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Conformal symmetry breaking and degeneracy ofhigh-lying unflavored mesonsTo cite this article Mariana Kirchbach et al 2012 J Phys Conf Ser 378 012036

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Conformal symmetry breaking and degeneracy of

high-lying unflavored mesons

Mariana Kirchbach1 Adrian Pallares-Rivera1 Cliffor Compean2 and

Alfredo Raya2

1Institute of Physics Autonomous University of San Luis Potosi Manuel Nava 6 San LuisPotosi SLP 98290 Mexico2Institute of Physics and Mathematics University of Michoacan San Nicolas de HidalgoBuilding C-3 University City Morelia Michoacan 58040 Mexico

E-mail marianaifisicauaslpmxpallaresifisicauaslpmxclifforifmimichmx

rayaifmumichmx

Abstract We show that though conformal symmetry can be broken by the dilaton such canhappen without breaking the conformal degeneracy patterns in the spectra Our argumentationgoes as follows We departure from the gauge-gravity duality which predicts on the boundaries ofthe AdS5 geometry a conformal theory associated with QCD at high temperatures and considerS1 times S3 slicing The inverse radius R of S3 relates to the temperature of the deconfinementphase transition and has to satisfy hcR ≫ ΛQCD On S3 whose isometry group is SO(4) wethen focus on the eigenvalue problem of the conformal Laplacian there given by 1

R2

(K2 + 1

)

with K2 standing for the Casimir invariant of the so(4) algebra This eigenvalue problemdescribes the spectrum of a scalar particle to be associated with a qq system Such a spectrumis characterized by a (K + 1)2-fold degeneracy of its levels with K isin [0infin) We then breakthe conformal S3 metric ds2 = dχ2 + sin2 χ(dθ2 + sin2 θdϕ2) -in polar χ θ and azimuthal ϕcoordinates- according to ds2 = eminusbχ

((1 + b24)dχ2 + sin2 χ(dθ2 + sin2 θdϕ2)

) and attribute

the symmetry breaking scale bh2c2R2 to the dilaton Next we show that the above metricdeformation is equivalent to a breaking of the conformal curvature of S3 by a term proportional

to b cot χ and that the perturbed conformal Laplacian is equivalent to(K2 + cK

) with cK

a representation constant and K2 being again an so(4) Casimir invariant but this time in arepresentation unitarily inequivalent to the 4D rotational one As long as the spectra before andafter the symmetry breaking happen to be determined each by eigenvalues of a Casimir invariantof an so(4) no matter whether or not in a representation that generates the orthogonal groupSO(4) as a subgroup of the conformal group SO(2 4) the degeneracy patterns remain unalteredthough the conformal symmetry breaks at the level of the representation of the algebra Wefit the S3 radius and the h2c2bR2 scale to the high-lying excitations in the spectra of theunflavored mesons as reported by the Crystal Barrel collaboration and observe the correcttendency of the hcR = 373 MeV value to notably exceed ΛQCD The size of the symmetrybreaking scale is calculated as hc

radicbR = 6737 MeV

1 Introduction and background

The five-dimensional Anti-de Sitter space AdS5 is presently one of the most intriguinggeometries in field theoretical studies due to its relevance in establishing the link between branetheory and QCD The fact is that according to the gauge-gravity duality conjecture [1] a string

XIII Mexican Workshop on Particles and Fields IOP PublishingJournal of Physics Conference Series 378 (2012) 012036 doi1010881742-65963781012036

Published under licence by IOP Publishing Ltd 1

theory at the boundary of AdS5 appears dual to a conformal field theory in (1+3) dimensionsassociated with QCD [2] The AdS5 geometry is defined as a surface embedded in a (2+4)-dimensional flat ambient space of two time-like (X0X5) and four space-like (X1X2X3X4)dimensions according to [3]

X20 +X2

5 minusX21 minusX2

2 minusX23 minusX2

4 =

(X0 +X4)(X0 minusX4) + (X5 +X3)(X5 minusX3) minus (X1 minus iX2)(X1 + iX2) = L2 (1)

where L2 is a scale In changing variables to

X0 +X4 = ev Xmicro = evxmicro micro = 5 1 2 3 (2)

the equation (1) becomes

ev(X0 minusX4) + e2v(x25 minus x2

1 minus x22 minus x2

3) = L2 (3)

This equation shows that the AdS5 space contains a flat Minkowski space-time warped by theexponential factor of exp(2v) Dividing now by that very warp factor and taking the v rarr infinlimit one encounters the Minkowskian light-cone C1+3 as

C1+3 limvrarrinfin

[X0 minusX4

ev+ x2


2 minus x23

]= lim

vrarrinfin

L2

e2vrarr 0 (4)

The v variable is known as the holographic variable Therefore the intersections of the AdS5

null-ray cone with X0 + X4 = ev hyperplanes produce slices Πv called branes which arecopies of flat Minkowski space-times in the xmicro coordinates with micro = 5 1 2 3 In terms of theholographic variable the conformal AdS5 boundary corresponds to the v rarr infin limit in whichcase the warp factor blows up QCD on C1+3 as its emerges at the AdS5 boundary is knownas holographic QCD and has produced a variety of interesting insights into the properties ofhadrons (see ref [4] for a recent review)

However flat Minkowski space-time in combination with an interaction potential of aninfinite range represent a significant obstacle toward the formulation of a finite-temperaturefield theory and the description of the deconfinement phase transition The reason is that theabsence of a scale presents a serious difficulty in defining the QCD chemical potential Thisdifficulty has been circumvented [2] [5] [6] in replacing the flat Minkowski space-time by thecompactified one S1timesS3 which equally well describes the conformal AdS5 boundary as visiblefrom casting eq (1) into the form [7]

detU minus detg = L2

U =

(X5 + iX0 0

0 X5 minus iX0

) g =

(X4 + iX3 iX1 minusX2

iX1 +X2 X4 minus iX3

)

X0 + iX5 = R0eiη X1 + iX2 = R sinχ sin θeiϕ

X3 = R sinχ cos θ X4 = R cosχ R20 minusR2 = L2 (5)

The null-ray AdS5 cone can now be approached for R0 rarr R The proof that S1 times S3 is aconformally compactified flat Minkowski space-time R1+3 has been elaborated in the literaturein great detail [8] [9] and the explicit map that takes the R1+3 metric to the metric of S1 timesS3

can be found for example in [7] The line interval on the unit hypersphere reads

ds2 = dχ2 + sin2 χ(dθ2 + sin2 θdϕ2) (6)


2

Particle motions on this space have been equally well studied the ref [7] providing valuableinsights in that regard In particular it has been found that the free geodesic motion of a scalarparticle is described in terms of the eigenvalue problem of the conformal Laplacian LS3

LS3 = minus 1

R2∆S3 + RS3 minus 1

R2∆S3 = minus 1

R2

(d2

dχ2+ 2cotχ

d

dχminus L2

sin2 χ

)=

1

R2K2 (7)

where ∆S3 is the Laplace-Beltrami operator on S3 RS3 stands for the conformal curvature andK2 is the operator of the squared four-dimensional (4D) angular momentum which acts as aCasimir invariant of the isometry algebra so(4) of S3 The so(4) algebra in the representationchosen in (7) has the property to generate the orthogonal group SO(4) On the unit hyperspherethe spectral problem of LS3 reads

LS3YKlm(~Ω) = (minus∆S3 + 1)YKlm(~Ω) =(K2 + 1

)YKlm(~Ω) = (K + 1)2YKlm(~Ω) ~Ω = (χ θ ϕ)

YKlm(~Ω) = S lK(χ)Y m

l (θ ϕ) S lK(χ) = sinl χGl+1

Kminusl(cosχ)

K isin [0infin) l isin [0K] m isin [minusl+l] (8)

Here YKlm(~Ω) are the hyper-spherical harmonics Gαn (cosχ) denote the Gegenbauer polynomials

and Y ml (θ ϕ) stand for the ordinary three-dimensional spherical harmonics The χ dependent

part of the hyperspherical harmonics ie the S lK(χ) function is sometimes referred to as ldquoquasi-

radialrdquo function [10] a notation which we occasionally shall make use of It is obvious that thespectrum in (8) is characterized by a (K+1)2 fold degeneracy of the levels Such a spectrum aswhole would fit into an infinite unitary representation of the conformal group SO(2 4) wherethe group SO(4) appears in the reduction chain

SO(2 4) sub SO(4) sub SO(3) sub SO(2) (9)

It is within this context that (K+1)2-fold degeneracy is associated with conformal symmetry inparallel to the H atom where (K + 1) acquires the meaning of the principal quantum number ofthe Coulomb potential problem However conformal symmetry can be at most an approximatesymmetry of QCD because it requires

bull massless quarks

bull scale independent strong coupling and

bull a massless dilaton

While the first two conditions seem to be reasonably respected within the unflavored sector ofQCD in which the u and d quark masses can be considered sufficiently small at the scale ofthe nucleon excitations which start around 1500 MeV and in the infrared regime where therunning coupling constant starts walking toward a fixed value [11] the dilaton mass can notbe disregarded so far The breaking of the conformal symmetry is frequently treated in termsof breaking the metric of the Minkowski space-time encountered at the AdS5 boundary ie bydeforming the warp factor (soft-walls) [4] Notice the dilaton can be introduced in the actionS as a background field in an overall exponential according to

S =

intdDXeminusΦ(z)radicminusgL Φ(z) = (microz)ν z = eminusv (10)

where micro sets a mass scale usually taken as ΛQCD for simplicity As shown in [12] [13] thedilaton modifies the particle motion and leaves a print in the meson spectra


3

We here adapt the essentials of the above scheme to the geometry of the compactifiedMinkowski space-time and design a pragmatic quark model of conformal symmetrybreaking on S1 timesS3 through global S3 metric deformation by an exponential factor ofthe second polar angle We then study the impact of this deformation on the spectraof the high-lying unflavored mesons reported by the Cristal Barrel collaboration in thecompilation of [14] Our aim is to reveal a possibility to deform the conformal curvaturein (6) by a scale and without removing the conformal degeneracy patterns from thespectra

The outline of the paper is as follows In the next section we present the mechanismof conformal symmetry breaking without removing the conformal degeneracy patterns in thespectra In section 3 we analyze the Crystal-Barrel data [14] on the high-lying unflavored mesonspectra with the aim to obtain an estimate for the S3 radius and the order of the scale of theconformal symmetry breaking which we attribute to the dilaton The paper ends with conciseconclusions

2 Breaking the conformal curvature of S3 by an interaction

Any perturbation of the quantum mechanical geodesic motion on S3 by an interaction potentialV (~Ω) breaks the conformal curvature as

h2c2RS3 minusrarr h2c2RS3 =h2c2

R2+ V (~Ω) (11)

and the perturbed Laplacian is no longer conformal neither is its spectral problem necessarilyexactly solvable Not so for the potential given by

V (χ) = minus2B cotχ (12)

with B = bh2c2R2 and a dimensionless b which gives rise to an exactly solvable(h2c2LS3 + V (χ)

)spectral problem and one which moreover happens to carry same (K + 1)2-

fold degeneracy patterns as the spectral problem of the intact conformal Laplacian in (8)The cotχ potential is special insofar as it is a harmonic function on S3 just as is theCoulomb potential in E3 and can be treated along the line of harmonic analysis and potentialtheory Using harmonic functions as potentials brings the advantage of minimizing the Dirichletfunctional integral Occasionally the cotangent potential on S3 is referred to as ldquocurvedrdquoCoulomb potential In the present section we wish to take a closer look on the quantummechanical cotangent potential problem on S3 from the perspective of conformal Laplacians[15 16] The cotangent potential problem on S3 first considered by Schrodinger [17 18]is equivalent to the relativistic four-dimensional rigid-rotator problem perturbed by a cotχinteraction

h2c2

R2[minus∆S3 + 1 minus 2b cotχ] Ψ

Klm(~Ω) =

h2c2

R2

[(K + 1)2 minus b2

(K + 1)2

]Ψ

Klm(~Ω) (13)

The solutions of equation (13) are well studied and known Specifically in [19] they have beenexpressed in terms of real non-classical Romanovski polynomials Rαβ

n (cotχ) (reviewed in [20])and in units of h = 1 = c R = 1 read

ΨKlm

(~Ω) = eαKχ

2 ψlK(χ)Y m

l(θ ϕ)

ψlK(χ) = sinK χRαK βKminus1

Kminusl(cotχ)

αK = minus 2b

K + 1 βK = minusK (14)


4

The Romanovski polynomials satisfy the following differential hypergeometric equation

(1 + x2)d2Rαβ

n

dx2+ 2

(α

2+ βx

)dRαβ

n

dxminus n(2β + nminus 1)Rαβ

n = 0 (15)

They are obtained from the following weight function

ωαβ(x) = (1 + x2)βminus1 exp(minusα cotminus1 x) (16)

by means of the Rodrigues formula

Rαβn (x) =

1

ωαβ(x)

dn

dxn

[(1 + x2)nωαβ(x)

] (17)

Recently it has been shown in [19] that the ψlK(χ) functions allow for finite decompositions in

the SO(4) basis of the quasi-radial functions SlK(χ) in (8) Namely the following relation has

been found to hold valid

ψlK(χ) =

Ksum

l

Cl S lK(χ) (18)

The explicit expressions for the coefficients Cl defining the full similarity transformation betweenthe invariant spaces of the perturbed and intact Laplacians on S3 can be consulted in [19] Uponintroducing the notations

ωK(χ) =αKχ

2equiv minus bχ

K + 1 b = minusωprime

K(χ)(K + 1) (19)

and insertion of (18) into (14) allows to rewrite (13) to

[LS3 + 2ωprime

K(χ)(K + 1) cotχ]eωK(χ)ψl

K(χ) =

eωK(χ)Ksum

l

Cl

[(K + 1)2 minus ωprime

K(χ) 2]Sl

K(χ) (20)

Dragging now the exponential factor in the lhs in (20) from the right to the very left recallingthat LS3 = K2 + 1 and making use of (18) amounts to

eωK(χ)[K2 + 1 minus ωprime

K(χ) 2 minus 2ωprime

K(χ)DK

] Ksum

l

Cl S lK(χ) =

eωK(χ)Ksum

l

Cl


K(χ) 2]Sl

K(χ)

DK =d

dχminusK cotχ (21)

Using in the rhs of (21) the relationship(K2 + 1

)Sl

K(χ) = (K + 1)2SlK(χ) known from (7)

allows to equivalently cast eq (21) as

(K2 + 1 minus ωprime


K(χ)eωK(χ)DKeminusωK(χ)

)eωK(χ)

Ksum

l

Cl S lK(χ)

=Ksum

l

(K2 + 1 minus ωprime(χ)2

)eωK(χ)Cl S l

K(χ) (22)


5

This equality implies the following important property of the DK operator

(K2 minus 2ωprime

K(χ)eωK (χ)DKeminusωK(χ)

)eωK(χ)

Ksum

l

Cl S lK(χ) =

Ksum

l

K2 eωK(χ)Cl S lK(χ)

K2 = eωK(χ)K2eminusωK(χ) (23)

Now the equivalence of eqs (20)ndash(21) and (22) allows to draw the final conclusion as

[K2 + 1 + 2ωprime

K(χ)(K + 1) cotχ]eω(χ)ψl

K(χ) =Ksum

l

Cl

[K2 + 1 minus ωprime(χ)2

]eωK(χ)S l

K(χ)

=[(K + 1)2 minus ωprime

K(χ) 2]eω(χ)ψl

K(χ) (24)

with ψlK from (18) and ωprime

K(χ) being a representation constant In this fashion the eigenvalueproblem of the cotangent-broken conformal Laplacian on S3 presents itself as the eigenvalueproblem of a Casimir invariant of the so(4) algebra in a representation unitarily inequivalent tothe 4D-rotational As long as the eigenvalues of any Casimir invariant remain unaltered underany well defined similarity transformations no matter unitary or not the degeneracy patternsin the spectral problems of the conformal and the cotangent-broken Lapalacians result sameThe Killing vectors of this new so(4) algebra

Jik = eωK(χ)JikeminusωK(χ) = eωK(χ)

(Xj

part

partXk

minusXkpart

partXj

)eminusωK(χ)

[Jik Jkj

]= Jji χ = cotminus1 X4radic

X21 +X2

2 +X23

(25)

refer to a surface (call it S3) which will differ in shape from S3 In this fashion the breakingof the conformal symmetry reveals itself through metric deformation Equivalently one can saythat it reveals itself at the level of the representation functions in reference to the differencebetween the hyper-spherical harmonics YKlm(~Ω) in (8) and Ψ

Klm(~Ω) in (14) Indeed the global

metric of the S3 ball is incorporated by the scalar of the 4D rotational isometry algebra so(4)

which in being the lowest LS3 eigenstate is the hyper-spherical harmonic Y000(~Ω)

S3 Y000(~Ω) =4sum

i=1

X2i = cos2 χ+ sin2 χ

(cos2 θ + sin2 θ

(cos2 ϕ+ sin2 ϕ

))= 1 (26)

Correspondingly the global metric of the S3 surface will be implemented by the scalar of thenon-unitary so(4) algebra in (23) (25) and is given by the exponentially rescaled hyper-sphere

S3 eminusbχY000(~Ω) = eminusbχ(cos2 χ+ sin2 χ

(cos2 θ + sin2 θ

(cos2 ϕ+ sin2 ϕ

)))= eminusbχ (27)

The line element on the deformed surface S3 is then easily calculated as

d2s = eminusbχ((1 + b24)d2χ+ sin2 χ

(dθ2 + sin2 θd2ϕ

)) (28)

In the following we conjecture that the h2c2bR2 scale of conformal symmetry breaking can beattributed to the dilaton the only significant source of such a breaking

Notice that the perturbance of the conformal Laplacian on S3 by a cotangent interactiondoes not even produce a conformal map [15] [16] as it could have happened by a different ingeneral gradient potential


6

In summary a subtle mode of symmetry breaking has been identified in which thesymmetry breaking happens at the level of the representation functions of the isometryalgebra of a given geometric manifold and reveals itself through a metric deformationwhile the spectral problems of the intact and broken Laplacians are characterized byidentical degeneracy patterns

It seems opportune to refer to this symmetry breaking mechanism as ldquosymmetry breakingcamouflaged by degeneracyrdquo or shortly ldquocamouflaged symmetry breakingrdquo

3 Degeneracy of high-lying unflavored mesons

In the following we employ the cotangent-broken conformal Laplacian on S3 in (13) in thedescription of the spectra of the high-lying unflavored mesons as reported by the Crystal Barrelcollaboration (data compilation from [14]) and adjust the R and b potential parameters Thecotangent potential derives its utility for employment in the description of hadron spectra fromits close relationship to the Cornell potential predicted by Lattice QCD [21] as visible from itsTaylor series decomposition

minus2b cot

r

R= minus2bR

r

+4b

3R

r + for χ =

r

R (29)

wherer denotes the geodesic distance on S3 and R was the S3 radius

The Crystal Barrel data on the spectra of the high-lying unflavored mesons show wellpronounced conformal degeneracy patterns in the region above asymp 1200 MeV Our case is that

bull these patterns emerge in consequence of the breaking of the conformal SO(2 4) groupsymmetry at the level of the representation of its algebra and in accordance with eqs (24)(28)

bull such a breaking may be attributed to the dilaton

In Figs (1ndash4) we present the spectra of the mesons of interest Tables 1 and 2 contain theresults of the least square parameter fit

Table 1 The potential parameters from the least square fit to the meson spectra by means ofeq (13)

Isospin Lowest spin R [fm] (h2c2b)R2[GeV2]

I=0 Jmin=0minus 05157279 0459666I=0 Jmin=1minus 05293829 0425489I=1 Jmin=0minus 05443879 0522783I=1 Jmin=1minus 05268850 0411833


7

Table 2 The meson masses from the least square fit to the spectra by means of eq (13)(last column) The label K stands for the four-dimensional angular momentum characterizinga degenerate level in accordance with eq (13) The fourth column contains the averaged massof the experimentally observed states as displayed in the respective plots on the Figs 1-4

I Jmin K experiment [MeV] fit [MeV]

0 0minus 1 127833 1349602 16925 1659483 197066 1961834 227422 22801

1 0minus 1 132833 1404892 169666 1708673 2031 1985044 225375 227369

0 1minus 1 1325 1416192 165333 1697433 199414 1979194 226977 227969

1 1minus 1 134333 1385852 16825 1670283 182585 1957164 225666 226275

4 Discussion and conclusions

The Tables 1 and 2 show that our least square fits to the Crystal Barrel data on the high-lying unflavored mesons by means of the cotangent broken conformal Laplacian on S3 ineq (13) provide quite a reasonable description of the observed degeneracies in the spectra underinvestigation and in support of our hypothesis that the breaking of the conformal symmetrycan occur at the level of the representation of the conformal algebra and without affecting thedegeneracies The b parameter is obtained as

b = 32793 plusmn 00697 (30)

We furthermore observe that to a very good approximation both the S3 radius and the h2c2bR2

scale are isospin and parity independent as should be for a conformal symmetry breaking scaledue to the dilaton The mean value of the inverse radius obtained from our fits corresponds toa temperature of T = hcR= 373 MeV which reasonably fulfills the requirement to be notablylarger than ΛQCD = 175 MeV In order to extract the scale parameter of the conformal symmetry

breaking within our approach we make use of eq (29) ie of χ =rR

and rewrite the exp(minusbχ)factor equivalently as

exp(minusbχ) = exp

minus

(radic

b

R

)2 (radicr R

)2 equiv exp(micror)

micro =

radicb

R r =

radicr R (31)

Evaluating this scale with the numbers listed in Table 1 amounts to a mean value of micro = 6737MeV which is closer to the dilaton mass than to ΛQCD = 175 MeV in (10) We conclude


8

Figure 1 High-lying excitations of the isosinglet pseudoscalar η meson according to the CrystalBarrel measurements in the compilation of [14] On the one side the quantum numbers of themesons in the four groups displayed here fit well into the respective K = 1 2 3 and 4 eigenstatesof the conformal Laplacian on S3 in (8) At the same time they fit equally well into theeigenstates of the broken Laplacian in (24) which shows that the breaking of the conformalgroup symmetry at the level of the representation of its algebra by the h2c2bR2 mass scaledoes not leave a print in the degeneracies in the spectrum Empty squares denote ldquomissingrdquoresonances These are additionally marked by a question-mark inside the round brackets whilethe accompanying number is our prediction for the respective mass Notice that not only theconformal degeneracy patterns are well pronounced also practically all the states with spinslower than the maximal in a level (Jmax = K) appear parity duplicated a phenomenon thatis indicative of a manifest realization of the chiral symmetry in the Wigner-Wyle mode at thatscale

that the Crystal Barrel data on high lying unflavored mesons support our model of conformalsymmetry breaking by a mass scale at the level of the representation of the underlying so(4)algebra and without loosing the degeneracies in the spectra


9

Figure 2 High-lying excitations of the isotriplet pseudoscalar π meson according to the CrystalBarrel measurements in the compilation of [14] For notations and clarifying comments see Fig 1

Figure 3 High-lying excitations of the isosinglet vector ω meson according to the CrystalBarrel measurements in the compilation of [14] For notations and clarifying comments seeFig 1


10

Figure 4 High-lying excitations of the isotriplet vector ρ meson according to the Crystal Barrelmeasurements in the compilation of [14] For notations and clarifying comments see Fig 1

References[1] Maldacena J 1998 Phys Rev Lett 80 4859[2] Witten E 1998 Adv Theor Math Phys 2 233[3] Moschella U 2005 Seminaire Poincare 1 1[4] de Teramond G F and Brodsky S J 2011 J Phys Conf Ser 287 012007[5] de Boer J 2002 Proceedings of the 10 Int Conf on Supersymmetry and Unification of Fund Interactions

(SUSY02) Hamburg Vol 1 512[6] Hands S Hollowood T J and Myers J C 2010 JHEP 1007 086[7] Gibbons G W and Steif A R 1995 Phys Lett B 346 255[8] Luscher M and Mack G 1975 Commun Math Phys 41 203[9] Gibbons G W 2011 Anti-de-Sitter space and its uses Preprint arxiv11101206v1 [hep-th]

[10] Pervushin V N Pogosyan G S Sissakian A N and Vinitsky S I 1993 Phys Atom Nucl 5b 1027[11] Deur A Burkert V Chen J P and Korsch W 2008 Phys Lett 665 349[12] Colangelo P De Fazio F Gianuzzi F Jugeau F and Nicotri S 2008 Phys Rev D 78 055009[13] Kelley T M Bartz S B and Kapusta J I Phys Rev D 83 016002 (2011)[14] Afonin S S 2006 Eur Phys J A 29 327[15] Rosenberg S 1998 The Laplacian on a Riemannian manifold (London London Mathematical Society

Student Text 31 Cambridge Univerisity Press)[16] Gurski M J 2006 Conformal invariants and non-linear elliptic equations Proc of the Int Congress of

Mathematicians Madrid Spain pp 203-212[17] Schrodinger E 1940 Proc Roy Irish Acad A 46 9[18] Schrodinger E 1941 Proc Roy Irish Acad 46 183[19] Pallares-Rivera A and Kirchbach M 2011 J Phys AMathTheor 44 445302[20] Raposo A Weber H J Alvarez-Castillo D E and Kirchbach M 2007 C Eur J Phys 5 253[21] Takahashi T T Suganuma H Nemoto Y and Matsufuru H 2002 Phys Rev D 65 114509


11

Conformal symmetry breaking and degeneracy of

high-lying unflavored mesons

Mariana Kirchbach1 Adrian Pallares-Rivera1 Cliffor Compean2 and

Alfredo Raya2

1Institute of Physics Autonomous University of San Luis Potosi Manuel Nava 6 San LuisPotosi SLP 98290 Mexico2Institute of Physics and Mathematics University of Michoacan San Nicolas de HidalgoBuilding C-3 University City Morelia Michoacan 58040 Mexico

E-mail marianaifisicauaslpmxpallaresifisicauaslpmxclifforifmimichmx

rayaifmumichmx

Abstract We show that though conformal symmetry can be broken by the dilaton such canhappen without breaking the conformal degeneracy patterns in the spectra Our argumentationgoes as follows We departure from the gauge-gravity duality which predicts on the boundaries ofthe AdS5 geometry a conformal theory associated with QCD at high temperatures and considerS1 times S3 slicing The inverse radius R of S3 relates to the temperature of the deconfinementphase transition and has to satisfy hcR ≫ ΛQCD On S3 whose isometry group is SO(4) wethen focus on the eigenvalue problem of the conformal Laplacian there given by 1

R2

(K2 + 1

)

with K2 standing for the Casimir invariant of the so(4) algebra This eigenvalue problemdescribes the spectrum of a scalar particle to be associated with a qq system Such a spectrumis characterized by a (K + 1)2-fold degeneracy of its levels with K isin [0infin) We then breakthe conformal S3 metric ds2 = dχ2 + sin2 χ(dθ2 + sin2 θdϕ2) -in polar χ θ and azimuthal ϕcoordinates- according to ds2 = eminusbχ

((1 + b24)dχ2 + sin2 χ(dθ2 + sin2 θdϕ2)

) and attribute

the symmetry breaking scale bh2c2R2 to the dilaton Next we show that the above metricdeformation is equivalent to a breaking of the conformal curvature of S3 by a term proportional

to b cot χ and that the perturbed conformal Laplacian is equivalent to(K2 + cK

) with cK

a representation constant and K2 being again an so(4) Casimir invariant but this time in arepresentation unitarily inequivalent to the 4D rotational one As long as the spectra before andafter the symmetry breaking happen to be determined each by eigenvalues of a Casimir invariantof an so(4) no matter whether or not in a representation that generates the orthogonal groupSO(4) as a subgroup of the conformal group SO(2 4) the degeneracy patterns remain unalteredthough the conformal symmetry breaks at the level of the representation of the algebra Wefit the S3 radius and the h2c2bR2 scale to the high-lying excitations in the spectra of theunflavored mesons as reported by the Crystal Barrel collaboration and observe the correcttendency of the hcR = 373 MeV value to notably exceed ΛQCD The size of the symmetrybreaking scale is calculated as hc

radicbR = 6737 MeV

1 Introduction and background

The five-dimensional Anti-de Sitter space AdS5 is presently one of the most intriguinggeometries in field theoretical studies due to its relevance in establishing the link between branetheory and QCD The fact is that according to the gauge-gravity duality conjecture [1] a string


Published under licence by IOP Publishing Ltd 1


X20 +X2

5 minusX21 minusX2

2 minusX23 minusX2

4 =







3) = L2 (3)


C1+3 limvrarrinfin

[X0 minusX4

ev+ x2


2 minus x23

]= lim

vrarrinfin

L2

e2vrarr 0 (4)





U =

(X5 + iX0 0

0 X5 minus iX0

) g =



)







2


LS3 = minus 1


R2∆S3 = minus 1

R2

(d2

dχ2+ 2cotχ

d

dχminus L2

sin2 χ

)=

1

R2K2 (7)






Kminusl(cosχ)












S =




3






R2+ V (~Ω) (11)






h2c2


Klm(~Ω) =

h2c2

R2

[(K + 1)2 minus b2

(K + 1)2

]Ψ

Klm(~Ω) (13)



ΨKlm

(~Ω) = eαKχ

2 ψlK(χ)Y m

l(θ ϕ)


Kminusl(cotχ)

αK = minus 2b



4


(1 + x2)d2Rαβ

n

dx2+ 2

(α

2+ βx

)dRαβ

n


n = 0 (15)




Rαβn (x) =

1

ωαβ(x)

dn

dxn

[(1 + x2)nωαβ(x)

] (17)




ψlK(χ) =

Ksum

l

Cl S lK(χ) (18)


ωK(χ) =αKχ

2equiv minus bχ


K(χ)(K + 1) (19)


[LS3 + 2ωprime


K(χ) =

eωK(χ)Ksum

l

Cl


K(χ) 2]Sl

K(χ) (20)




K(χ)DK

] Ksum

l

Cl S lK(χ) =

eωK(χ)Ksum

l

Cl


K(χ) 2]Sl

K(χ)

DK =d



)Sl






)eωK(χ)

Ksum

l

Cl S lK(χ)

=Ksum

l


)eωK(χ)Cl S l

K(χ) (22)


5


(K2 minus 2ωprime


)eωK(χ)

Ksum

l

Cl S lK(χ) =

Ksum

l




[K2 + 1 + 2ωprime


K(χ) =Ksum

l

Cl


]eωK(χ)S l

K(χ)


K(χ) 2]eω(χ)ψl

K(χ) (24)




(Xj

part

partXk

minusXkpart

partXj

)eminusωK(χ)

[Jik Jkj


X21 +X2

2 +X23

(25)





S3 Y000(~Ω) =4sum

i=1


(cos2 θ + sin2 θ

(cos2 ϕ+ sin2 ϕ

))= 1 (26)



(cos2 θ + sin2 θ

(cos2 ϕ+ sin2 ϕ

)))= eminusbχ (27)



(dθ2 + sin2 θd2ϕ

)) (28)




6





minus2b cot

r

R= minus2bR

r

+4b

3R

r + for χ =

r

R (29)










7



0 0minus 1 127833 1349602 16925 1659483 197066 1961834 227422 22801

1 0minus 1 132833 1404892 169666 1708673 2031 1985044 225375 227369

0 1minus 1 1325 1416192 165333 1697433 199414 1979194 226977 227969

1 1minus 1 134333 1385852 16825 1670283 182585 1957164 225666 226275



b = 32793 plusmn 00697 (30)





exp(minusbχ) = exp

minus

(radic

b

R

)2 (radicr R


micro =

radicb

R r =

radicr R (31)



8




9




10








11


X20 +X2

5 minusX21 minusX2

2 minusX23 minusX2

4 =







3) = L2 (3)


C1+3 limvrarrinfin

[X0 minusX4

ev+ x2


2 minus x23

]= lim

vrarrinfin

L2

e2vrarr 0 (4)





U =

(X5 + iX0 0

0 X5 minus iX0

) g =



)







2


LS3 = minus 1


R2∆S3 = minus 1

R2

(d2

dχ2+ 2cotχ

d

dχminus L2

sin2 χ

)=

1

R2K2 (7)






Kminusl(cosχ)












S =




3






R2+ V (~Ω) (11)






h2c2


Klm(~Ω) =

h2c2

R2

[(K + 1)2 minus b2

(K + 1)2

]Ψ

Klm(~Ω) (13)



ΨKlm

(~Ω) = eαKχ

2 ψlK(χ)Y m

l(θ ϕ)


Kminusl(cotχ)

αK = minus 2b



4


(1 + x2)d2Rαβ

n

dx2+ 2

(α

2+ βx

)dRαβ

n


n = 0 (15)




Rαβn (x) =

1

ωαβ(x)

dn

dxn

[(1 + x2)nωαβ(x)

] (17)




ψlK(χ) =

Ksum

l

Cl S lK(χ) (18)


ωK(χ) =αKχ

2equiv minus bχ


K(χ)(K + 1) (19)


[LS3 + 2ωprime


K(χ) =

eωK(χ)Ksum

l

Cl


K(χ) 2]Sl

K(χ) (20)




K(χ)DK

] Ksum

l

Cl S lK(χ) =

eωK(χ)Ksum

l

Cl


K(χ) 2]Sl

K(χ)

DK =d



)Sl






)eωK(χ)

Ksum

l

Cl S lK(χ)

=Ksum

l


)eωK(χ)Cl S l

K(χ) (22)


5


(K2 minus 2ωprime


)eωK(χ)

Ksum

l

Cl S lK(χ) =

Ksum

l




[K2 + 1 + 2ωprime


K(χ) =Ksum

l

Cl


]eωK(χ)S l

K(χ)


K(χ) 2]eω(χ)ψl

K(χ) (24)




(Xj

part

partXk

minusXkpart

partXj

)eminusωK(χ)

[Jik Jkj


X21 +X2

2 +X23

(25)





S3 Y000(~Ω) =4sum

i=1


(cos2 θ + sin2 θ

(cos2 ϕ+ sin2 ϕ

))= 1 (26)



(cos2 θ + sin2 θ

(cos2 ϕ+ sin2 ϕ

)))= eminusbχ (27)



(dθ2 + sin2 θd2ϕ

)) (28)




6





minus2b cot

r

R= minus2bR

r

+4b

3R

r + for χ =

r

R (29)










7



0 0minus 1 127833 1349602 16925 1659483 197066 1961834 227422 22801

1 0minus 1 132833 1404892 169666 1708673 2031 1985044 225375 227369

0 1minus 1 1325 1416192 165333 1697433 199414 1979194 226977 227969

1 1minus 1 134333 1385852 16825 1670283 182585 1957164 225666 226275



b = 32793 plusmn 00697 (30)





exp(minusbχ) = exp

minus

(radic

b

R

)2 (radicr R


micro =

radicb

R r =

radicr R (31)



8




9




10








11


LS3 = minus 1


R2∆S3 = minus 1

R2

(d2

dχ2+ 2cotχ

d

dχminus L2

sin2 χ

)=

1

R2K2 (7)






Kminusl(cosχ)












S =




3






R2+ V (~Ω) (11)






h2c2


Klm(~Ω) =

h2c2

R2

[(K + 1)2 minus b2

(K + 1)2

]Ψ

Klm(~Ω) (13)



ΨKlm

(~Ω) = eαKχ

2 ψlK(χ)Y m

l(θ ϕ)


Kminusl(cotχ)

αK = minus 2b



4


(1 + x2)d2Rαβ

n

dx2+ 2

(α

2+ βx

)dRαβ

n


n = 0 (15)




Rαβn (x) =

1

ωαβ(x)

dn

dxn

[(1 + x2)nωαβ(x)

] (17)




ψlK(χ) =

Ksum

l

Cl S lK(χ) (18)


ωK(χ) =αKχ

2equiv minus bχ


K(χ)(K + 1) (19)


[LS3 + 2ωprime


K(χ) =

eωK(χ)Ksum

l

Cl


K(χ) 2]Sl

K(χ) (20)




K(χ)DK

] Ksum

l

Cl S lK(χ) =

eωK(χ)Ksum

l

Cl


K(χ) 2]Sl

K(χ)

DK =d



)Sl






)eωK(χ)

Ksum

l

Cl S lK(χ)

=Ksum

l


)eωK(χ)Cl S l

K(χ) (22)


5


(K2 minus 2ωprime


)eωK(χ)

Ksum

l

Cl S lK(χ) =

Ksum

l




[K2 + 1 + 2ωprime


K(χ) =Ksum

l

Cl


]eωK(χ)S l

K(χ)


K(χ) 2]eω(χ)ψl

K(χ) (24)




(Xj

part

partXk

minusXkpart

partXj

)eminusωK(χ)

[Jik Jkj


X21 +X2

2 +X23

(25)





S3 Y000(~Ω) =4sum

i=1


(cos2 θ + sin2 θ

(cos2 ϕ+ sin2 ϕ

))= 1 (26)



(cos2 θ + sin2 θ

(cos2 ϕ+ sin2 ϕ

)))= eminusbχ (27)



(dθ2 + sin2 θd2ϕ

)) (28)




6





minus2b cot

r

R= minus2bR

r

+4b

3R

r + for χ =

r

R (29)










7



0 0minus 1 127833 1349602 16925 1659483 197066 1961834 227422 22801

1 0minus 1 132833 1404892 169666 1708673 2031 1985044 225375 227369

0 1minus 1 1325 1416192 165333 1697433 199414 1979194 226977 227969

1 1minus 1 134333 1385852 16825 1670283 182585 1957164 225666 226275



b = 32793 plusmn 00697 (30)





exp(minusbχ) = exp

minus

(radic

b

R

)2 (radicr R


micro =

radicb

R r =

radicr R (31)



8




9




10








11






R2+ V (~Ω) (11)






h2c2


Klm(~Ω) =

h2c2

R2

[(K + 1)2 minus b2

(K + 1)2

]Ψ

Klm(~Ω) (13)



ΨKlm

(~Ω) = eαKχ

2 ψlK(χ)Y m

l(θ ϕ)


Kminusl(cotχ)

αK = minus 2b



4


(1 + x2)d2Rαβ

n

dx2+ 2

(α

2+ βx

)dRαβ

n


n = 0 (15)




Rαβn (x) =

1

ωαβ(x)

dn

dxn

[(1 + x2)nωαβ(x)

] (17)




ψlK(χ) =

Ksum

l

Cl S lK(χ) (18)


ωK(χ) =αKχ

2equiv minus bχ


K(χ)(K + 1) (19)


[LS3 + 2ωprime


K(χ) =

eωK(χ)Ksum

l

Cl


K(χ) 2]Sl

K(χ) (20)




K(χ)DK

] Ksum

l

Cl S lK(χ) =

eωK(χ)Ksum

l

Cl


K(χ) 2]Sl

K(χ)

DK =d



)Sl






)eωK(χ)

Ksum

l

Cl S lK(χ)

=Ksum

l


)eωK(χ)Cl S l

K(χ) (22)


5


(K2 minus 2ωprime


)eωK(χ)

Ksum

l

Cl S lK(χ) =

Ksum

l




[K2 + 1 + 2ωprime


K(χ) =Ksum

l

Cl


]eωK(χ)S l

K(χ)


K(χ) 2]eω(χ)ψl

K(χ) (24)




(Xj

part

partXk

minusXkpart

partXj

)eminusωK(χ)

[Jik Jkj


X21 +X2

2 +X23

(25)





S3 Y000(~Ω) =4sum

i=1


(cos2 θ + sin2 θ

(cos2 ϕ+ sin2 ϕ

))= 1 (26)



(cos2 θ + sin2 θ

(cos2 ϕ+ sin2 ϕ

)))= eminusbχ (27)



(dθ2 + sin2 θd2ϕ

)) (28)




6





minus2b cot

r

R= minus2bR

r

+4b

3R

r + for χ =

r

R (29)










7



0 0minus 1 127833 1349602 16925 1659483 197066 1961834 227422 22801

1 0minus 1 132833 1404892 169666 1708673 2031 1985044 225375 227369

0 1minus 1 1325 1416192 165333 1697433 199414 1979194 226977 227969

1 1minus 1 134333 1385852 16825 1670283 182585 1957164 225666 226275



b = 32793 plusmn 00697 (30)





exp(minusbχ) = exp

minus

(radic

b

R

)2 (radicr R


micro =

radicb

R r =

radicr R (31)



8




9




10








11


(1 + x2)d2Rαβ

n

dx2+ 2

(α

2+ βx

)dRαβ

n


n = 0 (15)




Rαβn (x) =

1

ωαβ(x)

dn

dxn

[(1 + x2)nωαβ(x)

] (17)




ψlK(χ) =

Ksum

l

Cl S lK(χ) (18)


ωK(χ) =αKχ

2equiv minus bχ


K(χ)(K + 1) (19)


[LS3 + 2ωprime


K(χ) =

eωK(χ)Ksum

l

Cl


K(χ) 2]Sl

K(χ) (20)




K(χ)DK

] Ksum

l

Cl S lK(χ) =

eωK(χ)Ksum

l

Cl


K(χ) 2]Sl

K(χ)

DK =d



)Sl






)eωK(χ)

Ksum

l

Cl S lK(χ)

=Ksum

l


)eωK(χ)Cl S l

K(χ) (22)


5


(K2 minus 2ωprime


)eωK(χ)

Ksum

l

Cl S lK(χ) =

Ksum

l




[K2 + 1 + 2ωprime


K(χ) =Ksum

l

Cl


]eωK(χ)S l

K(χ)


K(χ) 2]eω(χ)ψl

K(χ) (24)




(Xj

part

partXk

minusXkpart

partXj

)eminusωK(χ)

[Jik Jkj


X21 +X2

2 +X23

(25)





S3 Y000(~Ω) =4sum

i=1


(cos2 θ + sin2 θ

(cos2 ϕ+ sin2 ϕ

))= 1 (26)



(cos2 θ + sin2 θ

(cos2 ϕ+ sin2 ϕ

)))= eminusbχ (27)



(dθ2 + sin2 θd2ϕ

)) (28)




6





minus2b cot

r

R= minus2bR

r

+4b

3R

r + for χ =

r

R (29)










7



0 0minus 1 127833 1349602 16925 1659483 197066 1961834 227422 22801

1 0minus 1 132833 1404892 169666 1708673 2031 1985044 225375 227369

0 1minus 1 1325 1416192 165333 1697433 199414 1979194 226977 227969

1 1minus 1 134333 1385852 16825 1670283 182585 1957164 225666 226275



b = 32793 plusmn 00697 (30)





exp(minusbχ) = exp

minus

(radic

b

R

)2 (radicr R


micro =

radicb

R r =

radicr R (31)



8




9




10








11


(K2 minus 2ωprime


)eωK(χ)

Ksum

l

Cl S lK(χ) =

Ksum

l




[K2 + 1 + 2ωprime


K(χ) =Ksum

l

Cl


]eωK(χ)S l

K(χ)


K(χ) 2]eω(χ)ψl

K(χ) (24)




(Xj

part

partXk

minusXkpart

partXj

)eminusωK(χ)

[Jik Jkj


X21 +X2

2 +X23

(25)





S3 Y000(~Ω) =4sum

i=1


(cos2 θ + sin2 θ

(cos2 ϕ+ sin2 ϕ

))= 1 (26)



(cos2 θ + sin2 θ

(cos2 ϕ+ sin2 ϕ

)))= eminusbχ (27)



(dθ2 + sin2 θd2ϕ

)) (28)




6





minus2b cot

r

R= minus2bR

r

+4b

3R

r + for χ =

r

R (29)










7



0 0minus 1 127833 1349602 16925 1659483 197066 1961834 227422 22801

1 0minus 1 132833 1404892 169666 1708673 2031 1985044 225375 227369

0 1minus 1 1325 1416192 165333 1697433 199414 1979194 226977 227969

1 1minus 1 134333 1385852 16825 1670283 182585 1957164 225666 226275



b = 32793 plusmn 00697 (30)





exp(minusbχ) = exp

minus

(radic

b

R

)2 (radicr R


micro =

radicb

R r =

radicr R (31)



8




9




10








11





minus2b cot

r

R= minus2bR

r

+4b

3R

r + for χ =

r

R (29)










7



0 0minus 1 127833 1349602 16925 1659483 197066 1961834 227422 22801

1 0minus 1 132833 1404892 169666 1708673 2031 1985044 225375 227369

0 1minus 1 1325 1416192 165333 1697433 199414 1979194 226977 227969

1 1minus 1 134333 1385852 16825 1670283 182585 1957164 225666 226275



b = 32793 plusmn 00697 (30)





exp(minusbχ) = exp

minus

(radic

b

R

)2 (radicr R


micro =

radicb

R r =

radicr R (31)



8




9




10








11



0 0minus 1 127833 1349602 16925 1659483 197066 1961834 227422 22801

1 0minus 1 132833 1404892 169666 1708673 2031 1985044 225375 227369

0 1minus 1 1325 1416192 165333 1697433 199414 1979194 226977 227969

1 1minus 1 134333 1385852 16825 1670283 182585 1957164 225666 226275



b = 32793 plusmn 00697 (30)





exp(minusbχ) = exp

minus

(radic

b

R

)2 (radicr R


micro =

radicb

R r =

radicr R (31)



8




9




10








11




9




10








11




10








11








11

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