Problem 1(a): In N N matrix form, the local SU(N) symmetry acts on the adjoint matter field Φ(x) and the gauge field A µ (x) according to

Transcription

1 PHY 396 K. Solutions for problem set #11. Problem 1: In N N mtrix form, the locl SUN symmetry cts on the djoint mtter field Φx nd the guge field A µ x ccording to Φ x = UxΦxU x, A µ x = UxA µxu x + i µ UxU x. S.1 Consequently, the covrint derivtives 4 become D µ Φx D µ Φ x = µ Φx + i A µ x,φ x ] S. where the first term on the RHS expnds to µ Φ = µ UΦU = µ UΦU + U µ ΦU + UΦ µ U = UU µ UΦU + U µ ΦU UΦU µ UU = U U µ UΦ + µ Φ ΦU µ U U = U µ Φ + Φ,U µ U ] U S.3 while the second term expnds to A µ,φ ] = UA µ U,UΦU ] + i µ UU,UΦU ] Aµ = U,Φ ] + iu µ U,Φ ] U S.4 Combining the two expnsions, we rrive t D µ Φ = U µ Φ + Φ,U µ U ] + i A µ,φ ] U µ U,Φ ] U = U D µ Φ U. S.5 Thus, the D µ Φx mtrix trnsforms exctly like the Φx mtrix itself, which mkes the D µ derivtive 4 covrint. Q.E.D. 1

2 Problem 1b: Let s strt with the second line of eq. 6. In the mtrix form, the djoint multiplet Φ is mtrix, the fundmentl multiplet Ψ is column vector, nd their mtrix product ΦΨ is lso column vector. The covrint derivtives cts on these mtrices nd vectors s D µ Φ = µ Φ + i A µ,φ ], D µ Ψ = µ Ψ + ia µ Ψ, S.6 while D µ ΦΨ def = µ ΦΨ + ia µ ΦΨ = µ ΦΨ + Φ µ Ψ + i A µ,φ ] Ψ + iφa µ Ψ = D µ ΦΨ + ΦD µ Ψ. S.7 Likewise, on the third line of eq. 6, Ξ is mtrix, Ψ is row vector, nd their mtrix product Ψ Ξ is lso row vector. Therefore D µ Ψ = µ Ψ iψ A µ, D µ Ξ = µ Ξ + i A µ,ξ ], S.8 while D µ Ψ Ξ def = µ Ψ Ξ iψ ΞA µ = µ Ψ Φ + Ψ µ Ξ iψ A µ Ξ + Ψ Ξ,A µ ] = D µ Ψ Ξ + Ψ D µ Ξ. S.9 Finlly, on the first line of eq. 6, Φ nd Ξ re both N N mtrices nd their product ΦΞ is lso mtrix. Consequently, D µ ΦΞ def = µ ΦΞ + i A µ,φξ ] = µ ΦΞ + Φ µ Ξ + i A µ,φ ] Ξ + iφ A µ,ξ ] S.10 = D µ ΦΞ + ΦD µ Ξ. Q.E.D.

3 Problem 1c: For the djoint multiplet of fields Φx, D µ D ν Φ = µ D ν Φ + i A µ,d ν Φ ] = µ ν Φ + i A ν,φ ] + i A µ, ν Φ + i A ν,φ ]] S.11 = µ ν Φ + i µ A ν,φ ] + i A ν, µ Φ ] + i A µ, ν Φ ] A µ, A ν,φ ]] where the three red terms re symmetric with respect to µ ν while the two blue terms re not symmetric. Consequently, when we ntisymmetrize in µ ν, we my ignore the red terms nd consider the blue terms only, thus ] Dµ,D ν Φ = i µ A ν,φ ] A µ, A ν,φ ]] i ν A µ,φ ] + A ν, A µ,φ ]] = i µ A ν ν A µ,φ ] ] ] S.1 A µ,a ν,φ where the second term follows from the Jcobi identity for the mtrix commuttor: A µ, A ν,φ ]] + A ν, A µ,φ ]] = + A µ, Φ,A ν ]] + Aν, A µ,φ ]] = Φ, A ν,a µ ]] = Aµ,A ν ],Φ ]. S.13 Altogether, we hve Dµ,D ν ] Φ = i µ A ν ν A µ + ia µ,a ν ],Φ ] i F µν,φ ] = ig F µν,φ ] S.14 where the second equlity follows from the definition of the non-belin F µν nd the third equlity from F µν = gf µν. Finlly, in components ig F µν,φ ] = igf b µν Φc λ b ], λc = igfµν b Φc bc λ if S.15 nd hence D µ,d ν ]Φ = gf bc F b µν Φc. 3

4 Problem 1d: In mtrix nottions, the non-belin guge symmetries ct on vector potentils A µ x ccording to A µx = UxA µ xu x + i µ UxU x. S.16 Tking F µν x def = µ A ν x ν A µ x + ia µ x,a ν x] S.17 s the definition of the tension fields F µν x, we then hve F µν x = µa ν x νa µ x + i A µ x,a ν x], S.18 whtever tht evlutes to. Specificlly, the first term here evlutes to µ A ν = µ UA ν U + i ν UU = U µ A ν U + where the second equlity follows from µ UU,UA ν U ] + i µ ν UU i ν UU µ UU µ UA ν U = U µ A ν U + S.19 µ UU,UA ν U ] S.0 cf. similr formul S.3 nd µ ν UU = µ ν UU + ν U µ U = µ ν UU ν UU µ UU. S.1 Likewise ν A µ = U νa µ U + ν UU,UA µ U ] + i ν µ UU i µ UU ν UU S. nd hence µ A ν νa µ = U µ A ν ν A µ U + µ UU,UA ν U ] i µ UU, ν UU ]. ν UU,UA µ U ] S.3 4

5 At the sme time, the commuttor prt of the tension field trnsforms into i A µ,a ] ν = i = i UA µ U + i µ UU, UA ν U + i ν UU ] UA µ U,UA ν U ] µ UU,UA ν U ] UA µ U, ν UU ] i µ UU, ν UU ], S.4 Combining eqs. S.3 nd S.4 leds to mssive cncelltion of 6 out of terms on the combined right hnd side. Only the first terms on right hnd sides of S.3 nd S.4 survive the cncelltion, thus µ A ν ν A µ + i A µ,a ν] = U µ A ν ν A µ U + i UA µ U,UA ν U ] = U µ A ν ν A µ + A µ,a ν ] U, S.5 or in other words, F µν x = UxF µνxu x. S.6 Q.E.D. Problem 1e: There re two wys to prove the non-belin Binchi identity: using prt b nd the Jcobi identity for the commuttors, or the hrd clcultion bsed directly on eq. S.17. Let me strt with the esier proof. In prt b we hve proved the Leibniz rules for the covrint derivtives of mtrix products of djoint nd ntifundmentl multiplets. In clss nd lso in prt d we sw tht the tension fields Fµν form n djoint multiplet. So if we tke ny fundmentl multiplet of some fields Ψ i x, then ccording to eq. 6.b D λ Fµν Ψ = D λ F µν Ψ +Fµν Dλ Ψ. S.7 But we lso sw in clss tht F µν Ψ = id µ,d ν ]Ψ, nd for the sme reson we should lso 5

6 hve F µν D λ Ψ = id µ,d ν ]D λ Ψ. Plugging these reltions into eq. S.7, we rrive t Dλ F µν Ψ = Dλ Fµν Ψ F µν Dλ Ψ = id λ Dµ,D ν ]Ψ + id µ,d ν ] D λ Ψ S.8 = id λ,d µ,d ν ]]Ψ. Now let s combine such formule for the 3 cyclic permuttions of the indices λ,µ,ν: Dλ F µν Ψ = idλ,d µ,d ν ]]Ψ, Dµ F νλ Ψ = idµ,d ν,d λ ]]Ψ, S.9 Dν F λµ Ψ = idν,d λ,d µ ]]Ψ, nd therefore Dλ F µν + D µ F νλ + D ν F λµ Ψ = i Dλ,D µ,d ν ]] + D µ,d ν,d λ ]] + D ν,d λ,d µ ]] Ψ. S.30 The right hnd side here must vnish by the Jcobi identity for the commuttors, hence on the left hnd side Dλ F µν + D µ F νλ + D ν F λµ Ψ = 0. S.31 Moreover, this must be true for ny fundmentl multiplet Ψx, which mens tht the mtrix in here must vnish, D λ F µν + D µ F νλ + D ν F λµ = 0. 8 Quod ert demonstrndum. The second proof of the Binchi identity follows directly from the definition S.17 of the non-belin tension fields nd the covrint derivtives 4. Let s spell out D λ F µν in detil: D λ F µν = λ F µν + ia λ,f µν ] = λ µ A ν ν A µ + ia µ,a ν ] + i A λ, µ A ν ν A µ + ia µ,a ν ] ] = λ µ A ν λ ν A µ + i λ A µ,a ν ] + ia µ, λ A ν ] = + ia λ, µ A ν ] ia λ, ν A µ ] A λ,a µ,a ν ]] λ µ A ν λ ν A µ + i λ A µ,a ν ] µ A ν,a λ ] + i A µ, λ A ν ] A λ, ν A µ ] A λ,a µ,a ν ]]. S.3 6

7 On the bottom two lines here I hve grouped terms in so tht fter summing over cyclic permuttions of the indices λ,µ,ν, we get zero sum seprtely for ech group. Indeed, λ µ A ν λ ν A µ + cyclic = λ µ A ν ν λ A µ + cyclic λ A µ,a ν ] µ A ν,a λ ] A µ, λ A ν ] A λ, ν A µ ] = 0 by inspection, + cyclic = 0 by inspection, + cyclic = 0 by inspection, nd S.33 A λ,a µ,a ν ]] + cyclic = 0 by Jcobi identity. Therefore, D λ F µν + cyclic D λ F µν + D µ F νλ + D ν F λµ = 0. S.34 Problem 1f: The Euler Lgrnge field equtions follow from requiring zero first vrition of the ction S = L under infinitesiml vrition of the independent fields Aµ x. Let s strt by clculting the vrition of the tension fields F µν : δf µν δ µ A ν ν A µ + ia µ,a ν ] = µ δa ν ν δa µ + i ] ] δa µ,a ν + i Aµ,δA ν = µ δa ν + i ] A µ,δa ν ν δa µ + i ] A ν,δa µ S.35 = D µ δa ν D ν δa µ where we tret the mtrix-vlued vritions δa ν x s djoint fields so their covrint derivtive work ccording to eq. 4, D µ δa ν µ δa ν +ia µ,δa ν ] nd likewise for the D ν δa µ. In light of eq. S.35, the trce in the Yng Mills Lgrngin 9 vries by δtr F µν F µν = tr F µν δf µν = tr F µν D µ δa ν D ν δa µ = 4tr F µν D µ δa ν since F µν = F νµ Dµ = 4tr F µν δa ν + 4 µ tr F µν δa ν S.36 where the lst equlity follows from the Leibniz rule for the two djoint fields Φ = F µν nd 7

8 Ξ = δa ν : tr D µ ΦΞ + tr ΦD µ Ξ = tr D µ ΦΞ = tr µ ΦΞ + itr A µ,φξ] = µ tr Φξ + 0 S.37 trce of commuttor is zero. Thus δl YM = g tr D µ F µν δa ν totl divergence S.38 so the net Yng-Mills ction vries by δs = g d 4 x tr D µ F µν xδa ν x = 1 g d 4 x D µ F µν x δa ν x. S.39 To mke this vrition vnish for ny infinitesiml δa νx we need D µ F µν x 0. Problem : In problem 1f we sw tht for infinitesiml vritions of the guge fields the YM Lgrngin vries by δl YM = 1 g D µ F µν δa ν + µ = D µ F µν δa ν + µ. S.40 Now let s dd the mtter Lgrngin L mt φ,dφ for some mtter fields in non-trivil multiplet or multiplets of the guge symmetry. When we vry the guge fields A νx while keeping the mtter fields φx fixed, the covrint derivtives Dφ vry due to iga ν t φ terms in D ν φ, which leds to non-trivil vrition δl mt = L mt A ν δa ν J ν δa ν. S.41 Altogether, the net ction of the theory vries by δs = d 4 x D µ F µν x J ν x δa ν x. S.4 Requiring this vrition to vnish for ny δa νx leds to the field equtions D µ F µν = J ν, S.43 or in mtrix nottions D µ F µν = J ν. This is the non-belin version of the Mxwell equtions 8

9 µ F µν = J ν. In the belin EM theory, the equtions µ F µν = J ν require the electric current to be conserved, ν J ν = ν µ F µν = 0 since F µν = F νµ nd the derivtives commute with ech other. The non-belin tension fields F µν re lso ntisymmetric in µ ν, but the covrint derivtives do not commute, D µ D ν D ν D µ. Therefore, D ν J ν = D ν D µ F µν = 1 D µ,d ν ]F µν = ig F µν,f µν ] S.44 where the lst equlity works exctly s in problem 1c the F µν fields form n djoint multiplet of fields, nd for ny such multiplet pcked into n hermitin N N mtrix Φ, D µ,d ν ]Φ = if µν,φ] = igf µν,φ]. However, unlike generic mtrix Φ which my commute or not commute with the F µν, for ny µ nd ν the F µν mtrix lwys commutes with itself. Thus, F µν,f µν ] = 0 even before summing over µ nd ν. S.45 Of course, fter the summing over µ nd ν we still hve zero, thus D ν D µ F µν x 0. Thus, consistency of the field equtions S.43 for the guge fields requires the non-belin currents J µ to be covrintly conserved: D ν J ν = D µ D ν F µν = 0, S.46 or in components ν J ν gf bc A b νj cν = 0. S.47 Note: becuse of the covrintizing term here, we do not hve conserved net chrges; ls, d d 3 xj 0 x,t 0. dt S.48 9

10 Problem b: The Lgrngin for fundmentl multiplet of Dirc fermions is spelled out in eq. 14. In components, λ L mt = Ψiγ µ D µ mψ = Ψ i iγ µ µ mψ i + Ψ i iγ µ iga µ hence J µ = L A µ = gψ i γ µ λ i j i j Ψ j, S.49 Ψ j gψγ µλ Ψ. S.50 To pck these currents into trceless hermitin mtrix J µ = Jµ 1 λ, we my use eq. 15. where the Ψ i ply the role of the ξ i while the Ψ j γ µ ply the role of the η j. Thus, J µ i j = on in mtrix nottions, J µ = gψγ µλ Ψ λ i J µ = I µ triµ N 1 N N where j = g Ψ jγ µ Ψ i g N Ψ kγ µ Ψ k δ i j, I µ i j = g Ψ jγ µ Ψ i. S.51 Note: the I µ mtrices re hermitin but not trceless, but the J µ mtrices re both hermitin nd trceless. Now consider the SUN trnsformtion rule for these mtrices. In components, Ψ i x = U i k xψk x, Ψ jx = Ψ l x U x lj S.5 hence or in mtrix nottions I µ x i j = g Ψ jxγ µ Ψ i x = g Ψ lx U x lj γµ U i k xψk x g = U i k x Ψ lxγ µ Ψ k x U x l j = U i k x I µ x k l U x lj I µ x = Ux I µ x U x. S.53 S.54 This trnsformtion leves the trce tri µ invrint, while its trceless prt J µ trnsforms 10

11 ccording to eq. 3, indeed J µ x = I µ x tri µ x N 1 = UxI µ xu x truxiµ xu x 1 N = UxI µ xu x triµ x Ux 1 U x N = Ux I µ x triµ x 1 U x N = Ux J µ x U x. S.55 In other words, the currents J µ x trnsform into ech other s members of n djoint multiplet, quod ert demonstrndum. Problem c: First, let ξx be fundmentl multiplet of some fields while η x is n ntifundmentl multiplet, nd let s prove the Leibniz rule for the djoint multiplet of the form Q = η λ ξ : D µ η λ ξ = D µ η λ Ψ + η λ D µ ξ. S.56 Proof: D µ η λ ξ µ η λ ξ gf bc A b µ η λ c ξ = µ η λ ξ + η λ µ ξ ga b µη f bc λ c = i λ,λ b ] ξ = µ η ig Ab µ η λ b λ ξ + η λ µ ξ + ig Ab µ λb ξ S.57 = D µ η λ ξ + η λ D µ ξ. Now tht we hve this generl formul, let s pply it to the non-belin currents J µ = g/ψγ µ λ Ψ. Using Ψ for ξ nd Ψγ µ for η, we rrive t D µ J µ = g/ D µ Ψ γ µ λ Ψ + g/ψλ γ µ D µ Ψ. S.58 Note: in this formul, the order of the λ nd γ µ mtrices is unimportnt since they ct on unrelted indices of the Dirc fields. 11

12 Now let s pply the Euler-Lgrnge equtions for the fermionic fields, which re simply the covrint Dirc eqution nd its conjugte, iγ µ D µ Ψ = mψ, id µ Ψγ µ = mψ. S.59 Plugging these equtions of motion into eq. S.58, we immeditely obtin D µ J µ = g/ D µ Ψγ µ = imψ λ Ψ+ g/ψλ γ µ D µ Ψ = imψ = img/ Ψλ Ψ img/ Ψλ Ψ S.60 = 0. Quod ert demonstrndum. Problem 3 : Regrdless of the nture of the mtter field multiplet m from the symmetry group s point of view it could be vector, tensor, spinor, whtever let me tret the Ψ α x s components of some big column vector Ψx of length sizem so I cn use mtrix nottions α for the Tm without bothering with indices α,β,... Thus insted of Tm β Ψβ x I will write simply Tm Ψx. A locl symmetry prmetrized by n infinitesiml Λ x cts on the guge nd mtter fields ccording to δa µ x = µλ x f bc Λ b xa c µ x, δψx = iλ xtm Ψx Consequently, the covrint derivtives D µ Ψx = µ Ψx + ia µ xt m Ψx. 19 1

13 chnge by δ D µ Ψx = D µ δψx + δd µ due to δa µ Ψx = µ δψx + ia µ xt m δψx + iδa µ xt m Ψx = i µ Λ xt m Ψx + iλ xt m µψx A b µ Tb m Λ xt m Ψx i µ Λ xt m Ψx ifbc Λ b xa c µ xt m Ψx S.61 relbel c b if cb Λ xa b µ xtc m Ψx = iλ xtm µψx Λ xa b µx Tm b T m +ifcb Tm c Ψx Now let s simplify the mtrix inside the big in the second term on the bottom line. The mtrices Tm represent the Lie lgebr of the locl symmetry, so they obey the sme commuttion reltions s the genertors ˆT, T m,tm b ] = if bc Tm c. S.6 Also, f bc is totlly ntisymmetric so f cb = +f bc. Consequently if cb T c m = if bc T c m = T m Tb m Tb m T m = T b m T m + ifcb T c m = T m Tb m, nd therefore S.63 δ D µ Ψx = iλ xtm µψx Λ xa b µ xt m Tb m Ψx = iλ xtm µ Ψx + ia b µ xtb m Ψx S.64 iλ xt m D µψx. Thus the D µ Ψx trnsforms under the infinitesiml locl symmetries exctly like the Ψx itself, which mkes the derivtive D µ covrint. Q.E.D. 13

14 Problem 4: When guge symmetry is spontneously broken, the guge fields cquire msses which come from the guge-covrint kinetic terms for the sclr fields with non-zero VEVs vcuum expecttion vlues. The simplest wy to seprte the vectors mss terms from the shifted sclrs kinetic energies nd from the sclr-vector interctions is to freeze the sclr fields to their VEVs. Indeed, let s freeze Φx Φ = C 1 N N. Then ccording to eq. 1, D µ Φ = ig B µ Φ + igl µ Φ ig Φ R µ = ig CB µ 1 N N + igc L µ R µ = ig CB µ 1 N N + igc L µ R µ λ, S.65 nd consequently Dµ tr Φ D µ Φ = Ng C B µ B µ + g C L µ R µ Lµ R µ. S.66 Thus, the belin B µ field hs mss M B = Ng C while the non-belin fields L µ nd R µ hve non-digonl mss terms. To digonlize those terms, let s mix the fields ccording to V µ = 1 L µ + R µ, X µ = 1 L µ R µ, S.67 where the 1/ coefficients mke the V µ nd X µ cnoniclly normlized, i.e. L kin L,R = 1 4 µl ν] + µ Rν] = 1 4 µxν] + µ Vν]. S.68 In terms of the V µ nd X µ, the mss terms for L µ nd R µ in eq. S.66 become L msses L,R = g C X µ Xµ. S.69 Thus, the V µ fields remin mssless while the X µ cquire common mss M X = g C. 14

15 Problem 4b: To write down n effective theory for the mssless fields, we simply freeze ll the mssive fields B µ, Xµ, ϕ 1, nd ξ 1 in the nottions of the previous homework, only the mssless Vµ remin un-frozen. In other words, we let Φx Φ = C 1 N N, B µ x 0, L µ x = R µ x = 1 Vµ x, S.70 nd then substitute these vlues into the Lgrngin 0. According to eq. S.65, for fields s in eq. S.70 D µ Φ = 0, so the only un-frozen terms in the Lgrngin re L unfrozen = 1 trl µνl µν 1 trr µνr µν for L µν = Rµν = trl µν L µν = 1 L µν S.71 = 1 V 4 µν which is precisely the Yng-Mills Lgrngin for the cnoniclly normlized tension fields V µν = L µν +R µν L µν when L µν = R µν. S.7 of the un-broken SUN V guge theory. Indeed, in terms of the cnoniclly normlized SUN V potentil fields V µ, Vµν = µ L ν νl µ gfbc L b µ Lc ν = Vν Vµ µ ν gf bc V µ b Vν c S.73 = µ V ν ν V µ g f bc V b µv c ν. The coefficient of the non-belin lst term on the bottom line is the guge coupling of the unbroken SUN V guge group g v = g. S.74 15

16 Problem 4 : For g L g R, the covrint derivtives of the sclr fields become D µ Φ = µ Φ + ig B µ Φ + ig L L µ Φ ig R ΦR µ. S.75 As in prt, the mss terms for the vector fields obtin from plugging Φ into these covrint derivtives nd then expnding the kinetic terms for the sclrs: nd hence D µ Φ = ig CB µ 1 N N + icg L L µ g R R µ λ S.76 L trd µ Φ D µ Φ Ng C B µ B µ + C g LL µ G RR µ g LL µ G R R µ. S.77 As inprt, thebelin gugefields gets mss M B = Ng C, while the non-belinvector mss is more tricky. Let s define the coupling g nd the mixing ngle θ ccording to g L = g cosθ, g R = g sinθ = g = g L + g R, tnθ = g R g L. S.78 Then the non-belin mss term in eq. S.77 becomes c g L µcosθ R µsinθ, S.79 which tells us which prticulr combintion of the non-belin guge fields become mssive. Indeed, if we let L kin L,R = 1 4 X µ = cosθ L µ sinθ R µ, Y µ = sinθ L µ + cosθ R µ, then both combintions of vector fields re cnoniclly normlized indeed, µl ν] + µ Rν] µxν] + while the mss term S.79 becomes = 1 4 Thus, the X µ fields hve mss M X = gc, while the Y µ µ Y ν] S.80, S.81 L mss L,R = C g X µx µ. S.8 fields remin mssless. 16

17 Now let s derive the effective Lgrngin for just the mssless vector fields Yµ x while freezing ll the other fields, i.e. setting Φx Φ, B µ x 0, nd Xµ x 0. In terms of the L µ nd R µ fields, this mens L µ = Y µ sinθ, R µ = Y µ cosθ, S.83 or in terms of the group-normlized guge fields L µ = g L L µ = g L sinθ Y µ, R µ = g R R µ = g R cosθ Y µ. S.84 However, for the mixing ngle θ relted to the couplings s in eq. S.78, we hve g L sinθ = g R cosθ = g Lg R g def = g Y, S.85 nd therefore L µ x = R µ x = g Y Y µ x def = Y µ x. S.86 In terms of the Y µ x guge field, the non-belin tension fields re L µν x = R µν x = Y µν x = µ Y ν x ν Y µ x + iy µ,y ν ] S.87 precisely s for group-normlized SUN connection Y µ x while the net YM Lgrngin is L YM = 1 g L tr L µν L µν 1 g R tr R µν R µν = 1 1 precisely s for the SUN YM theory with inverse guge coupling g L + 1 gr tr Y µν Y µν S.88 1 g Y = 1 g L + 1 gr. S.89 After bit of lgebr, this formul becomes g Y = g L g R, 4 gl +g R in perfect greement with Yµ x = g Y Yµ x for the cnoniclly normlized guge fields Y µx. 17