The quantization of the B=1 and B=2 Skyrmions
aa r X i v : . [ h e p - ph ] F e b The quantization of the B = 1 and B = 2 Skyrmions
Jimmy Fortier and Luc Marleau ∗ D´epartement de Physique, de G´enie Physique et d’Optique,Universit´e Laval, Qu´ebec, Qu´ebec, Canada G1K 7P4 (Dated: November 21, 2018)We propose to set the Skyrme parameters F π and e such that they reproduce the physical massesof the nucleon and the deuteron. We allow deformation using an axially symmetric solution andsimulated annealing to minimize the total energy for the B = 1 nucleon and B = 2 deuteron. Firstwe find that axial deformations are responsible for a significant reduction (factor of ≈
4) of therotational energy but also that it is not possible to get a common set of parameters F π and e whichwould fit both nucleon and deuteron masses simultaneously at least for m π = 138 MeV, 345 MeVand 500 MeV. This suggests that either m π >
500 MeV or additional terms must be added to theSkyrme Lagrangian.
PACS numbers: 12.39.Dc, 10.11Lm
I. INTRODUCTION
The Skyrme model [1] is a nonlinear theory of pions that admits topological solitons solutions called Skyrmions.These solutions fall into sectors characterized by an integer-valued topological invariant B . In its quantized version,a Skyrmion of topological charge B may be identified with a nucleus with baryon number B .Since the model is non-renormalizable, a canonical quantization is not possible and one has to resort to semiclassicalquantization of the zero modes of the Skyrmion. This method adds kinematical terms (rotational, vibrational,..) tothe total energy of the Skyrmion. The B = 1 Skyrmion was first quantized by Adkins, Nappi and Witten [2, 3]. Itprovided then a useful mean to set the parameters of the Skyrme model F π and e by fitting to the proton and deltamasses. The experimental value of the pion mass m π completes the set of input parameters when a pion mass term isadded to the Skyrme Lagrangian. The same calibration was assumed by Braaten and Carson [4] in their quantizationof the B = 2 Skyrmions and their predictions of a rather tightly bound and small sized deuteron. Further analysisby Leese, Manton and Schroers [5] considering the separation of two single Skyrmions in the most attractive channelled to more accurate predictions for the deuteron.Yet all these calculations have three caveats: First, they used a rigid-body quantization, i.e. the kinematic termis calculated from the solution which minimizes the static energy neglecting any deformation that could originateform the kinematical terms. This was pointed out by several authors [6, 7] who proposed to improve the solutions byallowing the B = 1 Skyrmion to deform within the spherically symmetric ansatz. Second, even with such deformationsit was noted that the set of parameters mentioned above does not allow for a stable spinning solution for the deltasince the rotational frequency Ω would not satisfy the constraint for stability against pion emission, Ω ≤ m π . Thesetwo problems were addressed recently in [8, 9]. Assuming an axial symmetry, the calculations were performed usinga simulated annealing algorithm allowing for the minimization of the total energy (static and kinetic). It was alsoshown that the stable spinning nucleon and delta masses could be obtained only if the pion mass is fixed at more thantwice its experimental value. One may argue that in this case m π could be interpreted as a renormalized pion masswhich could explain its departure form the experimental value. A third difficulty remains: Fixing the parameters ofthe Skyrme model still involve the delta which is interpreted as a stable spinning Skyrmion although physically itis an unstable resonance. Recently, Manton and Wood [10] took a different approach and chose data from B = 6Lithium-6 to set the Skyrme parameters. The purpose was to provide for a better description of higher B solutions.Unfortunately, their calculations were based on a two-fold approximation, the rational map ansatz and rigid-bodyquantization for values of B up to 8.In this work, we propose to set the Skyrme parameters such that they reproduce the physical masses of the nucleonand the deuteron. We allow deformation within an axially symmetric solution approximation and use simulatedannealing to minimize the total energy for the B = 1 nucleon and B = 2 deuteron. We argue that this procedureprovide a solution which is very close to if not the exact solution. Following the procedure in [8], we find the setsof parameters F π and e that are required to fit the nucleon and deuteron masses respectively for m π = 138 MeV, ∗ Electronic address: [email protected]
345 MeV and 500 MeV. The numerical calculations are compared to those obtained with the rational maps ansatzand rigid-body quantization. We find the latter approximation to be misleading since it suggests that it is possibleto simultaneously fit the nucleon and deuteron masses which is not the case when we perform an our numericalcalculation even for larger m π = 500 MeV. However for the solution to remains realistic with regard with the size ofthe nucleon or deuteron, lower values of F π and e are to be excluded.In section II, we introduce briefly the Skyrme model and find the static energy for the axially symmetric solutionproposed in [8]. The quantization of rotational and isorotational excitation using this solution leads to an expressionfor the kinetic energy terms at the end of section III. These expression suggest that the axial symmetry could bepreserved to a large extent. Finally we discuss and compare our numerical results from simulated annealing with anaxial solution on a two dimensional grid with that coming from rational maps with rigid-body approximation in thelast section. II. THE SKYRME MODEL
The SU (2) Skyrme Lagrangian density is L S = − F π
16 Tr ( L µ L µ ) + 132 e Tr [ L µ , L ν ] + m π F π U −
2) (1)where L µ is the left-handed chiral current L µ = U † ∂ µ U and the parameters F π and e are respectively the pion decayconstant and the dimensionless Skyrme constant. U is a SU (2) field associated to the pion field π by U = σ + i τ · π (2)where τ are the Pauli matrices, π = ( π , π , π ) is the triplet of pion fields and the scalar meson field σ satisfy σ + π · π = 1. The third term where m π is the pion mass was first added by Adkins and Nappi [2] to account forthe chiral symmetry breaking observed in nature.The field configurations that satisfy the boundary condition U ( r , t ) → for | r | → ∞ (3)fall into topological sectors labelled by a topological invariant B = 12 π Z d x det { L ai } = − ε ijk π Z d x Tr ( L i [ L j , L k ]) (4)taking integral values. Skyrme interpreted this topological invariant as the baryon number. The minimal staticenergy Skyrmion for B = 1 and B = 2 turns out to have spherical and axial symmetry respectively. Since we are onlyinterested by these values of B , the solution will be cast in terms of cylindrical coordinates ( ρ, θ, z ) in the form σ = ψ π = ψ cos nθ π = ψ sin nθ π = ψ (5)introduced in [11] where ψ ( ρ, z ) = ( ψ , ψ , ψ ) is a three-component unit vector that is independent of the azimuthalangle θ . The boundary conditions (3) implies that ψ → (0 , ,
1) as ρ + z → ∞ . Moreover, we must impose that ψ = 0 and ∂ ρ ψ = ∂ ρ ψ = 0 at ρ = 0. The B = 1 hedgehog solution appears as a special case of (5) having sphericalsymmetry and corresponds to ( ψ , ψ , ψ ) = (sin F sin θ, sin F cos θ, cos F )where F = F ( r ) is the profile or chiral angle.With the axial ansatz (5) and a appropriate scaling the expressions for the static energy and the baryon numberbecome E n = − Z d x L S = 2 π (cid:18) F π √ e (cid:19) Z d z d ρρ (cid:26) ( ∂ ρ ψ · ∂ ρ ψ + ∂ z ψ · ∂ z ψ ) (cid:18) n ψ ρ (cid:19) + 12 | ∂ z ψ × ∂ ρ ψ | + n ψ ρ + 2 β (1 − ψ ) (cid:27) (6) We have used 2 √ /eF π and F π / √ B = nπ Z d z d ρ ψ | ∂ ρ ψ × ∂ z ψ | (7)with β = √ m π eF π .However, to describe baryons, Skyrmions must acquire a well defined spin and isospin state. This is possible onlyupon proper quantization of the Skyrmions as it will be done in the next section. As we shall see in the next sectionadding (iso-)rotational energy will in general brake the axial symmetry manifest in B = 1 and B = 2 static solutions. III. QUANTIZATION
Since the Skyrme Lagrangian (1) is invariant under rotation and isorotation , the usual method of Skyrmionquantization consist of allowing the zero modes to depend on time and then quantize the resulting dynamical systemaccording to standard semiclassical methods. From this perspective, the dynamical ansatz is assumed to be e U ( r , t ) = A ( t ) U ( R ( A ( t ) r ) A † ( t ) (8)where A , A are SU (2) matrices and R ij ( A ) = Tr( τ i A τ j A † ) is the associated SO (3) rotation matrix. Introducingthis ansatz into the Skyrme Lagrangian (1) one gets the kinematical contribution to the total energy which can becast in the form T = 12 a i U ij a j − a i W ij b j + 12 b i V ij b j (9)where a j = − i Tr (cid:16) τ j A † ˙ A (cid:17) , b j = i Tr (cid:16) τ j ˙ A A † (cid:17) and the U ij , V ij and W ij are inertia tensors U ij = − √ e F π ! Z d x (cid:26) Tr ( T i T j ) + 18 Tr ([ L k , T i ] [ L k , T j ]) (cid:27) , (10) V ij = − √ e F π ! ǫ ikl ǫ jmn Z d xx k x m (cid:26) Tr ( L l L n ) + 18 Tr ([ L p , L l ] [ L p , L n ]) (cid:27) , (11) W ij = √ e F π ! ǫ jkl Z d xx k (cid:26) Tr ( T i L l ) + 18 Tr ([ L m , T i ] [ L m , L n ]) (cid:27) (12)where T i = iU † (cid:2) τ i , U (cid:3) . For the axial ansatz (5), these tensors are all diagonal and satisfy U = U , V = V , W = W and U = W n = V n . The components of these inertia tensors are U = 2 π √ e F π ! Z d z d ρρ (cid:26) ψ + 2 ψ + 12 (cid:20) (cid:18) ∂ ρ ψ · ∂ ρ ψ + ∂ z ψ · ∂ z ψ + n ψ ρ (cid:19) ψ + ( ∂ ρ ψ ) + ( ∂ z ψ ) + n ψ ρ (cid:21)(cid:27) , (13) U = 2 π √ e F π ! Z d z d ρρψ ( ∂ ρ ψ · ∂ ρ ψ + ∂ z ψ · ∂ z ψ + 2) , (14) Since we are interested only in the computation of the static properties, we will ignore translational modes and quantize the Skyrmionsin their rest frame. V = 2 π √ e F π ! Z d z d ρρ (cid:26) | ρ∂ z ψ − z∂ ρ ψ | (cid:18) n ψ ρ (cid:19) + z n ψ ρ + 12 (cid:0) ρ + z (cid:1) | ∂ ρ ψ × ∂ z ψ | (cid:27) , (15) W = 2 π √ e F π ! Z d z d ρρ (cid:26) [ ψ ( ρ∂ z ψ − z∂ ρ ψ ) − ψ ( ρ∂ z ψ − z∂ ρ ψ )] (cid:18) (cid:20) ( ∂ z ψ ) + ( ∂ ρ ψ ) + ψ ρ (cid:21)(cid:19) + ψ z∂ z ψ + ρ∂ ρ ψ ) [ ∂ ρ ψ ∂ z ψ − ∂ ρ ψ ∂ z ψ ]+ zψ ψ ρ (2 + ∂ ρ ψ · ∂ ρ ψ + ∂ z ψ · ∂ z ψ ) (cid:27) . (16)Let us note that W = 0 only for n = 1. In order to obtain the energy corresponding to the nucleons and thedeuteron, we must compute the Hamiltonian for the rotational and isorotational degrees of freedom in term of theinertia tensors (13) to (16) as well as its eigenvalues for the states corresponding to these particles.The body-fixed isospin and angular momentum canonically conjugate to a and b are respectively K i = ∂T∂a i = U ij a j − W ij b j , (17) L i = ∂T∂b i = − W T ij a j + V ij b j . (18)These momenta are related to the usual space-fixed isospin ( I ) and spin ( J ) by the orthogonal transformations I i = − R ( A ) ij K j , (19) J i = − R ( A ) T ij L j . (20)We can now write the Hamiltonian for the rotational and isorotational degrees of freedom as H = K · a + L · b − T where T is the kinematical energy (9).The quantization procedure consist of promoting four-momenta as quantum operators that satisfy each one the SU (2) commutation relations. According to (19) and (20), we see that the Casimir invariants satisfy I = K and J = L . Then the operators form a O (4) I , K ⊗ O (4) L , J Lie algebra. The physical states on which these operators actare the states formed in the base | Ψ i = | ii k i| jj l i where − i < i , − j < j and l < j that satisfy the constraintsformulated by Finkelstein and Rubinstein [12]. One of these constraints, namely, e πi n · L | Ψ i = e πi n · K | Ψ i = ( − B | Ψ i , (21)implies that the spin and isospin must be an integer for even B or a half-integer for odd baryon numbers. Anotherconstraint ( nK + L ) | Ψ i = 0 (22)comes from the axial symmetry imposed on the solution here.This formalism allows one to obtain the total energy in terms of the inertia tensors of (13-16) for the nucleon [9] E N = E + 14 (cid:16) − W U (cid:17) V − W U + 1 U + 12 U (23)and for the deuteron [4] E D = E + 1 V . (24)For the B = 1 solution the minimization of the static energy E leads to a spherically symmetric solution. Thenucleon mass in (23) however receives (iso-)rotational energy contributions from (iso-)rotations around the threeprincipal axis among which two are equivalent (direction 1 and 2). The deformation should then lead to a axiallysymmetric solution along the z -axis (direction 3) as argued in Refs.[8, 9].The situation is somewhat different for B = 2 which is known to have a toroidal solution upon minimization of thestatic energy E . In our scheme it corresponds to the axially symmetric solution with respect to the z -axis. Assumingsuch a solution the deuteron mass in (24) gets kinetic energy contributions which may be recast in the form E rot = 12 V + 12 V = 1 V , (25)which means that the contributions come from rotations perpendicular to the axis of symmetry. Unfortunately thereis no guaranty that in a non-rigid rotator approach the axial symmetry will be preserved when the rotational termsare added. However the deformations are not expected to be very large since the magnitude rotational energy onlyaccounts for less than 4% of the total mass of the deuteron in the rigid-body approximation [4]. Allowing deformationsshould increase the moments of inertia and bring the relative contribution of rotational energy to an even lower value.Clearly large changes in the configuration are prohibited by an eventual increase in the static energy E so, from thatargument alone we can infer that deformations from axial symmetry are bound to be contained into a 4% effect. Ourresults will show that non-axial deformations represents less than 1% of the deuteron mass if they contribute at all.This justifies the use of the axial solution in (5).In order to obtain the configurations ψ ( ρ, z ) = ( ψ , ψ , ψ ) that minimize the energies defined in (23) and (24),we use a two-dimensional version of a simulated annealing algorithm used in [13]. The minimization is carried outon a grid in a plane ( ρ, z ) made up of 250 ×
500 points with a spacing of 0.042 . The algorithm starts with aninitial configuration ψ ( ρ, z ) on the grid and evolves towards the exact solution. Here ψ ( ρ, z ) is generated using thesuitable rational map ansatz [14] to ensure that the initial solution have the appropriate baryon number, B = 1 or 2 , with a profile function of the form F ( r ) = 4 arctan (cid:0) e − αr (cid:1) (26)as inspired from Ref.[15]. Here α is a parameter chosen so that the entire baryon number density (such that B = 1or 2 respectively for the nucleon or the deuteron) fits into the grid. IV. RESULTS AND DISCUSSION
There are several ways to fix the parameters of the Skyrme model e , F π and m π and the authors of Ref. [8] showedthat one must take some precautions in order that the spinning solutions for the nucleon and the delta remain stableagainst pion emission. And indeed it was found that, in order to achieve a fit for the energies of spinning Skyrmionsto the masses of the nucleon and delta, one must impose a value for the pion mass that is larger than its experimentalvalue. Differences between the fitted and experimental values of F π and m π are not proscribed since after all, thevalues that enters the Lagrangian (1) are the unrenormalized parameters which could differ from the physical ones.Even so it remains that this procedure still assumes that a physically unstable particle, the delta resonance, isdescribed as a stable spinning Skyrmion and to be perfectly consistent one should instead rely on stable particles. Forthese reasons we chose to carry out calculations using data from two stable particles, the nucleon and the deuteron.We proceed as follows: Assuming a value for the pion mass and e , and an initial value for the Skyrme parameter F π we compute the lowest energy solution for the spinning Skyrmion corresponding to the nucleon using simulatedannealing which leads to a mass prediction. We iterate the procedure adjusting values of F π until the predicted massfits that of the nucleon. The same procedure is repeated for the B = 2 deuteron as well as for several values of e . Aset of points requires about two weeks of computer calculations on a regular PC. Performing an equivalent calculationon a 3D grid with similar spacing for a non-axial solution for example would be prohibitive. This explain in partswhy we use the axially symmetric ansatz. These parameters were adopted in order to be similar with those used in [8, 9]
Initially, the first set of calculations was performed with the pion mass equal to its experimental value m π = 138MeV. Since Ref. [8] suggests that a pion mass value of m π = 345 MeV or more is necessary to avoid instability due topion emission we repeated the calculations and evaluated the Skyrme parameters using m π = 345 MeV. The resultsare displayed in FIG. 1. Although it may seem interesting here to consider values of e larger than those illustratedon FIG. 1, the physical relevance of lower values of e is questionable. Below a certain value, the rotational energy ofthe Skyrmion is larger than the contribution coming from the pion mass term. As was highlighted in Refs. [6, 7], thisleads to an unstable Skyrmion with respect to emission of pions.
20 40 60 80 100 120 2.5 3 3.5 4 4.5 5 5.5eF π m π = 138 MeV m π = 345 MeV FIG. 1: F π as a function of e for which M + E rot is equal to the nucleon mass (circles) and that of the deuteron (squares).The solid and dashed lines correspond to the results obtained from the rigid body approach for the nucleon and deuteronrespectively. The set of data and lines at the upper left are for m π = 138 MeV whereas that at the bottom right are for m π = 345 MeV. Our results for B = 1 reproduce the same behavior as in Refs. [8, 9], i.e. the deformation of the Skyrmion becomesrelatively important only for larger values of m π and e . However, the results for B = 2 are far more interestingwith respect to deformation. Indeed, as we can see directly from FIG. 1, there is a noticeable difference between ournumerical results (squares) which allow for deformations as long as they preserve axial symmetry and that of therigid body approximation (dashed lines). Note that our implementation of the rigid body approximation shown hererelies on the rational map ansatz which is known to be accurate to a few percent and so it could not be responsiblealone such large difference in energy. This difference indeed corresponds to a surprisingly much smaller energy for thedeformed spinning Skyrmion than what is obtained from a the rigid body approximation wether it is based on therational map ansatz or not.To illustrate this difference, we carried out our minimization procedure using the set of parameters F π = 108 MeVand e = 4 .
84 for the deformed deuteron. The results, set SA, are listed in of Table I. For comparison we alsopresent three rigid body calculations: set SA-RB performed with our simulated annealing algorithm, set Ref.[4]-RBfrom Braaten et al and finally set RM-RB obtained through the rational maps approximation. Note that our resultsSA-RB agrees fairly well with that of Ref.[4]. Clearly the mass of the deuteron E D must be larger than E min2 = 1655 SA SA-RB Ref.[4]-RB RM-RB Exp. E D (MeV) 1679 1716 1720 1750 1876 E rot (MeV) 13.3 60.6 61.2 55.2 - h r i / D (fm) 0.94 0.93 0.92 0.94 2.095TABLE I: Total energy, rotational energy and charge radius of the deuteron using parameters F π = 108 MeV and e = 4 . . The results are presented for our simulated annealing calculations with axial deformation (SA) along with three rigid bodycalculations: with simulated annealing (SA-RB), from Braaten et al. (Ref.[4]-RB) and with the rational maps approximation(RM-RB) respectively. The last column (Exp.) shows the experimental values where the charge radius come from [16].
MeV , the minimum static energy for this choice of parameters. Computing the rotational energy for this solution leadsto the rigid body approximation result (SA-RB) for the deuteron mass of E D = 1716 MeV . As expected the resultsfor the deformed deuteron (SA) E D = 1679 MeV lies between these two values. Moreover our axial solution bringsthe relative contribution of the rotational term to about 1% of the minimum static energy E min2 . So allowing for axialdeformation reduces by about a factor of 4 the rotational term going from 60.6 MeV to 13.3 MeV. As for non-axialdeformations that might be present in a completely general solution, they must at the very most represent an 1%correction to the deuteron mass since it is bounded by E min2 (= 1655 MeV) < E exact D ≤ E D (= 1679 MeV). On theother hand they may still represent a significant portion of the remaining 13.3 MeV rotational energy. We concludenonetheless that the axial symmetry ansatz represents a very good approximation of the deuteron configuration.In addition, since these bounds are both based on axially symmetric solutions and their energy only differ by 1%,one can even contemplate the possibility that the exact solution may have axial symmetry and therefore is to oursolution, contrary to what physical intuition might suggest. This has yet to be proven and needless to say that sucha demonstration would require a 3D calculation with a level of precision less than 1%.Note that, both for m π = 138 MeV and 345 MeV, the solid and dashed lines intersect for a relatively low value of e. However despite what this rigid body calculations suggests, there is no set F π and e that leads simultaneously tothe masses of the nucleon and the deuteron. However, the data of FIG. 1 would indicate that the gap between thevalues of F π decreases for a larger pion mass. For that reason, we repeated the calculations with a relatively large m π = 500 MeV. Some of the computations required that we adjust the spacing of our grid to 0.0057 for smaller e since the configurations turned out to be much smaller in size. The results for m π = 500 MeV are presented in FIG.2. The main conclusion one can draw is that even for this pion mass, not common set of Skyrme parameters can befound.It is also interesting to note that the gap between the values of F π decrease as e decreases. This would suggest thatat very low values of e a fit is possible. However computing the charge radius, i.e. the square root of r n = 8 e F π π Z d z d ρ (cid:0) ρ + z (cid:1) ψ | ∂ ρ ψ × ∂ z ψ | (27)leads to an significant increase for the radius for small values of e and F π as illustrated in FIGS. 3 to 5. So lowervalues of e are incompatible with the physical size of the deuteron and nucleon and may be discarded. The resultsalso indicate that the best fit for the radius of the nucleon and deuteron would favor intermediate values of e around e ≃ . m π .To summarize, our calculations showed that the axial symmetry ansatz is a very good approximation of the exactsolution for the deuteron. This hints at the possibility that it may even represent the exact solution. This remainsto be proved with a general 3D calculation. We also found that allowing for axial deformation reduces the rotationalenergy by a significant factor. On the other hand we found that it is not possible to get a common set of parameters F π and e which would fit both nucleon and deuteron masses simultaneously at least for m π = 138 MeV, 345 MeV and500 MeV. This conclusion should hold even for the exact B = 2 solution since if the solution was allowed to adjustfree of any symmetry contraints it would achieve a configuration with lower total energy which would require largervalues of F π for the same set of e. This suggests that either m π >
500 MeV or additional terms must be added to theSkyrme Lagrangian. We also observed an increase in the deformations due to the spinning of the B = 2 Skyrmion(deuteron) especially for larger values of e and F π so the rigid body approximation may not be appropriate in thatcase.This work was were supported by the National Science and Engineering Research Council.
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