Problem.15: For example, P L u(+) m φ+ E E, so the rates for the wrong he- 0 licity are suppressed by m /4E. Problem.18: (a) with d σ d cos θ = 1 3πs M [ ( M = h 4 1 t µ ) ( 4m t ) ( + 1 u µ ) ( 4m u ) 1 ( ) ( ) 1 1 [(4m t µ u µ u ) ( + 4m t ) ( s 4m ) ]], where E = s s 4m, k =, t = k (1 cos θ), u = k (1 + cos θ). (b) d σ d cos θ = 3h4 3πs, σ = d cos θ d σ d cos θ = 3h4 3πs.

7 Review of Perturbative Field Theory 3 Problem.0: d σ p [ = 4πα s + (E + p cos θ) d cos θ st m µ(1 cos θ) ] Problem.1: d σ d cos θ = πα s ( 3 + cos ) θ, 1 cos θ which displays the singularity from the t-channel pole in the forward direction. Problem.: d σ d cos θ = πα s = πα s [ (1 + cos 4 θ ) sin 4 θ [ 16 sin 4 θ 8 sin θ (1 + sin4 θ ) + cos 4 θ ]. ] sin θ cos θ Problem.3: d σ d cos θ = πα βf 3 sin θ, 4s σ = πα β 3 f 3s Problem.4: d σ d cos θ = (Zα) π(1 β cos θ β p sin 4 θ ) β 1 Z α π β p sin 4 θ Problem.5: p f Γ = 8πMZ M m 0 M Z ( g 1π V + ga ). Problem.6: dγ d cos θ = p f [( gs + g P ) E p + ( g S g P ) m p 8πm Λ + Re (g P gs) p f cos θ]. Problem.7: σ(s) = 1π(s/M V ) Γ aā Γb b (s MV ) + MV, Γb b = g b M V Γ V 1π. Problem.9: M(, ) = M(+, +) = 8πα i ( 1 β π + 1 ) cos θ 1 cos θ.

8 4 The Standard Model and Beyond Problem.30: (a) eφ s (mã0( q ) ( p 1 + p ) ) A( q ) σ B( q ) [1 + F (0)] φ s1. (b) d e = ieg (0) m σ. Problem.31: τ 1 = 1 4 δ α m e, δ < Problem.3: p() H p(1) = p() d 3 x ej µ Q A µ p(1) [ = m p e φ G E (Q )Ã0( q ) G M (Q ) σ ] B( q ) m p φ 1 Problem.33: (a) (b) (c) Ṽ ( q ) = λ, V ( r ) = λδ 3 ( r ). Ṽ ( q ) = g 1g q + m, V (r) = g 1g φ 4π V (r) = + g 1g 4π (d) Sign change for vector, not for scalar. (e) V ( r ) = g πm 3 π 16πm p e M V r r. e m φr ( σ p x ) ( σ n x ) ( e x x r ).. Problem.34: Γ = e m 3 µ 8π ( A + B ).

9 C H A P T E R 3 Lie Groups, Lie Algebras, and Symmetries 3.1 PROBLEMS Problem 3.4: cos γ = cos α cos β sin α sin β ˆα ˆβ sin γ ˆγ = sin α cos β ˆα + cos α sin β ˆβ + sin α sin β ˆα ˆβ. Problem 3.13: π + p M π + p = M 3/, π p M π p = 1 3 M 3/ + 3 M 1/ π 0 n M π p = 3 M 3/ 3 M 1/, π 0 n M π 0 n = 3 M 3/ M 1/ Problem 3.14: M(Σ + π 0 ( ) p) = M3/ M 1/ 3 M(Σ + π + n) = 1 3 M 3/ + 3 M 1/ M(Σ π n) = M 3/. Problem 3.16: σ A+ p A + p = σ A 0 p A 0 p = 1 4 σ A + n A 0 p = λ 8πs (E 1 + m p), E 1 = s + m p m A s 5

10 6 The Standard Model and Beyond Problem 3.19: g K+ nσ = g π + pn + 6g K+ pλ Problem 3.0: M p = M n M N = m m β + M Ξ 0 = M Ξ M Ξ = m m β M Σ ± = M Σ 0 M Σ = m 0 + m β 3 m α 3 m α M Λ = m 0 m β 3 Problem 3.: (b) g p /g n = 3/ (exp: 1.46) (c) g p / = 3 (exp:.79); g n / = (exp: 1.91) Problem 3.3: ( ) [ gf g d M π0 p = ( ie)ū γ 5 i( k 1+ p 1 + m) (k 1 + p 1 ) m ɛ+ ɛ i( p ] 1 k + m) (p 1 k ) m γ5 u 1 ( ) [ M K + Σ 0 = gf + g d ( ie)ū γ 5 i( k ] 1+ p 1 + m) (k 1 + p 1 ) m ɛ + k i ɛ (k k 1 ) µ γ5 u 1 Problem 3.5: (a) (b) ( L i A )ab;cd = Li acδ bd + L i bdδ ac L i adδ bc L i bcδ ad, T (L A ) = m. ( L i S )ab;cd = c ab c cb [L i acδ bd + L i bdδ ac + L i adδ bc + L i bcδ ad ], T (L S ) = m +. For m =, ψ S 11, ψ S 1 and ψ S correspond to the J = 1 angular momentum representation with J 3 = +1, 0, and 1, respectively. For example, ( L 3 S )11;11 = 1 4 [4] = 1, ( L 3 S )1;1 = 1 4 [1 1] = 0, ( ) L 1 S 11;1 = 1 1 [ ] = 1. Problem 3.6: JLµ i = ψ ( ) L γ µ L i nψ L + itr φ L i n µ φ JRµ i = ψ ( R γ µ L i nψ R + itr φl i n µ φ ), and L i n I for the U(1) currents.

11 Lie Groups, Lie Algebras, and Symmetries 7 Problem 3.8: a = ν = µ /λ, b = µ / ( ) H(x) = µ4 µ x λ sech4, + H(x)dx = µ 3 3 λ Problem 3.30: (a) (b) M I = i ( mψ ν ) ū4 [ [ ( M = iλ 1 + 3m 1 1 t m + m 1 1 s + 1 )] u ] m η 1 m ψ t m + γ 5 1 γ 5 + γ 5 1 γ 5 η p 1 + p m ψ p p 3 m ψ u Problem 3.3: ψ = 1 {[ a b + a + b e iφ m ] + [ a b + a + b e iφm ] γ 5} ψ γ 5 ψ m = c + d a b, tan φ m = d c Problem 3.33: λ > 0 : φ 1 = ν b µ /λ, φ = 0, m 1 = λν b, m = λ ν b. φ φ survives. λ = 0 is similar except there is a circle of degenerate minima. For λ < 0, require λ + λ > 0. φ 1 = φ = ν c µ /(λ + λ ), m 1 = λ ν c, m = (λ + λ )ν c. φ 1 φ survives. Problem 3.34: λ 1 > 0, λ 3 > 0, µ µ 1 0, (flat for µ = µ 1 ). Problem 3.35: (a) µ I + µ II A > 0. (c) Minimum for µ I µ II A > 0; saddle point for µ I µ II A < 0. (d,e) µ A = µ I + µ II = A sin γ, 1 M Z = λν = µ I µ II tan γ tan. γ 1 (f) ( µ M Re = Z cos γ + µ A sin γ (MZ + ) µ A ) sin γ cos γ (MZ + µ A ) sin γ cos γ M Z sin γ + µ. A cos γ

12 8 The Standard Model and Beyond Problem 3.36: (b) (c) µ > 0 : m 1 = m = µ, m ψ = 0 µ < 0 : m 1 = µ, m = 0, m ψ = hν/, with ν = µ /λ µ > 0 : m 1 = µ + 3λν, m = µ + λν, m ψ = hν, with ν a/µ µ < 0 : m 1 = µ + 3 a a, m =, m ψ = h ν 0 + ɛ, ν 0 ν 0 with ν = ν 0 + ɛ, ν 0 = µ /λ, ɛ = a µ (e) µ J µ = i m L ( ψ L C ψ T L ψ T LCψ L )

13 C H A P T E R 4 Gauge Theories 4.1 PROBLEMS Problem 4.1: A µ = ( 0, nˆθ ) ( ), A ds rg = πn g. Problem 4.11: (a) gγ µ /, gγ µ /, igc µνσ ( p 3, p 3 + p 4, p 4 ) (b) M = ig v ɛ 1 ɛ+ u 1 p 1 p 3 m ψ + ig 1 s M A v [ ɛ + p 3 ɛ + ɛ + ɛ ( p 4 p 3 ) ɛ p 4 ɛ +] u 1. (c) M ig M A v p 4 u 1 ig MA v p 4 u 1 = 0 Problem 4.1: (a) A ijk = d ijk. (b) A D ijk = md ijk. One can also show that A A ijk = (m 4)d ijk and A S ijk = (m + 4)d ijk for the antisymmetric (A) and symmetric (D) products of two fundamentals. 9

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15 C H A P T E R 5 The Strong Interactions and QCD 5.1 PROBLEMS Problem 5.1: (b) b = 1, c = 1/ 3. (c) R = 4() for HN (QCD); Drell Yan 5/9 (1/3); anomaly coefficient = 4 for both. Problem 5.: M = 4 9 gs 4 [ (s m t c ) + (u m c) + m ct ] u m c = p(e + p cos θ), t = p (1 cos θ), E = s + m c s, p = s m c s Problem 5.3: (a) where (b,c) L φ =Tr [(D µ φ) D µ φ] µ Tr (φ φ) λ 1 ( Tr (φ φ) ) λ Tr (φ φφ φ) + q L hφq R + q R h φ q L, D µ φ = µ φ + ig L G µ L L φ ig R φ G µ R L. G i = sin δ G i L + cos δ G i R, G i A = cos δ G i L sin δ G i R, where tan δ g R /g L and M GA = g L + g R v φ. L q = g s q G L q gs cot δ q L GA L q L + g s tan δ q R GA L q R, where g s g L sin δ = g R cos δ; T i = T i L + T i R. Problem 5.4: σ = 8πα s 7s β Q ( 3 β Q σ = πα sβ 3 0 7s ) with β 0 = with β Q = 1 4m Q s 1 4m 0 s. 11

16 1 The Standard Model and Beyond Problem 5.5: F r=1 σ q0 q 0 q r q r β rel = n f 4πα s 7m 0 β i, σ q0 q 0 GG β rel = 8 7 παs m. 0 Problem 5.6: (a) M /gs 4 = 4 ( ( s + u m r m s + s t + u ) + m 4 r m s(s t + u) + ms) 4 9t (b) ( M 7m /gs 4 4 7m t 7m u + 4t tu + 4u ) = 4s (m t) (m u) ( 6m 8 3m 4 t 14m 4 tu 3m 4 u + m t 3 + 7m t u + 7m tu + m u 3 t 3 u tu 3) (d) Use the expressions for t and u in (.40) for m 1, = 0, m 3,4 = m t, and the cross section formula in (.57), integrating cos θ from 1 to σ t t _ (nb) qq tt GG t t s (TeV) Problem 5.8: (a) B = 7, α(m P ) 0.019; (b) B = 3, α(m P ) Problem 5.10: charmonium bottomonium exp model difference exp model difference 1S S % % 3S % 4S %

17 The Strong Interactions and QCD 13 for m c 1.7 GeV and m b 4.61 GeV. Problem 5.17: (b)tr ( M 1 M ) [ (, Tr M 1 M )][ ( )] ( ) Tr M a M b, Tr M 1 M M 1 M (c) Tr (M[ a M b ), ψ L ψ] R + ψ R ψ L (d) itr ˆn τm 1 M.

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19 C H A P T E R 6 Collider Physics 6.1 PROBLEMS Problem 6.3: dlij/ds dy (pb) LHC (14 TeV), s = 1 TeV uu uu dlij/ds dy (pb) LHC (14 TeV), s = 1 TeV uu Gu GG y y Problem 6.4: Cross sections in pb: p p ( TeV) pp (8 TeV) pp (14 TeV) σ q q (pb) σ GG (pb) σ tot (pb) pp tt ( TeV) pp tt ( TeV) dσ /dy (pb) total qq GG dσ/dp T (pb/gev) total qq GG y p T (GeV) 15

20 16 The Standard Model and Beyond Problem 6.5: λ GG = π 8 Γ R dl GG 130 fb, M R dŝ λ uū = 4π 9 Γ R dl uū+ūu 79 fb, M R dŝ λ d d 51 fb. Problem 6.6: 1, 1, 3

21 C H A P T E R 7 The Weak Interactions 7.1 PROBLEMS Problem 7.: (a) 1ɛ (1 ɛ) (b) ɛ [ (3 ɛ) + 6(1 ɛ) ], ρ = 3 8 (c): ν µ : ɛ (3 ɛ), ν e : 1ɛ (1 ɛ). Problem 7.10: (a), (b), (c) in text. (d): α eν = g V g A. gv +3g A Problem 7.11: dγ(πν τ ) d cos θ = M 3πm τ Γ(πν τ ) = G F cos θ c fπm 3 τ. 16π = G F cos θ c fπm 3 τ (1 + cos θ) 3π 17

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23 C H A P T E R 8 The Standard Electroweak Theory 8.1 PROBLEMS Problem 8.4: (a) V (φ, σ) = µ σσ σ + µ φφ φ + λ σ ( σ σ ) + λφ ( φ φ ) + λφσ σ σφ φ + κ φσ σφ φ + κ φσ σ φ φ. (c) M A cos θ W g ν σ (d) Γ = p ( ) f 6πMZ g sin θ W MA + E A MA, which survives for M A 0 because of the longitudinal mode. Problem 8.5: (b) L lh = 0. L l(w,a,z) = g ( Nγ µ E W µ + + Ēγµ N Wµ ) + e Ēγ µ EA µ g [ 1 cos θ W Nγ µ N + ( 1 ) ] Ēγµ + sin θ W E Z µ. Problem 8.6: A L MA R = diag( ) for i i i A L = i i i i i A R = i i i diag(e +3.14i e +0.14i e 0.67i ) Problem 8.7: m 1 = m = 3c, ψ L = ψ 0 L = ( ψ 0 1L ψ 0 L ), ψ R = i ( ψ 0 R ψ 0 1R ) 19

24 0 The Standard Model and Beyond Problem 8.8: (a) m 1 x, m B, A R I, ( ) A 1 y L = B y. B 1 (b) J µ Z = ē Lγ µ e L y B (ēl γ µ E L + ĒLγ µ e L ) ( y b ) ĒL γ µ E L + sin θ W (ēγ µ e + Ēγµ E ). Problem 8.9: M = g4 M Z (g V + g A ) cos 4 θ W 1 (q M Z ) d σ d cos θ = 1 3πs [ p 1 p 3 + p ] 1 p 4 p 3 p 4 MZ k 3 k 1 M. k 1 = s M H s, E = s + M H s, k 3 = s M Z s, E 4 = s + M Z s q = (p 3 p 1 ) = p 1 p 3 = k 1 k 3 (1 cos θ) p 1 p 4 = k 1 (E 4 + k 3 cos θ), p 3 p 4 = s M Z. Problem 8.10: L = Γ D κσ q L φd R /M D + h.c.. Problem 8.13: (a) d σ dz = s [( ɛll + ɛ RR ) (1 + z) + ( ɛ LR + ɛ RL ) (1 z) ] 384π z z for q r q r µ µ +. (b) For y > 0 d σ dydẑ =λ Z ɛ AB = Q re g s ZD(s)ɛ A (µ)ɛ B (r). [ ( Lr r G r N (1 + ẑ) + G r F (1 ẑ) ) r +L rr ( G r N (1 ẑ) + G r F (1 + ẑ) )], (ẑ ẑ for y < 0), where gz 4 λ Z, L r r dl q r q r, 384M Z Γ Z dŝdy ŝ=m L rr dl q rq r dŝdy ŝ=m Z Z G r N = ɛ L (µ) ɛ L (r) + ɛ R (µ) ɛ R (r), G r F = ɛ L (µ) ɛ R (r) + ɛ R (µ) ɛ L (r).

25 The Standard Electroweak Theory 1 (c) A F B (y) = 3 r L r r L rr (G r N Gr F ) 4 r (L r r + L rr ) (G r N + Gr F ) A similar expression holds for A F B (y 1, y ) except the numerator and denorminator are integrated ( y y y 1 dy, or y 1 + ) y 1 y dy. For more detail, see (Han et al., 013). (d) A F B (0, ln( s/m Z )) A FB (pp μ - μ + ), s = 14 TeV, s M Z 0.08 AFB (y) Problem 8.14: M νe [ ig = ɛ 3µɛ 4ν v 3 γ ν (1 γ 5 ) ) ( i M γ = ɛ 3µɛ 4ν v 3 [ieγ α ]u 1 s [ M Z = ɛ 3µɛ ig 4ν v 3 y ] ( i( p1 p 3 ) t ) [ ] ig γ µ (1 γ 5 ) u 1 ie C µαν ( p 3, p 1 + p, p 4 ) γ ( α 1 cos θ + sin θ W 1 )] γ5 u 1 W ( ) i e s MZ i C µαν ( p 3, p 1 + p, p 4 ) tan θ W

26 The Standard Model and Beyond 40 e + e - W + W - 30 σww (pb) only ν e ν e, γ ν e, γ, Z s (GeV) Problem 8.15: (a) dγ d cos θ = ĝ Eb [ ] (1 + cos θ) + m t 4πm t MW (1 cos θ) (c) m t F 0 = MW + m t 0.70, F = 1 F 0, F + = 0 Problem 8.0: (a) σ GGF pp H = π 8 σ W W F q rq s q rq sh(ŝ) = Γ H B(H GG) dl GG M H dŝ [( α3 g 16MW 1 + M H ) ln ŝ ŝ M H M H 11 pb + M H ] ŝ σ W W F pp H = q rq s dŝ ŝ dl qrq s dŝ ŝ σ W W F q rq s q rq sh(ŝ) 1.5 pb, σ ZZF pp H 0.5 pb σ qr q s W ± H(ŝ) = V rs πα g 36 p f p i 3M W + p f (ŝ M W ) (b) σ pp W ± H = q r q s dŝ ŝ dl qr q s dŝ ŝ σ qr q s W ± H(ŝ) 0.5 pb, σ pp ZH 0.3 pb. σ NNLO (pb) σ LO (pb) N γγ N 4l GGF VBF W H ZH total

27 The Standard Electroweak Theory 3 Problem 8.1: Γ(H GG) increases by a factor 9, while Γ increases by 1.8. B(H ZZ ) B(H ZZ 1 ) SM , B(H γγ).8/ , B(H γγ) SM 1.7 The rates σ GG B change by factors of and for ZZ and γγ, respectively. Problem 8.: (b) Define V (H + ν) c n= c nh n. Then MH = c = λν + ρν 4, c 3 = M ) H (1 + 8ρν4, c 4 = M ) H (1 ν 8ν + 16ρν4 c 5 = ρν, c 6 = ρ 6. 3M H (c) κ 1 4σν and M H / λν (1 4σν ). Interactions terms: M H σ ( νh + H ) ( µ H) M H H M H ( 1 σν ) H 3 M ( H 1 4σν ) ν 8ν H 4. Problem 8.3: M = i G F MH s s MH, σ = M 16πs Problem 8.4: M = 4ĝ ( m t M W ) mt E b (1 cos θ)

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29 C H A P T E R 9 Neutrino Mass and Mixing 9.1 PROBLEMS Problem 9.: m 1 = ɛ, ν 1L = ν µl ν τl m,3 = + ɛ (, ν,3l = ɛ ) ν el + ν µl + ν τl, Problem 9.3: m T γ µ (1 γ 5 )γ ν γ µ[ (1 γ 5 )m T + ɛ k ] γ ν + O(ɛ ) Problem 9.5: Dirac: σβ rel = G [ F π m ν ɛ L + ɛ ] R. Majorana: σβ rel = 8β 3 G [ F π m ν ɛ L + ɛ ] R. 5

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31 C H A P T E R 10 Beyond the Standard Model 10.1 PROBLEMS Problem 10.1: (a) (b) n AF = 11C (G) = 11m n fp = 17C (G) 5C (G) + 3C (L m ) = 34m3 13m 3 < n AF α = 4πˆb 1 (m, n f ) ˆb (m, n f ) = 4π m(11m n f ) 34m 3 13m n f + 3n f 3.5 Running Coupling (m=3) g(t) n f =1, α g (M P )=0.6 n f =13, α g (M P )=0.4 n f =14, α g (M P )= t=ln Q/M Z (c) α < α c for n f > n c. α = α c for n c = 5 50m 3 33m 5m. 3 7

32 8 The Standard Model and Beyond Problem 10.3: For infinitesimal transformation M σ 0 M σ 0 β i σ i, M σ i M σ i ɛ ijk ω j σ k β i σ 0 Problem 10.4: ψ 1M ψ M = ξ 1 ξ + ξ 1 ξ, ψ1m γ 5 ψ M = ξ 1 ξ ξ 1 ξ ψ 1M γ µ ψ M = ξ 1 σ µ ξ + ξ 1 σ µ ξ, ψ1m γ µ γ 5 ψ M = ξ 1 σ µ ξ ξ 1 σ µ ξ ψ 1M σ µν ψ M = ξ 1 s µν ξ + ξ 1 s µν ξ Problem 10.10: σ(ah) = cos (β α) g4 Z kah 3 96π k i σ(zh) = sin (β α) g4 Z kzh 3 96π k i ɛ L + ɛ R (s M Z ) ɛ L + ɛ R (s MZ ) [ 1 + 3M Z k Zh ] Problem 10.13:

33 Beyond the Standard Model 9 σ(pp q G ), s = 14 TeV 1 σ(pb) m (GeV) m. TeV M Z g Problem 10.14: Typeset by (a) FoilTEX θ = Q φ g 1. M 1 Z (b) Γ Z w + w = Γ Z zh = gq M Z φ 48π. ( Problem 10.15: B 3 W h0 d h0 u S Z ) basis: m B 0 g ν d g ν u 0 0 gν 0 m d W gνu 0 0 g ν d gν d 0 λ S s λ S ν u g Q d ν d g ν u gνu λ S s 0 λ S ν d g Q u ν u 0 0 λ S ν u λ S ν d 0 g Q S s 0 0 g Q d ν d g Q u ν u g Q S s m Z Decoupling limit: m B 0 g ν d g ν u 0 0 gν 0 m d W gνu 0 0 g ν d gν d 0 µ eff 0 0 g ν u, gνu µ eff g Q S s g Q S s 0 where µ eff = λ S s. Vector, Dirac fermion, and scalar masses are all g Q S s

34 30 The Standard Model and Beyond Problem 10.17: ζ for ω = 0; ζ 0.0 for ω = π/ (1 + ζ cos ω) (6) Problem 10.18: V = 15 4 µ ν Φ (15a + 7b)ν4 Φ