PHYSICS 218 PROBLEM SET #2 BEN LEHMANN

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1 PHYSICS 218 PROBLEM SET #2 BEN LEHMANN 1. Our Lagrangian is 1 4 F µνf a aµν, so let s start by working out the transformation law for Fµν a µ A a ν ν A a µ + gf ab A b µa ν. Taking A a µ A a µ + 1 g µα a f ab α b A µ, the field strength transforms as Fµν a Fµν a = µ A a ν + 1 g να a f ab α b A ν ν A a µ + 1 g µα a f ab α b A µ gf ab A b µ + 1 g µα b f bde α d A e µ A ν + 1 g να f gh α g A h ν Expanding in full, this is Fµν a = Fµν a + 1 g µ ν α a 1 g ν µ α a f ab µ α b A ν + f ab ν α b A µ f ab[ A ν µ α b + A b µ ν α + 1 g µα b ν α f gh α g A h ν µ α b gf gh α g A h νa b µ gf bde α d A e µa ν f bde α d A e µ ν α + gf bde f gh α d α g A e µa h ν We an remove two terms immediately from ommutativity of partial derivatives. Next, sine we are working with an infinitesimal transformation, we disard all terms of Oα 2. This leaves Fµν a = Fµν a +f ab[ ν α b A ] µ µ α b A ν +A ν µ α b A b µ ν α gf gh α g A h νa b µ gf bde α d A e µa ν 3 Expanding the derivatives and aneling, we have Fµν a = Fµν a + f ab α b ν A µ α b µ A ν gf gh α g A h νa b µ gf bde α d A e µa ν Now onsider the last two terms of eq. 4. We have a sum of two terms ontaining two fators of f ab, so we shall attempt to simplify to one term using the Jaobi identity. We have gα g f ab f gh A b µa h ν + gα d f ab f bde A e µa ν = gα d f ab f de A b µa e ν + gα d f ab f bde A e µa ν 5 ] 4 = gα d f ab f de A b µa e ν + gα d f ae f db A b µa e ν 6 = gα d f ab f de + f ae f db A b µa e ν 7 With some relabeling, we an apply the Jaobi identity to eq. 7, finding gα d f ab f de + f ae f db A b µa e ν = gα b f ab f de A d µa e ν 8 We now insert this result into eq. 4, whih yields a very satisfying transformation rule for the field strength: Fµν a = Fµν a + f ab α b ν A µ α b µ A ν gα b f de A d µa e ν 9 = F a µν f ab α b F µν 10 1

2 PHYSICS 218 PROBLEM SET #2 2 So equipped, let us now work out the transformation of the full kineti term. We have L 1 4 F a = 1 4 Fµν a f ab α b Fµν = L 1 4 µνf aµν 11 F aµν f ade α d F eµν 12 α b f ab F µνf aµν α d f ade F a µνf eµν + α b α d f ab f ade F µνf eµν 13 We disard the Oα 2 term, and the remaining two terms anel after interhanging indies. We are left with L L, so we have verified the gauge invariane of the Lagrangian. 2. a We use a defining relation of the fundamental representation: T a T b = 1 2n δab 1 n n if ab + d ab T 14 If we take the trae of this equation, we get tr T a T b = 1 2n δab tr 1 n n if ab + d ab tr T 15 Sine the T are the generators of speial unitary matries, we know that expit is unitary with determinant 1, i.e., det expit = 1. But det expit = exp trit, so exp trit = 1, whih is only possible if trit = 0. But then the T are all traeless, so eq. 16 beomes simply tr T a T b = 1 2n δab tr 1 n n = 1 2 δab 16 b Written out in full, we wish to show that i 2 f ad f bde i 2 f bd f ade = if abd if de, or f bd f ade f ad f bde = f abd f de 17 We shall see that this is nothing but a rearranged form of the Jaobi identity. Swap indies in the seond term so that the summation is arried out on the first index of one f ab and the last the other. With some additional swaps, our desired statement is f bd f ade + f ad f dbe + f bad f de = 0 18 Now ylially permute b, d in bad; our statement now reads f bd f ade + f ad f dbe + f dba f de = 0 19 Now swap d, a in the first term,, a in the seond term, and d, a in the third term. Caneling signs and rearranging, our desired statement is equivalent to But this is exatly the Jaobi identity. f abd f de + f bd f dae + f ad f dbe = a The kineti term is already gauge-invariant, so we onsider the gauge-fixing term on its own: L GF 1 2 µ A a µ + µ ɛdµ ab ω b 21 2a = 1 2 µ A a 2a µ 2 + 2ɛ µ A a µ ν Dν ab ω b + ɛ 2 µ Dµ ab ω b 22 The first term is just L GF. Working to Oɛ, we disard the third term, whih leaves L GF L GF ɛ a µ A a µ ν D ab ν ω b }{{} δl GF 23

3 = η a µ Dµ ab ω b }{{} L G PHYSICS 218 PROBLEM SET #2 3 We an t eliminate the remaining term, so the gauge-fixing term is not gauge-invariant. b To implement the gauge transformation of L G, we must note that Dµ ab itself transforms. We begin with L G = η a µ δ ab µ gf ab A µω b, and transform as [ L G η a + δη a µ δ ab µ gf ab A µ + δa ] µ ω b 24 = η a ɛ a ν A a ν µ Dµ ab ω b gf ab µ δa µω b 25 +gf ab η a µ δa ɛ µω b + a ν A a ν µ Dµ ab ɛg ω b a f ab ν A a ν µ δa µω b 26 } {{ } δl GF We an now read off the transformation of L GF + L G as δ L GF + L G = gf ab η a µ δa µω b ɛg a f ab ν A a ν µ δa µω b In partiular, note that this transformation would be trivial if we did not need to aount for δa a µ in the ovariant derivatives. Sine δa a µ is Oɛ, the seond term of eq. 27 is Oɛ 2, and we disard it. We are left with Our Lagrangian is now 27 δ L GF + L G = gɛf ab η a µ D d µ ω d ω b 0 28 L = L YM + L GF + L G = 1 4 F a µνf aµν 1 2a µ A a µ 2 η a µ D ab µ ω b 29 The kineti term remains gauge-invariant, and sine ω a does not expliitly appear in L GF, the only term whose transformation is affeted is L G. Before we implement the transformation in the Lagrangian, let us study the transformation of the ovariant derivative itself. The ovariant derivative only appears expliitly as Dµ ab ω b, whih transforms as Dµ ab ω b δ ab µ gf ab A µ + δa µ ω b + δω b 30 = D ab µ ω b + µ δω a gf ab A µ δω b gf ab δa µ ω b 31 Now we an write δ Dµ ab ω b as δ Dµ ab ω b = 1 2 ɛgf ab µ ω b ω 1 2 ɛg2 f ab f bde A µω d ω e ɛgf ab Dµ d ω d ω b 32 = 1 2 ɛgf ab µ ω b ω ɛgf ab ω b µ ω 1 2 ɛg2 f ab f bde A µω d ω e ɛgf ab D d µ ω d ω b 33 Now observe that sine the ω a antiommute, f ab µ ω b ω = f ab ω µ ω b = f ab ω µ ω b = f ab ω b µ ω. Thus we an ombine terms to get δ Dµ ab ω b = ɛgf ab µ ω b ω 1 2 ɛg2 f ab f bde A µω d ω e ɛgf ab Dµ d ω d ω b 34 Now we an use the Jaobi identity in the middle term: we have f ab f bde = f ab f bde = f abd f de = f bd f dae f ad f dbe 35 But observe that f bd f dae A µω d ω e = f ad f dbe A µω d ω e under a swap a and a relabeling b a. Making this substitution in eq. 34 gives δ Dµ ab ω b = ɛgf ab µ ω b ω ɛg 2 f bd f dae A µω d ω e ɛgf ab Dµ d ω d ω b 36

4 PHYSICS 218 PROBLEM SET #2 4 In the middle term, swapping and relabeling indies gives δ Dµ ab ω b = ɛgf ab δ d µ ω d ω b gf de A e µω d ω b ɛgf ab Dµ d ω d ω b 37 = ɛgf ab D d µ ω d ω b ɛgf ab D d µ ω d ω b = 0 38 So, with our new transformation rule, Dµ ab ω b is itself gauge-invariant. Now implementing the transformation in the Lagrangian is easy: we need only transform η a, whih gives δl = δl GF + δη a µ Dµ ab ω b 39 = ɛ a µ A a µ ν D ab ν ω b + ɛ a νa ν a µ D ab µ ω b = 0 40 Thus our Lagrangian is indeed BRST-invariant, and the day is saved! d First onsider δ 2 A a µ, whih is given by δ 2 A a µ = δ ɛ 1 Dµ ab ω b = ɛ 1 δ Dµ ab ω b But in part, we found that δ Dµ ab ω b = 0. Thus δ 2 A a µ = 0. We an do something similar for δ 2 η a, whih has the form δ 2 η a = δ ɛ 1 a µa µ a = ɛ 1ɛ 2 a µ D ab µ ω b 43 Generially, this is non-zero. But suppose we require that the Lagrange equations of motion are satisfied. Rewriting the Lagrangian in an equivalent form, we have L G = η a δ ab µ µ ω b gf ab µ A µω b 44 = η a µ µ ω a + gη a f ab µ A µω b 45 = µ η a µ ω a + gη a f ab µ A µω b 46 where we have integrated by parts in the last step. Now we an find the equations of motion: we have L = gf ab µ A L η µω b µ a µ η a = µ µ ω a = δ ab µ µ ω b 47 so the equations of motion are µ δ ab µ ω b gf ab A ω b = 0 48 but this is just µ D ab µ ω b = 0. Substituting in eq. 43 gives δ 2 η a = 0. e The equations of motion are easy to derive: L L µ B a µ B a = ab a + µ A aµ = 0 49 Thus B a = 1 a µa aµ, and if we substitute into the Lagrangian, we obtain L GF = 1 a µa µ a a µa µ a 2 = 1 2a µa µ a 2 50 So indeed, the auxiliary fields reprodue the gauge-fixing term we used before.

5 f Our Lagrangian is now PHYSICS 218 PROBLEM SET #2 5 L = L YM + B a µ A µ a ab ab a η a µ D ab µ ω b Clearly L YM will still be gauge-invariant, sine we have not hanged the presription for transforming A a µ. Under a gauge transformation, the remainder of the Lagrangian has δl = B a µ δa µ a ɛgb a f ab µ δa µ ωb + A µ δω b + δa µ δωb 52 where we have omitted δb a, sine δb a = 0. All terms but the first are Oɛ 2, and the first term is a total derivative; hene δl = 0, and the Lagrangian remains BRST-invariant. Now let us onsider δ 2 applied to the four fields of our theory: A a µ, η a, ω a, and B a. Sine δa a µ is independent of B a and η a, the argument of part d still holds to show that δ 2 A a µ = 0. That δ 2 B a = 0 is trivial, sine δb a = 0. We similarly have δ 2 η a = ɛδb a = 0. It remains only to show that δ 2 ω a = 0, and while this is not modified by the introdution of the auxiliary fields, we did not onsider this field in d. Expliitly, we have Now we an strategially relabel: 51 δ 2 ω a = 1 2 ɛ 1gf ab δ ω b ω 53 = 1 2 ɛ 1gf ab δω b ω + ω b δω 54 = 1 4 ɛ 1ɛ 2 g 2 f ab f bde ω d ω e ω + f de ω b ω d ω e 55 f ab f bde ω d ω e ω + f ab f de ω b ω d ω e = f ab f bde ω d ω e ω + f ab f de ω d ω e ω b 57 and thus δ 2 ω a = = f ab f bde ω d ω e ω + f ab f de ω d ω e ω b 58 = f ab f de ω d ω e ω b + f ab f de ω d ω e ω b = 0 59