Testing the Bethe ansatz with large N renormalons
PPrepared for submission to JHEP
Testing the Bethe ansatz with large N renormalons Marcos Mari˜no, Ramon Miravitllas Mas and Tom´as Reis
D´epartement de Physique Th´eorique et Section de Math´ematiquesUniversit´e de Gen`eve, Gen`eve, CH-1211 Switzerland
E-mail:
[email protected] , [email protected] , [email protected]
Abstract:
The ground state energy of integrable asymptotically free theories can be conjec-turally computed by using the Bethe ansatz, once the theory has been coupled to an externalpotential through a conserved charge. This leads to a precise prediction for the perturbativeexpansion of the energy. We provide a non-trivial test of this prediction in the non-linear sigmamodel and its supersymmetric extension, by calculating analytically the associated Feynman di-agrams at next-to-leading order in the 1 /N expansion, and at all loops. By investigating thelarge order behaviour of the diagrams, we locate the position of the renormalons of the theoryand we obtain an analytic expression for the large N trans-series associated to each. As a spin-off of our calculation, we provide a direct derivation of the beta function of these theories, atnext-to-leading order in the 1 /N expansion. a r X i v : . [ h e p - t h ] F e b ontents /N expansion of the effective potential 8 N renormalons and their trans-series 185 The supersymmetric non-linear sigma model 22 (cid:98) B m,(cid:96)
31B Analytic computation of the integrals 32C Asymptotic expansions of the discontinuities 34
In quantum field theory, the conventional perturbative series explores just the fluctuations aroundthe trivial vacuum. It is believed that a full picture should involve additional sectors, corre-sponding to non-trivial vacua, and leading to exponentially suppressed contributions to physicalobservables. The origin of these sectors might be additional semi-classical saddles, also calledinstantons, but in many cases, like in Yang–Mills theory, the most important non-perturbativeeffects are associated to elusive objects called renormalons [1].The contribution of instanton sectors can be obtained in principle by expanding the pathintegral around these non-trivial saddle-points. In the case of renormalons, we do not havesuch a semi-classical picture. It is known however that renormalons manifest themselves inthe large order behavior of perturbation theory [2, 3]. Therefore, one way to obtain precisequantitative information about renormalon sectors and their exponentially small corrections isto use conventional perturbation theory at large number of loops. This connection, in which anon-perturbative correction “resurges” in the perturbative series, is part of the general theoryof resurgence, which gives a systematic framework to understand non-perturbative sectors (see[4–8] for introductions). – 1 –n spite of some impressive lattice calculations [9] (see also [10]), unveiling the resurgentstructure of realistic quantum field theories remains a difficult problem. One could however tryto address these issues in more tractable quantum field theories. Among these, integrable fieldtheories in two dimensions play an important rˆole. On the one hand, they display many of thephysical phenomena of interest, like asymptotic freedom and the presence of renormalon sectors.On the other hand, they are much more tractable analytically. In particular, exact expressionsfor their S matrices have been conjectured in [11, 12].It was noted by Polyakov and Wiegmann in [13, 14] that, in integrable field theories, one cancalculate exactly the dependence of the ground state energy on a chemical potential coupled toa global conserved charge. This is done by combining the exact S -matrix with the Bethe ansatz,and the answer can be elegantly expressed in terms of a set of integral equations. One beautifulapplication of this observation was the determination of the exact mass gap of these theories,by comparing the Bethe ansatz to conventional perturbation theory (see [15–19] for various casestudies, and [20] for a review). The ground state energy as a function of the chemical potential,which we will call for short the free energy of the theory, is then an ideal observable to understandquantitatively the different sectors, perturbative and non-perturbative, of the theory.However, even extracting the perturbative expansion of the free energy from the Betheansatz answer remained a challenge for a long time (this is a generic difficulty of the Betheansatz, which goes back to the Lieb–Liniger solution of the interacting Bose gas in one dimension[21]). An algorithm to do this was finally found by Volin in [22, 23]. This opened the wayto a resurgent analysis of quantum integrable systems. The presence of renormalons in thenon-linear sigma model was tested numerically in [22]. In [24], this analysis was extended tomany two-dimensional asymptotically free theories, and some classical predictions of renormalonasymptotics were verified in detail. A similar resurgent study of various non-relativistic modelswas carried out in [25–27]. Very precise results on the resurgent structure of the free energy inthe O (4) sigma model have been recently obtained in [28, 29], where Volin’s method was used togenerate the perturbative series up to very large order.In this paper we study the free energy of the O ( N ) non-linear sigma model and its supersym-metric extension. Our first goal is a detailed test of the Bethe ansatz result against conventionalperturbation theory. So far, this has been verified up to two-loops in [30]. In order to do acomprehensive test, we calculate the free energy to all loops , at next-to-leading order in the 1 /N expansion. Our result matches the Bethe ansatz answer obtained in [22, 24] to all available or-ders, and provides very convincing evidence for its validity. It also provides additional evidencefor the conjectural S matrices (although these have been tested directly, up to next-to-leadingorder in the 1 /N expansion, in [31, 32]). We should note that, in order to make contact withperturbation theory, our 1 /N expansion is performed around the perturbative vacuum, and notaround the non-trivial large N vacuum where particles get a non-perturbative mass already attree level. In this sense, our calculation is very similar to what was done in [33]: we re-organizeperturbation theory in powers of 1 /N and keep the first two non-trivial contributions.An interesting spin-off of our calculation is a new derivation of the beta function of thenon-linear sigma model, at next-to-leading order in the 1 /N expansion. This is a known result,going back to [34]. However, in [34] the beta function was derived from the epsilon expansion ofthe critical exponents (see [35] for a review). Our derivation is a direct one, similar in spirit andin the details to the computation of Palanques-Mestre and Pascual of the QED beta functionin the limit of large number of fermions N f [36]. In the case of the supersymmetric non-linearsigma model, we verify the vanishing of the beta function at next-to-leading order in 1 /N , whichwas also derived from the epsilon expansion of critical exponents in [37].– 2 –n some cases, one can calculate the 1 /N expansion of the free energy directly from the Betheansatz equations [18, 38–40]. However, in the non-linear sigma model and its supersymmetricextension, the Bethe ansatz is not amenable to a 1 /N expansion, since the integral equationdetermining the free energy becomes singular when N → ∞ . Moreover, even if the Betheansatz result contains complete information about the ground state energy, including both theperturbative expansion and non-perturbative corrections, at present no known method exists toextract analytically the non-perturbative sectors, not even at large N . In contrast, our calculationof the perturbative series at next-to-leading order in 1 /N is fully analytic. By using techniquesdeveloped in [27, 33, 41], we can extract the trans-series at large N from our all-loop results.This trans-series signals the presence of an isolated infrared (IR) renormalon singularity , andan infinite sequence of ultraviolet (UV) renormalon singularities.In the supersymmetric extension of the non-linear sigma model, an important consequenceof our perturbative computation is an analytic proof, at next-to-leading order in 1 /N , that theIR singularity arising from bosonic diagrams does not cancel against the IR singularity arisingfrom fermionic diagrams. Therefore, the supersymmetric model exhibits IR renormalons . Thisis in contrast to the disappearance of leading IR renormalons in some supersymmetric theories,pointed out [42, 43].The organization of this paper is as follows. In section 2 we review basic aspects of thenon-linear sigma model, its 1 /N expansion, and its Bethe ansatz solution. Section 3 presents thecalculation of the effective potential at next-to-leading order in 1 /N . This result is then usedin section 4 to extract the large N renormalons and their trans-series. In 5 we extend all theseresults to the supersymmetric non-linear sigma model. Finally, in 6 we present our conclusionsand prospects for future developments. There are three Appendices with additional details andclarifications on our calculations. The O ( N ) non-linear sigma model is a quantum field theory in two dimensions for a vector field S ( x ) = ( S ( x ) , . . . , S N ( x )), satisfying the constraint S = 1 . (2.1)The Lagrangian density is L = 12 g ∂ µ S · ∂ µ S , (2.2)where g is the bare coupling constant. The non-linear sigma model is asymptotically free [44],and it can be regarded as a toy model for gauge theories. It also has many different applicationsin condensed matter physics, where it is used to model the low-energy dynamics of statisticalsystems with a global O ( N ) symmetry. We will write the beta function for the coupling constant g as β g ( g ) = µ d g d µ = − β g − β g − · · · , (2.3) In [22, 23] it was erroneously concluded, from the behavior of the very first terms of the perturbative series,that there are no IR renormalons at this order in 1 /N . In the first published version of [24], the numerical results for the Borel–Pad´e transform of the perturbativeseries of this model were interpreted as an indication that the first IR renormalon is absent. This is incorrect. Amore careful analysis of this series confirms the presence of this IR renormalon, also at finite N . – 3 –ith this convention, asymptotically free theories have β >
0. All of our perturbative calcula-tions will be done in the MS scheme. For the non-linear sigma model, the first two coefficientsof the beta function are [45] β = 14 π ∆ , β = 18 π ∆ , (2.4)where ∆ = 1 N − . (2.5)The beta function is known up to four loops [34, 46].The non-linear sigma model can be also studied in the limit in which the number of compo-nents of S is large and relevant quantities are computed in a large N expansion. In this setting,one introduces the ’t Hooft coupling λ = g π ∆ , (2.6)which is kept fixed in the large N limit. As we will see, it will be more natural to make theexpansion in powers of ∆, rather than in 1 /N .The large N expansion makes it possible to obtain non-perturbative results for this model(see e.g. [5]). In perturbation theory, one expands around an ordered vacuum with (cid:104) S (cid:105) (cid:54) = 0, inwhich the global O ( N ) symmetry is spontaneously broken. This leads to a perturbative spectrumconsisting of N − N , one finds a spectrum consistingof N massive particles in the fundamental representation of O ( N ), which is thought to be thetrue spectrum of the theory. It is important however to keep in mind that perturbation theoryaround the “false vacuum” gives the correct asymptotic expansion of O ( N ) invariant observables,as pointed out in [47, 48].Many quantities in the non-linear sigma model can be calculated systematically in a 1 /N expansion, like for example critical exponents. By using this method, one finds that the betafunction for the ’t Hooft coupling, defined as β ( λ ) = µ ∂λ∂µ , (2.7)has the following 1 /N expansion [34]: β ( λ ) = (cid:88) (cid:96) ≥ β ( (cid:96) ) ( λ )∆ (cid:96) , (2.8)where β (0) ( λ ) = − (cid:15)λ − λ , (2.9) β (1) ( λ ) = − λ (cid:90) λ d x sin( πx ) πx Γ(1 + x )Γ(1 + x ) x + 1 x + 2 , (2.10)and (cid:15) is related to the number of dimensions of the theory by d = 2 − (cid:15). (2.11)As a spin-off of the results in this paper, we will rederive the result for β (1) ( λ ) by a directcalculation in perturbation theory. – 4 – conjectural expression for the exact S -matrix of the two-dimensional non-linear sigmamodel was put forward in [11, 12]. This makes it possible the following exact computation [13].Let H be the Hamiltonian of the model, and let Q be a conserved charge, associated to a globalconserved current. Let h be an external field coupled to Q . h can be regarded as a chemicalpotential, and as usual in statistical mechanics we can consider the ensemble defined by theoperator H − h Q . (2.12)The corresponding free energy per unit volume is then defined by F ( h ) = − lim V,β →∞ V β log Tr e − β ( H − h Q ) , (2.13)where V is the volume of space and β is the total length of Euclidean time. This is the groundstate energy of the model in the presence of the additional coupling. As pointed out in [13], wecan compute δ F ( h ) = F ( h ) − F (0) (2.14)by using the exact S matrix and the Bethe ansatz. We will refer to (2.14) as the free energy. Afterturning on the chemical potential h beyond an appropriate threshold, there will be a density ρ of particles, charged under Q , and with an energy per unit volume given by e ( ρ ). These twoquantities can be obtained from the density of Bethe roots χ ( θ ). This density is supported onan interval [ − B, B ] and satisfies the integral equation m cosh θ = χ ( θ ) − (cid:90) B − B d θ (cid:48) K ( θ − θ (cid:48) ) χ ( θ (cid:48) ) . (2.15)In this equation, m is the mass of the charged particles, and with a clever choice of Q , it isdirectly related to the mass gap of the theory. The kernel of the integral equation is given by K ( θ ) = 12 π i dd θ log S ( θ ) , (2.16)where S ( θ ) is the S -matrix appropriate for the scattering of the charged particles. The energyper unit volume and the density are then given by e = m π (cid:90) B − B d θ χ ( θ ) cosh θ, ρ = 12 π (cid:90) B − B d θ χ ( θ ) . (2.17)Let us note that B is fixed by the value of ρ , and this leads implicitly to a function e ( ρ ). Finally,the free energy can be obtained as a Legendre transform of e ( ρ ): ρ = − δ F (cid:48) ( h ) ,δ F ( h ) = e ( ρ ) − ρh. (2.18)Note that the first equation defines ρ as a function of h .The above program can be implemented in a number of models. In the case of the non-linearsigma model, one considers the conserved currents associated to the global O ( N ) symmetry, J ijµ = S i ∂ µ S j − S j ∂ µ S i . (2.19)– 5 –e will denote by Q ij the corresponding charges. Usually [15, 16], one considers in (2.12) thequantum version of Q . The exact S matrix of the O ( N ) non-linear sigma model, for particlescharged under Q , is given by S ( θ ) = − Γ(1 + i x )Γ( + ∆ + i x )Γ( − i x )Γ(∆ − i x )Γ(1 − i x )Γ( + ∆ − i x )Γ( + i x )Γ(∆ + i x ) , x = θ π (2.20)and ∆ is given in (2.5).The perturbative series can be extracted from the Bethe ansatz solution by a method de-veloped by Volin [22, 23]. In his original work, the method was applied to the non-linear sigmamodel, but it was later extended to other quantum integrable models in [24–27]. It is convenientto use an expansion variable which can be connected directly to the perturbative answer. Sucha variable was introduced in [30] and is defined by α + ( ξ −
1) log α = log (cid:18) ρ β Λ (cid:19) , (2.21)where ξ = β β = ∆ (2.22)and Λ is the dynamically generated scaleΛ = µ (cid:0) β g (cid:1) − β / (2 β ) e − / (2 β g ) exp (cid:18) − (cid:90) g (cid:26) β g ( x ) + 1 β x − β β x (cid:27) d x (cid:19) . (2.23)This scale is proportional to the mass m appearing in the Bethe ansatz. In the case of thenon-linear sigma model, one has [15, 16]: m = (cid:18) (cid:19) ∆ . (2.24)In this way, we obtain a power series in α for the normalized energy density, e ( ρ ) ρ π ∆ = α (cid:88) n ≥ a n α n . (2.25)The result, up to order α , is e ( ρ ) ρ π ∆ = α + α α ∆2+ ∆32 α (cid:0) − (3 ζ (3) + 1) + 14∆(3 ζ (3) + 2) − ζ (3) + 8 (cid:1) + O ( α ) . (2.26)The free energy δ F ( h ) can also be calculated in perturbation theory from the effectivepotential. The coupling to the conserved charge Q leads to the modified Lagrangian [15, 16, 30] L h = 12 g (cid:26) ∂ µ S · ∂ µ S + 2i h ( S ∂ S − S ∂ S ) + h (cid:0) S + · · · + S N − (cid:1)(cid:27) . (2.27) A scheme for QCD closely related to (2.21) was studied in [49]. – 6 – F ( h ) was obtained at one-loop in [15, 16], and at two-loops in [30]. The result of the perturbativecalculation can be expressed in various convenient ways, depending on an appropriate choice ofcoupling. We can use the renormalization group (RG) to re-express the perturbative series interms of the coupling ¯ g ( µ/h, g ), defined bylog (cid:16) µh (cid:17) = − (cid:90) ¯ gg d xβ ( x ) , (2.28)or in terms of the dynamically generated scale defined in (2.23). In terms of ¯ g , the free energy is δ F ( h ) = − h (cid:26) g − β − β g + O (cid:0) ¯ g (cid:1)(cid:27) . (2.29)In order to compare with the result of the Bethe ansatz, we can use the Legendre transform of δ F ( h ) to calculate e ( ρ ), and then re-express the result in terms of the coupling α introduced in(2.21). We note that, at leading order in the coupling expansion,¯ g = 12 β α + O ( α ) . (2.30)One obtains in this way from perturbation theory, and up to two-loops [30], e ( ρ ) ρ π ∆ = α + α α ∆2 + O (cid:0) α (cid:1) . (2.31)This is in agreement with the result of the Bethe ansatz calculation (2.26), as verified in [22, 30].It is convenient to organize the free energy as a 1 /N expansion: δ F ( h ) = (cid:88) (cid:96) ≥ δ F ( (cid:96) ) ( h ) ∆ (cid:96) − . (2.32)Similarly, the normalized energy density in (2.25) can be organized as a double power seriesexpansion in α , ∆: e ( ρ ) ρ π ∆ = (cid:88) (cid:96) ≥ E ( (cid:96) ) ( α )∆ (cid:96) , (2.33)where E (cid:96) ( α ) are power series in α . One finds, at leading order, E (0) ( α ) = α + α , (2.34)and at subleading order in ∆ we have, for the very first terms [22, 24], E (1) ( α ) = α (cid:18) − ζ (3)32 (cid:19) α + (cid:18)
14 + 35 ζ (3)32 (cid:19) α + (cid:18) − ζ (3)512 − ζ (5)2048 (cid:19) α + O (cid:0) α (cid:1) . (2.35)The series E (1) ( α ) has been computed analytically up to order 44 in [24]. In the next section wewill compute E (1) ( α ) analytically and at all loops, directly in perturbation theory, and we willmatch the result (2.35) and up to order 44. – 7 – The 1 /N expansion of the effective potential To evaluate δ F ( h ) in perturbation theory we have to calculate the effective potential in thetheory with Lagrangian (2.27). As in other problems with an O ( N ) symmetry, it is convenientto reformulate the model in terms of a linear sigma model, by including an additional field X which implements the constraint (2.1). In this way we consider L h = 12 g (cid:26) ∂ µ S · ∂ µ S + 2i h ( S ∂ S − S ∂ S ) + h ( S + · · · + S N −
1) + X ( S − (cid:27) . (3.1)We expand around the following classical vacuum: σ ( x ) = (cid:16) σ, , . . . , (cid:17) + (cid:112) π ∆ λ (cid:16) ˜ σ ( x ) , ˜ σ ( x ) , η ( x ) , . . . , η N − ( x ) (cid:17) ,X ( x ) = χ + (cid:112) π ∆ λ ˜ χ ( x ) , (3.2)where σ and χ are constants that minimize the potential. Neglecting linear terms, the resultingLagrangian can be organized as L h = 12 π ∆ λ L tree + L G + (cid:112) π ∆ λ L int . (3.3)The tree-level Lagrangian is L tree = χ σ − − h , (3.4)while the quadratic or Gaussian part is given by L G = 12 η · ( − ∂ + χ + h ) η + 12 ˜ σ ˜ σ ˜ χ T − ∂ + χ ih∂ σ − ih∂ − ∂ + χ σ ˜ σ ˜ σ ˜ χ , (3.5)with η = ( η , . . . , η N − ). The interaction part is L int = 12 ˜ χ η · η + 12 ˜ χ (cid:16) ˜ σ + ˜ σ (cid:17) . (3.6)In the theory with h = 0, there is an ordered phase with χ = 0, σ (cid:54) = 0 in which ˜ σ , η areGoldstone bosons. This phase is not realized quantum-mechanically, as we explained in theprevious section. However, once h is turned on, these η bosons acquire a mass.After writing the quadratic terms L G in momentum space and inverting the matrix for thefields (˜ σ , ˜ σ , ˜ χ ), we obtain the propagators D ˜ σ ˜ σ = D ˜ σ ˜ σ = D ˜ σ ˜ σ = 0 , D ˜ σ ˜ σ = 1 k + χ ,D ˜ σ ˜ χ = D ˜ χ ˜ σ = 1 σ , D ˜ σ ˜ χ = − D ˜ χ ˜ σ = 2 hk σ ( k + χ ) ,D ˜ χ ˜ χ = − σ (cid:20) k + χ + 4 h k k + χ (cid:21) , D η i η j = δ ij k + h + χ , i, j = 1 , . . . , N − . (3.7)They are represented by the lines in figure 1.From the interaction terms in L int , we have three cubic vertices coupling the field ˜ χ with η and ˜ σ i , i = 1, 2. They are represented in figure 2.– 8 – σ ˜ χ η η ˜ σ ˜ χ ˜ χ ˜ χ ˜ σ ˜ σ Figure 1 : Propagators arising in the non-linear sigma model when its Lagrangian is expandedaround the classical vacuum. ˜ σ ˜ σ ˜ χ ˜ σ ˜ σ ˜ χ η η ˜ χ Figure 2 : Interaction terms arising in the non-linear sigma model when its Lagrangian is ex-panded around the classical vacuum.
The effective potential is a function of the vaccum expectation values (vev) σ and χ , and theparameter h . It has a 1 /N expansion in powers of ∆ given by V ( σ, χ ; h ) = (cid:88) (cid:96) ≥ V ( (cid:96) ) ( σ, χ ; h )∆ (cid:96) − . (3.8)The vevs σ and χ are obtained by extremizing the potential, and they can be also expanded in1 /N : σ = σ (0) + O (∆) , χ = χ (0) + O (∆) . (3.9)The leading order vevs σ (0) and χ (0) , which are the only ones needed in our calculation, areobtained as ∂V (0) ∂σ (0) = ∂V (0) ∂χ (0) = 0 . (3.10)At next-to-leading order in ∆, we have δ F ( h ) = 1∆ V (0) ( σ (0) , χ (0) ; h ) + V (1) ( σ (0) , χ (0) ; h ) + O (∆) . (3.11)Let us first compute V (0) ( σ, χ ; h ). It has contributions from the tree level Lagrangian L tree and from the 1 / ∆ = N − η at one-loop: V (0) ( σ, χ ; h ) = 14 πλ (cid:0) χ ( σ − − h (cid:1) + 12 (cid:90) d d k (2 π ) d log( k + h + χ ) . (3.12)By using dimensional regularization to evaluate the integral, we find V (0) ( σ, χ ; h ) = 14 πλ (cid:0) χ ( σ − − h (cid:1) + ( h + χ ) d/ (4 π ) d/ d Γ (cid:16) (cid:15) (cid:17) . (3.13)From here we can calculate σ (0) and χ (0) . There is a “disordered” non-perturbative vacuumwith χ (0) (cid:54) = 0 and an “ordered” perturbative vacuum with χ (0) = 0. As in [33], in order to– 9 –ake contact with conventional perturbation theory, we choose the perturbative vacuum, where χ (0) = 0. Imposing (3.10), we find σ λ = 1 λ − (cid:18) h π (cid:19) − (cid:15) Γ (cid:16) (cid:15) (cid:17) . (3.14)We can now introduce the renormalized coupling λ by the usual equation, λ = ν (cid:15) Zλ, (3.15)where ν = µ e γ − log(4 π ) , (3.16)parametrizes the scale choice µ and features the MS scheme, and γ is the Euler–Mascheroniconstant. Z is the renormalization constant, for which we consider the 1 /N expansion Z − = (cid:88) (cid:96) ≥ Z − (cid:96) ) ∆ (cid:96) . (3.17)Cancelation of singular terms in the r.h.s. of (3.14) fixes Z − = 1 + λ(cid:15) . (3.18)Now, the r.h.s. of (3.14) is manifestly finite as (cid:15) →
0, and we find σ λ = 1 λ + log (cid:18) hµ (cid:19) + O (∆ , (cid:15) ) . (3.19)By evaluating the leading order effective potential (3.13) at the critical point σ (0) , χ (0) = 0,we obtain the leading order free energy δ F (0) , defined in (2.32). After writing the resultingexpression in terms of the renormalized coupling λ and in the limit (cid:15) →
0, one finds δ F (0) ( h ) = − h π (cid:26) λ + log (cid:18) hµ (cid:19) − (cid:27) . (3.20) As explained in [50] (see also [33]), the next-to-leading correction to the effective potential in the1 /N expansion is given by a sum of ring diagrams. A ring diagram at m loops is constructedwith m bubbles of η particles successively connected by m propagators of ˜ χ particles until thediagram closes on itself. Each bubble comes with a factor 1 / ∆ = N − χ propagators. The sum of such ring diagrams is shown in figure 3. Followingthe structure used in [33], the contribution of these diagrams is the infinite sum − (cid:88) m ≥ m (cid:90) d d k (2 π ) d (cid:0) πλ Π( k , h + χ ) D ˜ χ ˜ χ (cid:1) m , (3.21)where Π( k , M ) = 12 (cid:90) d d q (2 π ) d q + M )(( k + q ) + M ) (3.22)– 10 – + λ + λ + · · · Figure 3 : Sum of the ring diagrams contributing to V (1) ( σ (0) , χ (0) ; h ).is the scalar polarization loop arising from the η bubbles.The momentum integrals in (3.21) are divergent. However, going back to (3.13), after renor-malization of the term with 1 /λ , the renormalization constant at next-to-leading order Z − provides additional divergent terms that should cancel the divergences of our momentum inte-grals. By evaluating at the critical point with χ (0) = 0, we find the renormalized free energy atnext-to-leading order δ F (1) ( h ) = − h ν − (cid:15) π Z − λ + 2 πν (cid:15) h (cid:88) m ≥ ( − m m (cid:32) πλ σ (cid:33) m I m . (3.23)The integrals I m are defined as I m = (cid:90) d d k (2 π ) d (cid:20) Π( k , h )( k + 4 h k ) k (cid:21) m . (3.24)By first expanding ( k + 4 h k ) m with the binomial theorem, the angular integral of the mo-mentum can be computed term by term. I m can then be expressed as a finite sum of rotationalinvariant integrals. Finally, to compute the integral over k , it will be crucial to write the scalarpolarization function in terms of a hypergeometric function:Π( k , h ) = h − − (cid:15) Γ(1 + (cid:15) )2(4 π ) d/ (cid:18) x + 44 (cid:19) − − (cid:15)/ F (cid:18) (cid:15) , ,
32 ; xx + 4 (cid:19) . (3.25)In this equation, x = k h . (3.26)The integrals I m then can be expressed as I m = h − (cid:15) ( m +1) S d π ) d (cid:20) Γ(1 + (cid:15) )2(4 π ) d/ (cid:21) m m (cid:88) (cid:96) =0 (cid:18) m(cid:96) (cid:19) (1 / (cid:96) ( d/ (cid:96) (cid:96) I m,(cid:96) , (3.27)where ( x ) n is the Pochhammer symbol, S d = 2 π d/ / Γ( d/
2) is the volume of a d -dimensionalsphere, and I m,(cid:96) = 4 d/ m − (cid:96) (cid:90) d z z m − (cid:96) − (cid:15)/ (1 − z ) − (cid:96) − ( m +1) (cid:15)/ (cid:20) F (cid:18) (cid:15) , ,
32 ; z (cid:19)(cid:21) m , (3.28)– 11 –hich we have appropriately expressed in terms of the integration variable z = x/ ( x + 4). It iseasy to see that, when (cid:96) ≥
2, the integrals I m,(cid:96) are finite as (cid:15) →
0. Thus, for now we focus onthe integrals with (cid:96) = 0, 1 and their singular part.It is convenient to re-express these integrals in yet another form, specially in order to extractthe singular part. We use fractional linear transformations of the hypergeometric function towrite F (cid:18) (cid:15) , ,
32 ; z (cid:19) = − (cid:15) Γ(1 − (cid:15) ) (cid:15) Γ(1 − (cid:15) ) z − / × (cid:20) − (cid:15) Γ(1 − (cid:15) )Γ(1 − (cid:15) ) (1 − z ) − (cid:15)/ F (cid:18) − (cid:15) , , − (cid:15) − z (cid:19)(cid:21) . (3.29)Similar manipulations of the polarization loop were done in [51, 52] to calculate critical exponentsin the 1 /N expansion. We can now write the integral I m,(cid:96) , up to overall factors, as I m,(cid:96) = 4 d/ m − (cid:96) (cid:18) − (cid:15) Γ(1 − (cid:15) ) (cid:15) Γ(1 − (cid:15) ) (cid:19) m B m,(cid:96) , (3.30) B m,(cid:96) = (cid:90) d z (1 − z ) m/ − (cid:96) − (cid:15)/ z − (cid:96) − ( m +1) (cid:15)/ (cid:20) − (cid:15) Γ(1 − (cid:15) )Γ(1 − (cid:15) ) z − (cid:15)/ F (cid:18) − (cid:15) , , − (cid:15) z (cid:19)(cid:21) m . (3.31)In writing this expression, we have changed the integration variable from z to 1 − z . The integral(3.31) can be written as an infinite sum which will be useful for our analysis. Let us define theexpansion coefficients (cid:20) F (cid:18) − (cid:15) , , − (cid:15) z (cid:19)(cid:21) s = (cid:88) k ≥ c ( s ) k z k . (3.32)Then, by expanding the binomial in (3.31) and using (3.32) for each of the hypergeometricfunctions arising in the binomial sum, we can integrate term by term in z , and we find: B m,(cid:96) = (cid:88) k ≥ B m,(cid:96),k , (3.33)where B m,(cid:96),k = m (cid:88) s =0 (cid:18) ms (cid:19)(cid:18) − (cid:15) Γ(1 − (cid:15) )Γ(1 − (cid:15) ) (cid:19) s Γ (cid:0) − (cid:96) + m − (cid:15) + 1 (cid:1) Γ (cid:0) k + (cid:96) + ( m − s + 1) (cid:15) − (cid:1) Γ (cid:0) k + m + ( m − s ) (cid:15) (cid:1) c ( s ) k . (3.34)In our result (3.34)–(3.33) we have successfully isolated the full singularity of I m,(cid:96) in the firsttwo terms k = 0, 1. From the factor 1 /(cid:15) m in (3.30), we note that the singular part of I m,(cid:96) corresponds to the Laurent expansion of B m,(cid:96) up to order (cid:15) m − . Thus, the first two terms k = 0,1 fully contain the expansion of B m,(cid:96) up to this order. We prove this statement in appendix A. The renormalized free energy (3.23) must be finite. By imposing cancellation of divergenceswe should be able to obtain an explicit expression for Z − , and thus, for the next-to-leadingorder of the beta function in the 1 /N expansion. This result, which we quoted in (2.10), hasbeen known for some time [34], based on the 1 /N calculation of critical exponents [51, 52]. Ourcalculation provides a direct derivation of the beta function, very similar to the calculation by– 12 –alanques-Mestre and Pascual in [36], where they studied the beta function of QED in the 1 /N f expansion.Let us write the next-to-leading correction to Z − in (3.17) as a Laurent expansion with yetundetermined coefficients B ( n ) i : Z − = (cid:88) n ≥ λ n n − (cid:88) i =0 B ( n ) i (cid:15) n − i . (3.35)Firstly we will relate the coefficients B ( n ) i directly with the coefficients of the beta function and,secondly, we will find an explicit expression for B ( n ) i by imposing cancellation of divergences in(3.23) and using our result in (3.34).The beta function for the ’t Hooft coupling can be obtained from the renormalization constant Z − as β ( λ ) = − (cid:15) λ − λ ∂∂λ log Z − . (3.36)By using the Laurent expansion (3.35) and the leading order result (3.18) for the renormalizedcoupling, the above equality can be written at order ∆ in the 1 /N expansion as β (1) ( λ ) = − λ (cid:88) n ≥ λ n ( n + 1) B ( n +1) n . (3.37)In addition, finiteness of the β function as (cid:15) → B ( n +1) i = − n − n + 1 B ( n ) i , i = 0 , . . . , n − , n ≥ . (3.38)Thus, the computation of β (1) ( λ ) now reduces to determining the coefficients B ( n ) i .As we prove in appendix A, the divergent part of the integrals I m,(cid:96) comes from a finitenumber of terms k in the sum of (3.33). To be more specific, B m,(cid:96),k leads to singularities only for k = 0, 1 when (cid:96) = 0, and for k = 0 when (cid:96) = 1. Thus, the divergent part of I m will be containedcompletely in the sum S m = B m, , + B m, , + m − (cid:15) B m, , , (3.39)where the factor in front of the (cid:96) = 1 term arises from the binomial coefficient and Pochhammersymbols in (3.27). By combining appropriately the Γ factors in (3.34), one finds S m = 2 (cid:15) Γ (cid:16) m − (cid:15) (cid:17) m (cid:88) s =0 (cid:18) ms (cid:19) ( − s m − s + 1 f m ( s(cid:15), (cid:15) ) , (3.40)where f m ( y, x ) = Γ( m +12 x − y + 1)Γ( m (1+ x )2 − y ) (cid:18) m − xx ( m + 1) − y − y ( m − x )2( x − m ( x + 1) − y ) + m − x (cid:19) e y g ( x ) , g ( x ) = 1 x log (cid:18) x Γ(1 − x )Γ(1 − x ) (cid:19) . (3.41)By Taylor expanding f m ( s(cid:15), (cid:15) ) in powers of the first variable and using properties of the binomialcoefficient, we can compute the sum in (3.40). We obtain, S m = ( − m m + 1 (cid:18) (cid:15) Γ(1 − (cid:15) )Γ(1 − (cid:15) ) (cid:19) m +1 (cid:15) − (cid:15) − − Γ (cid:16) m (cid:17) − m m + 1 (cid:20) d m +1 f m ( y, y m +1 (cid:21) y =0 (cid:15) m + O (cid:0) (cid:15) m +1 (cid:1) . (3.42)– 13 –e remark that the expression we have obtained is only valid up to order (cid:15) m , while the first termin the r.h.s. is equal to S m up to order (cid:15) m − (so it correctly encodes the singular part of theintegrals I m ). We will however need the second term to extract the finite part of the free energy.We are now ready to extract the singular part of the sum over ring diagrams in (3.23). First,let us introduce some notation. Let A ( (cid:15) ) be a Laurent series in (cid:15) . We will denote by div A ( (cid:15) ) thesingular or principal part of the series. Then, it is easy to show thatdiv πν (cid:15) h (cid:88) m ≥ ( − m m (cid:32) πλ σ (cid:33) m I m = div (cid:88) r ≥ λ r ν − r(cid:15) Π r ( h ; (cid:15) ) , (3.43)where Π r ( h ; (cid:15) ) = H − (cid:15) ( r +1) − (cid:15) (cid:15) r (cid:2) Γ(1 + (cid:15) ) (cid:3) r Γ (cid:0) − (cid:15) (cid:1) r − (cid:88) p =0 r − p (cid:18) r − p (cid:19)(cid:18) (cid:15) Γ(1 − (cid:15) )Γ(1 − (cid:15) ) (cid:19) r − p S r − p , (3.44)and we have denoted H = h πν . (3.45)The expression in (3.43) is obtained by using (3.14) in place of λ /σ and reexpanding in powersof λ . Replacing S m in (3.44) by the first term of (3.42), we finddiv (cid:2) Π r ( h ; (cid:15) ) (cid:3) = div (cid:20) r + 1) (cid:15) r P ( (cid:15), ( r + 1) (cid:15) ) (cid:21) , (3.46)where P ( x, y ) = − H − y e ( y − x ) j ( x ) Γ(1 − x ) Γ(1 + x )Γ(1 − x ) x − x − , j ( x ) = 1 x log Γ (cid:16) x (cid:17) . (3.47)We can now use a similar argument to the one in [36]. In the r.h.s. of (3.43), we replace λ bythe renormalized coupling at leading order in 1 /N : λ = ν (cid:15) λ λ(cid:15) . (3.48)We then obtain, (cid:88) r ≥ λ r (1 + λ(cid:15) ) r Π r ( h ; (cid:15) ) = (cid:88) m ≥ λ m (cid:26) ( − m +1 m ( m + 1) P ( (cid:15) ) (cid:15) m + ( m − P m ( (cid:15) ) + O ( (cid:15) ) (cid:27) , (3.49)where we used that m − (cid:88) s =0 (cid:18) m − s (cid:19) ( − s ( m + 1 − s ) j − = ( − m +1 m ( m + 1) , j = 0 , , j = 1 , . . . , m − , ( m − , j = m, (3.50)and we expanded P ( x, y ) = ∞ (cid:88) j =0 P j ( x ) y j . (3.51)– 14 –ecause the expansion functions P j ( x ) are regular at x = 0, it is obvious from (3.49) that only P ( (cid:15) ) contributes to the singular part of the sum over ring diagrams. By using the reflectionformula for the gamma function, we obtain the explicit form P ( (cid:15) ) = P ( (cid:15),
0) = − (cid:0) π(cid:15) (cid:1) π(cid:15) Γ(1 − (cid:15) )Γ(1 − (cid:15) ) (cid:15) − (cid:15) − , (3.52)which has a power series expansion at (cid:15) = 0: P ( (cid:15) ) = (cid:88) i ≥ P ,i (cid:15) i . (3.53)Requiring cancellation of divergences in (3.23) determines the expansion of (3.35), and we findthe values B ( m ) i = ( − m − m ( m − P ,i − , B ( m )0 = 0 , (3.54)which satisfy the constraint (3.38). The beta function at next-to-leading order in the 1 /N expansion can be now computed by going back to (3.37) and using our result for the coefficients B ( n ) i : β (1) ( λ ) = − λ (cid:88) m ≥ ( − m m P ,m − λ m = λ (cid:90) λ P ( − x )d x. (3.55)This result of course coincides with (2.10). Once the singularities have been canceled, we can focus on the finite part of the renormalizedfree energy. The finite part arises from the integrals I m,(cid:96) with (cid:96) ≥ I m,(cid:96) with (cid:96) = 0 ,
1. This last contribution comes from three types of terms. Two ofthem are already written down in the previous section, since they have their origin in S m : thelast term in the r.h.s. of (3.42), and the finite part in the r.h.s. of (3.49). In addition, we haveto take into account the contribution from the terms (cid:98) B m, = (cid:88) k ≥ B m, ,k , (cid:98) B m, = (cid:88) k ≥ B m, ,k , (3.56)which are not included in (3.39). It will be convenient to rewrite the series (3.56) in the followingintegral representation, which we derive in appendix A: (cid:98) B m,(cid:96) = (cid:15) m ( − m (cid:90) d z (1 − z ) m/ − (cid:96) z − (cid:96) (cid:20) d m g (cid:96) ( y ; z )d y m (cid:21) y =0 + O (cid:0) (cid:15) m +1 (cid:1) , (3.57)where g (cid:96) ( y ; z ) = (cid:18) √ z (cid:19) − y (cid:20)(cid:18) √ − z (cid:19) y − δ (cid:96) zy (cid:21) , (cid:96) = 0 , . (3.58)The derivation of this result relies on the same tricks that we used to obtain (3.42) (similarmanipulations can also be found in [53]).We can now write the next-to-leading correction to the renormalized free energy as δ F (1) ( h ) = − h (cid:26) W (cid:18) λ ; hµ (cid:19) + X (cid:18) λ ; hµ (cid:19) + Y (cid:18) λ ; hµ (cid:19) + Z (cid:18) λ ; hµ (cid:19)(cid:27) , (3.59)– 15 –here W (cid:18) λ ; hµ (cid:19) = 12 π (cid:88) m ≥ (cid:20) ( − m +1 m ( m + 1) P ,m + ( m − P m (0) (cid:21) λ m ,X (cid:18) λ ; hµ (cid:19) = 1 π (cid:88) m ≥ m lim (cid:15) → (cid:15) m (cid:16) (cid:98) B m, + m (cid:98) B m, (cid:17)(cid:18) λ λ log( h/µ ) (cid:19) m ,Y (cid:18) λ ; hµ (cid:19) = − π (cid:88) m ≥ ( − m m ( m + 1) Γ (cid:16) m (cid:17)(cid:20) d m +1 f m ( y, y m +1 (cid:21) y =0 (cid:18) λ λ log( h/µ ) (cid:19) m ,Z (cid:18) λ ; hµ (cid:19) = 14 π (cid:88) m ≥ ( − m m m (cid:88) (cid:96) =2 (cid:18) m(cid:96) (cid:19) (1 / (cid:96) (cid:96) ! 4 (cid:96) − m I m,(cid:96) (cid:18) λ λ log( h/µ ) (cid:19) m . (3.60)Many of the ingredients appearing in these formulae have been already spelled out in detail. Thecoefficients P ,m can be read from (3.52). The coefficients P m (0) follow from (3.47) and (3.51): P m (0) = ( − m +1 m ! log m (cid:18) hµ (cid:19) . (3.61)The function f m ( y, x ) is given in (3.41). It remains to compute the integrals (cid:98) B m, , (cid:98) B m, and I m,(cid:96) , (cid:96) = 2 , . . . , m , for arbitrary m, (cid:96) . This can be done analytically, and the results are presented inappendix B. This allows us to compute δ F (1) ( h ) at any given order. The very first terms read δ F (1) ( h ) = − h π (cid:20)(cid:18) −
14 + log( h/µ ) (cid:19) λ + (cid:18) −
724 + 21 ζ (3)32 + log( h/µ )2 − log ( h/µ )2 (cid:19) λ + (cid:18) − − ζ (3)96 + (8 − ζ (3)) log( h/µ )16 − log ( h/µ )2 + log ( h/µ )3 (cid:19) λ + (cid:18) − ζ (3)2560 + 3 ζ (4)320 + 4185 ζ (5)2048 + (24 + 105 ζ (3)) log( h/µ )32+ ( −
24 + 63 ζ (3)) log ( h/µ )32 + log ( h/µ )2 − log ( h/µ )4 (cid:19) λ + O (cid:0) λ (cid:1)(cid:21) . (3.62)In order to compare with the Bethe ansatz solution, we have to rexpress this result in terms ofthe coupling constant α , defined in (2.21). The first step is to set λ to the running couplingconstant at the scale µ = h , λ = λ ( µ = h ). λ is related to the dynamically generated scale Λ and h by log (cid:18) h Λ (cid:19) = 1¯ λ + ξ log(¯ λ ) + (cid:90) ¯ λ (cid:20) β ( u ) + 1 u − ξu (cid:21) d u, (3.63)where ξ is defined in (2.22). At this scale, δ F (1) ( h ) simplifies greatly to δ F (1) ( h ) = − h π (cid:88) m ≥ v m λ m = − h π (cid:20) − λ (cid:18) −
724 + 21 ζ (3)32 (cid:19) λ + · · · (cid:21) . (3.64)This defines the coefficients v m , m ≥
1. We can now use the Legendre transform (2.18) to obtainthe normalized energy density (2.25). In order to do that, it is useful to introduce yet anothercoupling 1˜ α + ξ log ˜ α = log (cid:18) h Λ (cid:19) , (3.65)– 16 –hich was first considered in [30] and is related to ¯ λ by¯ λ = ˜ α − ∆ (cid:18) ˜ α (cid:90) ˜ α (cid:20) u + β (1) ( u ) u (cid:21) d u (cid:19) + O (cid:0) ∆ (cid:1) . (3.66)We can use ˜ α to write the free energy as δ F ( h ) = h (cid:18)
1∆ Σ ( ˜ α ) + Σ ( ˜ α ) + O (∆) (cid:19) , (3.67)where Σ ( ˜ α ) = − π (cid:18) α − (cid:19) , Σ ( ˜ α ) = − π (cid:88) m ≥ v m ˜ α m − π (cid:90) ˜ α (cid:20) u + β (1) ( u ) u (cid:21) d u. (3.68)The Legendre transform gives, ρh = 12 π ∆ ˜ α + ˜ α π − ( ˜ α ) + ˜ α Σ (cid:48) ( ˜ α ) + O (∆) ,eh = 1∆ (cid:18) π ˜ α + 18 π (cid:19) + ˜ α π − Σ ( ˜ α ) + ˜ α Σ (cid:48) ( ˜ α ) + O (∆) . (3.69)The final step is to relate ˜ α to the coupling α defined in (2.21):˜ α = α + 2 πα ∆ α + 1 (cid:16) α π − ( α ) + α Σ (cid:48) ( α ) (cid:17) + O (cid:0) ∆ (cid:1) , (3.70)which leads to a remarkably simple expression for the normalized energy density: eρ π ∆ = α + α − ∆ α (cid:88) m ≥ v m α m + α (cid:90) α (cid:20) u + β (1) ( u ) u (cid:21) d u + O (cid:0) ∆ (cid:1) . (3.71)Expanding β (1) ( u ) with (3.55), we can read the following result for the series E (1) ( α ), defined in(2.33), in terms of the perturbative coefficients v m : E (1) ( α ) = − α (cid:88) m ≥ (cid:18) v m − ( − m +1 m ( m + 1) P ,m (cid:19) α m . (3.72)It follows from this expression that E (1) ( α ) = − πα (cid:2) X ( α ; 1) + Y ( α ; 1) + Z ( α ; 1) (cid:3) , (3.73)where the functions in the r.h.s. were defined in (3.60).One can now compare the expression (3.72) with the result of the Bethe ansatz (2.35). Wefind perfect agreement up to order α . Up to this order, we find that every coefficient is the sumof a rational number plus a linear combinations of odd Riemann zeta functions . In appendix Bwe prove to all orders that Z ( α ; 1) can be written as linear combination of ζ (2 k + 1) (in fact, withno rational term). On the other hand, X ( α ; 1) and Y ( α ; 1) do not satisfy this transcendentalityproperty when alone, but their combination indeed does, up to order α (in this case, a rationalterm has to be included with the linear combination of odd zetas), although we do not have aproof of this statement to all orders. From the Bethe ansatz result, we also know that higher corrections in ∆ can be written as polynomials inmultiple variables of odd Riemann zeta functions. – 17 –
Large N renormalons and their trans-series In [22, 24], numerical evidence was given for the factorial growth of the perturbative series for(2.31), at fixed N . This was interpreted as a signature of renormalons [1–3]. In [24], the con-tribution of UV and IR renormalons was disentangled, and detailed evidence was given that thelarge order behavior of the perturbative series is in agreement with the predictions of renormalonphysics. In particular, it was shown that the next-to-leading behavior of the asymptotics involvesthe first two coefficients β and β of the beta function. The evidence for these effects was basedon a numerical study of the perturbative series and it focused on the leading singularities in theBorel plane.At large N , the ring diagrams studied in the previous section should give the leading renor-malon behavior. One advantage of having explicit results for these diagrams is that we can obtainfrom them analytic results on the large order behavior of the perturbative series. Equivalently,we can find explicit results for the exponentially suppressed trans-series associated to each Borelsingularity. These can be obtained without even calculating the loop integrals. As shown in[27, 33, 41], it is enough to write the generating functions (3.60) in integral form and study theirimaginary parts (or, equivalently, their imaginary discontinuities).We will now present the integral forms for the series X ( λ ; 1), Y ( λ ; 1) and Z ( λ ; 1) appearingin (3.73). We note that the coefficients of the series W ( λ ; 1) do not grow factorially. This is easilyobserved from (3.61) and the fact that P ,m are the Taylor coefficients of an analytic function at (cid:15) = 0.Let us start with X ( λ ; 1). The integral form in this case is easily obtained by using theexplicit expression (3.58) and Laplace transforms. We find X ( λ ; 1) = 1 π (cid:90) d zz X (cid:0) λ √ − z, z (cid:1) + 12 π (cid:90) d zz (1 − z ) X ( λ √ − z, z ) , (4.1)where X ( y, z ) = − yz − y log (cid:16) √ z (cid:17) + log (cid:20) − y log (cid:18) √ z (cid:19)(cid:21) − log (cid:20) − y log (cid:18) √ z √ − z + 1 (cid:19)(cid:21) , X ( y, z ) = 11 − y log (cid:16) √ z √ − z (cid:17) − − y log (cid:16) √ z (cid:17) . (4.2)In the case of Y ( λ ; 1) we use the explicit expression (3.41) and we write the Euler betafunctions that appear in the resulting expression as integrals over z . In this way we find Y ( λ ; 1) = 14 π (cid:90) d zz (1 − z ) (cid:88) m ≥ ( − λ √ − z ) m m + 1 d m +1 d y m +1 (cid:20) y yz − y/ (cid:18) yy + 2 + z (cid:19)(cid:21) y =0 . (4.3)We can use again Laplace transforms to sum up this series and we eventually find Y ( λ ; 1) = 14 π (cid:90) d zz (1 − z ) Y ( λ √ − z, z ) , (4.4)where Y ( y, z ) = − z + z e − /y y E (cid:20) − y (cid:18) − y log (cid:18) √ z (cid:19)(cid:19)(cid:21) + z + 21 − y log (cid:16) √ z (cid:17) (4.5)– 18 –nd E ( z ) is the exponential integral.Finally, after using the identity (cid:88) m ≥ (cid:96) m (cid:18) m(cid:96) (cid:19) x m = x (cid:96) (cid:96) (1 − x ) (cid:96) , (4.6)the last series can be written as Z ( λ ; 1) = 1 π (cid:90) d z (1 − z ) (cid:34) Z ( z, λ ) − − (cid:32) (cid:112) Z ( z, λ )2 (cid:33)(cid:35) , (4.7)where Z ( z, λ ) = 1 + F ( z ) λ zF ( z ) λ , F ( z ) = F (cid:18) , ,
32 ; z (cid:19) = tanh − ( √ z ) √ z . (4.8)The advantage of the representations (4.1), (4.4) and (4.7) is that they lead to explicit, expo-nentially small imaginary terms. These are precisely the trans-series associated to the renormalonsingularities.Let us first consider the function X ( λ ; 1). It has discontinuities when λ <
0, due to thepoles and logarithmic branch cut in the integrands of (4.1). The singularities of the integrandare located at z and z , which are defined by1 λ = √ − z log (cid:18) √ z (cid:19) , λ = √ − z log (cid:18) √ z √ − z (cid:19) . (4.9)We finddisc X ( λ ; 1) = 2 π i (cid:34) π (cid:90) z z d zz − π Res (cid:18) z X (cid:0) λ √ − z, z (cid:1) , z = z (cid:19) − (cid:88) i =1 , π Res (cid:18) z (1 − z ) X (cid:0) λ √ − z, z (cid:1) , z = z i (cid:19)(cid:35) . (4.10)This discontinuity can be computed term by term as a trans-series in λ , i.e. as a power series inboth e /λ and λ . See appendix C for details of this computation. We finddisc X ( λ ; 1) = − i (cid:20)(cid:18) λ − (cid:19) e /λ + (cid:18) λ + 48 λ (cid:19) e /λ + (cid:18) λ + 1188 λ + 306 λ (cid:19) e /λ + (cid:18) λ + 634883 λ + 11264 λ + 1536 λ (cid:19) e /λ + (cid:18) λ + 330000 λ + 8020003 λ + 80600 λ + 7100 λ (cid:19) e /λ + (cid:18) λ + 238878725 λ + 5197824 λ + 2489472 λ + 501696 λ + 31632 λ (cid:19) e /λ + O (cid:16) e /λ (cid:17)(cid:21) . (4.11)Let us now consider the function Y ( λ ; 1). By investigating the function (4.5), we see that Y ( λ ; 1) has discontinuities both for positive and negative λ . When λ <
0, there is a pole at z = z , where z was defined in (4.9), and a discontinuity due to the exponential integral. For– 19 – >
0, we have a discontinuity due again to the exponential integral, and one findsdisc Y ( λ >
0; 1) = i e − /λ , disc Y ( λ <
0; 1) = i (cid:20)(cid:18) λ + 10 λ (cid:19) e /λ + (cid:18) λ + 128 λ + 64 λ (cid:19) e /λ + (cid:18) λ + 1584 λ + 1476 λ + 330 λ (cid:19) e /λ + (cid:18) λ + 573443 λ + 25600 λ + 12032 λ + 1568 λ (cid:19) e /λ + (cid:18) λ + 6800003 λ + 11800003 λ + 284000 λ + 82200 λ + 7140 λ (cid:19) e /λ + (cid:18) λ + 2654208 λ + 5640192 λ + 5501952 λ + 2536704 λ + 504576 λ + 31680 λ (cid:19) e /λ + O (cid:16) e /λ (cid:17)(cid:21) . (4.12)Finally, we consider the discontinuity of Z ( λ ; 1), which arises from two sources. The firstone is a pole of the integrand, which appears for λ <
0. This occurs at a z ∈ (0 ,
1) satisfying1 + λz F ( z ) = 0 . (4.13)There is another source of discontinuity due to the square root inside the logarithm, which occurswhen Z ( z, λ ) <
0. The discontinuity of this source is given by4 π (cid:90) z z d z (1 − z ) tanh − (cid:32)(cid:115) F ( z ) λ zF ( z ) λ (cid:33) , (4.14)where ( z , z ) is the subinterval of (0 ,
1) where Z ( z, λ ) is negative. The value z is the polepreviously discussed in (4.13), while z satisfies1 + λF ( z ) = 0 . (4.15)The trans-series obtained in this way isdisc Z ( λ ; 1) = − i (cid:20)(cid:18) λ + 6 λ (cid:19) e /λ + (cid:18) λ + 64 λ + 16 λ (cid:19) e /λ + (cid:18) λ + 756 λ + 288 λ + 24 λ (cid:19) e /λ + (cid:18) λ + 8192 λ + 133123 λ + 768 λ + 32 λ (cid:19) e /λ + (cid:18) λ + 2900003 λ + 1932503 λ + 503003 λ + 1600 λ + 40 λ (cid:19) e /λ + (cid:18) λ + 53084165 λ + 43130885 λ + 304128 λ + 47232 λ + 2880 λ + 48 λ (cid:19) e /λ + O (cid:16) e /λ (cid:17)(cid:21) . (4.16)We can now put all these results together and calculate the trans-series associated to E (1) ( α ).It is given by 1 / (2i) times the discontinuity, and readsIm E (1) ( α ) = π (cid:20) − α e − /α + (cid:0) − α (cid:1) e /α + (cid:18) α + 72 α (cid:19) e /α + (cid:18) α + 200003 α + 65003 α + 200 α (cid:19) e /α + O (cid:16) e /α (cid:17)(cid:21) . (4.17)– 20 – · · Figure 4 : Location of the renormalon singularities in the Borel plane. There is a single singu-larity in the positive real axis for ζ = 2, corresponding to the IR renormalon, and and infiniteseries of UV renormalon singularities at ζ = − k − k = 0 , , . . . .Let us analyze this result. The first term in the r.h.s. of (4.17) corresponds to a Borel singularityon the positive real axis at ζ = 2. It is an IR renormalon, which has been identified in [22, 24]at finite N and more recently in [28, 29] in the O (4) model. At this order in the 1 /N expansionthere are no additional IR renormalons. The next terms correspond to singularities in the Borelplane on the negative real axis, and they are UV renormalons. We conjecture that they arelocated at ζ = − k − , k ∈ Z ≥ . (4.18)See figure 4 for a representation of the renormalon singularities in the Borel plane.From the discontinuity in (4.17) we can derive the large order behavior of the coefficients ofthe series E (1) ( α ) = (cid:88) m ≥ e (1) m α m . (4.19)The derivation follows from the general theory of resurgence (see e.g. [5, 7]), which relates thetrans-series of a function to the singularities of its Borel transform. In turn, one can extract thelarge order behavior of the perturbative coefficients directly from the Borel singularities. We find e (1) m = − m )( − − m (cid:20) − m − m − (cid:21) + 2Γ( m − − m −
32 Γ( m + 2)( − − m (cid:20) − m + 1 (cid:21) + O (cid:0) m !( − − m (cid:1) . (4.20)The first term in the r.h.s. gives the leading order asymptotics, which is due to the first UVrenormalon singularity at ζ = −
2. The second term, with fixed sign, is due to the IR renormalonsingularity at ζ = 2; while the third term, which is exponentially subleading with respect to thefirst two, is due to the next UV renormalon at ζ = − N .In general, one has to exercise care in this comparison, since some of the contributions to theasymptotic behavior detected at finite N might be suppressed at large N . In [24] it was foundthat the series (2.25) at finite N has the following asymptotics: a n ∼ C IR − n Γ( n + 2∆ −
1) + C UV ( − − n Γ( n −
2∆ + 1) , n (cid:29) , (4.21)where the first and second term in the r.h.s. correspond respectively to the IR and UV singu-larities at ζ = ±
2, and C IR , C UV are in principle functions of N . After taking into account thedifference in the labelling, m = n + 1, we find that (4.21) leads to the behavior (4.20) at large– 21 – , since ∆ → C IR , C UV areof order ∆.In [24] it was argued that the IR renormalon at ζ = 2 is associated to the condensate ofthe operator O = ∂ µ S · ∂ µ S , and that the large order behavior (4.21) is compatible with theexpectations of renormalon physics. It follows that our large N result for this IR renormalonmight be also explained by the contribution of this condensate.The calculation above determines the functional form of the trans-series associated to thedifferent singularities. An additional, interesting question is the “semiclassical decoding” of thenormalized density (2.33) at large N , i.e. its expression as a Borel–Ecalle resummation of thesetrans-series and the perturbative series. At finite N = 4, this has been recently done in [28, 29] forthe full energy density, by using trans-series at finite N . However, it is not clear to us what is thenon-perturbative definition of the functions E ( (cid:96) ) ( α ), since as we mentioned in the Introduction,the Bethe ansatz for the non-linear sigma model is not amenable to a 1 /N expansion. We can extend all of the above results to the supersymmetric version of the non-linear sigmamodel considered in [54]. This model consists of the vector field S ( x ) of the purely bosonicversion, satisfying as well the constraint (2.1), and an N -uple of two-component Majorana spinors Ψ = (Ψ , . . . , Ψ N ), satisfying the constraint S · Ψ = 0 . (5.1)The Lagrangian density is L susy = 12 g (cid:26) ∂ µ S · ∂ µ S + Ψ · i /∂ Ψ + 14 (cid:0) Ψ · Ψ (cid:1) (cid:27) , (5.2)where we follow the conventions of [54] for the gamma matrices. This model is asymptoticallyfree and its beta function is of the form (2.3), with β = 14 π ∆ , β = 0 , (5.3)and ∆ is again given by (2.5). The model can also be studied in the large N expansion, whereone finds a non-perturbative mass gap and dynamical breaking of the discrete chiral symmetry Ψ → γ Ψ [55]. The beta function in the 1 /N expansion has the structure (2.8), where β (0) ( λ )is given again by the expression in (2.9), while β (1) ( λ ) = 0 due to cancellations between bosonsand fermions [37]. We will rederive this result in section 5.2.As in the non-linear sigma model, we couple the present model to an external potential byusing again the conserved charge Q associated to the global O ( N ) symmetry. The dependenceof the ground state energy on the external potential can be obtained from the Bethe ansatzand the exact S -matrix conjectured in [56]. The resulting integral equation was written downexplicitly in [19], where it was used to obtain the exact mass gap of the model. In [24] the groundstate energy was computed as a power series in the coupling α , defined in (2.21) (although ξ = 0in this case). At leading order in the 1 /N expansion, one obtains for E (0) ( α ) the same result thatwe presented in (2.34). At next-to-leading order in 1 /N one finds E susy(1) ( α ) = − ζ (3)32 α + 35 ζ (3)32 α − (cid:18) ζ (3)512 + 4185 ζ (5)2048 (cid:19) α + O (cid:0) α (cid:1) , (5.4)– 22 – σ ˜ χ η η ˜ σ ˜ χ ˜ χ ˜ χ ˜ σ ˜ σ ˜ τ ˜ τ Figure 5 : Propagators for bosonic fields in the supersymmetric non-linear sigma model. ν λ ψ ψ ν λλ λ Figure 6 : Propagators for fermionic fields in the supersymmetric non-linear sigma model.which is available up to order α in [24]. Interestingly, this expansion is almost identical to thebosonic result (2.35), but it only keeps its transcendental part.Our goal in this section is to test the result (5.4) against a perturbative calculation. Likebefore, it is more convenient to use the linearized version of the model, which is obtained byintroducing three auxiliary fields: one scalar field X to impose the constraint (2.1), a Majoranafermion λ to impose the constraint (5.1), and a Hubbard–Stratonovich scalar field τ to integrateout the quartic fermionic term in the Lagrangian (5.2). The resulting Euclidean Lagrangian,which includes the coupling to the chemical potential h for both S and Ψ fields, is given by L susy h = 12 g (cid:26) ∂ µ S · ∂ µ S + 2i h ( S ∂ S − S ∂ S ) + h ( S + · · · + S N −
1) + X ( S − Ψ · /∂ Ψ + 2 λ S · Ψ + τ + τ Ψ · Ψ − i h (cid:0) Ψ γ Ψ − Ψ γ Ψ (cid:1)(cid:27) . (5.5)We now expand the Lagrangian around the following classical vacuum, S ( x ) = (cid:16) σ, , . . . , (cid:17) + (cid:112) π ∆ λ (cid:16) ˜ σ ( x ) , ˜ σ ( x ) , η ( x ) , . . . , η N − ( x ) (cid:17) , Ψ ( x ) = (cid:16) , . . . , (cid:17) + (cid:112) π ∆ λ (cid:16) ν ( x ) , ν ( x ) , ψ ( x ) , . . . , ψ N − ( x ) (cid:17) ,X ( x ) = χ + (cid:112) π ∆ λ ˜ χ ( x ) , τ ( x ) = τ + (cid:112) π ∆ λ ˜ τ ( x ) , λ ( x ) = 0 + (cid:112) π ∆ λ λ ( x ) . (5.6)Up to linear terms in the fields, one finds that the Lagrangian can be written as12 π ∆ λ L susytree + L susyG + (cid:112) π ∆ λ L susyint . (5.7)The tree-level Lagrangian is given by L susytree = χ σ −
1) + τ − h . (5.8) In Euclidean signature, the gamma matrices satisfy the anti-commutator relation { γ µ , γ ν } = 2 η µνE I , where η µνE is the Euclidean metric. After a Wick rotation of the action, the Eulidean gamma matrices γ E that enter in theEuclidean Lagrangian (5.5) are connected to the Minskowki matrices γ M by γ = γ M and γ j = i γ jM , for j (cid:54) = 0. – 23 –he quadratic terms are given by L susyG = 12 η · ( − ∂ + χ + h ) η + 12 ˜ τ + 12 ˜ σ ˜ σ ˜ χ T − ∂ + χ h∂ σ − h∂ − ∂ + χ σ ˜ σ ˜ σ ˜ χ + 12 ψ · ( /∂ + τ ) ψ + 12 ν ν λ T /∂ + τ − i hγ σ i hγ /∂ + τ σ ν ν λ , (5.9)where ψ = ( ψ , . . . , ψ N − ). Finally, the interaction terms are given by L susyint = 12 ˜ χ η · η + 12 ˜ χ (cid:16) ˜ σ + ˜ σ (cid:17) + λ (cid:16) ˜ σ ν + ˜ σ ν + η · ψ (cid:17) + 12 ˜ τ (cid:16) ν ν + ν ν + ψ · ψ (cid:17) . (5.10)From (5.9) we can compute the propagators in momentum space. The propagators of the bosonfields η , ˜ σ , ˜ σ and ˜ χ were already obtained in (3.7). There is, however, an additional boson ˜ τ with propagator D ˜ τ ˜ τ = 1 . (5.11)The fermion propagators are S ν ν = S ν ν = S ν ν = 0 , S ν ν = 1i /k + τ ,S ν λ = S λν = 1 σ , S ν λ = i hσ γ /k + τ , S λν = − i hσ /k + τ γ ,S λλ = − σ (cid:20) i /k + τ − γ h i /k + τ γ (cid:21) , S ψ i ψ j = δ ij /k + τ , i, j = 1 , . . . , N − . (5.12)The propagators of the bosonic and the fermionic fields are represented diagrammatically as infigure 5 and figure 6. The interaction terms are represented by the vertices in figure 7.We can now calculate the effective potential in an expansion in powers of ∆, as in (3.8). Theleading order term comes from the tree-level Lagrangian together with the one-loop contributionsof the η bosons and ψ fermions, for which there are 1 / ∆ = N − V (0) ( σ, χ, τ ) = 14 πλ (cid:2) χ ( σ −
1) + τ − h (cid:3) + 12 (cid:90) d d k (2 π ) d log (cid:18) k + h + χk + τ (cid:19) . (5.13)By using dimensional regularization, we obtain V (0) ( σ, χ, τ ; h ) = 14 πλ (cid:2) χ ( σ −
1) + τ − h (cid:3) + ( h + χ ) d/ (4 π ) d/ d Γ (cid:16) (cid:15) (cid:17) − ( τ ) d/ (4 π ) d/ d Γ (cid:16) (cid:15) (cid:17) . (5.14)The extremization procedure for χ and σ is identical to the non-supersymmetric case. As for τ ,much like χ , there is the non-perturbative choice τ (cid:54) = 0 and the trivial one τ = 0. We choosethe latter to connect our result with perturbation theory. After renormalizing the coupling likein the non-supersymmetric case, we find δ F susy(0) ( h ) = − h π (cid:26) λ + log (cid:18) hµ (cid:19) − (cid:27) . (5.15)– 24 – σ ˜ σ ˜ χ ˜ σ ˜ σ ˜ χ η η ˜ χν ν ˜ τ ν ν ˜ τ ψ ψ ˜ τλ ˜ σ ν λ ˜ σ ν λ η ψ Figure 7 : Interaction terms in the supersymmetric non-linear sigma model.
There are three types of ring diagrams that contribute to the effective potential at next-to-leadingorder in ∆: ring diagrams with bosonic η bubbles, ring diagrams with fermionic ψ bubbles, andring diagrams with mixed ψ - η bubbles. The first type of ring diagrams are the same ones of thebosonic sigma model, so we do not have to compute them again. Since the propagators for ψ and˜ τ are both h independent, fermionic ring diagrams do not contribute to δ F ( h ) after subtractionat h = 0. Thus we only need to consider mixed ring diagrams.The ψ - η bubbles are connected by λ lines. Then the contribution of mixed ring diagrams tothe effective potential is (cid:88) m ≥ m (cid:90) d d k (2 π ) d Tr (cid:104)(cid:16) πλ S λλ ( k )Π µψη ( k, h + χ, τ ) γ µ (cid:17) m (cid:105) . (5.16)A minus sign arises from the single fermion loop that runs across the entire diagram. This signgets canceled by an additional sign that appears in the computation of the effective potential.The polarization loop is in this caseΠ µψη ( k, M , τ ) = (cid:90) d d q (2 π ) d q + M − i( k µ + q µ ) + τ ( k + q ) + τ . (5.17)This integral can be computed with standard Feynman techniques and, after evaluation at thevacuum χ (0) = τ (0) = 0, we obtainΠ µψη ( k, h ,
0) = − i k µ Π ψη ( k , h ) , Π ψη ( k , h ) = Γ(1 + (cid:15) )(4 π ) d/ (cid:90) d y y − (cid:15)/ (cid:2) h + (1 − y ) k (cid:3) (cid:15)/ = h − − (cid:15) Γ(1 + (cid:15) )(4 π ) d/ (1 + x ) − − (cid:15)/ − (cid:15)/ F (cid:18) (cid:15) , − (cid:15) , − (cid:15) xx + 1 (cid:19) . (5.18)– 25 – a) Bosonic (b) Fermionic (c) Mixed Figure 8 : The three types of ring diagrams that appear at order ∆ . (a) Ring diagram fromthe regular non-linear sigma model. (b) Ring diagram with pure fermion bubbles. These type ofdiagrams do not depend on h . (c) Ring diagrams with mixed boson-fermion bubbles, which wecompute in this section.Again, as in the non-supersymmetric case, we have expressed the polarization loop in terms of ahypergeometric function. The free energy at next-to-leading order in ∆ from bosonic plus mixedring diagrams can now be written as δ F susy(1) ( h ) = − h ν − (cid:15) π πν (cid:15) h (cid:88) m ≥ ( − m m (cid:32) πλ σ (cid:33) m ( I m − I m ) . (5.19)The integrals I m , corresponding to bosonic diagrams, were already defined in (3.24), and we havea set of new integrals from the mixed diagrams given by I m = (cid:90) d d k (2 π ) d (cid:2) Π ψη ( k , h ) (cid:3) m Tr (cid:20)(cid:18) k + k h γ /kγ /k (cid:19) m (cid:21) . (5.20)The front factor in (5.19) has been extracted for better comparison with the non-supersymmetricresult of (3.23). As we already mentioned, one important difference in the present model is thatwe do not need a renormalization constant to cancel the divergences of the ring diagrams. Instead,there is a total cancellation of divergences between bosonic and mixed diagrams. We will see thisexplicitly in section 5.2, thus proving that the beta function at subleading order is β (1) ( λ ) = 0.At this stage, it is convenient to compute the trace in (5.20). Expanding with the binomialtheorem, we obtainTr (cid:20)(cid:18) k + k h γ /kγ /k (cid:19) m (cid:21) = m (cid:88) (cid:96) =0 (cid:18) m(cid:96) (cid:19) ( h ) (cid:96) k m − (cid:96) Tr (cid:104) ( γ /k ) (cid:96) (cid:105) . (5.21)We can calcuate the trace recursively in arbitrary dimension d :Tr (cid:104) ( γ /k ) (cid:96) (cid:105) = 2 (cid:98) d (cid:99) (cid:34) (cid:18) k + i (cid:113) k − k (cid:19) (cid:96) + 12 (cid:18) k − i (cid:113) k − k (cid:19) (cid:96) (cid:35) . (5.22)As is the standard procedure, we take the dimension of the spinor representation to be a fixedinteger 2 (cid:98) d (cid:99) = 2. In d -dimensional spherical coordinates we can always pick k = k cos θ . Then– 26 –e obtain Tr (cid:104) ( γ /k ) (cid:96) (cid:105) = 2 k (cid:96) cos(2 (cid:96)θ ) , (5.23)and the momentum angular integral can now be computed term by term in (5.21), yielding I m = h − (cid:15) ( m +1) S d (2 π ) d (cid:20) Γ(1 + (cid:15) )(4 π ) d/ (cid:21) m m (cid:88) (cid:96) =0 (cid:18) m(cid:96) (cid:19) ( − (cid:96) Γ( d ) Γ( d − (cid:96) )Γ( d + (cid:96) ) I m,(cid:96) , (5.24)where we singled out the integrals I m,(cid:96) = 1(1 − (cid:15) ) m (cid:90) d z z m − (cid:96) − (cid:15)/ (1 − z ) − (cid:96) − ( m +1) (cid:15)/ F (cid:16) (cid:15) , − (cid:15) , − (cid:15) z (cid:17) m . (5.25)For (cid:96) ≥ (cid:15) = 0. Since the factor 1 / Γ( d/ − (cid:96) ) vanishes in the sum of(5.24) for (cid:15) = 0, none of the terms with (cid:96) ≥ The goal in this section is to check that the subleading free energy (5.19) in the supersymmetricmodel is already finite without the need of renormalization. For that, we follow similar techniquesto those in section 3.2 and 3.3. That is, we want to isolate the singular part of the integrals I m,(cid:96) ,as we did in (3.33)–(3.34). We start by expressing the hypergeometric function in (5.25) as asum of two terms, by using fractional linear transformations: F (cid:16) (cid:15) , − (cid:15) , − (cid:15) z (cid:17) = − − (cid:15)(cid:15) Γ(1 − (cid:15) ) Γ(1 − (cid:15) ) z − (cid:15)/ × (cid:20) − Γ(1 − (cid:15) )Γ(1 − (cid:15) ) (1 − z ) − (cid:15)/ F (cid:16) (cid:15) , − (cid:15) , − (cid:15) − z (cid:17)(cid:21) . (5.26)We now go back to (5.25), plug in (5.26) and change the variable of integration from z to 1 − z .We get, I m,(cid:96) = (cid:18) − (cid:15) Γ(1 − (cid:15) ) Γ(1 − (cid:15) ) (cid:19) m B m,(cid:96) , (5.27) B m,(cid:96) = (cid:90) d z (1 − z ) − (cid:96) +( m − (cid:15)/ z − (cid:96) − ( m +1) (cid:15)/ (cid:20) − Γ(1 − (cid:15) )Γ(1 − (cid:15) ) z − (cid:15)/ F (cid:16) (cid:15) , − (cid:15) , − (cid:15) z (cid:17)(cid:21) m . (5.28)By expanding the square bracket with the binomial theorem and integrating term by term withthe Euler beta function, we obtain B m,(cid:96) = (cid:88) k ≥ B m,(cid:96),k , (5.29) B m,(cid:96),k = m (cid:88) s =0 (cid:18) ms (cid:19)(cid:18) − Γ(1 − (cid:15) )Γ(1 − (cid:15) ) (cid:19) s Γ (cid:0) − (cid:96) + ( m − (cid:15) + 1 (cid:1) Γ (cid:0) k + (cid:96) + ( m − s + 1) (cid:15) − (cid:1) Γ (cid:0) k + (2 m − s ) (cid:15) (cid:1) d ( s ) k . (5.30)The coefficients d ( s ) k are defined by the Taylor expansion (cid:104) F (cid:16) (cid:15) , − (cid:15) , − (cid:15) z (cid:17)(cid:105) s = (cid:88) k ≥ d ( s ) k z k . (5.31)– 27 –he Laurent expansion of B m,(cid:96) up to order (cid:15) m is obtained by summing only the terms k = 0, 1for (cid:96) = 0 and k = 1 for (cid:96) = 1. This follows from a computation similar to the one in appendixA. We then find, B m, = B m, , + B m, , + O (cid:0) (cid:15) m +1 (cid:1) , B m, = B m, , + O (cid:0) (cid:15) m +1 (cid:1) . (5.32)This result is slightly different to the non-supersymmetric case (compare (5.32) to (A.1)), and itgreatly simplifies the computation of mixed ring diagrams, as it means that the terms in (5.32)already incorporate the singular part plus the finite part of I m,(cid:96) .Equation (5.30) gives a divergent result for m = 1, (cid:96) = 1 (even for arbitrary (cid:15) ). In thisspecial case we have to compute the integral (5.25) explicitly. Instead of using the hypergeometricrepresentation, we use the integral representation of (5.18) and commute the x and y integrals.Then we can compute both integrals analytically in terms of the Euler beta function B ( x, y ).We obtain, I , = B (cid:16) − (cid:15) , (cid:15) (cid:17) B (cid:16) − (cid:15) , (cid:15) (cid:17) = 2 (cid:15) + π O (cid:0) (cid:15) (cid:1) . (5.33)We are now ready to compute the divergent part arising from mixed ring diagrams in (5.19).We start by considering the sum S m = B m, , + B m, , + m(cid:15) − (cid:15) B m, , , (5.34)which can be written in the form S m = Γ (cid:16) ( m − (cid:15) (cid:17) m (cid:88) s =0 (cid:18) ms (cid:19) ( − s m − s + 1 f m ( s(cid:15), (cid:15) ) , (5.35)where f m ( y, x ) = Γ (cid:0) ( m + 1) x − y + 1 (cid:1) Γ (cid:0) mx − y (cid:1) (cid:26) m − m + 1) x − y − yx ( m − x − mx − y ) + m − x (cid:27) e y h ( x ) , h ( x ) = 1 x log (cid:18) Γ(1 − x )Γ(1 − x ) (cid:19) . (5.36)We can now perform the sum in s , as we did in (3.42), and we obtain the result S m = 2( − m m + 1 (cid:18) Γ(1 − (cid:15) )Γ(1 − (cid:15) ) (cid:19) m +1 (cid:15) − (cid:15) − − − m ( m + 1)( m − (cid:20) d m +1 f m ( y, y m +1 (cid:21) y =0 (cid:15) m + O (cid:0) (cid:15) m +1 (cid:1) . (5.37)The first term in the r.h.s. is equal to S m up to order (cid:15) m − , so that term is sufficient to computethe singular part of the integrals I m . The second term, contributing to order (cid:15) m , will be neededfor the computation of the finite part of I m .At this point, it becomes a simple exercise to verify that the singular parts of I m and I m cancel each other in (5.19). We first observe that S m = 2 · − (cid:15) ( m +1) S m + O ( (cid:15) m ) . (5.38)If we now rewrite the sum in (3.27) in terms of S m , and the sum in (5.20) in terms of S m , weobtain I m = I m + O (cid:0) (cid:15) (cid:1) . (5.39)– 28 –his proves that the Laurent expansion of (5.19) starts at the constant term. In other words,there is a complete cancellation of divergent terms between bosonic and mixed ring diagrams,thus showing that β (1) ( λ ) = 0 from a direct computation in perturbation theory. We can also organize the finite parts of the mixed ring diagrams contributing to the free energyin a manner similar to what we did in (3.60) for bosonic ring diagrams. We note, however, thatthere are a few simplifications in this case. The analogue of Z , corresponding to the sum of theintegrals I m,(cid:96) for (cid:96) ≥
2, is equal to 0. The analogue of X , corresponding to the contributionfrom the terms we missed in the sum (5.34), is also 0, as is easily inferred from the result (5.32).Moreover, the analogue of W is present in the mixed ring diagrams, but it exactly cancels withthat of the bosonic diagrams (this can be deduced from the derivation that led to (5.38)).In short, we only need to consider the analogue of Y , which arises from the second term inthe r.h.s. of (5.37). After simplifying f m ( y,
0) we find f m ( y,
0) = y (cid:26) m − y/ − m (cid:27) , (5.40)and the computation of the ( m + 1)-th derivative follows naturally from the geometric series.Then, the contribution to the free energy is simply Y (cid:18) λ ; hµ (cid:19) = 12 π (cid:88) m ≥ ( m − m (cid:18) λ λ log( h/µ ) (cid:19) m . (5.41)Putting all our results together we obtain δ F susy(1) ( h ) = − h (cid:26) X (cid:18) λ ; hµ (cid:19) + Y (cid:18) λ ; hµ (cid:19) + Z (cid:18) λ ; hµ (cid:19) + Y (cid:18) λ ; hµ (cid:19)(cid:27) , (5.42)Thus, the ground state energy is E susy(1) ( α ) = − πα (cid:2) X ( α ; 1) + Y ( α ; 1) + Z ( α ; 1) + Y ( α ; 1) (cid:3) . (5.43)Here, α is defined by the equation (2.21) with β given in (5.3) and ξ = 0. We find perfect agree-ment between (5.43) and the coefficients obtained from the Bethe ansatz, which were calculatedup to order α in [24].As we previously observed in (5.4), the perturbative expansion of E susy(1) ( α ) does not containany rational term (at least to the available order). In our perturbative computation, we verifiedthat Y ( α ; 1) cancels the rational part of X ( α ; 1) + Y ( α ; 1), so that the coefficients are linearcombinations of Riemann zeta functions evaluated at odd arguments.As we did in the bosonic case, we can obtain from these results the location of the renormalonsingularities and their associated trans-series. The mixed diagrams (5.41) contribute only to theIR renormalon singularity at ζ = 2, and we findIm E susy(1) ( α ) = π (cid:20) α e − /α + (cid:0) − α (cid:1) e /α + (cid:18) α + 72 α (cid:19) e /α + (cid:18) α + 200003 α + 65003 α + 200 α (cid:19) e /α + O (cid:16) e /α (cid:17)(cid:21) , (5.44)– 29 –here only the sign of the IR term differs from the regular case. The asymptotic behaviourextracted from this discontinuity matches the coefficients with the expected precision. Let us notethat, although divergences cancel between bosonic and mixed ring diagrams, the IR renormalonat ζ = 2 does not cancel. This is in contrast to the cancelation of leading IR renormalons thatoccurs in some supersymmetric theories according to [42, 43]. The Bethe ansatz calculation of the free energy in two-dimensional integrable models is one ofthe most interesting exact results in quantum field theory. It makes it possible to understandquantitatively many important aspects of asymptotically free theories, like dynamical mass gen-eration and the presence of renormalons and condensates. For this reason, it is important to testthe predictions of the Bethe ansatz against more conventional methods in quantum field theory.This has been done in the past, in particular in the case of the Gross–Neveu model [17, 18].However, a direct test against perturbation theory was limited until very recently by the diffi-culty of extracting perturbative series from the Bethe ansatz. The results of [22–24] have openedthe possibility of such a test, and this has been the first goal of this paper. We have performeda calculation of the free energy at next-to-leading order in the 1 /N expansion and at all loops,and we have verified the predictions of the Bethe ansatz up to very high order in the couplingconstant, both in the non-linear sigma model and its supersymmetric extension. The second goalof this paper has been to use this analytic, all-orders result to obtain detailed information aboutrenormalon singularities and their associated trans-series, at leading order in the 1 /N expansion.There are many problems open by our investigation and by closely related efforts. First of all,one could extend the tests presented here to other integrable field theories. In [24], explicit resultsfor the perturbative series of the free energy have been obtained for the Gross–Neveu model andthe principal chiral field with different choices of chemical potentials. There are also resultsfor integrable, non-relativistic models, like the Gaudin–Yang model [25]. All these perturbativeexpansions, obtained from the Bethe ansatz, could be tested by conventional techniques, andthese tests would provide additional insights. It should be mentioned that, in the case of theprincipal chiral field, which is a matrix model, a direct perturbative calculation at higher loops ismore challenging than for the vector models that we studied in this paper. Another interestingproblem is to extend our results to higher orders in the 1 /N expansion, even for the modelsconsidered here.In our calculation of the free energy we did an expansion around the trivial large N vacuum,in order to make contact with perturbation theory. However, it is well-known that both thenon-linear sigma model and its supersymmetric version have a non-trivial large N saddle point,which leads to a non-perturbative mass gap. In fact, previous analysis of renormalons in thenon-linear sigma model have been based on expansions around the non-trivial large N saddlepoint [57–59]. A preliminary calculation of the free energy considered in this paper around thisnon-trivial vacuum seems to lead to a purely non-perturbative result, at next-to-leading order inthe 1 /N expansion. It would be very interesting to understand more precisely the relationshipbetween the expansions around these two very different vacua, and the rˆole of renormalons ineach of them. Note that similar issues are raised in the two-dimensional linear O ( N ) model.There, the ground state energy can be calculated in the trivial large N vacuum, similarly towhat we did in this paper [33], but it can be also calculated in the non-trivial large N vacuum[60]. – 30 –he exact Bethe ansatz solution contains in principle all the information in the problem, bothperturbative and non-perturbative. One of the goals of the resurgence program is to “semiclas-sically decode” this exact answer, by writing it as a Borel–´Ecalle resummation of a trans-series.Substantial evidence that this can be done was obtained recently in [28, 29], in the case of the O (4) sigma model, by an impressive calculation. Additional evidence has been given in the con-text of integrable many-body systems, in [25, 27]. The first step in the semiclassical decoding isto obtain the explicit form of the trans-series. So far, in all the problems solved by the Betheansatz, this has been done by looking at the large order behavior of the perturbative sector. Amore challenging problem is to extract the trans-series directly from the integral equation defin-ing the free energy. This would require an extension of Volin’s method by explicitly includingthe exponentially suppressed corrections.Another important open question is to provide a physical interpretation of the trans-seriesthat we have obtained. It was argued in [24] that the IR renormalon at ζ = 2 in the non-linearsigma model might be explained by the condensate of the operator O = ∂ µ S · ∂ µ S . Is it possibleto devise some sort of perturbation theory in the background of the condensate which allows usto reproduce analytically the trans-series? Such a generalized perturbation theory, akin to theone used in the calculation of OPEs [61], would provide one of the missing ingredients in ourunderstanding of quantum field theory. Acknowledgements
We would like to thank Lorenzo di Pietro, Marco Serone and Giacomo Sverbeglieri for usefuldiscussions. This work has been supported in part by the Fonds National Suisse, subsidy 200020-175539, by the NCCR 51NF40-182902 “The Mathematics of Physics” (SwissMAP), and by theERC-SyG project “Recursive and Exact New Quantum Theory” (ReNewQuantum), which re-ceived funding from the European Research Council (ERC) under the European Union’s Horizon2020 research and innovation program, grant agreement No. 810573.
A Integral representation of the series (cid:98) B m,(cid:96) In this appendix we will prove that the integrals (3.31) for (cid:96) = 0 , B m, = B m, , + B m, , + O ( (cid:15) m ) , B m, = B m, , + O ( (cid:15) m ) . (A.1)where the coefficients B m,(cid:96),k are given in (3.34). In particular, this means that the singular partof the integrals I m can be extracted from the combination in (3.39).To prove (A.1), we start from (3.31) and we expand the binomial. We then subtract theterms k = 0 and/or k = 1 to obtain the following expression for the sums in (3.56): (cid:98) B m,(cid:96) = (cid:90) d z (1 − z ) m/ − (cid:96) − (cid:15)/ z − (cid:96) − ( m +1) (cid:15)/ m (cid:88) s =0 (cid:18) ms (cid:19) ( − s g (cid:96) ( s(cid:15), (cid:15) ; z ) . (A.2)In this equation, g (cid:96) ( y, (cid:15) ; z ) = z − y/ e y g ( (cid:15) ) (cid:26) e y Φ( (cid:15),z ) − yz − (cid:15) ) δ (cid:96), (cid:27) , (A.3)– 31 –here g is given in (3.41), and Φ( (cid:15), z ) is defined byΦ( (cid:15), z ) = 1 (cid:15) log (cid:20) F (cid:18) − (cid:15) , , − (cid:15) z (cid:19)(cid:21) = log (cid:20) √ − z (cid:21) + O ( (cid:15) ) . (A.4)We now Taylor expand g (cid:96) ( y, x ; z ) in its first argument g (cid:96) ( y, x ; z ) = (cid:88) r ≥ ( s(cid:15) ) r r ! (cid:20) d r g (cid:96) ( y, x ; z )d y r (cid:21) y =0 . (A.5)The binomial identity m (cid:88) s =0 (cid:18) ms (cid:19) ( − s s r = (cid:40) , ≤ r < m, ( − m m ! , r = m, (A.6)allows us to perform the sum over s in (A.2) for each term of the Taylor expansion. Every termwith r < m vanishes after the binomial sum, and we obtain (cid:98) B m,(cid:96) = (cid:15) m ( − m (cid:90) d z (1 − z ) m/ − (cid:96) − (cid:15)/ z − (cid:96) − ( m +1) (cid:15)/ (cid:20) d m g (cid:96) ( y, (cid:15) ; z )d y m (cid:21) y =0 + O (cid:0) (cid:15) m +1 (cid:1) . (A.7)The O (cid:0) (cid:15) m +1 (cid:1) contains further terms of the Taylor expansion that we have ignored. Becausethe above integral is convergent at (cid:15) = 0, we can set (cid:15) = 0 in the integrand of (A.7) (furthercorrections will be of order (cid:15) m +1 ). This completes the proof of (A.1) and yields the result in(3.57), where g (cid:96) ( y ; z ) ≡ g (cid:96) ( y, z ). B Analytic computation of the integrals
In this appendix we derive and present analytic results for the integrals I m,(cid:96) and (cid:98) B m,(cid:96) appearingin (3.60).Let us start with the integrals (cid:98) B m,(cid:96) , defined in (3.57). One way to compute them is to obtaina closed expression for the generating functional g m,(cid:96) ( y ) ≡ (cid:90) d z (1 − z ) m/ − (cid:96) z − (cid:96) g (cid:96) ( y, z ) , (cid:96) = 0 , , (B.1)and then extract the desired integral by taking the m -th Taylor coefficient of g m,(cid:96) ( y ) around y = 0.First we fix (cid:96) = 1, in which case we have the closed expression g m, ( y ) = 2 Γ( m + 1)Γ( − y ) F (cid:0) m, − y ; m − y ; − (cid:1) m Γ( m − y ) − y +1 Γ( m + 1)Γ( − y ) m Γ( m − y ) . (B.2)This expression is not suited for a Taylor expansion around y = 0 (because of the hypergeometricfunction), but it can be put in a more appropriate form by observing that F (cid:16) m, − y m − y − (cid:17) = 12 F (cid:16) − y , m − − y − ( m −
1) + (2 m − − (cid:17) . (B.3)– 32 –e then use Theorem 3 in [62] and expand the singular gamma functions around their poles,which leads to the expression F ( a, j, a − j + m ; −
1) = 2 − a √ π Γ( a − j + m )Γ( j )Γ( m − j ) × m − (cid:88) k =0 ( − k + j (cid:18) m − k (cid:19) − k Γ( a + k )Γ( a )Γ( a + k +12 ) ψ (0) ( a + k − j + 1)Γ( a + k − j + 1) , (B.4)where ψ (0) is the polygamma function. After some additional massaging, we obtain the followingalternative expression for (B.2), in terms of Pochhammer symbols: g m, ( y ) = ( − m m − m − (cid:88) k =0 ( − k +1 (cid:18) m − k (cid:19)(cid:18) k − m − y (cid:19) m − ψ (0) (cid:18) k − m − y (cid:19) + 2 y Γ( m )Γ( − y )Γ( m − y ) . (B.5)For (cid:96) = 0 and m ≥
2, a similar calculation leads to g m, ( y ) = 18Γ( m − m (cid:88) k =0 ( − k (cid:18) mk (cid:19)(cid:18) − k y (cid:19) m − ψ (0) (cid:18) k − m − y (cid:19) − y (4 m + ( y − y )Γ( m + 1)Γ( − y − m − y + 1) . (B.6)The expressions (B.5) and (B.6) are now ready to be Taylor expanded around y = 0. From theTaylor coefficients we can then extract analytic expressions for the integrals in (3.57).Let us now consider the integrals I m,(cid:96) , defined in (3.28), with 2 ≤ (cid:96) ≤ m . We first focuson the particular case (cid:96) = m : I m,m . Knowing that the integrals converge without the need ofregularisation, we can set (cid:15) = 0, and the hypergeometric function in the integrand becomes F (cid:18) ,
12 ; 32 ; z (cid:19) = tanh − ( √ z ) √ z . (B.7)Then, after the change of variable x = tanh − ( √ z ), the integral (3.28) at (cid:15) = 0 can be writtenas I m,m = 12 (cid:90) ∞ x m e − ( m − x/ (1 − e − x ) m − d x. (B.8)By expanding the denominator and integrating term by term, we obtain a sum of Hurwitz zetafunctions: I m,m = m !2 m − (cid:88) i =0 b i ζ (cid:18) m − i + 1 , m − (cid:19) , (B.9)where b i = 1( m − m − (cid:88) j = i (cid:18) ji (cid:19)(cid:18) m − (cid:19) j − i S ( j ) m − (B.10)– 33 –nd S ( i ) m are the Stirling numbers of the first kind. The Hurwitz zeta function appears with integervalues in the first argument and both integer and half-integer values in the second argument. Inthese cases, the zeta function reduces to ζ ( n, k ) = ζ ( n ) − H ( n ) k − , k ∈ Z , (B.11) ζ ( n, k ) = (2 n − ζ ( n ) − n − k − / (cid:88) q =0 q + ) n , k + 12 ∈ Z , (B.12)where H ( n ) k is the k -th harmonic number of order n , and ζ ( z ) is the Riemann zeta function.This completes the analytic computation of I m,m . However, our result (B.9) can be written in acompact form by using the conventions of umbral calculus: I m,m = m ( m − ζ m +1 m − (cid:89) i =0 (cid:18) ζ − m −
32 + i (cid:19) , ζ n (cid:55)→ ζ (cid:18) n, m − (cid:19) . (B.13)In these conventions, we first expand the product as a finite polynomial in an abstract variable ζ , and then we replace each power of ζ by the Hurwitz zeta function, according to the mappingin (B.13).By using similar tricks, we can also compute I m,(cid:96) , with (cid:96) = 2 , . . . , m −
1, in closed form.However, by integrating (3.28) by parts, one finds the following recursion formula: I m,(cid:96) = 2 (cid:96) − (cid:104) ( m − (cid:96) ) I m,(cid:96) +1 + m I m − ,(cid:96) (cid:105) , m ≥ , (cid:96) ≥ . (B.14)Using this recursion and the values I m,m for m ≥ I m,(cid:96) .One can use this derivation to obtain results on the transcendentality of the integrals. Byobserving that the function m − (cid:88) i =0 b i x i = − sin (cid:0) πx + ( m − π (cid:1) π ( m − (cid:18) m − x − (cid:19) Γ (cid:18) m x − (cid:19) (B.15)is odd for odd m and even for even m , one shows that the b i vanish for m − i + 1 even. Thus, onlyHurwitz zeta functions ζ ( n, m − ) where n is odd show up in (B.9). Furthermore, the rationalparts in (B.12) are a finite sum of rational numbers, for both m odd and even. By exchangingthese sums with the sum over i in (B.9), one can easily show that the total rational part vanishes.Thus I m,m is a linear combination of the values ζ (2 k + 1), with k ∈ N and 1 ≤ k ≤ m/
2. Thanksto the recursion relation (B.14), the same statement trivially applies to all I m,(cid:96) with 2 ≤ (cid:96) ≤ m . C Asymptotic expansions of the discontinuities
To obtain the discontinuity of X ( λ ; 1), as presented in (4.11), we first have to express the residuesin (4.10) in terms of the position of the poles z and z . This can be easily done with L’Hˆopital’srule. Then it suffices to compute z and z perturbatively in powers of e /λ . These perturbativesolutions can be conveniently obtained by defining variables v and ξ such that z = 1 − (1 − ξv ) , ξ = e /λ , (C.1)– 34 –nd define v i such that z i = 1 − (1 − ξv i ) . The variable ξ in (C.1) should not be confused withthe parameter introduced in (2.22). From (4.9) we obtain v = (cid:18) ξv + e ξv − ξv log ξ (cid:19) , v = (cid:18) ξ + e − ξv − ξv log ξ (cid:19) − . (C.2)We now write the perturbative solution as v i = 1 + v (1) i ξ + v (2) i ξ + . . . , (C.3)If we plug this ansatz in (C.2) we can calculate the coefficients v ( j ) i recursively, and we find v = 1 + ξ (2 log ξ + 1) + ξ (cid:0) ξ + 10 log ξ + 2 (cid:1) + O (cid:0) ξ (cid:1) ,v = 1 + ξ (2 log ξ −
1) + ξ (cid:0) ξ − ξ + 1 (cid:1) + O (cid:0) ξ (cid:1) . (C.4)To compute the discontinuity of Y ( λ ; 1) presented in (4.12), one distinguishes the case λ > λ <
0. We recall that the exponential integral E ( x ) can be written as an entirefunction plus − log( x ). Thus the discontinuity of E is − π i when its argument is negative. Inthe case of λ >
0, we find that the argument of E is negative for all z ∈ (0 , Y ( λ >
0; 1) = − π (cid:90) d zz (1 − z ) z e λ √ − z λ √ − z ( − π i) = ie − /λ . (C.5)When λ < X ( λ ; 1). We have a logarithmic discontinuitywhen z ∈ (0 , z ), and a pole at z = z . In total we finddisc Y ( λ <
0; 1) = − π i (cid:34) π (cid:90) z d zz (1 − z ) z e − λ √ − z λ √ − z + 14 π Res (cid:0) Y ( λ √ − z, z ) , z = z (cid:1)(cid:35) = − i2 (cid:34) e − /λ − e − λ √ − z − z + 2 λz (1 − z ) (cid:20) dd z √ − z log (cid:18) √ z (cid:19)(cid:21) − (cid:35) , (C.6)where we have to plug the z solution in (C.1) and (C.4), leading to (4.12).For the discontinuity of Z ( λ ; 1), it is convenient to change variables as z = (1 − ξv ) . (C.7)From (4.13) we find that z = (1 − ξv ) , (C.8)where v is the same as in (C.2) and (C.4). The asymptotic approximation to the residue of thepole in (4.7) now follows in the same way as for the pole of Y ( λ ; 1). Meanwhile, from (4.15) weobtain a much more simple equation for v , v = e − ξ log ξv (1 − ξv ) , (C.9)which is the same equation found in the closely related calculations of [27, 41]. Thus, we cantake the same solution v = 1 − ξ (2 log( ξ ) + 1) + ξ (cid:0) ξ + 6 log ξ + 1 (cid:1) + O (cid:0) ξ (cid:1) . (C.10)– 35 –e are left with computing the discontinuity of the logarithm term in (4.7). This discontinuity,given by (4.14), comes from the branch cut of the square root when Z ( z, λ ) <
0. Following thestrategy of [27, 41], we define the functions A ( v ) = 2( v − v ) ξ log ξ + 2 log ξ + log (cid:18) v − ξvv v − ξvv (cid:19) + 4 ξ (cid:20) ( v − ξv ) log (cid:18) − ξv ξv (cid:19) − ( v − ξv ) log (cid:18) − ξvξv (cid:19)(cid:21) ,A ( v ) = 2( v − v ) ξ log ξ + log (cid:18) v − ξvv v − ξvv (cid:19) . (C.11)By construction we have that A i ( v i ) = 0 for i = 2, 4. In terms of these functions, the discontinuityin (4.14) can be written as12 πξ (cid:90) v v d v − ξvv (1 − ξv ) log (cid:32) (cid:112) A ( v ) + (cid:112) A ( v ) (cid:112) A ( v ) − (cid:112) A ( v ) (cid:33) . (C.12)It is useful to introduce one further variable v = 1 + ξw, (C.13)and v i = 1 + ξw i . The advantage of using A , A and w is that we can expand the integrand of(C.12) as a power series in ξ , and we find the following structure: (cid:88) i ≥ (cid:34) P ( i ) l ( w, w , w ) log (cid:18) √ w − w + √ w − w √ w − w − √ w − w (cid:19) + i P ( i ) r ( w, w , w ) (cid:112) ( w − w )( w − w )( w − w ) i (cid:35) ξ i , (C.14)where P ( i ) l,r are polynomials. For example, up to order O (cid:0) ξ (cid:1) , we find, (cid:0) − ξw + O (cid:0) ξ (cid:1)(cid:1) log (cid:18) √ w − w + √ w − w √ w − w − √ w − w (cid:19) + i ξ (8 log ξ + w − w + 8)2( w − w ) (cid:112) ( w − w )( w − w ) + O (cid:0) ξ (cid:1) . (C.15)The coefficients of the ξ series in (C.14) are functions of w which are integrable on the interval[ w , w ]. After integration, we replace w and w by their respective perturbative solutionsthrough (C.4), (C.10) and (C.13). Summing up the result of this computation with the residueof the pole at z = z leads to (4.16). References [1] M. Beneke,
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