Holographic Nuclear Physics with Massive Quarks
Salvatore Baldino, Lorenzo Bartolini, Stefano Bolognesi, Sven Bjarke Gudnason
IIFUP-TH-2021
Holographic Nuclear Physics with Massive Quarks
Salvatore Baldino (1) , Lorenzo Bartolini (2) ,Stefano Bolognesi (3) and Sven Bjarke Gudnason (4) (1)
Center for Mathematical Analysis, Geometry and Dynamical Systems,Department of Mathematics, Instituto Superior T´ecnico,Universidade de Lisboa, Av. Rovisco Pais, 1049-001 Lisboa, Portugal (2)
Institut f¨ur Theoretische Physik, Technische Universit¨at Wien,Wiedner Hauptstrasse 8-10, A-1040 Vienna, Austria (3)
Department of Physics “E. Fermi”, University of Pisa and INFN Sezione di PisaLargo Pontecorvo, 3, Ed. C, 56127 Pisa, Italy (4)
Institute of Contemporary Mathematics, School of Mathematics and Statistics,Henan University, Kaifeng, Henan 475004, P. R. China emails: salvatore.baldino(at)tecnico.ulisboa.pt, lorenzobartolini89(at)gmail.com,stefano.bolognesi(at)unipi.it, gudnason(at)henu.edu.cn
February 2021
Abstract
We discuss nuclear physics in the Witten-Sakai-Sugimoto model, in the limit oflarge number N c of colors and large ’t Hooft coupling, with the addition of a finitemass for the quarks. Individual baryons are described by classical solitons whose sizeis much smaller than the typical distance in nuclear bound states, thus we can usethe linear approximation to compute the interaction potential and provide a naturaldescription for lightly bound states. We find the classical geometry of nuclear boundstates for baryon numbers up to B = 8. The effect of the finite pion mass – inducedby the quark mass via the GMOR relation – is to decrease the binding energy of thenuclei with respect to the massless case. We discuss the finite density case with aparticular choice of a cubic lattice, for which we find the critical chemical potential,at which the hadronic phase transition occurs. a r X i v : . [ h e p - t h ] F e b ontents The holographic model of Witten-Sakai-Sugimoto (WSS) [1–3] is the top-down holographictheory closest to QCD to date. The model is based on a D4–D8 brane setup in type IIAstring theory and the flavor dynamics is encoded in the low-energy action for the gaugefields on the D8 flavor branes in the geometry left by the gauge D4-branes. The modelat low-energy reproduces a large- N c SU( N c ) gauge theory with N f massless quarks plus atower of massive adjoint matter fields. Being a top-down model, it has very few parameters: N c , N f , the ’t Hooft coupling λ and the mass scale M KK . In this paper, we will also includea quark mass term. The model shares all the important features with QCD, in particularconfinement and chiral symmetry breaking as well as the existence of a low-energy chiralLagrangian, which is of a Skyrme-type theory coupled to an infinite tower of vector mesons.Baryons in the WSS model are identified with instantons of the gauge theory describingthe flavor branes [4–7], just like baryons can be seen as solitons of the Skyrme model [8, 9].Quantization of low-energy instantonic degrees of freedom provides quantum numbers for1he corresponding nucleons. Many techniques developed in the context of quantization ofzeromodes of Skyrmions as nuclei have been used in our approach, see e.g. Refs. [10–17].Composite nuclei are described by multi-instantons configurations. We used this ap-proach to describe composite nuclei from a “solitonic perspective” in a previous work [18],where we restricted to the case of massless quarks. This approach can be viewed as com-plimentary to other approaches to holographic nuclear physics, see e.g. Refs. [19–26]. Inthe limit which we are considering, the instanton radius scales as λ − and the distancesbetween individual nuclei in the bound state configuration scale as λ and so a linear ap-proach can be used for the computation of the dominant two-body potential between thenuclei as an infinite sum of monopole and dipole interactions. In this limit the instan-tons become point-like but with an SU(2) orientation, which is very similar to the modelsconsidered in 3+1 dimensions in Refs. [27, 28]. In this work we include the quark masses,which via the Gell-Man–Oakes–Renner (GMOR) relation induces a pion mass, which inturn is felt by the solitonic (pionic) degrees of freedom. In Ref. [18], we found bound statesin the large N c and large λ limit and computed their respective nuclear binding energies.In distinction to the previous work (i.e. the massless case), the inclusion of a pion massmakes the nuclear bound states larger and the corresponding binding energies smaller.The paper is organized as follows. In section 2 we give an overview of the model anddiscuss the introduction of the quarks mass term in the framework. In section 3 we computethe nucleon-nucleon potential. In section 5 we discuss the quantization for the deuteronstate. In section 4 we present the numerical results for the nuclear bound states. In section6 we discuss the case of finite density and the hadronic phase transition. We conclude insection 7 with a discussion and outlook. The model encodes color degrees of freedom in a background metric from type IIA stringtheory [1]: ds = (cid:16) uR (cid:17) (cid:0) η µν dx µ dx ν + f ( u ) dx (cid:1) + (cid:18) Ru (cid:19) (cid:18) du f ( u ) + u d Ω (cid:19) ,e φ = g s (cid:16) uR (cid:17) , F = dC = 2 πN c Vol (cid:15) , f ( u ) = 1 − u u , (2.1)where φ is the dilaton, C n , F n +1 indicate the Ramond-Ramond n -form and its correspond-ing field strength, Vol and (cid:15) stand for the volume of a unit radius 4-sphere ( S ) andits volume form, respectively. The function f ( u ) makes the geometry terminate at a fixedcoordinate u KK , so in order to avoid singularities, the coordinate x has to be periodic with2eriod δx = 4 π R u KK ≡ πM KK . (2.2)Throughout this article we will work in units of the intrinsic mass scale of the theory, thatis, we set: M KK = u KK = 1 . (2.3)The inclusion of flavor degrees of freedom is performed via insertion of a couple of N f stacked D8/D8-branes (with N f being the number of light quark flavors), transverse tothe color branes in the x direction (that is, localized on the circle) [2, 3]. We will workin the setup with antipodal flavor branes, in which case the two stacks merge at the cigartip labeled by u = u KK : this is in every sense a geometrical realization of the spontaneousbreaking of chiral symmetry. A rigorous treatment should include also the backreactionon the geometry due to the presence of these stacks of branes, but since we will only beconsidering the presence of two light flavors, N f = 2, the effect of the modified geometrycan be neglected as a first approximation (see Ref. [29] for a treatment of the backreactionof the flavor branes and the implications).We will be interested in the theory on the D8/D8-branes, so we employ for the cigarsubspace bulk coordinates, defined by u = u + u KK r ,x = 2 R u KK θ, ⇒ (cid:40) y = r cos θ,z = r sin θ, (2.4)with z running on the curve that defines the embedding of the flavor branes, and y beingits transverse coordinate. The action on the D8-brane world volume is composed by twoterms: a Yang-Mills action in warped spacetime arising from the truncation of the DBIaction to quadratic terms, and a Chern-Simons term, originating from the coupling of theD8-branes to the Ramond-Ramond 3-form C : S = S YM + S CS ,S YM = − κ Tr (cid:90) d x dz (cid:20) h ( z ) F µν + k ( z ) F µz (cid:21) ,S CS = N c π (cid:15) α α α α α (cid:90) d x dz (cid:98) A α (cid:104) (cid:0) F aα α F aα α (cid:1) + 2 tr (cid:0) (cid:98) F α α (cid:98) F α α (cid:1)(cid:105) , (2.5)with the warp factors k ( z ) , h ( z ) and κ given by3 ( z ) = 1 + z , h ( z ) = (1 + z ) − , κ ≡ N c λ π . (2.6)The notation for the indices we use is as follows: α , α , ... = { , , , , z } , M, N, ... = { , , , z } ,i, j, k, ... = { , , } , µ, ν, ... = { , , , } . (2.7)In continuity with Ref. [18], it is useful to define a new coupling Λ, and a unique warpfactor H ( z ) as: Λ = 8 λ π , H ( z ) = (cid:0) z (cid:1) . (2.8)Moreover, we rescale the action by: S = κ − S , (2.9)so that the full rescaled action in the new notation reads: S = − (cid:90) d x dz H (cid:18) (cid:98) F α α (cid:98) F α α + tr ( F α α F α α ) (cid:19) + 1Λ (cid:15) α α α α α (cid:90) d x dz (cid:98) A α (cid:18) F α α F α α + 16 (cid:98) F α α (cid:98) F α α (cid:19) . (2.10) The addition of a quark mass to the model is performed by the insertion of a Wilson lineoperator in the dual QCD-like theory: this is the best we can do given the geometry, sinceflavor and anti-flavor degrees of freedom are not localizable at the same position (theywill always remain separated along the x direction). Hence, a quark mass term will benonlocal and take the form: δS ∝ (cid:90) d x N f (cid:88) i =1 OW ii ( x ) + h . c . , (2.11)where OW ji is the open Wilson line operator: OW ji ≡ ψ † j L (cid:16) x µ , x = − π R (cid:17) P exp (cid:20)(cid:90) dx (i A + Φ) (cid:21) ψ i R (cid:16) x µ , x = + π R (cid:17) . (2.12)4o obtain this object, we insert an open string stretching between the flavor branes, andprovide the action with the Aharony-Kutasov term [30]: S AK = κ (cid:48) Vol (cid:90) d x d Ω (cid:88) i e − S i str + h . c . , (2.13)with S str being the action of the stretched string. The action S str is composed of twoterms: the Nambu-Goto action for the free string ( S NG ) and an interaction term of thestring endpoints with the flavor branes to which they are attached: N (cid:48) (2 π ) g s l s (cid:16) u KK R (cid:17) (cid:90) d x e − S NG tr (cid:104)(cid:16) P e − i (cid:82) ∂ WS A − (cid:17) + h . c . (cid:105) , (2.14)where the subtraction of the identity matrix accounts for the subtraction of the vacuum.After factorizing away the contribution from the Nambu-Goto term in S str , we are left withan action that includes the interaction of the endpoint of the string with the D c and in the quark massmatrix M . Finally, since the endpoints are forced to move on the y = 0 curve in the cigarsubspace, the action in terms of the gauge fields is: S AK = c (cid:90) d x tr P (cid:104)(cid:16) M e − i (cid:82) + ∞−∞ dz A z − (cid:17) + h . c . (cid:105) , c = λ π . (2.15)At this point it is useful to evaluate c in terms of Λ: c = Λ √ π , (2.16)and to perform the rescaling (2.9): S AK = 2 (2Λ π ) N c (cid:90) d x tr P (cid:104) M (cid:16) e − i (cid:82) + ∞−∞ dz A z + e i (cid:82) + ∞−∞ dz A z − (cid:17)(cid:105) . (2.17) The additional term in the Lagrangian induces two effects: it obviously changes the massof the 1-instanton configuration, but could also potentially modify the size of the instanton.The classical YM instanton possesses a modulus ρ , that is interpreted as the size of theinstanton and does not affect the energy. As explored in Ref. [5], the combined effect ofthe Chern-Simons term (that tends to dilate the instanton) and of the gravitational field(that tends to shrink it) is reflected by the fact that ρ ceases to be a modulus, and theenergy gains a ρ dependence. The size of the instanton is then determined by minimizingthe energy with respect to ρ . Here we will determine the contribution of the mass term5o the total mass, and size of the instanton. We will follow the computations made inRef. [31], adapting them to the gauge that was used in Ref. [7].The energy of the static model is obtained by computing the opposite of the staticaction. This action is obtained by considering A I and ˆ A as the only non vanishing fields.The energy then reads E = (cid:90) d x dz (cid:18) tr (cid:18) H ( z ) − F ij + H ( z ) F zi (cid:19) − H ( z ) − ( ∂ i ˆ A ) − H ( z ) ( ∂ z ˆ A ) (cid:19) − (cid:90) d x dz ˆ A tr( F IJ F KL ) (cid:15) IJKL − π ) N c (cid:90) d x tr P (cid:104) M (cid:16) e − i (cid:82) + ∞−∞ dzA z + e i (cid:82) + ∞−∞ dzA z − (cid:17)(cid:105) . (2.18)We will use the same field configuration as was used in Ref. [7]. In fact, as argued inRef. [31], modifications to the field profile are subleading in Λ, and the first-order effectsonly affect the energy. The field configuration we use is A I = − σ IJ x J b ( ρ ) , ˆ A = a ( ρ ) , (2.19)where ρ = √ x I x I and the profiles a and b are given by a ( r ) = 8Λ r + 2 ρ ( r + ρ ) , b ( r ) = 1 r + ρ , (2.20)where r = √ x i x i . With this configuration, we can perform the integrals. The integral ofthe first two lines of Eq. (2.18) is done in great detail in Ref. [7]: resulting in:8 π (cid:18) ρ ρ (cid:19) . (2.21)For the third line of Eq. (2.18), using the arguments of Ref. [31] to get the leading ordercontribution in Λ, we can write the integral as (cid:90) d x tr (cid:0) M ( U + U † − ) (cid:1) . (2.22)To evaluate this integral in a divergenceless fashion, we must perform a gauge transforma-tion to avoid singularities: we will thus work in the gauge A z = (cid:18) r + ρ − ρ (cid:19) x i σ i . (2.23)6erforming the integration, the matrix exponentiation and the trace, we gettr (cid:0) M ( U + U † − ) (cid:1) = − m (cid:32) π (cid:115) r r + ρ (cid:33) , (2.24)where we assumed the quark masses to be degenerate, and thus M = m , with m u = m d = m . This term cannot be integrated in closed form. The integral can be written as (cid:90) d x tr (cid:0) M ( U + U † − ) (cid:1) = − mπρ I , (2.25)with I ≡ (cid:90) + ∞ x (cid:18) cos π √ x − + 1 (cid:19) ≈ . . (2.26)The total energy of the system is then E ( ρ ) = 8 π (cid:18) ρ ρ (cid:19) + 32 mπ (2Λ π ) N c Iρ . (2.27)As expected, the mass term works as an inertia factor, favoring configurations of smallerradius.Minimization of Eq. (2.27) cannot be done analytically. In order to obtain some insightof the deformed quantities, we proceed by writing ρ = ρ + x , where ρ = 4 √ Λ (cid:18) (cid:19) , (2.28)and making a power series expansion of the derivative of Eq. (2.27) around x = 0, trun-cating at first order and then finding the zero of the derivative (ramification point). Then,the result can also be expanded in a power series using m or N c . The deformation is ρ = 4 √ Λ (cid:18) (cid:19) (cid:32) − (cid:18) (cid:19) I √ π mN c + O (cid:16)(cid:0) mN c (cid:1) (cid:17)(cid:33) . (2.29)The mass of the system is equal to the energy at this value of ρ : we get (after rescalingthe energy) M B = N c (cid:32) Λ8 + (cid:114) (cid:33) + 16 I √ π (cid:18) (cid:19) m . (2.30)We see that the corrections brought by the mass term can be written as a power series in mN c (rescaled by an initial power of N c ), so the first correction given by the mass is subleading7n N c . We will eventually be interested in the configuration with (cid:98) A M = 0, since those componentsarise as N c − corrections (induced via the Chern-Simons term by the presence of a slowmotion, be it rotational or translational), so we neglect the factor exp (cid:0) − i (cid:82) dz (cid:98) A z (cid:1) . Wealso still impose the condition that the masses of the two light flavors are degenerate. TheAharony-Kutasov action contributes to the equations of motion with: δ S AK δA z = − i (2Λ π ) N c m (cid:16) e i (cid:82) + ∞−∞ dzA z − e − i (cid:82) + ∞−∞ dzA z (cid:17) = − π ) N c m sin (cid:18)(cid:90) + ∞−∞ dzA z (cid:19) . (2.31)Keeping only the first order of the sine power series, we obtain the new equation of motionfor the component A z (i.e. it is a set of three equations, since A z = A az T a ): H ( z ) ( ∂ i ∂ i A z − ∂ i ∂ z A i ) − π ) N c m (cid:90) + ∞−∞ dz (cid:48) A z ( x, z (cid:48) ) = Source terms , (2.32)while the other equations are unaltered (to be precise, the equation of motion for (cid:98) A z wouldreceive a correction if we go beyond the static approximation, but that is beyond ourgoal to include time derivatives only). Since the holonomy of the A z field is dual to thepion field, and since the first-order term in the equations of motion arise from a Lagrangianterm quadratic in it, the modification we performed to the equations of motion correspondscompletely with the relaxation of the massless pion regime, in favor of a finite pion mass.This is fully consistent with the quark description: by giving finite mass to single quarks,pions are now pseudo-Goldstone bosons, hence massive. However, since the quark massesare degenerate, we still have a residual symmetry, reflected in the degenerate masses of thepion triplet. For completeness and later convenience, we provide once again the full set of equations It could look like the mass of the η (cid:48) meson is also bound to be the same as that of the pion triplet (inthis limit with only two flavors), since the corresponding mass term has to originate from this very sameAharony-Kutasov action, and the fields are equally normalized. However, the Chern-Simons term for theRamond-Ramond form, C , induces the holographic realization of the Witten-Veneziano mechanism, thusremoving this degeneracy. See Ref. [2], and Refs. [32, 33] for more detailed discussions
8f motion restoring the parity indices and the sources: H − ∂ i ∂ i (cid:98) A + ∂ z (cid:16) H ∂ z (cid:98) A (cid:17) = − π Λ δ ( x ) δ ( z ) , (2.33) H − ∂ j ∂ j A + i + ∂ z (cid:16) H ∂ z A + i (cid:17) = − π ρ (cid:15) ijk σ k ∂ j δ ( x ) δ ( z ) , (2.34) H (cid:0) ∂ i ∂ i A + z − ∂ i ∂ z A − i (cid:1) − a (cid:90) + ∞−∞ dz (cid:48) A z ( x, z (cid:48) ) = − π ρ σ i ∂ i δ ( x ) δ ( z ) , (2.35) ∂ j ∂ j A − i − ∂ j ∂ i A − j H − ∂ z (cid:16) H (cid:0) ∂ i A + z − ∂ z A − i (cid:1)(cid:17) = 2 π ρ σ i δ ( x ) ∂ z δ ( z ) , (2.36)where we have defined the new parameter a ≡ π ) N c m . (2.37)We define the fields in terms of Green’s functions G ( x, z, x (cid:48) , z (cid:48) ) and L ( x, z, x (cid:48) , z (cid:48) ) as: (cid:98) A = − π Λ G ( x, z, , ,A + i = − π ρ (cid:15) ijk σ k ∂ j G ( x, z, , ,A − i = − π ρ σ i ∂ z (cid:48) G ( x, z, , z (cid:48) ) (cid:12)(cid:12) z (cid:48) =0 ,A + z = − π ρ σ i ∂ i L ( x, z, , , (2.38)and we take the functions G ( x, z, x (cid:48) , z (cid:48) ) and L ( x, z, x (cid:48) , z (cid:48) ) to obey H − ∂ i ∂ i G ( x, z, ,
0) + ∂ z (cid:16) H ∂ z G ( x, z, , (cid:17) = δ ( x − x (cid:48) ) δ ( z − z (cid:48) ) , (2.39) H ( ∂ i ∂ i L ( x, z, x (cid:48) , z (cid:48) ) − ∂ i ∂ z G ( x, z, x (cid:48) , z (cid:48) )) − a (cid:90) + ∞−∞ dz (cid:48) L ( x, z, x (cid:48) , z (cid:48) ) = δ ( x − x (cid:48) ) δ ( z − z (cid:48) ) , (2.40) H − ∂ z (cid:48) G ( x, z, x (cid:48) , z (cid:48) ) + ∂ z ( H L ( x, z, x (cid:48) , z (cid:48) )) = 0 , (2.41)with G ( x, z, x (cid:48) , z (cid:48) ) and L ( x, z, x (cid:48) , z (cid:48) ) given as in Ref. [6, 18] by G ( x, z, x (cid:48) , z (cid:48) ) = − π ∞ (cid:88) n =1 ψ n ( z ) ψ n ( z (cid:48) ) c n e − k n | x − x (cid:48) | | x − x (cid:48) | , (2.42) L ( x, z, x (cid:48) , z (cid:48) ) = − π ∞ (cid:88) n =0 φ n ( z ) φ n ( z (cid:48) ) d n e − k n | x − x (cid:48) | | x − x (cid:48) | . (2.43)Our goal is to derive the values of k n dual to the meson masses again, to check how thepresence of finite quark masses influences the masses of the bound states. To start off, we9efine a scalar product as in Ref. [18]:( f, g ) ≡ (cid:90) + ∞−∞ dz H − ( z ) f ( z ) g ( z ) . (2.44)The Ans¨atze (2.42)-(2.43) and Eq. (2.39) give us the usual condition for the profile functions ψ n ( z ): H ( z ) ∂ z (cid:16) H ( z ) ∂ z ψ n ( z ) (cid:17) = − k n ψ n ( z ) , (2.45)so that ψ n ( z ) are found as solutions to an eigenvalue problem of an Hermitian (with respectto Eq. (2.44)) operator. They must then obey a completeness relation given by ∞ (cid:88) n =1 ψ n ( z ) ψ n ( z (cid:48) ) H ( z ) c n = δ ( z − z (cid:48) ) , c n = ( ψ n , ψ n ) , (2.46)so that at this stage there are no differences compared to the massless quarks problem.The pion profile function corresponds to φ so let us turn our attention to that: Eq. (2.41)is solved by imposing, for every n , the conditions ∂ z ( H ( z ) φ n ( z )) φ n ( z (cid:48) ) d n + ψ n ( z ) ( ∂ z (cid:48) ψ n ( z (cid:48) )) H ( z ) c n = 0 , (2.47) ∂ z ( H ( z ) φ ( z )) φ ( z (cid:48) ) d = 0 . (2.48)Eq. (2.47) is solved by choosing φ n ( z ) = ∂ z ψ n ( z ) and d n = k n c n : note however that d is notdetermined by this condition, since there is no normalizable mode ψ . Finally, Eq. (2.48)imposes the shape of the pion wave function φ ( z ) = H − ( z ).Now we substitute into Eq. (2.35) obtaining H ( z ) ∞ (cid:88) n =0 φ n ( z ) φ n ( z (cid:48) ) d n δ ( x − x (cid:48) ) + a π ∞ (cid:88) n =0 (cid:82) + ∞−∞ dzφ n ( z ) φ n ( z (cid:48) ) d n e − k n | x − x (cid:48) | | x − x (cid:48) |− H ( z ) π φ ( z ) φ ( z (cid:48) ) d k e − k | x − x (cid:48) | | x − x (cid:48) | = δ ( x − x (cid:48) ) δ ( z − z (cid:48) ) . (2.49)Here all contributions to the sum in the second term vanish due to the relation φ n ( z ) = ∂ z ψ n ( z ) and the normalizability of the eigenfunctions ψ n ( z ): (cid:90) + ∞−∞ dzφ n ( z ) = ψ n (+ ∞ ) − ψ n ( −∞ ) , n ≥ , (2.50)so that only the n = 0 contribution survives. In the first term we can make use of a10ompleteness relation for the functions φ n ( z ): We define a new inner product (cid:104) f, g (cid:105) ≡ (cid:90) + ∞−∞ dz H ( z ) f ( z ) g ( z ) , (2.51)which satisfies (cid:104) φ n , φ n (cid:105) = d n = k n c n for all n but n = 0. Extending the notation to φ andperforming the integration we identify d = π and obtain the completeness relation ∞ (cid:88) n =0 H ( z ) φ n ( z ) φ n ( z (cid:48) ) d n = δ ( z − z (cid:48) ) , d n = (cid:104) φ n , φ n (cid:105) . (2.52)Exploiting this new relation, we can check that the first term of Eq. (2.49) cancels out withthe deltas on the right hand side. Hence all we are left with is the following equation: a π φ ( z (cid:48) ) e − k | x − x (cid:48) | | x − x (cid:48) | − H ( z ) π φ ( z ) φ ( z (cid:48) ) d k e −| x − x (cid:48) | | x − x (cid:48) | = 0 , (2.53)that we can further simplify by recalling that φ ( z ) = H − ( z ). After doing so, it is possibleto identify the pion mass by recalling that the k n are dual to the masses of the mesons, sowe obtain k = πa = (cid:112) (2 π ) Λ N c m . (2.54)An alternative, perhaps more intuitive way to obtain the same result is to obtainthe effective Lagrangian for the mesons by expanding the gauge fields: doing so for theAharony-Kutasov action, and remembering that the holonomy of the A z field is related tothe pseudoscalars as U ≡ P exp (cid:18) − i (cid:90) dz A z (cid:19) = exp (cid:18) i f π ( π a τ a + S ) (cid:19) , (2.55)we can easily derive the Gell-Man–Oakes–Renner relation for this model:4 mc = f π m π , (2.56)from which the pion mass squared can be read off. Making use of Eq. (2.8) and theholographic formula for the pion decay constant f π = 4 κπ , (2.57)it is easy to verify that m π indeed coincides with k of Eq. (2.54).11 Nucleon-Nucleon potential
We now turn to computing the interaction potential between two nucleons: this is donealong the lines of the previous work (i.e. Ref. [18]). As a first step we build a two-instantonconfiguration by positioning two instantons at a distance far larger than their size andgiving them arbitrary orientations: (cid:98) A = (cid:98) A p + (cid:98) A q , A = BA p B † + CA q C † , (3.1)with (cid:0) X ( p ) , Z ( p ) (cid:1) = (0 , , ,
0) and (cid:0) X ( q ) , Z ( q ) (cid:1) = ( r , r , r , R (cid:29) ρ with R being the distance between the instanton centers.This is a viable approximation scheme since the holographic model extrapolates from thelarge λ regime, and the size of the single instantons is of order ρ − while R (cid:39) λ . Space isthen divided into three regions: two balls of radius ρ centered at (cid:126)X = (cid:126)X ( p ) and (cid:126)X = (cid:126)X ( q ) ,where the corresponding field is strong and the other is in its linear approximation, andthe rest of space where both fields can be approximated by their linear form.To compute the (static) energy, we employ the definition S = − (cid:82) dt E so that E = κ − E = − κ − (cid:90) d x dz L , (3.2)is the rescaled energy, while E is the energy in physical units ( L is the Lagrangian densityin physical units). Part of the integral in Eq. (3.2) will account for the self energies ofthe solitons (the masses of the single baryons), so to compute the interaction potential,we have to subtract off these self energy terms: the Ansatz (3.1) allows us to easily do soby keeping just the cross-terms that involve both fields A p,q . The full rescaled energy isobtained by the on-shell action as: E = (cid:90) d x dz (cid:32) H tr (cid:0) F ij (cid:1) + H tr (cid:0) F iz (cid:1) + 12 H (cid:0) ∂ i (cid:98) A (cid:1) + H (cid:0) ∂ z (cid:98) A (cid:1) (cid:33) + a (cid:90) d x dz tr (cid:18)(cid:90) z −∞ dz (cid:48) A z ( x, z ) A z ( x, z (cid:48) ) (cid:19) , (3.3)where we have made use of the equations of motion to trade the Chern Simons term for achange in sign in the (cid:98) A terms of the Yang-Mills action.Now it is sufficient to insert Eq. (2.38) into the above expression and exploit Eqs. (2.39),(2.41) and (2.40) together with the Ans¨atze (2.42) and (2.43) to be able to perform integra-tions with Dirac deltas. The only difference that emerges with respect to Ref. [18], is thepresence of the last term in Eq. (3.3), and that now k (cid:54) = 0. We introduce the symmetric12ensor: P ij ( r, k ) = δ ij (cid:0) ( rk ) + rk + 1 (cid:1) − r i r j r (cid:0) ( rk ) + 3 rk + 3 (cid:1) , (3.4)as well as the rotation matrix M ij ( G ) = 12 tr (cid:0) σ i Gσ j G † (cid:1) , (3.5)which gives the spatial rotation corresponding to an SU(2) rotation implemented via thematrix G ∈ SU(2). The full potential at the end of a somewhat lengthy (but analogous tothat of section 3.1 of Ref. [18]) is given in physical units by: V ( r, B † C ) = 4 πN c Λ (cid:32) ∞ (cid:88) n =1 (cid:18) c n − e − k n − r r + 65 1 c n − M ij ( B † C ) P ij ( r i , k n − ) e − k n − r r −
65 1 d n e − k n r r M ij ( B † C ) P ij ( r i , k n ) (cid:19) − π e − k r r M ij ( B † C ) P ij ( r i , k ) (cid:33) . (3.6)In this formula, r i is the relative position of the single soliton cores, while r = r i r i . Wesee immediately that the difference with respect to the potential obtained in Ref. [18] isentirely contained in the last term, corresponding to the contribution of the n = 0 mode,that is, the pion: here we have simply another short-range interaction (actually absorbablein the previous term by extending the sum to n = 0 and remembering that d = π ),mediated by a particle of mass k , as can be argued by the exponential decay. In Ref. [18]we had a long-range interaction, as appropriate for a massless mediating boson.Here we assumed fixed a position in the z coordinate and fixed the instanton sizes:It is possible to consider these quantities as massive moduli as long as we consider smalloscillations around their equilibrium value, so that they will actively enter in the expressionof the potential to give the more general formula V ( r, B † C, ρ i , Z i )= 4 πN c Λ ∞ (cid:88) n =1 ψ n ( Z ) ψ n ( Z ) c n e − k n − r r + π N c Λ ρ ρ φ ( Z ) φ ( Z ) π e − k r r M ij ( B † C ) P ij ( r i , k )+ π N c Λ ρ ρ (cid:32) ∞ (cid:88) n =1 (cid:18) ψ n ( Z ) ψ n ( Z ) c n + φ n ( Z ) φ n ( Z ) d n (cid:19) e − k n r r P ij ( r i , k n ) (cid:33) M ij ( B † C ) . (3.7)As we can see now, every sum runs over all values of n , not alternating between even andodd values: It is easy to verify that this generalization is due to the presence of Z i (cid:54) = 0,and that in the case Z i = 0 the appropriate terms vanish because of the parity propertiesof the functions ψ n ( z ), and φ n ( z ). 13 Numerical evaluations
We now proceed to the numerical evaluation of the various physical quantities that wehave found. In our model, we have three free parameters , M KK , Λ and m , of which M KK is the only dimensional quantity. We calibrate the parameters to the following values: • For the mass parameter M KK , we choose the value 949 MeV. This is chosen to fitthe mass of the ρ meson, that is the second-lightest massive meson in our model.Even if the usual practice is to fit the mass scale using the lightest massive particlein the model, we choose to use the ρ meson for the fit, following Ref. [18]. Thisway the introduction of the pion mass will appear in numerically in the model as aperturbation of the massless model. • For the coupling Λ, we choose the value 1 . • The new parameter m is interpreted as the quark mass. We have chosen the value3 . × − for this parameter. This yields a quark mass of 2 .
910 MeV, which isbetween the masses of the u and d quarks.With the above parameter values, we get the following values for the observables: • Pion mass: Evaluating Eq. (2.54), we get a pion mass of 134 . . π )and 139 . π ± ). This result confirms that the model workswell at low energies. • Baryon mass: Evaluating Eq. (2.30), we get a baryon mass of 1 .
628 GeV for the lightbaryons (neutrons and protons). The baryons in our model are significantly heavierthan the physical baryons, whose masses sit around 938 MeV for the proton andneutron. This is to be expected, as the model is expected to break down at energyscales larger around 949 MeV.
We will now turn to the minimization of the 2-body potential acting on B nuclei as V B tot = B (cid:88) i,j =1 i (cid:54) = j V ( r ij , B † i B j ) , (4.1) In principle, the number of colors, N c , is also a free parameter, but we will make the obvious choice N c = 3. r ij ≡ | x i − x j | and B i is the SU(2) rotation matrix of the i -th nucleon (instanton).The numerical methods we will use are a simple random walk algorithm (only movingforward when the potential energy is lowered) a Metropolis algorithm allowing for statisticalrandom movement (with some probability of moving “uphill”) as well as a simple gradientflow method.Figure 1: Geometric configurations for stable and metastable nuclei up to B = 8. Thestable (energetically favorable) configurations are labeled with the baryon number B andthe metastable states have an added suffix in form of a Latin letter; the further in thealphabet the higher the energy. Bonds are shown for short enough distances between thenuclei as black solid lines. The color scheme is described in the text.In Fig. 1 we present the numerical results for classical composite nuclei, stable andmeta-stable, up to baryon number B = 8. The stable configurations (ground states) arelabeled with their corresponding baryon number B , while metastable states have a Latinletter added to the label. With increasing letter in the alphabet, the higher is the energy.This bond between two nuclei symbolizes the 2-body potential, which is used to find theseconfigurations and the bond is displayed only for distances shorter than 1 . R with R being the optimal distance of two nucleons in deuterium ( B = 2) for m = 0 . B = . All points on the color sphere for a nonstandard orientation of theinstanton are then rotated by the rotation matrix M ij ( B ) of Eq. (3.5). BR B Bm = 0 m = 0.142 m = 0.251 0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 2 3 4 5 6 7 8 Figure 2: Binding ratios for the stable nuclei (ground states) up to B = 8, for variousvalues of m . In our calibration, the physical pion mass is m = 0 .
142 (dimensionless units).We will now calculate the binding ratios of the B nuclear bound states, defined byBR B ≡ BE − E B BE . (4.2)The result is shown in Fig. 2. Qualitatively, the bound states show the same spatial/geome-trical distributions in the case of a finite pion mass as compared to the massless case, seeFig. 1 and Fig. 5 of Ref. [18]. The optimal distance between two nucleons in the 2-bodypotential grows with an increasing pion mass and this effectively enlarges all the sizes of thenuclear bound states, but does not affect the orientations of the instantons in SU(2) space.In particular, the optimal distance of the 2-body potential for m = 0 in dimensionless unitsis R = 2 .
06 and for m = 0 .
142 it is R = 2 .
23. The binding energies and thus the bindingratios, similarly decrease with an increasing pion mass.On the other hand, the presence of the pion mass removes the long-range force of thepotential, so even though the minimum is pushed a bit away in the 2-body potential,the attraction is also shorten somewhat. This has the consequence that some states aredifferent in the massive case, with respect to the massless case. In particular, a metastablestate in the B = 5 sector made of a tetrahedron with a satellite exists in the masslesscase [18], but has disappeared in the massive case. Similarly, the ground state of the B = 6 state has changed by moving the left-most and the top most nucleon in Fig. 1 closerto the hexagon in the massive case, than in the massless case where it was more compact.16he details of the solutions in the massive case with m = 0 .
142 are shown in Tab. 1. B shape details V min BR2 line bond length = 2.23 − . . − . . − . . − . . − . . − . . − . . − . . − . . − . . − . . B = 2 , . . . V min is the potential in dimensionless units and BR is the binding ratio. The quantum description of the deuteron is obtained by quantization of the collective modesstarting from the B = 2 classical system that we examined in section 3. The procedure hasalready been described in detail in Ref. [18]: Here we will review the analysis and obtainthe binding energy of the deuteron in the system at hand.A generic field configuration in the two-instanton sector is written by considering onlythe degrees of freedom that do not change the potential computed in section 3, and fixingall other degrees of freedom to have the potential as attractive as possible. The unfixeddegrees of freedom form the zeromode manifold, and can be interpreted in the followingway: • Overall position of the center of mass of the system, that we denote as x = ( x , x , x ).Due to the gravitational field in the z -direction, we do not allow motion along the z -axis. Quantizing those degrees of freedom gives the system an overall momentum.17 Isospin of the system: This is represented by an SU(2) matrix called U . • Spin of the system: This is represented by an SU(2) matrix called E and the corre-sponding SO(3) matrix, M ij ( E ) is given in Eq. (3.5). • Parity: We can always switch the two components of the system. As this is a discretesymmetry, it will not induce a momentum. We will indicate parity with a binaryvariable P = 0 , R = ( R , ,
0) (with both nuclei centered at the origin of theholographic coordinate) with phase opposition, B † C = i σ . The numerical value of R isthe value at which the potential (3.7) is minimized.A configuration A I in the two-instanton sector belongs to the zeromode manifold, if itcan be written as A I ( x, z ) = U E † A I (cid:16) x − ( − ) P M ( E ) r , z (cid:17) ( U E † ) † + U i σ E † A I (cid:16) x + ( − ) P M ( E ) r , z (cid:17) ( U i σ E † ) † . (5.1)The kinetic energy is computed by starting from the kinetic energy of two baryons, com-puted in Ref. [2], and then comparing configuration (5.1) with configuration (3.1) to un-derstand how the collective coordinates are related to the coordinates describing the singlebaryons. After performing the transformation and freezing the coordinates that are notzeromodes, we obtain the kinetic energy for the zeromode manifold. Since the derivationis the same as in Ref. [18], we will just cite the result here. In terms of the angular left-invariant velocities Ω i = − i tr( U † ˙ U σ i ) and ω i = − i tr( E † ˙ Eσ i ), we have the kinetic energy T = 12 M B (cid:18) ρ ω + (cid:18) ρ + R (cid:19) ω + R ω + ρ (cid:0) Ω + Ω + (Ω − ω ) (cid:1)(cid:19) . (5.2)The mass M and radius ρ have been computed in subsection 2.2, and the parameter R is obtained from the potential (3.7) in the attractive channel, by finding the minimum.The Hamiltonian for the deuteron is computed straightforwardly. First, the momenta aredefined as L = M B ρ ω , L = M B (cid:18) ρ + R (cid:19) ω , L = M B (cid:18) R ρ (cid:19) ω − M ρ Ω ,K = M B ρ Ω , K = M B ρ Ω , K = M B ρ (Ω − ω ) . (5.3)In terms of those momenta, the Hamiltonian reads H = 12 M B ( X ij L i L j + Y ij K i K j + 2 Z ij L i K j ) + V min + 2 M B , (5.4)18here we have included rest mass and defined the inertia tensors X, Y, Z as X ≡ ρ ρ + R
00 0 R , (5.5) Y ≡ ρ ρ
00 0 R + ρ , (5.6) Z ≡ R . (5.7)Quantum states are expressed as | ψ (cid:105) = | k, k , i , l, l , j (cid:105) , (5.8)where k and l are eigenvalues for K and L , of values k ( k + 1) and l ( l + 1) respectively, k and l are eigenvalues for K and L , and i and j are eigenvalues for the right-invariantmomenta I and J , that commute with the (left-invariant) momenta K and L and havethe properties I = K and L = J . The right-invariant momenta can be obtained fromthe left-invariant momenta as J i = − M ( E ) ij L j and I i = − M ( U ) ij K j (where E and U arepromoted to position operators).Due to the fact that the configuration space has discrete symmetries (as an example,multiplying U and E by i σ on the right leaves the configuration invariant) we have toimpose Finkelstein-Rubenstein constraints. The analysis in Ref. [18] is still valid: Theonly states that are compatible with the quantization of the single baryons as fermions andthat include the constraints are | D (cid:105) = | , , , , , j (cid:105) , | I (cid:105) = | , , i , , , (cid:105) , | I (cid:105) = 1 √ | , , i , , , (cid:105) + | , − , i , , , (cid:105) ) . (5.9)Of those states, | D (cid:105) is the only one having the quantum numbers of the deuteron (isospin0 and spin 1). The energies of the states are computed by acting on them with the19amiltonian: the final result is H | D (cid:105) = ρ M B R ρ + 2 M B + V min | D (cid:105) = E D | D (cid:105) ,H | I (cid:105) = (cid:18) ρ M B + 2 M B + V min (cid:19) | I (cid:105) = E I | I (cid:105) ,H | I (cid:105) = (cid:18) ρ M B (cid:18) ρ R (cid:19) + 2 M B + V min (cid:19) | I (cid:105) = E I | I (cid:105) . (5.10)It is evident that E D is always the smallest energy of the three ones above, so the deuteronstate is effectively the ground state of the system. Although the result is identical to theresult in Ref. [18] in form, we emphasize that in this setting the values of M B , ρ, V min and R are different, so the final numerical result will be different.With E D obtained from the first line of (5.10), the most relevant quantity we cancompute is the binding energy E D − M B . Various parameters are needed to obtain thisvalue. The new parameters R and V min are not free, but they are fixed by looking forthe minimum of the classical potential (3.7) in the attractive channel ( B † C = σ , r =( R , , , ρ i = ρ, Z i = 0). In the massive model, we have found R = 2 .
23 and V min = − .
220 by including 40 massive mesons in the potential computation. Furthermore, theinstanton size can be computed from Eq. (2.29), obtaining a value of ρ = 2 . − V min = 208 . E D − M B = 208 . .
224 MeV found in experiments. As in our previouswork, the binding energy is two orders of magnitude larger than the expected bindingenergy. The pion mass improves the prediction, as the binding energy of the masslessmodel is 275 . N c and corrections is required to understand if theprediction for the binding energy can be improved, as the parameters N c and Λ are notthat large. The attractive channel of the potential between two nucleons is obtained by employing arelative rotation between their SU(2) orientations: The iso-rotation must act by an angle π with an axis orthogonal to the relative position vector. Using the axis-angle notation,the matrix M ij ( B † C ) that describes the relative orientation is given by M ij (ˆ u , α ) = δ ij cos α + (1 − cos α ) ˆ u i ˆ u j + (cid:15) ijk ˆ u k sin α . (6.1)It is possible to arrange nucleons in a cubic lattice in such a way that the interaction20igure 3: The fundamental cell of the infinite lattice. Careful inspection of the figurereveals that every pair of nucleons, connected with a solid black line, face each other witha matching color, representing the attractive channel described in the text.between all nearest neighbors is in the attractive channel, see fig. 3. We make the followingchoices: The relative iso-orientation B † C between every soliton and its nearest neighborsis given by • ± i σ for r = ( ± R, , α = π, ˆ u = ˆ x , • ± i σ for r = (0 , ± R, α = π, ˆ u = ˆ x , • ± i σ for r = (0 , , ± R ), corresponding to α = π, ˆ u = ˆ x .This choice for the nearest neighbors fixes completely the lattice in a self-consistent way(since σ σ = i σ ), so it is now possible to compute the energy density associated to it.Before moving to the computation, let us make some useful considerations: The relativeorientation enters the potential formula via the combination M ij P ij ( r , k ), which reads M ij (ˆ u , α ) P ij ( r , k ) = (cid:18) α − (1 − cos α ) (ˆ u · r ) r (cid:19) ( kr ) − (1 − cos α ) (cid:18) u · r ) r − (cid:19) ( rk + 1) , r ≡ √ r · r , (6.2)so that we have two classes of iso-orientations: α = 0 and α = π : M ij (ˆ u , P ij ( r , k ) = 2( rk ) M ij (ˆ u , π ) P ij ( r , k ) = 2 (ˆ u · r ) r (cid:2) − rk + 1) − ( rk ) (cid:3) + 2( rk + 1) . (6.3)21o compute the full energy density it is sufficient to evaluate the potential between asingle chosen nucleon and all the others: Then translational symmetry ensures that thefull energy is given by multiplying this contribution by the number of nucleons (which isinfinite) with a factor of one half to avoid double counting. The nucleon number can thenbe traded for the number of cells, as a function of the lattice volume V L , so the full energywill be computed as E = 12 (cid:88) p (cid:88) p (cid:48) (cid:54) = p V ( r pp (cid:48) , B † p C p (cid:48) ) = 12 V L R (cid:88) p (cid:48) (cid:54) = p V ( r pp (cid:48) , B † p C p (cid:48) ) , r pp (cid:48) ≡ r p (cid:48) − r p , (6.4)with p indicating a lattice site.To make the calculation more straightforward, we choose our reference frame in such away that r p = (0 , ,
0) and B p = , so we are left with the following formula for the energydensity: EV L = 12 R (cid:88) p (cid:48) V ( r p (cid:48) , C p (cid:48) ) = 12 R (cid:88) p (cid:48) V ( r p (cid:48) , α p (cid:48) , ˆ u p (cid:48) ) , (6.5)where p (cid:48) runs over all lattice sites with the exception of (0 , , r = R ( n , n , n ) , n i ∈ Z . (6.6)The position r not only enters the expression of M ij P ij , but it also determines whether α = 0 or α = π : We must therefore be careful and classify iso-orientations with respectto ( n , n , n ). There are four possible scenarios, depending on how many even or oddinstances of n i are present in the position vector: • all n i are EVEN ⇒ C p (cid:48) = , • all n i are ODD ⇒ C p (cid:48) = , • one n i is EVEN ⇒ C p (cid:48) = ± i σ j , • one n i is ODD ⇒ C p (cid:48) = ± i σ j ,where in the last two lines σ j indicates a rotation around the axis determined by our rulefor the rotation of nearest neighbors (that is, j = 2 for i = 1, j = 3 for i = 2 and j = 1for i = 3). For example, a nucleon sitting at R (2 b − , b , b ) with b ∈ Z will have anorientation of ± i σ , i.e. the same as a nucleon sitting at R (2 b , b − , b − α = π , and the axis of rotation enters Eq. (6.2) onlyvia (ˆ u · r ), which selects the component of r along the axis ˆ u . But since ˆ u is a unit vectorcorresponding to one of the coordinate axes, this simply selects the j -th component Rn j .22ow we will compute the contributions to the energy density: Let us start with thefirst two possibilities in the classification, corresponding to all even or all odd n i , for whichwe obtain E , V L = 4 πN c R Λ (cid:88) b ,b ,b (cid:34) ∞ (cid:88) n =1 (cid:18) c n − e − k n − r r + 12 k n − c n − e − k n − r r − k n d n e − k n r r (cid:19) − k π e − k r r (cid:35) , (6.7)with r = 4 R ( b + b + b ) for the even case ( E ) and r = R [(2 b − +(2 b − +(2 b − ]for the odd case ( E ). We now turn to the more difficult scenarios: We start by considering r = (2 b − , b , b ) R with b ∈ Z . In this case the energy density becomes E (1)3 V L = E p , (4 b ) V L , (6.8)where we have defined E p , ( b ) V L ≡ πN c R Λ (cid:88) b ,b ,b (cid:34) ∞ (cid:88) n =1 (cid:20) c n − e − k n − r r ++ 65 2 c n − (cid:18) br (cid:0) − rk n − + 1) − ( rk n − ) (cid:1) + ( rk n − + 1) (cid:19) e − k n − r r −−
65 2 d n (cid:18) br (cid:0) − rk n + 1) − ( rk n ) (cid:1) + ( rk n + 1) (cid:19) e − k n r r (cid:21) −− π e − k r r (cid:18) rk − bR r (cid:0) k r + k r (cid:1)(cid:19) (cid:35) , (6.9)with r = R [(2 b − + 4 b + 4 b ]. We can see that the other terms labeled by E (2 , willhave the same form except for the substitution of every instance of b with respectively b , b : Since the sum runs over all three parameters, then each of these terms will givecontribution equal to the result E V L = 3 E (1)3 V L . (6.10)We are thus only left with the terms of the case in which one coordinate is an evenmultiple of the lattice spacing, while the others are odd. As before we begin by analyzingthe situation in which this coordinate is r , so that r = R (2 b , b − , b − j -th componentof r due to the rotation axis will now pick up an odd value instead of an even one, so that E (1)4 V L = E p , ((2 b − ) V L . (6.11)23nd r is now given by r = R [4 b + (2 b − + (2 b − ]. Again, as in the previous case,the sum of all three possible combinations with only one even coordinate will just resultin a factor of three, so we get E V L = 3 E (1)4 V L . (6.12)The total energy density is then given by the sum of the contributions from all four possi-bilities in the classification, giving E tot V L = 1 V L ( E + E + E + E ) . (6.13) E R L max = 1 L max = 2 L max = 3 L max = 5 L max = 15 L max = 25 L max = 50−0.0015−0.001−0.0005 0 0.0005 0.001 0.0015 0.002 2 3 4 5 6 7 8 9 10 Figure 4: The shape of the potential energy density around its minimum. L max is the cutoffof lattice sites: | b j | ≤ L max .The total lattice potential can then be computed numerically and plotted as a functionof the lattice spacing R : the sum over lattice sites converges with good precision ( ∼ − )for b j ∈ [ − L max , L max ] with L max = 15, ∀ j = 1 , ,
3. Notice that the cutoff in terms oflattice sides is 2 L max in each direction (i.e. both in the x + direction an the x − direction)and the factor of two is due to b j being an integer parametrizing even or odd n j ’s. Theresulting potential density has a shallow minimum at R = 4 .
86 (see Fig. 4).It is not straightforwardly clear that the quadruple sum in Eqs. (6.7) and (6.9) areconvergent in the limit of L max → ∞ , i.e. in the limit of summing over all lattice sites b j and all massive vectors n (being the index of k n ). The converges of the quadruple sumwould prove that the density in the given form is finite. We give a mathematical proof ofthe convergence of each term in the sums separately in appendix A.24 .1 Hadronic phase transition We will now use the baryon lattice to study the presence of a hadronic phase transition:To do so we need to compute the free energy of the configuration and look for a criticaldensity at which it becomes negative.We can do this without relying on holography, simply by calculating the Legendretransform of the energy density. First of all, we recover the total energy density from theinteraction potential density by adding the mass density terms: the cubic lattice cell hasunit net baryon number, so the mass density as a function of the lattice spacing R is simply M ( R ) = M B R − , (6.14)with M B given by Eq. (2.30).The chemical potential is defined as the derivative of the energy density with respectto the baryon number density (which we can trade for a derivative with respect to R , sincethe baryon number density is d B = R − ): µ = − R ∂∂R (cid:0) EV − L + M B R − (cid:1) , (6.15)with EV − L given by Eq. (6.13).The free energy density F is then obtained from the energy density as: F = EV − L + M ( R ) − µR − = EV − L + R ∂∂R EV − L . (6.16)In Fig. 5, we plot the free energy density as a function of R , showing that it becomesnegative at a finite density, signaling the presence of a first order hadronic phase transition. In this paper, we have included the quark mass term using the Aharony-Kutasov action inour solitonic approach to holographic nuclear physics. We work in the limit Λ → ∞ , wherethe size of the instantons is much smaller than the typical separation distance betweennuclei in a nuclear bound state. This allows us to calculate multi-instanton solutions bygluing together two overlapping instantons in the linear regime; in particular, we computethe 2-body potential which now has acquired a mass term for the pions. This inductionof the pion mass from the quark mass term is in line with the GMOR relation. Usingnumerical methods we find nuclear bound states – stable and metastable – with baryonnumbers B = 2 through B = 8. 25 RL max = 1 L max = 2 L max = 3 L max = 5 L max = 15 L max = 25 L max = 50−0.0002−0.0001 0 0.0001 0.0002 0.0003 2 3 4 5 6 7 8 9 10 Figure 5: The free energy density as a function of the lattice spacing R . As the densityincreases (i.e. R decreases), the free energy becomes negative, signaling a phase transition. L max is the cutoff of lattice sites: | b j | ≤ L max .We find that the main difference by having massive pions in the 2-body potential isthat the nuclear bound states grow slightly in size and correspondingly reduce their bindingenergy. Similar effect apply also the deuteron bound state as to the other light nuclei.Using the 2-body potential, we consider the case of an infinite crystal of nucleons ina cubic lattice with each nucleon oriented in SU(2) space so as to put it in the attractivechannel with respect to nearest neighbors. We calculate the potential energy density andprove in appendix A that it is finite, and finally use it to calculate the free energy. The freeenergy becomes negative at a critical lattice spacing R crit ∼ .
9, which signals a hadronicphase transition of first order. The presence of the hadronic phase transition was alsostudied in the same model in Refs. [34–37]: The difference with our approach lies both inthe description of the solitons and in the setup of the flavor branes. In the cited works thebranes’ position at infinity are taken to be close enough to allow the use of the deconfinedgeometry up to arbitrarily low temperatures, thus being far from the antipodal branes’regime: the baryonic matter is described with various levels of accuracy, starting fromexactly pointlike instantons up to an infinite lattice of finite size instantons interactingwith the nearest neighbors via the ADHM construction. We work instead with antipodalbranes at zero temperature, using the confined geometry, and we employ the flat spaceinstanton approximation, then take into account every single interaction (up to a cutoff inlattice size) with every soliton seeing each other’s core as pointlike. The presence of thefirst order (except for exactly pointlike instantons) phase transition remains a feature ofthe model in both regimes.The work carried out in this paper has been done in the large Λ and large N c limit, whichis of course none other than approximations to real world physics, as neither parameterstake (that) large phenomenological values (i.e. Λ SS = 1 .
569 and N c = 3). It would be26nteresting to calculate, for instance, the leading Λ − correction to the 2-body potentialand see if this could improve some phenomenological properties of our results, for examplethe large binding energies. Of course, this would turn the problem into a nonlinear oneand make it severely more difficult that the one we have solved here. Acknowledgments
S. Baldino wishes to dedicate this work to the memory of Federico Tonielli, to rememberan extraordinary person and an excellent scientist who left us prematurely. The workof S. Baldino is supported by FCT - Fundac˜ao para a Ciˆencia e a Tecnologia, throughthe PhD fellowship SFRH/BD/130088/2017. The work of L. Bartolini is supported bythe “Fondazione Angelo della Riccia”. The work of S. Bolognesi is supported by theINFN special project grant “GAST (Gauge and String Theories)”. S. B. Gudnason thanksthe Outstanding Talent Program of Henan University for partial support. The work ofS. B. Gudnason is supported by the National Natural Science Foundation of China (GrantsNo. 11675223 and No. 12071111).
A Proof of convergence
We now give a proof that the energy density for the infinite lattice of baryons (6.13) is aconvergent sum, by giving an upper bound for each term.
Lemma 1
The following quadruple sum obeys the inequality ∞ (cid:88) n =0 b ,b ,b = −∞ ( b ,b ,b ) (cid:54) =0 e − k n R √ b + b + b c n R p ( b + b + b ) p ≤ A p ( m, R ) G (cid:0) ηR (cid:1) + B p ( m, R ) G (cid:0) ηR (cid:1) min( { c n } ) , (A.1) provided k n obeys k n ≥ m + ηn with m > , η > positive constants, and p ∈ Z > .Proof : Using √ A + B + C ≥ | A | + | B | + | C | , we have for the triple sum ∞ (cid:88) b ,b ,b = −∞ ( b ,b ,b ) (cid:54) =(0 , , e − k √ b + b + b ≤ ∞ (cid:88) b ,b ,b = −∞ ( b ,b ,b ) (cid:54) =(0 , , e − k | b | e − k | b | e − k | b | = 3 e k + e − k (cid:0) k (cid:1) . (A.2)27t this point we include the fourth sum over n : ∞ (cid:88) n =0 b ,b ,b = −∞ ( b ,b ,b ) (cid:54) =(0 , , e − k n R √ b + b + b ≤ ∞ (cid:88) n =0 b ,b ,b = −∞ ( b ,b ,b ) (cid:54) =(0 , , e − ( m + ηn ) R √ b + b + b ≤ ∞ (cid:88) n =0 b ,b ,b = −∞ ( b ,b ,b ) (cid:54) =(0 , , e − ( m + ηn ) R ( | b | + | b | + | b | ) = ∞ (cid:88) n =0 e ( m + ηn ) R + e − ( m + ηn ) R (cid:0) ( m + ηn ) R (cid:1) ≤ e mR G (cid:0) ηR (cid:1) + e − mR G (cid:0) ηR (cid:1) (cid:0) mR (cid:1) , (A.3)where we have used thatsinh (cid:18) mR ηnR (cid:19) ≥ sinh (cid:18) mR (cid:19) e ηnR , m > , R > , n ∈ Z ≥ , (A.4)and we have defined G ( x ) ≡ − e − x , x > . (A.5)Next we need the Yukawa-like sum ∞ (cid:88) b ,b ,b = −∞ ( b ,b ,b ) (cid:54) =(0 , , e − kR √ b + b + b R p ( b + b + b ) p = (cid:90) ∞ k d m · · · (cid:90) ∞ m p − d m p ∞ (cid:88) b ,b ,b = −∞ ( b ,b ,b ) (cid:54) =(0 , , e − m p R √ b + b + b ≤ (cid:90) ∞ k d m · · · (cid:90) ∞ m p − d m p (cid:20) coth (cid:18) m p R (cid:19) − (cid:21) , (A.6)for p ∈ Z > and k >
0. Combining this result with the fourth sum of Eq. (A.3), we obtainthe following result ∞ (cid:88) n =0 b ,b ,b = −∞ ( b ,b ,b ) (cid:54) =(0 , , e − k n R √ b + b + b R p ( b + b + b ) p ≤ ∞ (cid:88) n =0 b ,b ,b = −∞ ( b ,b ,b ) (cid:54) =(0 , , e − ( m + ηn ) R √ b + b + b R p ( b + b + b ) p = (cid:90) ∞ m d m · · · (cid:90) ∞ m p − d m p ∞ (cid:88) n =0 b ,b ,b = −∞ ( b ,b ,b ) (cid:54) =(0 , , e − ( m p + ηn ) R √ b + b + b ≤ (cid:90) ∞ m d m · · · (cid:90) ∞ m p − d m p e mR G (cid:0) ηR (cid:1) + e − mR G (cid:0) ηR (cid:1) (cid:0) mR (cid:1) = A p ( m, R ) G (cid:18) ηR (cid:19) + B p ( m, R ) G (cid:18) ηR (cid:19) , (A.7)28ith the functions A ( m, R ) = (cid:0) − e mR (cid:1) R (cid:0) e mR − (cid:1) , B ( m, R ) = R (cid:34) mR + − e mR (cid:0) e mR − (cid:1) − (cid:0) e mR − (cid:1)(cid:35) , A ( m, R ) = R (cid:104) mR + e mR − − (cid:0) e mR − (cid:1)(cid:105) , B ( m, R ) = R (cid:104) − mR + e mR − + 6 log (cid:0) e mR − (cid:1) + 4Li (cid:0) e − mR (cid:1)(cid:105) , A ( m, R ) = R (cid:104) mR − log (cid:0) e mR − (cid:1) + Li (cid:0) e − mR (cid:1)(cid:105) , B ( m, R ) = R (cid:20) mR − log (cid:0) e mR − (cid:1) − (cid:0) e − mR (cid:1) + 2Li (cid:0) e − mR (cid:1)(cid:21) , (A.8)where Li is the dilogarithm or Spence function and Li is the trilogarithm. Finally, wecan write ∞ (cid:88) n =0 b ,b ,b = −∞ ( b ,b ,b ) (cid:54) =(0 , , e − k n R √ b + b + b c n R p ( b + b + b ) p ≤ ∞ (cid:88) n =0 b ,b ,b = −∞ ( b ,b ,b ) (cid:54) =(0 , , e − k n R √ b + b + b min( { c n } ) R p ( b + b + b ) p = A p ( m, R ) G (cid:0) ηR (cid:1) + B p ( m, R ) G (cid:0) ηR (cid:1) min( { c n } ) . (A.9) (cid:3) Lemma 2
The following quadruple sum obeys the inequality ∞ (cid:88) n =0 b ,b ,b = −∞ ( b ,b ,b ) (cid:54) =0 k n e − k n R √ b + b + b c n R p − ( b + b + b ) p − ≤ m + A p − ( m − , R ) G (cid:16) η − R (cid:17) + B p − ( m − , R ) G (cid:16) η − R (cid:17) min( { c n } ) − η + R A p ( m − , R ) G (cid:48) (cid:16) η − R (cid:17) + B p ( m − , R ) G (cid:48) (cid:16) η − R (cid:17) min( { c n } ) , (A.10) provided k n obeys k n ≥ m − + η − n and k n ≤ m + + η + n with m ± > , η ± > positiveconstants, and p ∈ Z > . roof : Following the lines of the proof of Lemma 1, we have ∞ (cid:88) n =0 b ,b ,b = −∞ ( b ,b ,b ) (cid:54) =0 k n e − k n R √ b + b + b c n R p − ( b + b + b ) p − ≤ ∞ (cid:88) n =0 b ,b ,b = −∞ ( b ,b ,b ) (cid:54) =0 ( m + + η + n ) e − ( m − + η − n ) R √ b + b + b c n R p − ( b + b + b ) p − ≤ ∞ (cid:88) n =0 b ,b ,b = −∞ ( b ,b ,b ) (cid:54) =0 m + e − ( m − + η − n ) R √ b + b + b c n R p − ( b + b + b ) p − − ∞ (cid:88) n =0 b ,b ,b = −∞ ( b ,b ,b ) (cid:54) =0 η + dd η − e (cid:16) − ( m − + η − n ) R √ b + b + b (cid:17) c n R p ( b + b + b ) p ≤ m + A p − ( m − , R ) G (cid:16) η − R (cid:17) + B p − ( m − , R ) G (cid:16) η − R (cid:17) min( { c n } ) − η + R A p ( m − , R ) G (cid:48) (cid:16) η − R (cid:17) + B p ( m − , R ) G (cid:48) (cid:16) η − R (cid:17) min( { c n } ) , (A.11)where k n ≥ m − + η − n and k n ≤ m + + η + n and the derivative with respect to the argumentof G is G (cid:48) ( x ) = − e − x (1 − e − x ) . (A.12) (cid:3) Lemma 3
The following quadruple sum obeys the inequality ∞ (cid:88) n =0 b ,b ,b = −∞ ( b ,b ,b ) (cid:54) =0 k n e − k n R √ b + b + b c n R p − ( b + b + b ) p − ≤ m A p − ( m − , R ) G (cid:16) η − R (cid:17) + B p − ( m − , R ) G (cid:16) η − R (cid:17) min( { c n } ) − m + η + R A p − ( m − , R ) G (cid:48) (cid:16) η − R (cid:17) + B p − ( m − , R ) G (cid:48) (cid:16) η − R (cid:17) min( { c n } )+ η R A p ( m − , R ) G (cid:48)(cid:48) (cid:16) η − R (cid:17) + B p ( m − , R ) G (cid:48)(cid:48) (cid:16) η − R (cid:17) min( { c n } ) , (A.13) provided k n obeys k n ≥ m − + η − n and k n ≤ m + + η + n with m ± > , η ± > positiveconstants, and p ∈ Z > . roof : Following the lines of the proof of Lemmata 1 and 2, we have ∞ (cid:88) n =0 b ,b ,b = −∞ ( b ,b ,b ) (cid:54) =0 k n e − k n R √ b + b + b c n R p − ( b + b + b ) p − ≤ ∞ (cid:88) n =0 b ,b ,b = −∞ ( b ,b ,b ) (cid:54) =0 ( m + + η + n ) e − ( m − + η − n ) R √ b + b + b c n R p − ( b + b + b ) p − ≤ ∞ (cid:88) n =0 b ,b ,b = −∞ ( b ,b ,b ) (cid:54) =0 m e − ( m − + η − n ) R √ b + b + b c n R p − ( b + b + b ) p − − ∞ (cid:88) n =0 b ,b ,b = −∞ ( b ,b ,b ) (cid:54) =0 m + η + dd η − (cid:32) e − ( m − + η − n ) R √ b + b + b c n R p − ( b + b + b ) p − (cid:33) + ∞ (cid:88) n =0 b ,b ,b = −∞ ( b ,b ,b ) (cid:54) =0 η d d η − (cid:32) e − ( m − + η − n ) R √ b + b + b c n R p ( b + b + b ) p (cid:33) ≤ m A p − ( m − , R ) G (cid:16) η − R (cid:17) + B p − ( m − , R ) G (cid:16) η − R (cid:17) min( { c n } ) − m + η + R A p − ( m − , R ) G (cid:48) (cid:16) η − R (cid:17) + B p − ( m − , R ) G (cid:48) (cid:16) η − R (cid:17) min( { c n } )+ η R A p ( m − , R ) G (cid:48)(cid:48) (cid:16) η − R (cid:17) + B p ( m − , R ) G (cid:48)(cid:48) (cid:16) η − R (cid:17) min( { c n } ) , (A.14)where k n ≥ m − + η − n and k n ≤ m + + η + n and the derivative with respect to the argumentof G is given by Eq. (A.12) and the double derivative yields G (cid:48)(cid:48) ( x ) = e − x + e − x (1 − e − x ) . (A.15) (cid:3) Observation 1
The infinite sum (cid:80) ∞ n =0 a n with a n ≥ remains finite by restricting to oddor even n . Corollary 1
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