Antisymmetric Wilson loops in N = 4 SYM: from exact results to non-planar corrections
AAntisymmetric Wilson loops in N = 4 SYM: from exact results tonon-planar corrections Anthonny F. Canazas Garay , Alberto Faraggi , and Wolfgang M¨uck Instituto de F´ısica, Pontificia Universidad Cat´olica de ChileCasilla 306, Santiago, Chile Departamento de Ciencias F´ısicas, Facultad de Ciencias Exactas, Universidad Andr´esBelloSazie 2212, Piso 7, Santiago, Chile Dipartimento di Fisica “Ettore Pancini”, Universit`a degli Studi di Napoli “Federico II”Via Cintia, 80126 Napoli, Italy Istituto Nazionale di Fisica Nucleare, Sezione di NapoliVia Cintia, 80126 Napoli, Italy
Abstract
We consider the vacuum expectation values of 1 / N = 4super Yang-Mills theory in the totally antisymmetric representation of the gauge group U ( N ) or SU ( N ). Localization and matrix model techniques provide exact, but rather formal, expressionsfor these expectation values. In this paper we show how to extract the leading and sub-leadingbehavior in a 1 /N expansion with fixed ’t Hooft coupling starting from these exact results. Thisis done by exploiting the relation between the generating function of antisymmetric Wilson loopsand a finite-dimensional quantum system known as the truncated harmonic oscillator. Sum andintegral representations for the 1 /N terms are provided. a r X i v : . [ h e p - t h ] J u l ontents I ( y, z ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62.3 Truncated harmonic oscillator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72.4 Generating Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 I ( g, g ) and their traces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174.3 Generating function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 194.4 Holographic regime . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20 He N to an integral 26D Proof of (4.16)
28E Proof of (4.28) Wilson loop operators have played a central role in the development of gauge/gravity dualities [1,2].In this context, the -BPS circular loop in N = 4 super Yang-Mills theory with U ( N ) and SU ( N )gauge groups and ’t Hooft coupling λ has received special attention, mainly due to the conjectureput forward in [3, 4] that its expectation value is captured exactly, to all orders in N and λ , by aGaussian matrix model. This conjecture was later proved in [5] using supersymmetric localization,a technique that has since provided theorists with a several other exact results in supersymmetricgauge theories. These include, for example, Wilson loops preserving less supersymmetry [6–10],correlators of Wilson loops with chiral primary operators [11–18], correlators between Wilson loops[19], as well as Wilson loops and their correlators in N = 2 super Yang-Mills theory [20–26].2 key ingredient in the description of Wilson loop operators is the representation of the gaugegroup, typical ones for U ( N ) and SU ( N ) being the fundamental, totally symmetric and totallyantisymmetric representations. Depending on the rank of the representation, the holographic dualcorresponds to probe strings or D-branes [27, 28] propagating on AdS × S , or to fully back-reacted bubbling geometries [29–32]. In all cases, the on-shell action of the gravitational objectagrees perfectly with the matrix model calculation [33] at leading order in 1 /N and 1 / √ λ . Weexpect, however, that a thorough examination should yield a match at the next-to-leading order aswell. It is then crucial to systematically extract these corrections on both sides of the duality.From the gravitational perspective the calculation of sub-leading corrections amounts to analyz-ing semi-classical fluctuations of the background configurations and computing the correspondingone-loop partition functions. There has been considerable effort in this direction [34–44], althoughthe calculations are plagued with ambiguities inherent to string theory in curved spaces.On the gauge theory side, solving the matrix model is a non-trivial task, and computing sub-leading corrections is a conceptually clear, albeit technically difficult procedure. Recently, sub-leading corrections in 1 /N to the Wilson loop in the totally antisymmetric representation have beenfound using loop equation techniques [45] and topological recursion in the Gausssian matrix model[46] (see also [47] for the case of symmetric representations). A particularly interesting developmentwas reported in [48], where the authors managed to compute the exact vacuum expectation valueof the circular Wilson loop in arbitrary irreducible representations using the method of orthogonalpolynomials. In other words, they solved the matrix model exactly. While useful for some purposes,their results are rather formal, and it is not at all obvious how to extract the large- N (and large- λ )limit from them, much less any sub-leading corrections, except for simple cases. So far, they haveonly been used for numerical comparison with other approaches.In this paper, we address and solve the problem of extracting the leading and first sub-leadingterms in the 1 /N expansion of the generating function of totally antisymmetric Wilson loops startingfrom the exact results in [48]. Our motivation is two-fold. First, it is obviously interesting to seehow it can be done. Second, the techniques used in [45,46] solve the matrix model in the continuumlimit, in which the discrete poles of the matrix model resolvent give rise to a cut singularity anda continuous eigenvalue density. In other words, the analyticity properties of the resolvent changein this limit, and it would be interesting to see whether and where this has any implications. Weanticipate that we do not find any at order 1 /N .The paper is organized as follows. In section 2, we review the Gaussian matrix model andreproduce the formal result of [48] for the generating function of antisymmetric Wilson loops. Atthe same time, we establish a connection with the algebra of the truncated harmonic oscillator,which will be crucial for our developments. In section 3, we extract the leading large- N behaviour(fixed λ ) of the generating function from the formal solution of the matrix model. The mostsubstantial part of the paper is section 4, which contains the calculation of the 1 /N terms of the3ilson loop generating function, both for general λ and in the holographic regime of large λ . Ourresults are shown to agree with [45, 46], but turn out to be somewhat simpler in their final form.Section 5 contains the conclusions. Some technical details are deferred to the appendices. Localization techniques [5] map the expectation value of the circular Wilson loop in N = 4 SYMwith gauge group U ( N ) to an expectation value in a Gaussian matrix model. We begin this sectionby reviewing this model and developing some results that will be relevant for what follows. The Gaussian matrix model is defined by the partition function Z = (cid:90) [ dX ] exp (cid:18) − Nλ Tr (cid:0) X (cid:1)(cid:19) , (2.1)where X is a N × N hermitian matrix. If the gauge group is SU ( N ) the matrices are also traceless,condition that can be implemented with a Lagrange multiplier. The expectation value of anyquantity F ( X ) in the Hermitian ensemble is then given by (cid:104) F ( X ) (cid:105) = 1 Z (cid:90) [ dX ] F ( X ) exp (cid:18) − Nλ Tr (cid:0) X (cid:1)(cid:19) . (2.2)When F is invariant under similarity transformations, i.e., F ( V XV − ) = F ( X ), one can diagonalizethe matrix X in terms of its eigenvalues x n , n = 1 , . . . , N , and integrate out the remaining “angular”variables. This yields (cid:104) F ( X ) (cid:105) = 1 Z (cid:90) (cid:34) N (cid:89) m =1 dx m (cid:35) ∆ ( x ) F ( x ) exp (cid:32) − Nλ N (cid:88) n =1 x n (cid:33) , (2.3)where the transformation Jacobian∆( x ) = det (cid:2) x n − m (cid:3) = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) x x · · · x N − x x · · · x N − ... ... ... . . . ...1 x N x N · · · x N − N (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (2.4)is known as the Vandermonde determinant.The Vandermonde determinant enjoys several properties that we can exploit to our advantage.For example, one can show that∆( x ) = (cid:89) ≤ m 2. Some basic properties of the Hermite polynomials are listed in appendix A. I mn ( y, z ) = (cid:115) ( n − m − y m − n L ( m − n ) n − ( − yz ) = (cid:115) ( m − n − z n − m L ( n − m ) m − ( − yz ) , (2.15)where L ( α ) n ( x ) are Laguerre polynomials. Owing to the properties of these polynomials, the distinc-tion between the cases m ≥ n and m < n is not necessary. I ( y, z ) We now proceed to discuss some noteworthy attributes of the matrix (2.15). First, one easilyverifies that I (0 , 0) = ,I T ( y, z ) = I ( z, y ) ,I ( ξy, ξ − z ) = P ( ξ ) I ( y, z ) P − ( ξ ) , P mn ( ξ ) = ξ m δ mn . (2.16)(2.17)(2.18)Moreover, I ( y, 0) and I (0 , z ) are lower and upper triangular matrices, respectively, with unit di-agonal entries. Thus, det I ( y, 0) = det I (0 , z ) = 1. Some algebra shows that I ( y, z ) has the LUdecomposition I ( y, z ) = I ( y, I (0 , z ) . (2.19)It immediately follows that det I ( y, z ) = 1 . (2.20)The matrix also satisfies I ( y , I ( y , 0) = I ( y + y , , (2.21)which, combined with (2.17) and (2.19), implies I ( y + y , z + z ) = I ( y , I ( y , z ) I (0 , z ) . (2.22)We also deduce that I − ( y, z ) = I (0 , − z ) I ( − y, . (2.23)Furthermore, I ( y, z ) is similar to its inverse, because I − ( y, z ) = [ P ( − I (0 , z )] I ( y, z ) [ P ( − I (0 , z )] − , (2.24)as can be shown using (2.18), (2.19) and (2.23). This implies that the eigenvalues of I ( y, z ) comein reciprocal pairs and that its characteristic polynomial is either palindromic or anti-palindromic.The properties (2.21) are reminiscent of an exponential behavior. This is no coincidence. Indeed,one can verify the remarkable relation I (0 , z ) = e zA ⇒ I ( y, z ) = e yA T e zA , (2.25)6here A is the matrix given by A n,n +1 = √ n , (2.26)and all other entries vanishing. More explicitly, A = √ · · · 00 0 √ · · · · · · √ N − 10 0 0 · · · . (2.27)Note that any power of A is traceless and that A N = 0. The same is true for A T , of course. We promptly notice that A and A T are nothing more than the matrix representation of the ladderoperators of the harmonic oscillator truncated to the first N energy eigenstates. Surely, the numberoperator N = A T A = · · · 00 1 0 · · · 00 0 2 · · · · · · N − (2.28)is diagonal and satisfies [ N , A ] = − A , [ N , A T ] = A T . (2.29)However, A and A T themselves do not fulfill the Heisenberg algebra. Instead, their commutatorreads [ A, A T ] = − N diag (0 , . . . , , , (2.30)which is traceless, as must be for finite-dimensional operators.The N -dimensional quantum mechanical system formed by the operators A and A T , chieflycalled the truncated harmonic oscillator, is well known in quantum optics. We recall here some ofits properties, following the recent account [50] and refer the interested reader to more references inthat paper. In the next sections we will exploit this connection to the truncated harmonic oscillatorin order to extract the leading and sub-leading large N behavior of the circular Wilson loops.In a fairly obvious notation, we denote by | n (cid:105) , n = 0 , , . . . , N − n + 1-th position and zeros elsewhere. These states are eigenstates of the number operator7 and whence are called the number basis. According to (2.27), the ladder operators act uponthem as A | n (cid:105) = √ n | n − (cid:105) n = 0 , , . . . , N − ,A T | n (cid:105) = √ n + 1 | n + 1 (cid:105) n = 0 , , . . . , N − , A T | N − (cid:105) = 0 . (2.31)(2.32)Of course, the matrix elements (2.15) are I mn ( y, z ) = (cid:104) m | e yA T e zA | n (cid:105) . (2.33)Actually, this same expression can be obtained by computing the matrix elements of the displace-ment operator D ( y, z ) = e ya † e za = e yz e ya † + za in the full, infinite-dimensional, quantum harmonicoscillator [51], and then truncating to the first N number states.We will now construct a basis for the N -dimensional Hilbert space which is more suited to ourpurposes. Consider the states | ζ (cid:105) = N − (cid:88) n =0 He n ( ζ ) √ n ! | n (cid:105) , (2.34)where ζ is an arbitrary real parameter. By virtue of the recursion relations for the Hermite poly-nomials, the actions of A and A T on | ζ (cid:105) can be formally represented as A | ζ (cid:105) = (cid:18) ζ − ∂∂ζ (cid:19) | ζ (cid:105) − He N ( ζ ) (cid:112) ( N − | N − (cid:105) ,A T | ζ (cid:105) = ∂∂ζ | ζ (cid:105) . (2.35)(2.36)We see that | ζ (cid:105) is an approximate eigenstate of A + A T . It becomes an exact eigenstate if ζ is aroot of He N , (cid:0) A + A T (cid:1) | ζ (cid:105) = ζ | ζ (cid:105) , if He N ( ζ ) = 0 . (2.37)This construction makes explicit the fact that A + A T is actually the companion matrix of theHermite polynomial of degree N , namely,det (cid:2) ζ − (cid:0) A + A T (cid:1)(cid:3) = He N ( ζ ) . (2.38)Now, the Hermite polynomial He N ( ζ ) has precisely N distinct roots, which we denote in in-creasing order by ζ i , i = 1 , , . . . , N . One can show that the corresponding states | ζ i (cid:105) are linearlyindependent by computing their inner product with the aid of the Christoffel-Darboux formula, (cid:104) η | ζ (cid:105) = N − (cid:88) n =0 He n ( η ) He n ( ζ ) n ! = 1( N − N ( η ) He N − ( ζ ) − He N ( ζ ) He N − ( η ) η − ζ . (2.39)Then, (cid:104) ζ i | ζ j (cid:105) = c i δ ij , c i = He N +1 ( ζ i ) √ N ! , (2.40)8nd the vectors | ζ i (cid:105) form an orthogonal basis of the Hilbert space called the position basis.The matrix elements of I ( y, z ) in the position basis are I ij ( y, z ) = 1 c i c j (cid:104) ζ i | e yA T e zA | ζ j (cid:105) . (2.41)We will use this expression extensively in the next sections. Before we move on, we would liketo point out that it is possible to write an exact expression for I − ( y, z ) that takes a particularlysimple form. Indeed, since A T acts on | ζ (cid:105) as a derivative, we have I − ij ( y, z ) = 1 c i c j (cid:104) ζ i | e − zA e − yA T | ζ j (cid:105) = 1 c i c j (cid:104) ζ i − z | ζ j − y (cid:105) . (2.42)Even though we will not use this formula in what follows, we believe that it could lead to somesimplifications in the analysis of sub-leading corrections to the expectation value of the circularWilson loop. The precise relation between the circular Wilson loop and the Gaussian matrix model is (cid:104) W R (cid:105) U ( N ) = 1dim[ R ] (cid:10) Tr R (cid:2) e X (cid:3)(cid:11) , (2.43)where R denotes the representation of the U ( N ) gauge group and Tr R the corresponding trace.The vacuum expectation value on the left is defined on the N = 4 SYM theory; the right hand sidecorresponds to an insertion in the matrix model.In this paper we will be concerned with the totally antisymmetric representation of rank k defined by the Young diagram A k = ... k . k ≤ N . (2.44)It has dimension dim[ A k ] = (cid:18) Nk (cid:19) . (2.45)The structure of the trace Tr A k can get increasingly complicated as the rank k grows. Indeed, forany N × N matrix X we haveTr A X = Tr X , Tr A X = 12 (Tr X ) − 12 Tr (cid:0) X (cid:1) , Tr A X = 16 (Tr X ) − 12 Tr (cid:0) X (cid:1) Tr X + 13 Tr (cid:0) X (cid:1) , (2.46)(2.47)(2.48)9nd so forth. Happily, there is a natural way to encode this structure into the generating function F A ( t ; X ) ≡ det [1 + tX ] = N (cid:88) k =0 t k Tr A k [ X ] . (2.49)A straightforward calculation using (2.9), (2.10) and (2.12) reveals that the matrix model ex-pectation value of this generating function is (cid:10) F A ( t ; e X ) (cid:11) U ( N ) = det (cid:20) I ( y, z ) + t e g I ( g + y, g + z ) (cid:21) = det (cid:20) t e g I ( g, g ) (cid:21) , (2.50)where the matrix I ( g, g ) was introduced in (2.15). The dependence on y and z disappears as aconsequence of (2.21) and (2.20). This is essentially the answer reported by Fiol and Torrentsin [48], except that they use a slightly different matrix, which we call ˜ I ( g, g ), obtained from I ( g, g )by multiplying the m -th row by (cid:112) ( m − g − m and dividing the m -th column by the same factor.The determinant in (2.50) does not change under these operations, so (cid:10) F A ( t ; e X ) (cid:11) U ( N ) = det (cid:104) t e g ˜ I ( g, g ) (cid:105) , ˜ I mn ( g, g ) = L ( m − n ) n − ( − g ) , (2.51)which is the expression given in [48]. Note also that our generating functional differs by a factor of t N from theirs.In the answer (2.50) we recognize the form of the generating function of antisymmetric traces(2.49) for the matrix I ( g, g ), (cid:10) F A ( t ; e X ) (cid:11) = F A (cid:16) t e λ N ; I ( g, g ) (cid:17) , (2.52)which allows us, using (2.49) and (2.43), to obtain the formal yet remarkably simple result (cid:104) W A k (cid:105) U ( N ) = 1dim[ A k ] e λk N Tr A k [ I ( g, g )] . (2.53)For the gauge group SU ( N ), the matrix model must be restricted to traceless matrices. This canbe achieved either by using a Lagrange multiplier or by explicitly isolating the trace component inthe matrix model integral, as was done in [45]. The outcome is (cid:104) W A k (cid:105) SU ( N ) = (cid:104) W A k (cid:105) U ( N ) e − λk N = 1dim[ A k ] e λk ( N − k )8 N Tr A k [ I ( g, g )] . (2.54)Because the representations A k and A N − k are conjugate to each other for the gauge group SU ( N ),we expect (cid:104) W A k (cid:105) SU ( N ) = (cid:10) W A N − k (cid:11) SU ( N ) . This can be demonstrated by noting that, for an invert-ible matrix X , (2.49) implies Tr A N − k [ X ] = det[ X ] Tr A k [ X − ] . (2.55)10ow, since I ( g, g ) is similar to its inverse, and det I ( g, g ) = 1, this givesTr A N − k [ I ( g, g )] = Tr A k [ I ( g, g )] , (2.56)which proves the assertion. Equation (2.56) also shows that the generating function F A ( t ; I ( g, g ))is a palindromic polynomial.The main object of study in the remainder of this paper is the function F ( t ) = 1 N ln F A ( t ; I ( g, g )) = 1 N Tr ln[1 + tI ( g, g )] , (2.57)from which the traces Tr A k [ I ( g, g )] can be calculated byTr A k [ I ( g, g )] = (cid:73) dt πit e N [ F ( t ) − κ ln t ] , (2.58)where we have introduced the ratio κ = kN . (2.59)We are interested in the large N regime, keeping the ’t Hooft coupling λ and the ratio κ fixed. Inthis regime, the integral in (2.58) is dominated by the saddle point value, Tr A k [ I ( g, g )] = e N [ F ( t ∗ ) − κ ln t ∗ ] − ln[2 πN ( κ + t ∗ F (cid:48)(cid:48) ( t ∗ ))] , (2.60)where t ∗ satisfies the saddle point equation t ∗ F (cid:48) ( t ∗ ) = κ . (2.61)Moreover, F ( t ) admits an asymptotic expansion in 1 /N , F ( t ) = ∞ (cid:88) n =0 F n N − n . (2.62)Our aim is to calculate the terms F ( t ) and F ( t ) from the exact, but formal, expression (2.57). In this section, we shall calculate the leading order term F ( t ) of the generating function. Ofcourse, F ( t ) is known both from a matrix model calculation and from the holographic dual [28,33](the latter implying large λ in addition to large N ). Here, we reproduce it starting from the exactexpression (2.57) by means of two different calculations, which exploit the relation of the matrix I to the truncated harmonic oscillator. To obtain (2.60), introduce t = t ∗ e iz , expand the integrand in z up to second order and evaluate the Gaussianintegral. The second term in the exponent comes from the Gaussian integral. F ( t ) thatOkuyama [46] attributes to an unpublished note by Beccaria. Starting from (2.57) and Taylor-expanding the logarithm we may write F ( t ) = − N ∞ (cid:88) n =1 ( − t ) n n Tr[ I ( g, g ) n ] . (3.1)In order to compute the powers of the matrix I ( g, g ) we resort to the representation (2.25). Toleading order in 1 /N , commutators of the matrices A and A T can be neglected, so that we canessentially normal order the product in the trace. This leads to F ( t ) = − ∞ (cid:88) n =1 ( − t ) n n N Tr (cid:16) e ngA T e ngA (cid:17) + O (1 /N ) . (3.2)We can now expand the exponentials and calculate the traces. We defer this little calculation toappendix B. With the results (B.3) and (B.5), this yields F ( t ) = − ∞ (cid:88) n =1 ( − t ) n n N − (cid:88) k =0 ( ng ) k k !( k + 1)! ( N − k ) k +1 N . (3.3)The leading term in 1 /N is found to be F ( t ) = − ∞ (cid:88) n =1 ( − t ) n n ∞ (cid:88) k =0 ( n √ λ/ k k !( k + 1)! = − √ λ ∞ (cid:88) n =1 ( − t ) n n I ( n √ λ ) , (3.4)where I ( x ) denotes a modified Bessel function. This is precisely J in (2.21) of [46].A slightly simpler way of arriving at (3.4) is to recognize (3.2) as F ( t ) = − ∞ (cid:88) n =1 ( − t ) n n N Tr[ I ( ng, ng )] + O (1 /N ) . (3.5)The trace follows easily from (2.15) and the identity [52] N (cid:88) m =1 L ( α ) m − ( x ) = L ( α +1) N − ( x ) . (3.6)Hence, (3.5) becomes F ( t ) = − ∞ (cid:88) n =1 ( − t ) n n N L (1) N − ( − n g ) + O (1 /N ) . (3.7)For large order, the Laguerre polynomials satisfy [52]lim N →∞ N α L ( α ) N (cid:16) − zN (cid:17) = 1 z α I α (cid:16) z (cid:17) , (3.8)with I α ( x ) being a modified Bessel function. Recalling that g = (cid:113) λ N , applying the limit (3.8) to(3.7) gives (3.4). 12or a comparison with the matrix model calculation in the saddle point approximation [33], onecan use the following integral representation of the modified Bessel function [52],I ( z ) = zπ π (cid:90) e z cos θ sin θ d θ . (3.9)After substituting (3.9) into (3.4), the summation can be carried out to give F = 2 π π (cid:90) dθ sin θ ln (cid:16) t e √ λ cos θ (cid:17) . (3.10)A simple change of variables transforms this integral into the one in (2.26) of [33].The computation of F ( t ) turns out to be considerably simpler in the position basis. The reasonfor this is that, to leading order in 1 /N , we can also write I ( g, g ) = e g ( A + A T ) + O (1 /N ) , (3.11)where all commutators in the Baker-Campbell-Hausdorff formula contribute to the O (1 /N ) terms.According to (2.37), the operator in the exponential is diagonal in the position basis, so that (2.57)simply becomes F ( t ) = 1 N N (cid:88) i =1 ln (cid:16) t e gζ i (cid:17) + O (1 /N ) . (3.12)The sum is taken over the roots of the Hermite polynomial He N ( ζ ). For large N , the distributionof zeros becomes dense and is described by the Wigner semi-circle law. Using formulas (A.11) and(A.12) to determine the integral measure and dropping the 1 /N contributions, we get F ( t ) = 2 π (cid:90) π dθ sin θ ln (cid:16) t e √ λ cos θ (cid:17) , (3.13)which agrees with (3.10). In this section, we extend the calculation of the generating function F ( t ) to order 1 /N , i.e., weobtain the subleading term F ( t ) in the expansion (2.62). Our results agree with [46], althoughwe also managed to make a slight simplification. In the holographic regime of large λ , we will findagreement with [45].Exploiting the relation between the matrix I ( g, g ) in the position basis and the zeros of Hermitepolynomials appears to be the most promising path to compute the subleading corrections. Startingagain from (3.1), we will proceed as follows. In subsection 4.1, we compute the matrix elements of I ( g, g ) in the position basis up to order 1 /N , which results in an expression like I ( g, g ) = I (0) + 1 N I (1) + O (1 /N ) . (4.1)13he leading term I (0) is known from the calculation at the end of the previous section and isdiagonal. The main effort in subsection 4.1 consists in finding I (1) . We then need to computepowers of I ( g, g ). This is not as straightforward as one might think because I (1) is a dense matrix.Consider, for example, I ( g, g ) = [ I (0) ] + 1 N (cid:104) I (0) I (1) + I (1) I (0) (cid:105) + 1 N [ I (1) ] . (4.2)The last term, which naively seems to be of order 1 /N , is, in fact, of order 1 /N , because the matrixmultiplication contains a sum over N terms of order 1. The same happens for higher powers. Thecalculation of the powers of I ( g, g ) and their traces will be done in subsection 4.2. Another sourceof 1 /N contributions is the conversion of the sum over the roots of Hermite polynomials, whichlabel the position basis elements, into an integral. We present the details of this conversion inappendix C. The generating function F ( t ) will be calculated in subsection 4.3, where we will alsoprovide its integral representation. Finally, in subsecion 4.4, we consider F ( t ) in the holographicregime of large λ . Consider the matrix I ( g, g ) = e gA † e gA . With | ζ (cid:105) defined in (2.34), we find by direct application ofthe properties (2.31) and (2.32)e gA † e gA | ζ (cid:105) = N − (cid:88) k =0 N − (cid:88) l =0 min( N − ,N − − l + k ) (cid:88) n =0 g k + l k ! l ! (cid:18) nk (cid:19) k ! He n + l − k ( ζ ) √ n ! | n (cid:105) . (4.3)Reordering the summations such that the sum over states stays on the left, one obtainse gA † e gA | ζ (cid:105) = N − (cid:88) n =0 n (cid:88) k =0 N − − n + k (cid:88) l =0 g k + l l ! (cid:18) nk (cid:19) He n + l − k ( ζ ) √ n ! | n (cid:105) . (4.4)Note that the binomial coefficient restricts the sum over k . The next step consists in evaluating n N − − n ∞ N − k ↓ l → Figure 1: Illustration of the domain of summations over k and l in (4.4). The domain to be summedover is the dark shaded area. Instead, we sum over the semi-infinite rectangle and subtract the sumover the light shaded area. 14he sums over k and l . These sums, in the k, l plane, extend over a domain that can be representedby the dark shaded trapezoid area in Fig. 1. The trick is to extend the summation domain toa semi-infinite stripe until l = ∞ and subtract the sum over the domain illustrated by the lightshaded area. In the sum over the semi-infinite stripe, it comes handy to introduce r = k + l and tosum over r instead of l . This yields n (cid:88) k =0 ∞ (cid:88) l =0 g k + l l ! (cid:18) nk (cid:19) He n + l − k ( ζ ) = ∞ (cid:88) r =0 n (cid:88) k =0 g r r ! (cid:18) rk (cid:19)(cid:18) nk (cid:19) k ! He n + r − k ( ζ ) . (4.5)Here, we recognize the linearization formula (A.6) and, subsequently, the generating function forthe Hermite polynomials (A.5). Therefore, the sum over the semi-infinite stripe yields the simpleexpression n (cid:88) k =0 ∞ (cid:88) l =0 g k + l l ! (cid:18) nk (cid:19) He n + l − k ( ζ ) = e gζ − g He n ( ζ ) . (4.6)In the sum that must be subtracted (the light shaded area), we introduce the variable s = l − k + n − N and get n (cid:88) k =0 ∞ (cid:88) l = N − n + k g k + l l ! (cid:18) nk (cid:19) He n + l − k ( ζ ) = ∞ (cid:88) s =0 n (cid:88) k =0 g N − n + s +2 k ( N − n + s + k )! (cid:18) nk (cid:19) He N + s ( ζ ) . (4.7)Notice that, until now, it was not necessary to require that ζ be a root of He N . From now on, wewill assume that it is. This implies that the s = 0 term in the sum in (4.7) vanishes. Furthermore,it allows us to express the Hermite polynomials He N + s ( ζ ) as a sum of the polynomials He n ( ζ ) with n < N . This can be shown by means of the linearization formula (A.6), setting n = N . Moreprecisely, we can writeHe N + s ( ζ ) = − [ s − ] (cid:88) l =0 a s,l He N − s +2 l ( ζ ) , for He N ( ζ ) = 0 , (4.8)with coefficients a s,l , which satisfy m (cid:88) k =0 (cid:18) sk (cid:19)(cid:18) Nk (cid:19) k ! a s − k,m − k = (cid:18) sm (cid:19)(cid:18) Ns − m (cid:19) ( s − m )! . (4.9)The coefficients a s,l can be determined applying (4.9) recursively for increasing m . The first stepsof this recursion yield a s, = (cid:18) Ns (cid:19) s ! , a s +2 , = − (cid:18) Ns (cid:19) s ! s ( s + 2) , . . . . (4.10)We shall not investigate this recursion further, because it will turn out that only a s, is relevant forour purposes. After substituting (4.8) into (4.7), we can reorder the summations over s and l by15ntroducing r = s − l and obtain − ∞ (cid:88) s =0 n (cid:88) k =0 g N − n + s +2 k ( N − n + s + k )! (cid:18) nk (cid:19) [ s − ] (cid:88) l =0 a s,l He N − s +2 l ( ζ ) = − n (cid:88) k =0 N (cid:88) r =1 (cid:18) nk (cid:19) He N − r ( ζ ) ∞ (cid:88) l =0 g N − n + r +2 l +2 k ( r + N − n + 2 l + k )! a r +2 l,l . (4.11)With these results, let us now consider the matrix element of e gA † e gA between two (non-normalized) position eigenstates. Remember that the sum over k and l in (4.4) is given by (4.6)minus (4.11). We find (cid:104) ζ i | e gA † e gA | ζ j (cid:105) = e gζ j − g (cid:104) ζ i | ζ j (cid:105) (4.12)+ N (cid:88) r,s =1 He N − s ( ζ i ) He N − r ( ζ j ) ∞ (cid:88) k =0 ∞ (cid:88) l =0 g r + s +2( l + k ) a r +2 l,l k !( N − s − k )!( s + r + 2 l + k )! . where we have used the definition of the states | ζ (cid:105) (2.34), extended the summation over k (this canbe done because of the binomial) and let s = N − n . Furthermore, rewriting the sums over l and k as a sum over n = l + k and k gives (cid:104) ζ i | e gA † e gA | ζ j (cid:105) = e gζ j − g (cid:104) ζ i | ζ j (cid:105) (4.13)+ N (cid:88) r,s =1 He N − s ( ζ i ) He N − r ( ζ j ) ∞ (cid:88) n =0 g r + s +2 n ( r + s + 2 n )! n (cid:88) k =0 (cid:18) r + s + 2 nk (cid:19) a r +2( n − k ) ,n − k ( N − s − k )! . In the first term on the right hand side, one can recognize the leading order result, corrected bythe exponential e − g . Therefore, the second term is of order 1 /N , so that leading order relationscan be used to manipulate it. A short inspection of the sum over k in (4.13) for some low values of n shows that the leading order contribution in 1 /N comes from the term with k = n . Therefore,using (4.10), we get n (cid:88) k =0 (cid:18) r + s + 2 nk (cid:19) a r +2( n − k ) ,n − k ( N − s − k )! ∼ (cid:18) r + s + 2 nn (cid:19) N !( N − s − n + 1) n ( N − r )!( N − s )! ∼ (cid:18) r + s + 2 nn (cid:19) N ! N n ( N − r )!( N − s )! , (4.14)where subleading terms in 1 /N have been omitted. Then, after normalizing the matrix elements(4.13) by means of (2.40) and substituting (4.14), we obtain I ij ( g, g ) = e gζ j − g δ ij (4.15)+ N (cid:88) r,s =1 ( N !) ( N − r )!( N − s )! He N − s ( ζ i ) He N − r ( ζ j )He N +1 ( ζ i ) He N +1 ( ζ j ) ∞ (cid:88) n =0 g r + s +2 n N n ( r + s + n )! n ! + O (cid:18) N (cid:19) . 16t is possible to obtain simple expressions for the ratios He N − s ( ζ i ) / He N +1 ( ζ i ). We include thecalculation in Appendix D, where we show that, to leading order in 1 /N ,He N − s ( ζ )He N +1 ( ζ ) ∼ − ( N − s )! N ! N s − U s − (cos θ ) , (4.16)where cos θ is determined by ζ ∼ √ N cos θ , and U s denote the Chebychev polynomials of thesecond kind [49], U s (cos θ ) = sin[( s + 1) θ ]sin θ . (4.17)Hence, we obtain our final result for the matrix I ( g, g ) in the position basis, I ij ( g, g ) = e gζ j − g δ ij + 1 N N (cid:88) r,s =1 sin( sθ i ) sin( rθ j )sin θ i sin θ j ∞ (cid:88) n =0 ( √ λ/ r + s +2 n ( r + s + n )! n ! + O (cid:18) N (cid:19) = e − λ N e √ λ (cid:113) N cos θ i δ ij + 1 N ∞ (cid:88) r,s =1 sin( sθ i ) sin( rθ j )sin θ i sin θ j I r + s ( √ λ ) + O (cid:18) N (cid:19) . (4.18)In the step from the first to the second line we have recognized the modified Bessel functions in thesums over n and expressed the roots ζ i in the diagonal term in terms of cos θ i being careful to usethe expression (A.11), which is exact to order 1 /N , . Moreover, we have extended the summationsto infinity in the sense of an asymptotic expansion. I ( g, g ) and their traces In this subsection, we return to calculate the quantities needed in the generating function (3.1),namely the powers of the matrix I ( g, g ) and their traces. As mentioned at the beginning of thissection, calculating the powers of I ( g, g ) to order 1 /N is not straightforward. Fortunately, aftercalculating a few powers, e.g., I ( g, g ) and I ( g, g ), one recognizes a pattern that can then be provenby induction for all powers. The calculation is somewhat tedious and involves the facts that themodified Bessel functions have the generating functione x cos θ = ∞ (cid:88) k = −∞ I k ( x ) cos( kθ ) , (4.19)and that they satisfy the addition theorem [52]I n ( x + y ) = ∞ (cid:88) k = −∞ I n − k ( x ) I k ( y ) . (4.20)Dropping 1 /N terms, the result is summarized in the formula I mij ( g, g ) = e − mλ N e m √ λ (cid:113) N cos θ i δ ij + 1 N ∞ (cid:88) r,s =1 sin( rθ i ) sin( sθ j )sin θ i sin θ j (cid:104) I r + s ( m √ λ ) + M ( m ) rs (cid:105) . (4.21) In the second term, which is already of order 1 /N , this care was not needed. M ( m ) rs denote some (infinite) matrices, which satisfy the recursion relations M ( m +1) rs = ∞ (cid:88) t =1 M ( m ) rt I s − t ( √ λ ) − ∞ (cid:88) t =0 I r + t ( m √ λ ) I s + t ( √ λ ) (4.22)and M (1) rs = 0.Let us evaluate the trace of (4.21). This calculation is nearly identical to the one resulting in(C.6), in particular with respect to the cancellation of various 1 /N contributions, and yields1 N Tr I m ( g, g ) = 2 m √ λ I ( m √ λ ) e − mλ N + 1 N ∞ (cid:88) r =1 M ( m ) rr . (4.23)The first term on the right hand side reproduces the leading order expression (3.4), with an expo-nential correction that can be traced back to the exponential in (2.52), which is missing in (2.57).The second term, which is the substantial part of the 1 /N contributions, will be evaluated in theremainder of this subsection.Henceforth, let z = √ λ in order to simplify the notation. For m = 2, we simply have from(4.22) ∞ (cid:88) r =1 M (2) rr = − ∞ (cid:88) r =1 ∞ (cid:88) t =0 I r + t ( z ) I r + t ( z ) = − ∞ (cid:88) v =1 v (cid:88) r =1 I v ( z ) I v ( z ) , where we have re-ordered the summations over r and t . For m > 2, using the same reordering,(4.22) leads to the following pattern, ∞ (cid:88) r =1 M ( m ) rr = − m − (cid:88) a =1 ∞ (cid:88) v =1 I v (( m − a ) z ) S a ( v ; z ) , (4.24)where S a ( v ; z ) stands for S ( v ; z ) = v I v ( z ) (4.25)and S a ( v ; z ) = v (cid:88) r =1 ∞ (cid:88) t =1 · · · ∞ (cid:88) t a − =1 I v − r + t ( z ) I t − t ( z ) · · · I r − t a − ( z ) , a = 2 , . . . , m − . (4.26)Let us calculate (4.26) for a = 2, where there is only one t -summation. The simplest way toproceed is to use the invariance of the summand under the transformation r → v − r + 1, t → − t .This yields S ( v ; z ) = 12 v (cid:88) r =1 ∞ (cid:88) t = −∞ I v − r + t ( z ) I t − r ( z ) = 12 v I v (2 z ) , (4.27)where we have recognized the summation formula (4.20).Unfortunately, the same trick does not suffice to easily obtain the nested sums for a > S a ( v ; z ), we prove in appendix E that S a ( v ; z ) = va I v ( az ) . (4.28)18his is a remarkable result, which we have not found in the literature. With (4.28), we can returnto (4.24), which simplifies to ∞ (cid:88) r =1 M ( m ) rr = − m − (cid:88) a =1 ∞ (cid:88) v =1 va I v (( m − a ) z ) I v ( az )= − z m − (cid:88) a =1 ∞ (cid:88) v =1 I v (( m − a ) z ) [I v − ( az ) − I v +1 ( az )]= − z m − (cid:88) a =1 ∞ (cid:88) v =1 [I v (( m − a ) z ) I v − ( az ) − I v +1 (( m − a ) z ) I v ( az )]= − z m − (cid:88) a =1 I ( az ) I [( m − a ) z ] . (4.29) Putting together (3.1), (4.23) and (4.29), we obtain the generating function F ( t ) to order 1 /N , F ( t ) = − ∞ (cid:88) m =1 ( − t ) m (cid:34) m √ λ I ( m √ λ ) e − mλ N − N √ λ m m − (cid:88) a =1 I ( a √ λ ) I (( m − a ) √ λ ) (cid:35) + O (1 /N ) . (4.30)Let us compare (4.30) with Okuyama’s result, which is (2.21) of [46]. The first term in the bracketsreproduces J of [46], except for the exponential factor, which arises from the fact the our F ( t ) is notexactly the generating function of U ( N ) Wilson loops, but is defined in terms of the palindromicpolynomial F A ( t ; I ( g, g )), c.f. (2.57). The second term in the brackets can easily be shown toreproduce J of [46], because ∂∂λ m − (cid:88) a =1 (cid:104) √ λ I ( a √ λ ) I (( m − a ) √ λ ) (cid:105) = 12 m − (cid:88) a =1 (cid:104) a I ( a √ λ ) I (( m − a ) √ λ ) + ( m − a ) I ( a √ λ ) I (( m − a ) √ λ ) (cid:105) = m m − (cid:88) a =1 (cid:104) I ( a √ λ ) I (( m − a ) √ λ ) + I ( a √ λ ) I (( m − a ) √ λ ) (cid:105) , (4.31)which appears in the integrand in J . The step from the second to the third line consists insymmetrizing the summands with respect to a → m − a . We note that our result for the 1 /N termis slightly simpler than Okuyama’s, because it does not involve an integral.In the remainder of this section, we will find an integral representation of F ( t ). Consider thefirst term in brackets in (4.30). Because it contains F ( t ), but also corrections in 1 /N , we willdenote it by (cid:102) F ( t ). After using the integral representation of the modified Bessel function (3.9), The integral representation of F ( t ) is (3.10). m can be performed, which yields (cid:102) F ( t ) = 2 π π (cid:90) d θ sin θ ln (cid:16) t e √ λ cos θ − λ N (cid:17) = F − λ πN π (cid:90) d θ sin θ t e √ λ cos θ t e √ λ cos θ + O (1 /N ) . (4.32)Now, consider the second term in brackets in (4.30), which we shall denote by (cid:102) F ( t ) /N . Usingthe standard integral representation for I and I [52], the sum over a can be done, which gives (cid:102) F ( t ) = √ λ π ∞ (cid:88) m =1 ( − t ) m m π (cid:90) d θ π (cid:90) d φ cos φ e √ λ (cos θ + m cos φ ) − e m √ λ ( m cos θ +cos φ ) e √ λ cos φ − e √ λ cos θ . (4.33)Lets us rewrite the fraction in the integrand ase √ λ (cos θ + m cos φ ) − e m √ λ ( m cos θ +cos φ ) e √ λ cos φ − e √ λ cos θ = e √ λ (cos θ − cos φ ) − e √ λ (cos θ − cos φ ) (cid:16) e m √ λ cos φ − e m √ λ cos θ (cid:17) − e m √ λ cos θ The last term on the right hand side, which does not depend on φ , integrates to zero in the φ -integral. For the remaining term, performing the sum over m in (4.33) leads to (cid:102) F ( t ) = − √ λ π π (cid:90) d θ π (cid:90) d φ cos φ e √ λ (cos θ − cos φ ) − e √ λ (cos θ − cos φ ) ln 1 + t e √ λ cos φ t e √ λ cos θ . (4.34)Thus, combining (4.32) with (4.34), we obtain the integral representation for F ( t ) as F ( t ) = − √ λ π π (cid:90) d θ π (cid:90) d φ cos φ e √ λ (cos θ − cos φ ) − e √ λ (cos θ − cos φ ) ln 1 + t e √ λ cos φ t e √ λ cos θ (4.35) − λ π π (cid:90) d θ sin θ t e √ λ cos θ t e √ λ cos θ . Our aim in this subsection is to evaluate the saddle point value of (4.35) in the regime of large λ .We remind the reader that the saddle point, in this regime, is given by t = e −√ λ cos θ ∗ , (4.36)where the angle θ ∗ satisfies κ ≡ kN = 1 π (cid:18) θ ∗ − 12 sin 2 θ ∗ (cid:19) . (4.37)20 πθ ∗ θ ∗ θ φAA BC Figure 2: Division of the integration domain for the double integral in (4.35). On the dashed line,where θ = φ , both, the denominator and the logarithm in the integral vanish.Let us start with integral on the second line of (4.35). Dropping the terms that are exponentiallysuppressed for large λ , we easily obtain − λ π π (cid:90) d θ sin θ t e √ λ cos θ t e √ λ cos θ = − λ π θ ∗ (cid:90) d θ sin θ = − λκ . (4.38)To evaluate the double integral on the first line of (4.35), we divide the integration domain intothe four regions A , A (cid:48) , B and C illustrated in Fig. 2. What differs between these four regions isthe approximation of the logarithm in the integrand. In the region A ( θ < θ ∗ < φ ), we have A : ln 1 + e √ λ (cos φ − cos θ ∗ ) √ λ (cos θ − cos θ ∗ ) ≈ −√ λ (cos θ − cos θ ∗ ) , (4.39)while the fraction in front of the logarithm is approximately − 1. Therefore, the contribution of theregion A is A = − λ π θ ∗ (cid:90) d θ π (cid:90) θ ∗ d φ cos φ (cos θ − cos θ ∗ ) = λ π (cid:18) sin θ ∗ − θ ∗ sin 2 θ ∗ (cid:19) . (4.40)In the region A (cid:48) ( φ < θ ∗ < θ ), we have A (cid:48) : ln 1 + e √ λ (cos φ − cos θ ∗ ) √ λ (cos θ − cos θ ∗ ) ≈ √ λ (cos φ − cos θ ∗ ) , (4.41)but the fraction in front of the logarithm is exponentially suppressed. Hence, A (cid:48) = 0 . (4.42)21ext, look at region B , where θ > θ ∗ and φ > θ ∗ . Here, B : ln 1 + e √ λ (cos φ − cos θ ∗ ) √ λ (cos θ − cos θ ∗ ) ≈ e √ λ (cos φ − cos θ ∗ ) (cid:16) − e √ λ (cos θ − cos φ ) (cid:17) . (4.43)The term in parentheses precisely cancels the denominator of the term in front of the logarithm,and what remains of the integrand is again exponentially suppressed for large λ . Thus, B = 0 . (4.44)In the remaining region C , where θ < θ ∗ and φ < θ ∗ , we have C : ln 1 + e √ λ (cos φ − cos θ ∗ ) √ λ (cos θ − cos θ ∗ ) ≈ √ λ (cos φ − cos θ ) . (4.45)Therefore, we can write the contribution from the region C to the integral as C = − λ π θ ∗ (cid:90) d θ θ ∗ (cid:90) d φ cos φ (cos φ − cos θ ) e √ λ cos θ e √ λ cos φ − e √ λ cos θ . (4.46)Because the integration domain in (4.46) is symmetric with respect to φ and θ , we can symmetrizethe integrand and rewrite (4.46) as C = λ π θ ∗ (cid:90) d θ θ ∗ (cid:90) d φ (cid:34) (cos φ − cos θ ) + (cos θ − cos φ ) e √ λ cos φ + e √ λ cos θ e √ λ cos φ − e √ λ cos θ (cid:35) . (4.47)The first term is readily integrated and yields C = λ π θ ∗ (cid:90) d θ θ ∗ (cid:90) d φ (cos φ − cos θ ) = λ π (cid:18) θ ∗ + 12 θ ∗ sin 2 θ ∗ − θ ∗ (cid:19) . (4.48)In the second term in (4.47), we write cos θ − cos φ = sin φ − sin θ and realize that, using thesymmetry of the integrand and the integration domain, we can write C = λ π θ ∗ (cid:90) d θ θ ∗ (cid:90) d φ sin φ e √ λ cos φ + e √ λ cos θ e √ λ cos φ − e √ λ cos θ . (4.49)Here, the φ -integral must be interpreted as the principle value because of the pole for φ = θ ,whereas there was no pole with the symmetric integrand. However, after rewriting (4.49) as C = − √ λ π θ ∗ (cid:90) d θ θ ∗ (cid:90) d φ sin φ ∂ φ (cid:104) ln (cid:12)(cid:12)(cid:12) e √ λ (cos θ − cos φ ) − (cid:12)(cid:12)(cid:12) + ln (cid:12)(cid:12)(cid:12) e √ λ cos φ − e √ λ cos θ (cid:12)(cid:12)(cid:12)(cid:105) , (4.50)the φ -integral is easily done using integration by parts, which also takes care of the principal value.The result is C = λ π (cid:18) − sin θ ∗ + 12 θ ∗ sin 2 θ ∗ (cid:19) . (4.51)22umming the results (4.40), (4.42), (4.44), (4.48) and (4.51), we obtain the first line of (4.35), (cid:102) F ( t ) = λ π (cid:0) θ ∗ − θ ∗ sin 2 θ ∗ + sin θ ∗ (cid:1) . (4.52)This agrees with (3.53b) of [45]. Finally, after adding (4.38) to (4.52) and using (4.37), we end up with the final result for F ( t ), F ( t ) = λ (cid:20) − κ (1 − κ ) + 1 π sin θ ∗ (cid:21) . (4.53)We note that (4.53) is symmetric under θ ∗ → π − θ ∗ , which lets κ → − κ . This reflects, of course,the palindromic property of F A ( t ; I ( g, g )). We also note that the first term in brackets in (4.53)cancels against the exponential factor in (2.54), i.e., for the SU ( N ) Wilson loop. This reproduces(3.56) of [45]. In this paper we tackled the problem of non-planar corrections to the expectation value of -BPSWilson loops in N = 4 SYM with gauge group U ( N ) or SU ( N ). More precisely, we extractedthe leading and sub-leading behaviours in the 1 /N expansion at fixed ’t Hooft coupling λ of theWilson loop generating function. Unlike previous works, which had addressed this issue using loopequation techniques and topological recursion, our starting point was the exact solution of thematrix model, which had been known for some time. Our results for the 1 /N term of the Wilsonloop generating function agree with previous calculations, but appear to be somewhat more explicit.We have provided both sum and integral representations of the 1 /N terms and have evaluated themexplicitly in the holographic large- λ regime, which allows for easier comparison with the holographicdual picture. This term should match with the gravitational backreaction of the D-brane on thegravity side. A particularly interesting observation is the connection between the Wilson loopgenerating function and the finite-dimensional quantum system known as the truncated harmonicoscillator. This system, which is familiar to the Quantum Optics community, provides a descriptionof the problem that seems to be more amenable to an asymptotic 1 /N expansion. En route, weobtained interesting mathematical relations and sum rules involving the Hermite polynomials. Itwould be interesting to see these formulas, which we proved in appendices D and E, in differentapplications.One can envisage two main lines of generalization of the present work. First, it would beinteresting to extend the methods developed here to other representations of the gauge group.A particularly interesting case is the totally symmetric representation S k (see also [47]), whosegenerating function is slightly more complicated than the antisymmetric one but still quite simple, The difference in the sign stems from the different definitions of F ( t ). 23o the problem seems tractable. Second, one may investigate how the present approach extendsto higher orders in 1 /N . At first sight, there are a number of technical obstacles that must beovercome, because order-1 /N terms have been neglected at many points of the calculation. So,the question whether our approach lends itself to a systematic 1 /N expansion is highly non-trivial.This problem is closely related to the fact that the large, but finite- N matrix model solution differsin its analyticity properties from the continuum limit. Moreover, corrections to the saddle pointcalculation of the Wilson loop expectation values become relevant at order 1 /N . We leave theseinteresting questions for the future. Acknowledgements A.C. and A.F. were supported by Fondecyt A Some properties of the Hermite polynomials In this appendix, we list a number of formulae regarding the (probabilists’) Hermite polynomials,which are useful for the analysis in the main text. These relations can be found in standardreferences [49, 52]. Sometimes a translation from the physicists’ version of the polynomials isnecessary. They are related by He n ( x ) = 2 − n H n ( x/ √ . (A.1)The Hermite polynomials He n ( x ) satisfy the differential equationHe (cid:48)(cid:48) n − x He (cid:48) n + n He n = 0 (A.2)as well as the recurrence relations He (cid:48) n = n He n − , He n +1 = x He n − n He n − . (A.3)(A.4)The generating function is e xt − t = ∞ (cid:88) n =0 t n n ! He n ( x ) . (A.5)Another useful property is the linearization formulaHe m ( x ) He n ( x ) = m (cid:88) k =0 (cid:18) mk (cid:19)(cid:18) nk (cid:19) k ! He m + n − k ( x ) . (A.6)24 main ingredient in our analysis is the location of the N roots of He N for large N . It can beobtained from the relation of He N to the parabolic cylinder functionHe N ( x ) = e x / U (cid:18) − N − , x (cid:19) (A.7)and the asymptotic expansion of the parabolic cylinder functionU (cid:18) − µ , √ µt (cid:19) ∼ g ( µ )(1 − t ) / (cid:34) cos κ ∞ (cid:88) s =0 ( − s u s ( t )(1 − t ) s µ s − sin κ ∞ (cid:88) s =0 ( − s u s +1 ( t )(1 − t ) s +3 / µ s +2 (cid:35) , (A.8)where κ = µ η − π , η = 12 (cid:16) arccos t − t (cid:112) − t (cid:17) , (A.9)and u s ( t ) are polynomials u = 1 , u = 124 t ( t − , · · · (A.10)The function g ( µ ) is irrelevant for our purposes. Setting µ = √ N + 1 and t = cos θ , these relationsimply that the N roots of He N are given approximately by ζ i = 2 (cid:114) N + 12 cos θ i , (A.11)with (cid:18) N + 12 (cid:19) (cid:18) θ i − 12 sin 2 θ i (cid:19) − π = (cid:18) i − (cid:19) π ( i = 1 , , . . . , N ) . (A.12) B Trace of normal ordered products In this appendix, we calculate the trace of the normal ordered product A T m A n . First, consider A T n A n . Using the definition of the number operator (2.28) and the commutators (2.29), we get A T n A n = A T n − N A n − = A T n − A n − ( N − n + 1) = ( N − n + 1) n , (B.1)where we have iterated the first two steps to arive at the final expression. ( a ) n = Γ( a + n ) / Γ( a )denotes the Pochhammer symbol. This allows us to write immediately A T m A n = A T m − n ( N − n + 1) n for m ≥ n ,( N − m + 1) m A n − m for m < n . (B.2)Because N is diagonal and any power of A is off-diagonal, the trace of this quantity is non-zeroonly, if m = n , Tr (cid:0) A T m A n (cid:1) = δ mn Tr( N − n + 1) n . (B.3)25he trace in (B.3) can be calculated starting with the expansion of the Pochhammer symbol interms of Stirling numbers [52]. This givesTr( N − n + 1) n = n (cid:88) l =0 s ( n, l ) Tr N l = n (cid:88) l =0 s ( n, l ) N − (cid:88) k =0 k l = n (cid:88) l =1 s ( n, l ) l (cid:88) j =0 j ! (cid:18) Nj + 1 (cid:19) S ( l, j ) , (B.4)where S ( l, j ) denote the Stirling numbers of the second kind. After rearranging the sum and usingthe properties of the Stirling numbers, (B.4) becomesTr( N − n + 1) n = n ! (cid:18) Nn + 1 (cid:19) = ( N − n ) n +1 n + 1 . (B.5)Finally, expressing the Pochhammer symbol in terms of Stirling numbers of the first kind yields anexpansion in 1 /N , 1 N n +1 Tr( N − n + 1) n = 1 n + 1 n (cid:88) l =0 s ( n + 1 , n + 1 − l ) N − l . (B.6) C Conversion of the sum over the roots of He N to an integral Consider a sum of the form 1 N N (cid:88) i =1 f ( θ i ) , (C.1)where θ i are defined in (A.11) in terms of the zeros of the Hermite polynomial He N ( ζ ). In the large N limit, it is justified to convert such a sum into an integral. Here we describe here how to do itcorrectly to order 1 /N .We start by setting x = i − and use the Euler-Maclaurin formula in mid-point form [53].The mid-point form has the advantages that the integration domain lies manifestly symmetricwithin the inveral (0 , N ) and that the boundary terms in the Euler-Maclaurin formula contain onlyderivatives of the integrand. The latter turn out to contribute at least of order 1 /N and are,therefore, irrelevant for our purposes. Hence, we have1 N N (cid:88) i =1 f ( θ i ) = 1 N N (cid:90) d xf ( θ x +1 / ) + O (cid:18) N (cid:19) = 2 π (cid:18) N (cid:19) π − θ / (cid:90) θ / d θ sin θf ( θ ) + O (cid:18) N (cid:19) . (C.2)In the second equality, we have changed the integration variable using (A.11). The angle θ / ,which marks the tiny edges missing from the interval (0 , π ), also follows from (A.11), (cid:18) N + 12 (cid:19) (cid:18) θ / − 12 sin 2 θ / (cid:19) = 14 π ⇒ θ / ≈ π N . (C.3)26herefore, (C.2) becomes1 N N (cid:88) i =1 f ( θ i ) = 2 π (cid:18) N (cid:19) π (cid:90) d θ sin θf ( θ ) − N [ f (0) + f ( π )] + O (cid:18) N (cid:19) . (C.4)To obtain the second term on the right hand side we have assumed that the function f is regularat 0 and π .Let us check the formula (C.4) by calculating the trace of the matrix I ( g, g ) in the position basis(4.18) and compare the result with the exact expression, which can be calculated in the numberbasis as follows, N Tr (cid:16) e gA † e gA (cid:17) = N − (cid:88) k,l =0 g k + l k ! l ! 1 N Tr (cid:16) A † k A l (cid:17) = N − (cid:88) k =0 g k ( k !) N Tr( N − k + 1) k = N − (cid:88) k =0 g k N k k !( k + 1)! k (cid:88) l =0 s ( k + 1 , k + 1 − l ) N − l = (cid:18) − λ N (cid:19) √ λ I ( √ λ ) + O (cid:18) N (cid:19) . (C.5)In the position basis, we have from (4.18) and (C.4)1 N Tr I ( g, g ) = 2 π (cid:18) N (cid:19) π (cid:90) d θ sin θ e g (cid:113) N + cos θ − g + 1 N ∞ (cid:88) r,s =1 I r + s ( √ λ ) 2 π π (cid:90) d θ sin( rθ ) sin( sθ ) − N cosh √ λ = (cid:18) N (cid:19) e − λ N √ λ (cid:113) N I (cid:32) √ λ (cid:114) N (cid:33) + 1 N ∞ (cid:88) r =1 I r ( √ λ ) − N cosh √ λ = 2 √ λ I ( √ λ ) e − λ N + 1 N (cid:34) 12 I ( √ λ ) + ∞ (cid:88) r =1 I r ( √ λ ) − 12 cosh √ λ (cid:35) = 2 √ λ I ( √ λ ) e − λ N . (C.6)We especially point out the presence of the last term on the second line, which comes from the edgeterms of (C.4) and is crucial for cancelling other 1 /N contributions. The bracket on the penultimate This calculation appeared already in section 3 between (3.2) and (3.3) and makes use of the calculation inappendix B. /N . Obviously (C.6) agrees with(C.5) to order 1 /N . D Proof of (4.16) In this appendix, we provide a proof of the asymptotic formulaHe N − s ( ζ )He N +1 ( ζ ) ∼ − ( N − s )! N ! N s − U s − (cos θ ) , (D.1)which is (4.16) in the main text. This formula holds for s (cid:28) N , and ζ ∼ √ N cos θ is a root ofHe N .The proof is done by induction using the recursion formula for the Hermite polynomials (A.4).First, consider s = 1 and s = 2. Because ζ is a root of He N , we haveHe N − ( ζ ) = − N He N +1 ( ζ ) , (D.2)He N − ( ζ ) = 1 N − ζ He N − ( ζ ) = − N ( N − N θ He N +1 ( ζ ) , (D.3)so (D.1) obviously holds for s = 1 and s = 2. Now, assume that (D.1) holds for He N − s +1 . Then,from the recursion formula (A.4) we getHe N − s ( ζ ) = 1 N − s + 1 [ ζ He N − s +1 ( ζ ) − He N − s +2 ( ζ )] , He N − s ( ζ )He N − ( ζ ) = − N − s + 1 (cid:20) √ N cos θ ( N − s + 1)! N ! N s − U s − (cos θ ) − ( N − s + 2)! N ! N s − U s − (cos θ ) (cid:21) = − ( N − s )! N ! N s − (cid:20) θ U s − (cos θ ) − N − s + 2 N U s − (cos θ ) (cid:21) ∼ − ( N − s )! N ! N s − U s − (cos θ ) , where we have dropped the term s of order 1 /N in front of U s − (cos θ ) and used the recursionrelation for the Chebychev polynomials [49] in the last step. E Proof of (4.28) In this appendix, we shall prove the remarkable summation formula v (cid:88) r =1 ∞ (cid:88) t =1 · · · ∞ (cid:88) t a − =1 I v − r + t ( z ) I t − t ( z ) · · · I r − t a − ( z ) = va I v ( az ) , a = 2 , , . . . . (E.1)28or a = 2, we have established the result in (4.27). Our starting point is the generating functionof the modified Bessel functions (4.19),e z cos θ = I ( z ) + 2 ∞ (cid:88) k =1 I k ( z ) cos( kθ ) . (E.2)Differentiating it with respect to θ shows that z θ e az cos θ = ∞ (cid:88) k =1 k I k ( az ) a sin( kθ ) . (E.3)Therefore, we can prove (E.1) by showing that ∞ (cid:88) v =1 S a ( v ; z ) sin( vθ ) = z θ e az cos θ , (E.4)where S a ( v ; z ) is, as defined in the main text, the left hand side of (E.1). Since S ( v ; z ) is knownfrom (4.27), (E.4) trivially holds for a = 2 by virtue of (E.3).The left hand side of (E.4) is explicitly ∞ (cid:88) v =1 S a ( v ; z ) sin( vθ ) = ∞ (cid:88) v =1 sin( vθ ) v (cid:88) r =1 ∞ (cid:88) t =1 · · · ∞ (cid:88) t a − =1 I v − r + t ( z ) I t − t ( z ) · · · I r − t a − ( z ) . (E.5)We rearrange the sums over v and r by introducing s = v − r and t a = r , such that ∞ (cid:88) v =1 S a ( v ; z ) sin( vθ ) = ∞ (cid:88) s =0 ∞ (cid:88) t a =1 · · · ∞ (cid:88) t =1 sin[( s + t a ) θ ] I t a − t a − ( z ) · · · I t − t ( z ) I s + t ( z ) . (E.6)Now, we can extend the summation over t a to all integers without changing the value of the sum.The reason for this is that ∞ (cid:88) s =0 ∞ (cid:88) r =0 sin[( s − r ) θ ] M rs = 0 , (E.7)if M rs is symmetric, by the antisymmetry of the sine. It is easy to see that this is the case for thematrix involving the rest of the t -summations and the modified Bessel functions in (E.6). Therefore, ∞ (cid:88) v =1 S a ( v ; z ) sin( vθ ) = ∞ (cid:88) s =0 ∞ (cid:88) t a = −∞ ∞ (cid:88) t a − =1 · · · ∞ (cid:88) t =1 sin[( s + t a ) θ ] I t a − t a − ( z ) · · · I t − t ( z ) I s + t ( z )= ∞ (cid:88) s =0 ∞ (cid:88) t a − =1 · · · ∞ (cid:88) t =1 ∞ (cid:88) t a = −∞ { sin[( s + t a − ) θ ] cos( t a θ ) + cos[( s + t a − ) θ ] sin( t a θ ) }× I t a ( z ) I t a − − t a − ( z ) · · · I t − t ( z ) I s + t ( z )= e z cos θ ∞ (cid:88) v =1 S a − ( v ; z ) sin( vθ ) . 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