PHY–396 K. Solutions for problem set #4. Problem 1(a): First of all, a ...

PHY–396 K. Solutions for problem set #4.Problem 1(a):First of all, a point of notation: In class, I used denoted the original scalar fields ϕ i (x) (fori = 1, . . . , N), and σ(x) was the shifted field σ(x) = ϕ N (x) − v. But in this problem, thefirst N − 1 = 3 scalar fields are denoted π i (x) while σ(x) is the un-shifted field playing therole of π 4 (x). The shifted field σ(x) − 〈σ〉 will be called δσ(x).And now, the solution. The linear sigma model has scalar potentialV (σ, π˜) = λ 8(σ 2 + π˜2 − f 2) 2− λf 2 β × σ. (S.1)All stationary points of this potential are solutions to equationsand∂V∂π˜= λ 2(σ 2 + π˜2 − f 2) × = 0 (S.2)π˜∂V∂σ = λ (σ 2 +2π˜2 − f 2) × σ − λf 2 β = 0. (S.3)For f ≠ 0, the second equation (S.3) requires σ 2 + π˜2 − f 2 ≠ 0 and then the first equation(S.2) immediately leads to π˜= 0 — all stationary points are along the σ axis of the R 4 fieldspace. Plugging π˜= 0 back into the eq. (S.3), we obtain a cubic equation for the σ,σ 3 − f 2 σ − 2f 2 β = 0. (S.4)For small β > 0 this equation has three real solutions, namelyσ 1 ≈ −2β, σ 2 ≈ −f + β, σ 3 ≈ +f + β. (S.5)Now let’s find out which of this stationary points is a minimum.At π˜= 0 the second1

derivatives of the potential are∂ 2 V∂σ 2 = λ(3σ 2 − f 2 ),∂ 2 V∂π i ∂π j= λ(σ 2 − f 2 )δ ij ,∂ 2 v∂σ∂π i= 0. (S.6)Evaluating these second derivatives for the three stationary points (S.5), we find@ σ 1 : V ′′σσ < 0 and V ′′ππ < 0 =⇒ maximum,@ σ 2 : V ′′σσ > 0 and V ′′ππ < 0 =⇒ saddle point,@ σ 3 : V ′′σσ > 0 and V ′′ππ > 0 =⇒ minimum.(S.7)Thus, V (σ, π˜) has a unique minimum at = 0, σ = σ 3 ≈ f + β, exactly as in eq. (6.2).π˜Q.E.D.Problem 3(b):Let us shift the σ field by its vacuum expectation value 〈σ〉 = σ 3 ,σ(x) = 〈σ〉 + δσ(x).(S.8)Expanding the scalar potential into powers of the π˜and δσ fields, we haveV (π˜, δσ) = λ 8(π˜2 + ( σ 2 = 〈σ〉 2 + 2 〈σ〉 δσ + δσ 2) − f 2) 2− λf 2 β ( σ = 〈σ〉 + δσ )= const + V 1 + V 2 + V 3 + V 4(S.9)whereV 1 = λ 2 × ( 〈σ〉 2 − f 2) × 〈σ〉 × δσ − λβf 2 × δσ = 0 (S.10)(cf. eq. (S.3)) since we are expanding around the minimum V ,V 2 = λ (〈σ〉 2 − f 2) × ( 4π˜2 + δσ 2) + λ 2 〈σ〉2 × δσ 2 ,= λ (〈σ〉 2 − f 2) ×4π˜2 + λ (3 〈σ〉 2 − f 2) × δσ 2 ,4andV 3V 4= λ 〈σ〉 × δσ × ( π˜2 + δσ 2) (S.11),2= λ 8 × ( π˜2 + δσ 2) 2 .2

Now let’s expand the whole Lagrangian into powers of π˜(x) and δσ(x). ClearlyL = L kin − V = L free + L int (S.12)whereL free = L kin − V 2 (π˜, δσ)[ (∂µ δσ) 2=− λ (3 〈σ〉 2 − f 2) ]× δσ 22 4+[ (∂µ π˜) 22− λ 4(〈σ〉 2 − f 2) ]× π˜2(S.13)describes the free π˜and δσ fields, andL int = −V 3 (π˜, δσ) − V 4 (π˜, δσ) (S.14)describes their interactions.(S.13): the pion mass 2 isThe particles’ masses follow from the quadratic LagrangianM 2 π = λ 2(〈σ〉 2 − f 2) = λβf 2〈σ〉≈ λβf(S.15)— where the second equality follows from 〈σ〉 satisfying eq. (S.3), and the last, approximateequality follows from 〈σ〉 = σ 3 ≈ f, — while the σ’s mass 2 isM 2 σ = λ 2(3 〈σ〉 2 − f 2) = λ 〈σ〉 2 + M 2 π. (S.16)For β ≪ f we have 〈σ〉 ≈ f and hence M 2 σ ≈ λf 2 ≫ M 2 π.Q.E.D.Problem 3, the non-linear limit:In the strong-coupling limit λ → ∞, the first term in the potential (S.1) becomes a constraintσ 2 (x) + π˜2 (x) ≡ f 2 . (S.17)Solving the constraint eliminates the independent σ field from the theory asσ(x) =√f 2 − π˜2 (x) , (S.18)and we are left with the non-linear sigma model for the three π i (x) fields.The kinetic3

Lagrangian for this model follows fromL kin = 1 2 (∂ µπ˜) 2 + 1 2 (∂ µσ) 2 = 1 2 G ij(π˜) ∂ µ π i ∂ µ π j (S.19)(implicit sum over i and j) whereG ij (π˜) = δ ij + π i π jf 2 − 2 π˜. (S.20)As to the potential, the SO(4)–symmetric leading term vanishes, while the perturbationyieldsV= −λf 2 β × σ = −λf 2 β × √ f 2 − π˜2 = const + 1 2λfβ × π˜2 + O(π˜4 ). (S.21)Extracting the quadratic terms from eqs. (S.19) and (S.21), we obtain the free LagrangianL free = 1 2 (∂ µπ˜) 2 − 1 2 λfβ × π˜2 , (S.22)which clearly describes 3 fields π i of common mass M 2 π = λfβ.Problem 2(a):given Φ → e +iθ U L ΦU † R , (6.5)we have Φ † → e −iθ U R Φ † U † L , (S.23)hence Φ † Φ → U R (Φ † Φ)U † R , (S.24)(Φ † Φ ) 2 → UR Φ † ΦU † R U RΦ † ΦU † R = U R(Φ † Φ ) 2 U†R , (S.25)(likewise Φ † Φ ) n (→ UR Φ † Φ ) n U†R∀n = 1, 2, 3, . . . , (S.26)and thereforeall traces tr( (Φ † Φ ) n ) are invariant under (6.5), (S.27)thanks to the cyclic invariance rule for traces, tr ( U R XU † ) (R = tr XU†R U ( )R)= tr X for anyX = ( Φ † Φ ) n . Consequently, the scalar potential (6.4) is invariant under symmetries (6.5).4

For the global symmetries where e iθ , U L , and U R do not depend on x, the kinetic termin (6.3) is also invariant. Indeed,for constant e iθ , U L , U R ,∂ µ Φ → e +iθ U L (∂ µ Φ)U † R ,∂ µ Φ † → e −iθ U R (∂ µ Φ † )U † L ,(S.28)∂ µ Φ † ∂ µ Φ → U R (∂ µ Φ † ∂ µ Φ)U † R ,and tr ( ∂ µ Φ † ∂ µ Φ ) is invariant.Altogether, the whole Lagrangian (6.3) is invariant,Q.E.D.Problem 2(⋆):The kinetic term in (6.3) and the last two terms in the potential (6.4) have a much biggersymmetry than G, namely SO(2N 2 ), which does not care for the matrix structure of theΦ(x) and treats it as 2N 2 real component fields. Indeed,(tr Φ † Φ) = ∑ i,j∣∣Φ ji∣∣ 2= ∑ i,j( ( ) 2 ( ) ) 2Re Φ ji + Im Φ ji(S.29)is invariant under all SO(2N 2 ) “rotations” of the components, and so is the kinetic term.On the other hand, the tr ( Φ † Φ Φ † Φ ) in the potential does depend on the packing of2N 2 real components into a complex N × N matrix, and it is this term which reduces theinternal symmetry group of the theory to G = SU(N) × SU(N) × U(1).Proving that all the SO(2N 2 )/G symmetries are broken by the quartic trace term is anon-trivial exercise in group theory rather than field theory. You do not have to do it aspart of this homework set, and I am not writing down the proof here.Problem 2(b):Given the eigenvalues (κ 1 , . . . , κ N ) of the Φ † Φ matrix, the invariant traces (S.27) obtain astr( (Φ † Φ ) n ) =N∑κ n i .i=1(S.30)5

Consequently, the scalar potential isV = α 2(∑κ 2 i + β ∑2iiκ i) 2+ m 2 ∑ iκ i .(S.31)Now let’s minimize this potential. Since the matrix Φ † Φ cannot have any negativeeigenvalues, we are looking for a minimum of V (κ 1 , . . . , κ N ) under constraints κ i ≥ 0. Thisrequires∀i = 1, . . . , N,either κ i ≥ 0 and ∂V∂κ i= 0, or else κ i = 0 and ∂V∂κ i> 0, (S.32)where∂V∂κ i= ακ i + m 2 + β ∑ jκ j .(S.33)The solutions of these equations are as follows:• For α > 0, β > 0, and m 2 > 0, the only solution is κ 1 = · · · κ N = 0 and hence 〈Φ〉 = 0.• For α > 0, β > 0, and m 2 < 0, the only solutions is (6.6) and hence 〈Φ〉 = C × aunitary matrix. We shall focus on this regime in parts (c–e) of this problem and inproblem 3.• For α < 0 or β < 0, the situation is more complicated:— When α + β < 0 or α + Nβ < 0, the scalar potential (6.4) is unbounded frombelow and the theory is sick.— When α > 0 and β < 0 but α + Nβ > 0, the solutions are similar to the β > 0case: For m 2 > 0 all κ i = 0, while for m 2 < 0 the κ i are as in eq. (6.6).— When β > 0 and α < 0 but α + β > 0, there is only one 〈φ〉 = 0 solution form 2 > 0, but there are N distinct solutions for m 2 < 0, namely for k = 1, 2, . . . , N,κ 1 = · · · = κ k = −m2α + kβ > 0, κ k+1 = · · · = κ N = 0 (S.34)(modulo permutations of the eigenvalues). The deepest minimum of V correspondsto the k = 1 solution.6

Problem 2(c):Let’s act with some SU(N) L × SU(N) R × U(1) symmetry (6.5) on the vacuum expectationvalues (6.7):〈Φ〉 = C × 1 N×N → e iθ U L 〈Φ〉 U † R = C × eiθ U L U † R . (S.35)Clearly, to keep the VEVs 〈Φ〉 invariant, we neede iθ U L U † R = 1 N×N (S.36)and henceU R = e iθ × U L . (S.37)Moreover, since the U L and U R matrices have unit determinants, this meanse iθ = 1 and U L = U R ∈ SU(N). (S.38)In other words, the unbroken symmetry group is SU(N) which acts on the scalar fields asΦ(x) → UΦ(x) U † , U ∈ SU(N). (S.39)Problem 2(d):In terms of the shifted fields (6.8),∂ µ Φ = 1 √2(∂ µ ϕ 1 + i∂ µ ϕ 2 ) , ∂ µ Φ † = 1 √2(∂ µ ϕ 1 − i∂ µ ϕ 2 ) ,(S.40)hence the kinetic term in the Lagrangian becomestr ( ∂ µ Φ † ∂ µ Φ ) = 1 2 tr(∂ µϕ 1 ∂ µ ϕ 1 ) + 1 2 tr(∂ µϕ 2 ∂ µ ϕ 2 ).(S.41)As to the potential terms, we haveΦ † Φ = C 2 × 1 N×N+ C × ( δΦ † + δΦ) + δΦ † δΦ= C 2 × 1 N×N + √ 2C × ϕ 1 + 1 2 ϕ2 1 + 1 2 ϕ2 2 + i 2 [ϕ 1, ϕ 2 ](S.42)and consequentlytr ( Φ † Φ ) = NC 2 + √ 2C tr(ϕ 1 ) + 1 2 tr(ϕ2 1) + 1 2 tr(ϕ2 2), (S.43)7

tr 2( Φ † Φ ) = N 2 C 4 + 2 √ 2NC 3 tr(ϕ 1 ) + 2C 2 tr 2 (ϕ 1 ) + NC 2 ( tr(ϕ 2 1) + tr(ϕ 2 2) )+ √ 2C tr(ϕ 1 ) × ( tr(ϕ 2 1) + tr(ϕ 2 2) ) + 1 (4 tr(ϕ21 ) + tr(ϕ 2 2) ) 2(S.44)( (Φtr † Φ ) ) 2= NC 4 + 2 √ 2C 3 tr(ϕ 1 ) + 3C 2 tr(ϕ 2 1) + C 2 tr(ϕ 2 2)+ √ 2C tr(ϕ 3 1) + √ 2C tr(ϕ 1 ϕ 2 2)+ 1 4 tr(ϕ4 1) + 1 4 tr(ϕ4 2) +2 3 tr(ϕ2 1ϕ 2 2) − 1 2 tr(ϕ 1ϕ 2 ϕ 1 ϕ 2 ). (S.45)Plugging all these formulae into the potential (6.4) and expanding in powers of ϕ 1 and ϕ 2 ,we obtainV = const + V 1 + V 2 + V 3 + V 4 (S.46)whereV 1 = √ 2C × (m 2 + βNC 2 + αC 2 ) × tr(ϕ 1 ) = 0 (S.47)〈〈 because m 2 + (α + Nβ)C 2 = 0 〉〉V 2 = βC 2 × tr 2 (ϕ 1 ) + 1 2 (m2 + βNC 2 + 3αC 2 ) × tr(ϕ 2 1)+ 1 2 (m2 + βNC 2 + αC 2 ) × tr(ϕ 2 2)= βC 2 × tr 2 (ϕ 1 ) + αC 2 tr(ϕ 2 1) + 0, (S.48)V 3= βC()√ × tr(ϕ 1 ) × tr(ϕ 2 1) + tr(ϕ 2 2)2+ αC ()√ × tr(ϕ 3 1) + tr(ϕ 1 ϕ 2 2) , (S.49)2V 4 = β () 2tr(ϕ 281) + tr(ϕ 2 2)+ α ()tr(ϕ 481) + tr(ϕ 4 2) + 6 tr(ϕ 2 1ϕ 2 2) − 2 tr(ϕ 1 ϕ 2 ϕ 1 ϕ 2 ) . (S.50)Combining the quadratic part (S.48) of this potential with the kinetic terms (S.41), wearrive atL 2 = 1 2 tr(∂ µϕ 1 ∂ µ ϕ 1 ) − βC 2 × tr 2 (ϕ 1 ) − αC 2 tr(ϕ 2 1) + 1 2 tr(∂ µϕ 2 ∂ µ ϕ 2 ) (S.51)which gives us the mass spectrum of the theory: the ϕ 1 (x) matrix of fields is massive and the8

ϕ 2 (x) matrix is massless. Each matrix is N ×N and hermitian, so it contains N 2 independentreal scalar fields, which give rise to N 2 particles. Altogether, the spectrum comprises:• N 2 massless particles from the ϕ 2 (x) matrix.• N 2 − 1 massive particles with M 2 = 2αC 2 from the traceless part of the ϕ 1 (x) matrix.• One more massive particle with M 2 = 2(α + Nβ)C 2 = −2m 2 from the trace of ϕ 1 (x).To see where the values of the masses come from, let’s decompose the ϕ 1 matrix intothe pure trace plus the traceless part,ξ(x) def = tr(ϕ 1(x))√Nand ˜ϕ 1 (x) def = ϕ 1 (x) − ξ(x) √N× 1 N×N =⇒ tr( ˜ϕ 1 ) ≡ 0. (S.52)Consequently,tr 2 (ϕ 1 ) = N × ξ 2 , tr(ϕ 2 1) = ξ 2 + tr( ˜ϕ 2 1), (S.53)likewisetr ( ∂ µ ϕ 1 ∂ µ ϕ 1)= ∂µ ξ ∂ µ ξ + tr ( ∂ µ ˜ϕ 1 ∂ µ ˜ϕ 1),so the free Lagrangian (S.51) becomesL 2 = 1 2 (∂ µξ) 2 − 2(βC 2 N + αC 2 ) × 1 2 ξ2+ 1 2 tr( (∂ µ ˜ϕ 1 ) 2) − 2αC 2 × 1 2 tr( ˜ϕ2 1)(S.54)+ 1 2 tr( (∂ µ ϕ 2 ) 2) − 0 × 1 2 tr(ϕ2 2)where all the masses are manifest.Problem 2(e):The unbroken SU(N) symmetry acts on the scalar fields according to eq. (S.39), or in termsof the shifted fields,ϕ 1 (x) → Uϕ 1 (x) U † , ϕ 2 (x) → Uϕ 2 (x) U † . (S.55)To make the SU(N) multiplet structure manifest, let’s decompose both ϕ 1 and ϕ 2 into their9

traceless parts and pure traces along the lines of eq. (S.52),ϕ 1 (x) = ξ 1(x)√N×1 N×N + ˜ϕ 1 (x), ϕ 2 (x) = ξ 2(x)√N×1 N×N + ˜ϕ 2 (x), tr( ˜ϕ 1 ) ≡ tr( ˜ϕ 2 ) ≡ 0.(S.56)With this decomposition, the ξ 1 and the ξ 2 are both invariant under the SU(N) — whichputs each of them into its own singlet multiplet — while the each of the traceless parts ˜ϕ 1and ˜ϕ 2 makes each own adjoint multiplet.This multiplet structure agrees with the masses we obtained in part (d). Indeed, allN 2 − 1 members of the adjoint multiplet ˜ϕ 1 have the same mass 2αC 2 , while the singlet ξ 1has a different mass 2(α + Nβ)C 2 .On the other hand, both the adjoint multiplet ˜ϕ 2 and the singlet ξ 2 are massless. Thereason for this degeneracy goes beyond the un-broken SU(N) symmetry; instead, both the˜ϕ 2 and the ξ 2 are Goldstone bosons of the spontaneously broken symmetries inG/H =() /SU(N) L × SU(N) R × U(1) SU(N) .(S.57)Specifically, the singlet ξ 2 is the Goldstone boson of the broken U(1) symmetry. Indeed, theU(1)’s generator commutes with all the other generators, so it belongs in its own singlet ofthe symmetry, and the corresponding Goldstone particle should also be a singlet.Now consider the non-abelian generators. Generators TL a of the SU(N) L form an adjointmultiplet of the SU(N) L , but are invariant under the SU(N) R . Likewise, generators TR a ofthe SU(N) R form an adjoint multiplet of the SU(N) R , but are invariant under the SU(N) L .In other words, under an (U L , U R ) ∈ SU(N) L × SU(N) R they transform asT a L → U LT a L U † L , T a R → U RT a R U † R . (S.58)When the SU(N) L × SU(N) R is broken down to a single SU(N) spanning U L = U R = U,both TL a and T R a transform asT a L → UT a L U † , T a R → UT a R U † , (S.59)which puts them into two adjoint multiplets of the unbroken SU(N). Equivalently, we may10

form two adjoint multiplets out ofT a V = T a L + T a R and T a A = T a L − T a R , (S.60)which act on the scalar fields according toT a V Φ = i 2 [λa , Φ], T a A Φ = i 2 {λa , Φ}. (S.61)The TV a generate the unbroken SU(N) symmetry, cf. eq. (S.39). The T A a generators arespontaneously broken, hence there should be an adjoint multiplet of massless Goldstonebosons. And indeed there is — the ˜ϕ 2 .Problem 3(a):When a gauge symmetry is spontaneously broken, the gauge fields acquire masses — whichcome from the gauge-covariant kinetic terms for the scalar fields with non-zero VEVs (vacuumexpectation values). The simplest way to separate the vector mass terms from shiftedscalars’ kinetic energies and scalar-vector interactions is to freeze the scalar fields to theirVEVs. Indeed, let’s freeze Φ(x) ≡ 〈Φ〉 = C × 1 N×N . Then according to eq. (9),D µ 〈Φ〉 = ig ′ B µ 〈Φ〉 + igL µ 〈Φ〉 − ig 〈Φ〉 R µ= ig ′ C B µ × 1 N×N + igC × (L µ − R µ )= ig ′ C B µ × 1 N×N + igC ∑(L a µ − R a2µ) × λ a ,a(S.62)and consequently( (Dµtr 〈Φ〉 †)( D µ 〈Φ〉 )) = Ng ′2 C 2 × B µ B µ + g2 C 22∑(L a µ − Rµ)(L a aµ − R aµ ). (S.63)aThus, the abelian B µ field has mass M 2 B = 2Ng′2 C 2 while the non-abelian fields L a µ and R a µ11

have non-diagonal mass terms. To diagonalize those terms, let’s mix the fields according toVµ a = √ 1 (Laµ + R a )µ , Xaµ = 1 (√ Laµ − Rµ) a , (S.64)2 2where the 1/ √ 2 coefficients make the V a µ and X a µ canonically normalized, i.e.L kinL,R = −1 4∑a( ( 2 ( ) ) 2∂ [µ Lν]) a + ∂ [µ Rν]a= − 1 ∑( ( ) 2 ( ) ) 2∂4 [µ Vν]a + ∂ [µ Vν]a . (S.65)aIn terms of the V a µ and X a µ, the mass terms for L a µ and R a µ in eq. (S.63) becomeL massesL,R = g 2 C 2 × X a µX aµ . (S.66)Thus, the V a µ fields remain massless while the X a µ acquire common mass M 2 X = 2g2 C 2 .Problem 3(b):To write down an effective theory for the massless fields, we simply freeze all the massivefields B µ , X a µ, ˜ϕ 1 , and ξ 1 ; only the massless V a µ remain un-frozen. In other words, we letΦ(x) ≡ 〈Φ〉 = C × 1 N×N , B µ (x) ≡ 0, L a µ(x) = R a µ(x) = 1 √2V a µ (x),(S.67)and then substitute these values into the Lagrangian (4.9). According to eq. (S.62), for fieldsas in eq. (S.67) D µ Φ = 0, so the only un-frozen terms in the Lagrangian areL unfrozen = −2 1 tr(L µνL µν ) −2 1 tr(R µνR µν ) 〈〈 for L a µν = Rµν a 〉〉= − tr(L µν L µν ) = − 1 ∑( )La 22 µνa(S.68)= − 1 ∑( )Va 24 µνa— which is precisely the Yang-Mills Lagrangian for the canonically normalized tension fieldsV a µν = La µν + R a µν√2→ √ 2L a µν when L a µν = R a µν . (S.69)of the un-broken SU(N) V gauge theory. Indeed, in terms of the canonically normalized12

SU(N) V potential fields V a µ ,Vµν a = √ ()2 ∂ µ L a ν − ∂ ν L a µ − gf abc L b µL c ν= √ (Vνa Vµa 2 ∂ µ √ − ∂ ν √ − gf abc V )µb Vνc√ √2 2 2 2(S.70)= ∂ µ V aν − ∂ ν V a µ − g √2f abc V b µ V cν .The coefficient of the non-abelian last term on the bottom line is the gauge coupling of theunbroken SU(N) V gauge groupg v = g √2. (S.71)Problem 3(c):First, let me check that any Φ(x) is gauge-equivalent to a hermitianΦ ′ (x) ≡ e iθ(x) U L (x)Φ(x)U † R (x) = Φ′† (x)(S.72)for some SU(N) matrices U L (x) and U R (x) and phase θ(x). Furthermore, we may realizethis equivalence using only the broken part of the symmetry, namely the phase θ and thedifference between the U L and the U R ; together, they may be combined into a single unitarymatrixW (x) = e iθ(x) U L (x)U † R (x).(S.73)Indeed, in terms of W (x), the gauge transform (S.72) works asΦ ′ (x) = W (x) × U R (x)Φ(x)U † R (x),(S.74)and by the polar decomposition theorem, any complex matrix U R ΦU † Rproduct of a unitary matrix W −1 and a hermitian matrix H,can be written as aU R ΦU † R = W −1 H =⇒ Φ ′ = H = Φ ′† . (S.75)Moreover, this transform is non-singular for any invertible matrix Φ, i.e. for det(Φ) ≠ 0 —which is certainly true for Φ(x) = 〈Φ〉 + a small perturbation. To see how this works, let13

˜Φ = U R ΦU † R and note that ˜Φ†˜Φ = UR Φ † ΦU † R(S.76)is a hermitian, positive matrix (for det Φ ≠ 0) so it has a well defined, non-singular squarerootH =) 1/2. (˜Φ†˜Φ(S.77)Now let’s defineW = H × ˜Φ −1 −→ U R ΦU † R ≡ ˜Φ = W −1 H, (S.78)and check the unitarity of W :W † W =(˜Φ−1 ) †HH ˜Φ−1 =(˜Φ−1 ) †˜Φ†˜Φ˜Φ−1 = 1,so W is indeed a unitary matrix. And by construction, both H and W are non-singularfunctions of the Φ matrix when det Φ ≠= 0.Now let’s see which gauge symmetries preserve the hermiticity of the Φ(x). Clearly, theunbroken gauge symmetries with θ ≡ 0 and U L ≡ U R ≡ U turn any hermitian Φ(x) into ahermitianΦ ′ (x) = U(x)Φ(x)U † (x),(S.79)so the un-broken gauge symmetries are NOT fixed by the Φ ≡ Φ † gauge. On the other hand,the broken gauge symmetriesΦ ′ (x) = W (x)Φ(x)(S.80)generally do not preserve the hermiticity of Φ. Moreover, no infinitesimal symmetry of theform (S.80) can preserve the hermiticity of Φ(x) that has positive eigenvalues — which shouldbe true for Φ(x) = C × 1 N×N + a small perturbation — so all the infinitesimal broken gauge14

symmetries are indeed fixed by the condition Φ † ≡ Φ. To see that, let W (x) = 1 + iΛ(x) forinfinitesimal hermitian Λ(x). ThenΦ ′† − Φ ′ = W Φ − ΦW † = Φ + iΛΦ − Φ + iΦΛ = i{Λ, Φ}, (S.81)so to preserve the hermiticity of Φ(x), the Λ(x) should anticommute with it. But in thebasis where Φ is diagonal, the anti-commutator has matrix elements{Λ, Φ} ij = (φ i + φ j ) × Λ ij , (S.82)and since the eigenvalues φ i are all positive, all φ i + φ j ≠ 0, so the anti-commutator {Λ, Φ}cannot vanish unless Λ = 0. Thus, no infinitesimal symmetry W (x) = 1+iΛ(x) can preservethe hermiticity of Φ(x), so the gauge condition Φ † ≡ Φ indeed fixes all the broken gaugesymmetries.Problem 3(d):Let’s start with the non-abelian vector fields. In terms of V µ and X µ ,L µ = 1 √2(V µ + X µ ) ,L µν = 1 √2V µν + 1 √2(∂ µ X ν − ∂ ν X µ ) + ig 2 ([V µ, X ν ] − [V ν , X µ ]) + ig 2 [X µ, X ν ]where= 1 √2V µν + 1 √2(ˆDµ X ν − ˆD ν X µ)+ ig 2 [X µ, X ν ]V µνdef[ ]= ∂ µ V ν − ∂ ν V µ + ig v Vµ , V ν(S.83)for g V = g √2(S.84)in accordance with eq. (S.70), while ˆD µ are derivatives covariant with respect to the unbrokenSU(N) V ,ˆD µ X = ∂ µ X + ig V [V µ , X] for any adjoint X. (S.85)Similarly,R µν = 1 √2V µν − 1 √2(ˆDµ X ν − ˆD ν X µ)+ ig 2 [X µ, X ν ]. (S.86)15

Combining eqs. (S.83) and (S.86) together, we findL YM= − 1 2 tr( L µν L µν) − 1 2 tr( R µν R µν)= − 1 2 tr ( (Vµν+ ig V [X µ , X ν ] ) 2 ) − 1 2 tr ( ( ˆDµ X ν − ˆD ν X µ) 2)after a bit of algebra( (= −2 1 tr (V µνV µν )() − tr ˆDµ X ˆDµ ν X ν)) ( (+ tr ˆDµ X µ) ) 2− 2ig V tr ( V µν [X µ , X ν ] ) −2 1g2 V tr( −[X µ , X ν ] [X µ , X ν ] )= − 1 4∑a+ g V2(Vaµν) 2 −12∑∑f abc VµνX a bµ X cνabca( ( ˆDµ Xνa ) 2 ( )− ˆDµ Xµ) a 2+ g2 V4∑f abc f ade XµX b νX c dµ X eν .abcde(S.87)Altogether, we have covariant kinetic energy terms for the V and X vector fields as well asinteractions between V ’s and X’s and also among X’s themselves.Next, considerD µ Φ = ∂ µ Φ + ig V (V µ + X µ )Φ − ig V Φ(V µ − X µ ) + ig ′ B µ Φ= ∂ µ Φ + ig V [V µ , Φ] + ig V {X µ , Φ} + ig ′ B µ Φ(S.88)= ˆD µ Φ + ig V {X µ , Φ} + ig ′ B µ Φ.In the unitary gauge where the Φ(x) matrix is always hermitian, the first term on the lastline of (S.88) is hermitian while the other two terms are anti-hermitian. Consequently,()tr (D µ Φ † )(D µ Φ)(= tr ( ˆD µ Φ) 2) ( (gV+ tr {X µ , Φ} + g ′ B µ Φ ) ) 2. (S.89)The first term on the right hand side here provides SU(N) V –covariant kinetic energies forthe scalar fields, while the second terms contains scalar-dependent masses for the vectorfields B µ and Xµ. a To make this manifest, we shift Φ(x) by 〈Φ〉 and then expand into real(in the unitary gauge) components according toΦ(x) = C × 1 N×N +ξ(x) √2N× 1 N×N + 1 2∑ϕ a (x) × λ aa(S.90)16

In terms of these components,(tr ( ˆD µ Φ) 2) = 1 2 (∂ µξ) 2 + ∑ a12 ( ˆD µ ϕ a ) 2 (S.91)while( (gVtr {X µ , Φ} + g ′ B µ Φ ) ) 2= 1 2 M BB(ξ, ϕ) × B µ B µ + ∑ a+ 1 2M a BX (ξ, ϕ) × Xa µB µ∑a,bM abXX (ξ, ϕ) × Xa µX bµ (S.92)whereM BB (ξ, ϕ) = g ′2 (( √2NC + ξ) 2(M a BX (ξ, ϕ) = g′ g V(2 C + √ ξ )× ϕ a + ∑2Nbc)∑+ (ϕ a ) 2 , (S.93)ad abc ϕ b ϕ c ), (S.94)M abXX (ξ, ϕ) = g2 V⎛⎜⎝(C +ξ √2N) 2× δ ab +(C + √ ξ )× ∑2Nc+ 12N ϕa ϕ b+ ∑ cde⎞d abc ϕ c⎟d acd ϕ d d bce ϕ e ⎠ .(S.95)In the last two formulae here, I’ve used{λ a , λ b } = 2 N δab + 2 ∑ cd abc λ c(S.96)where d abc constants are totally symmetric in all three adjoint indices a, b, c. They exist forSU(N) theories with N ≥ 3, but not for the SU(2).Finally, there is the scalar potential (3). It’s decomposition in terms of Φ(x) − 〈Φ〉 worksexactly as in problem 2(d), except that in the unitary gauge there is no ϕ 2 . Expanding ϕ 1 (x)17

into components ξ(x) and ϕ a (x), we getV = const + V 2 + V 3 + V 4 ,V 2 = (α + Nβ)C 2 × ξ 2 + αC 2 × ∑ a(ϕ a ) 2 ,V 3= α + Nβ √2NC × ξ 3 +d abc ϕ a ϕ b ϕ c ,(S.97)V 4= α + Nβ8N × (ξ 2 + ∑ a3α + Nβ√2NC × ∑ a(ϕ a ) 2 ) 2+ α 4 × ∑ aξ(ϕ a ) 2 + αC 2 × ∑ abc(√2N ξϕa + ∑ bcd abc ϕ b ϕ b ) 2.Altogether, expanding the Lagrangian (8) in terms of fields of definite mass and theirSU(N) V –covariant derivatives yieldsL = (S.87) + (S.91) + (S.92) + (S.97).(S.98)18