The Bethe/Gauge Correspondence

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30 Chapter 2. BetheWe’ll get back to the m n momentarily. The potential Y (λ) bears the name Yang-Yang functionor Yang-Yang action after its discoverers Yang and Yang [48, §4] (see also [49]). See also §I.2,II.1 and the introduction to Chapter X of [25].Nekrasov and Shatashvili give two equivalent expressions for Y (λ). The first one looks a bitcomplicated [1]: 4Y (λ) = i πL∑N∑l=1 n=1( λn − ν)ls l ˆx − is l 2πˆx(λ) := λ arctan 1 λ + 1 2 log(1 + λ2 ) .N∑ˆx(λ n − λ m ) − iN 2π ϑ ∑λ n ,n,m=1n=1(2.27)Since the λ n are complex, the functions ˆx(λ) and Y (λ) are multivalued. The integers m n in(2.26) label the different branches of the complex logarithms. They can be absorbed by shiftingthe Yang-Yang function to Y ⃗m (λ) := Y (λ) − i ∑ λ n m n , for which (2.26) really corresponds tocritical points. Alternatively, we can get rid of the m n by exponentiation:(exp 2π ∂ Y (λ) )= 1 . (2.28)∂λ nBefore we check that these equations give rise to the bae (2.25) let’s work on (2.27) for abit. First notice thatˆx ′ (λ) = arctan 1 λ = 1 2i log λ + iλ − i ,where the second equality can be checked by solving z = tan w for w. 5Defineϖ 1 (λ) := (λ + i) log(λ + i) − (λ − i) log(λ − i)(the notation will make more sense soon). Since 2i ˆx ′ (λ) = ϖ ′ 1(λ), ˆx and ϖ 1 differ by at mostan additive constant. To find this constant, let’s evaluate both expressions at e.g. λ = 0. Sincethe arctangent is bounded, ˆx(λ) → 0 as λ → 0. But ϖ 1 (λ) → 0 in that limit too, so they’reactually equal: 2i ˆx(λ) = ϖ 1 (λ). (This can also be checked directly.)We can use this to rewrite the first part of Y (λ) in (2.27):2i s ˆx(λ/s) = (λ + i s) log(λ/s + i) − (λ − i s) log(λ/s − i)= (λ + i s) log(λ + i s) − (λ − i s) log(λ − i s) − 2i s log s . (2.29)Here the 1/s can be extracted from the logarithm because s ∈ 1 2N is a positive real number.The resulting last term in (2.29) is a constant which is not relevant for the potential Y (λ), andcan safely be dropped.For the next part of Y (λ) further note that ˆx is an even function, so thatN∑n,m=1ˆx(λ n − λ m ) = 2N∑ˆx(λ n − λ m ) , (2.30)n
2.3. Yang-Yang function 31To see that Y (λ) really is the potential of the bae we compute∂∂λ jN ∑n
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: 2.3. Yang-Yang function 292.2.3 Fur
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 54 and 55: 46 Chapter 3. Gaugeunder U(1) G as
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

The Bethe/Gauge Correspondence

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