The Bethe/Gauge Correspondence

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46 Chapter 3. Gaugeunder U(1) G as ¯Φ¯f ↦→ ¯Φ¯f e −iΛ , with the vector superfield acting from the right. In the presentcontext, with G = U(1), this distinction is of course a bit pedantic: ¯Φ¯f e −iΛ is the same ase −iΛ ¯Φ¯f in abelian theory. However, our convention pays out when we look at nonabelian gaugegroups, where the distinction does matter.Thus, the Lagrangian for two-dimensional sqed is given byL sqed = L kin,f + L kin,¯f + L gauge + L W∫= d 4 θ(¯Φf e 2V Φ f + ¯Φ¯f e −2V Φ¯f − 14 e 2 ¯Σ Σ) ( ∫ )+ d 2 θ m Φ f Φ¯f + h.c. .where m ∈ C is a complex mass. (For ‘realistic’ sqed we should of course take m ∈ R >0 .)The component expansion again involves gauge covariant derivatives, ∇ µ = ∂ µ ± i A µ , andYukawa-type interactions.Notice that gauge invariance requires the superpotential to involve an equal numbers offundamental and anti-fundamental chiral superfields. Now we also understand why the gaugecoupling appears in the denominator: it is the result of the rescaling of the vector superfield.Adding flavour. For the Bethe/gauge correspondence, we need a theory that is richer thansqed. We want to include several flavours of charged matter.Since choosing different charges for some of the fields breaks down the flavour symmetrygroup to a subgroup, flavour symmetry requires all flavours to have the same charge. Take L fflavours of fundamental fields Φ l fwith corresponding antichiral fields ¯Φ f,l = (Φ l f )† , 1 ≤ l ≤ L f .Because of our convention to treat anti-fundamental as antichiral superfields, it makes sense todenote them with a lower flavour index: ¯Φ¯f,k . We take L¯f of them and denote the conjugatefields by Φ k¯f , with 1 ≤ k ≤ L¯f.The flavour symmetry group is H max = ( U(L f ) × U(L¯f) ) /U(1) G . Here we have to take thequotient by the ‘diagonal’ action of the gauge group which simultaneously transforms the matterfields by Φ l f ↦→ eiΛ Φ l f and Φk¯f ↦→ Φ k¯f e −iΛ [17, p. 5].The total Lagrangian is a generalization of L sqed , where we also include the twisted superpotential:L = L kin,f + L kin,¯f + L gauge + L W + L∫˜W= d 4 θ(¯Φf,l e 2V Φ l f + ¯Φ¯f,k e −2V Φ k¯f − 14 e ¯Σ)2 Σ( ∫+ d 2 θ W (Φ f , ¯Φ¯f)∫∣+ d 2 ϑ ˜W )(Σ) ∣ + h.c.∣¯θ± =0 ϑ± =0(3.41)The superpotential W has to be a holomorphic gauge-invariant function. Of particular interestis the generalization of the complex mass term (3.15), which reads∫L m = d 2 θ ¯Φ¯f,k m k l Φ l f + h.c. , (3.42)with m k l ∈ C. This term is gauge invariant (provided L f , L¯f > 0) but generically breaks theglobal flavour symmetry group H max down to a subgroup H. To see what this residual flavoursymmetry group is, consider a flavour symmetry transformationΦ l f ↦→ (U f ) l l ′Φl′ f , Φ k¯f ↦→ Φ¯f,k ′(U¯f) k′ kwhere U f ∈ U(L f ) and U¯f ∈ U(L¯f). This transformation leaves (3.42) invariant if the equalitym k l = (U¯f) k k ′ mk′ l ′ (U f) l′ l (3.43)holds for generic complex masses. This requires(U¯f) k k ′ ∝ δk k ′ , (U f) l l ′ ∝ δl l ′ ,
3.3. Supersymmetric abelian gauge theory 47so both of the matrices have to be diagonal. Unitarity allows the diagonal elements to be acomplex phase. To satisfy (3.43) we can pair up phase factors of U f with complex conjugatefactors for U¯f. In this way, we are free to choose min(L f , L¯f) phases in total. Thus, the residualflavour symmetry group H is given by a subgroup of the maximal torus of H max isomorphic toU(1) min(L f,L¯f) /U(1) G∼ = U(1)min(L f ,L¯f)−1 .Twisted masses revisted. A manifestly supersymmetric way to introduce the twisted massterms (3.21) is as follows [17, §2]. Consider again sqed with L f fundamental matter fields andL¯f anti-fundamentals. In the absence of a superpotential this theory has a global symmetry withsymmetry group H max as above. We can gauge this group as well, and introduce a correspondingnonabelian vector superfield V ′ . The idea is to ‘weakly gauge’ H max , which means that we‘freeze’ V ′ to a background expectation value.We don’t have to know much about the nonabelian case to see what will happen. It isagain possible to go to the Wess-Zumino gauge (3.28). Since the vector component field A ′ µtransforms in the adjoint representation, it takes values in the Lie algebra h of H max . Supersymmetryrequires the other component fields to be h-valued as well, so we may think of themas square matrices of size L f + L¯f − 1. To see which constant values of the component fields ofV ′ are compatible with supersymmetry we examine the supersymmetry transformations of thecomponent fields for nonabelian gauge groups [14, §4.1]:δ ε A ′ 0 = i (¯ε − λ ′ − + ¯ε + λ ′ +) + h.c. ,δ ε A ′ 1 = i (¯ε − λ ′ − − ¯ε + λ ′ +) + h.c. ,δ ε σ ′ = − √ 2i (¯ε + λ ′ − + ε −¯λ′ + ) ,(δ ε λ ′ + = i ε + D ′ + i F 01 ′ +i )2 √ 2 [σ′ , ¯σ ′ ] + √ 2 ε − (∇ 0 + ∇ 1 ) ¯σ ′ ,(δ ε λ ′ − = i ε − D ′ − i F 01 ′ −i )2 √ 2 [σ′ , ¯σ ′ ] + √ 2 ε + (∇ 0 − ∇ 1 ) σ ′ ,δ ε D ′ = ε − (∇ 0 + ∇ 1 ) ¯λ ′ − + ε + (∇ 0 − ∇ 1 ) ¯λ ′ + + ε − [¯σ ′ , ¯λ ′ +] + ε + [σ ′ , ¯λ ′ −] + h.c.(3.44)The first three lines are the same as in the abelian case (3.29), with ‘¯λ ′ +’ now denoting theHermitian conjugate of λ ′ + in h. The last three lines contain covariant derivatives for H andadditional commutators in h.We want to make all fields constant, so the derivatives drop out and Lorentz invariancerequires A ′ µ = 0. Thus we want to solve:0 = δ ε A ′ 0 = i (¯ε − λ ′ − + ¯ε + λ ′ +) + h.c. ,0 = δ ε A ′ 1 = i (¯ε − λ ′ − − ¯ε + λ ′ +) + h.c. ,0 = δ ε σ ′ = − √ 2i (¯ε + λ ′ − + ε −¯λ′ + ) ,(0 = δ ε λ ′ ± = i ε ± D ′ ±i )2 √ 2 [σ′ , ¯σ ′ ] ,0 = δ ε D ′ = ε − [¯σ ′ , ¯λ ′ +] + ε + [σ ′ , ¯λ ′ −] + h.c.As the variations of A ′ µ, σ ′ and D ′ are all linear in λ ′ , we have to take λ ′ ± = ¯λ ′ ± = 0. Thisrequires D ′ = [σ ′ , ¯σ ′ ] = 0. We see that only σ ′ can be nonzero, provided that the matrix of σ ′is normal (commutes with its Hermitian conjugate), which is equivalent to requiring σ ′ to bediagonalizable.Thus, instead the entire flavour group, we only weakly gauge its maximal torus( )U(1)L f× U(1) L¯f /U(1)G ⊆ H . (3.45)This yields U(1)-vector superfields which we denote by Ṽ fl and Ṽ ¯f k . Their only non-vanishingcomponent fields are the complex scalars, fixed to the background values ˜σf l = ˜m l f∈ C and
Page 1 and 2:
Institute for Theoretical PhysicsMa
Page 4 and 5: 4 Bethe/gauge 614.1 The corresponde
Page 6 and 7: viPrefaceA note on notationMany asp
Page 9 and 10: Chapter 1OverviewIn quantum field t
Page 11 and 12: 1.1. Bethe: quantum integrable mode
Page 17 and 18: 1.2. Gauge: supersymmetry 9providin
Page 19 and 20: 1.2. Gauge: supersymmetry 11Because
Page 21 and 22: 1.2. Gauge: supersymmetry 13The def
Page 23 and 24: 1.3. Physics in two dimensions 15be
Page 25 and 26: 1.4. Bethe/gauge: the general idea
Page 27 and 28: Chapter 2BetheIn this chapter we ta
Page 29 and 30: 2.1. Algebraic Bethe Ansatz 21on ou
Page 31 and 32: 2.1. Algebraic Bethe Ansatz 23To ge
Page 33 and 34: 2.1. Algebraic Bethe Ansatz 25Accor
Page 35 and 36: 2.2. Generalizations of the xxx s m
Page 37 and 38: 2.3. Yang-Yang function 292.2.3 Fur
Page 39: 2.3. Yang-Yang function 31To see th
Page 42 and 43: 34 Chapter 3. GaugeThe Poincaré gr
Page 44 and 45: 36 Chapter 3. Gaugeexpansion we wil
Page 46 and 47: 38 Chapter 3. Gaugewhere we can in
Page 48 and 49: 40 Chapter 3. GaugeHere K is a real
Page 50 and 51: 42 Chapter 3. GaugeVector superfiel
Page 52 and 53: 44 Chapter 3. GaugeThus, L r is pre
Page 56 and 57: 48 Chapter 3. Gauge˜σ k¯f= ˜m k
Page 58 and 59: 50 Chapter 3. GaugeVacuum manifolds
Page 60 and 61: 52 Chapter 3. GaugeThis (almost) is
Page 62 and 63: 54 Chapter 3. GaugeLooking at the t
Page 64 and 65: 56 Chapter 3. Gaugethe effective tw
Page 66 and 67: 58 Chapter 3. GaugeMore matter repr
Page 69 and 70: Chapter 4Bethe/gaugeNow that we hav
Page 71 and 72: 4.1. The correspondence 63the spin
Page 73 and 74: 4.2. Further topics 654.1.3 Diction
Page 75: 4.2. Further topics 67N = 2 symin f
Page 78 and 79: 70 Chapter 4. Bethe/gauge• descri
Page 80 and 81: [17] N. Dorey, T. Hollowood, and D.
Page 82: [56] U. Lindström, “Supersymmetr
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The Bethe/Gauge Correspondence

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