Center of mass motion in bag model

Despite the great success on the mass spectra, the reputation of the bag model has been closely followed by the embarrassment from the center of mass motion. It leads to severe theoretical inconsistencies. For instance, the masses and the decay constants would no longer be independent of the momentum. In this work, we provide a systematical approach to resolve this problem. The meson decay constants as well as the baryon transition form factors can be computed consistently in our framework. Notably, the form factors in the neutron $\beta$ decays do not depend on any free parameters, and are determined to be $F^V_1 =1 $ and $F^A_1 = 1.31$ or $F_1^A/F_1^V= 1.31$, which is close the experimental value of $F^A_1/F^V_1 = 1.27$. In addition, we find that ${\cal B} (\Lambda_b \to \Lambda \gamma) = (6.8 \pm 3.3 ) \times 10^{-6} $, which agrees to the experimental value of $(7.1\pm 1.7)\times 10^{-6}$.


I. INTRODUCTION
The Massachusetts Institute of Technology (MIT) bag model describes a hadron as a bag which contains various quarks, antiquarks and perhaps gluons [1].It was proposed to reconcile two very different ideas, behind the quantum chromodynamics (QCD) with a single parameter of the bag radius (R), as follows: • Due to the asymptotic freedom, quarks move freely inside the bag; • Quarks are not allowed to penetrate the bag because of the QCD confinement.
Expressing the ideas mathematically, we start with the free Dirac equations, given by along with the boundary condition, where ψ q stands for the quark wave function, M q represents the quark mass with q being the quark flavor, r = | x|, γ = (γ 1 , γ 2 , γ 3 ), and R is the bag radius.The description is similar to the infinite square well in quantum physics.In practice, "bag" can be understood as the abbreviation for "infinite spherical well".The higher order mass corrections can be done by taking ψ q as an unperturbed state.
Thanks to its simplicity, the bag model can easily cooperate with various QCD systems.
In astrophysics, under the framework of the bag model, the strange stars, neutron stars and quark cores are viewed as large bags containing countless quarks [11][12][13][14][15].In spite of these great progresses, the bag model has little application in the particle decay system, mainly due to the center of mass motion (CMM) [16][17][18].This study is devoted to introduce the homogeneous bag model to provide a consistent framework to deal with the CMM.
The paper is organized as follows.We give a brief review of the MIT bag model in Sec.II, where we concentrate on the problem of the CMM rather than computational details.In Sec.III, we display some of the methods tackling with the CMM given in the literature.In Sec.IV, we present a systematic framework to deal with the CMM, and show the numerical results of the neutron β decay and Λ b → Λγ.We conclude the study in Sec.V.

II. THE MIT BAG MODEL
Since we only consider the low-lying hadrons, the quark wave function of ψ q can be safely taken to be isotropic.Consequently, Eq. ( 1) with J z = 1/2 can be easily solved as where φ q ( x) corresponds to the spatial part of the quark wave function with E k q = p 2 q + M 2 q , ω q± = E k q ± M q , p q and E q are the magnitudes of the quark 3-momentum and energy, respectively, N is the normalization constant, χ ↑ and χ ↓ stand for (1, 0) T and (0, 1) T , respectively, and j 0,1 are the zeroth and the first spherical Bessel functions, which are basically the cosine and sine functions in the spherical coordinate.For the absence of energy corrections, we have that E q = E k q .In analogy to the case of the infinite square well, imposing the boundary condition would quantize p q .Combining Eqs. ( 2) and (3), we find that p q must satisfy the relation [3] tan At the massless and heavy quark limits, we obtain lim MqR→0 p q R = 2.043, 5.396 respectively, which is valid for all hadrons.These results are handy for quick estimations as M u,d can be taken as massless, whereas M c,b as infinite in practice.It states that the magnitude of the 3-momentum grows along with the quark mass, approaching π/R.
At this stage, one can readily give the estimations of the hadron masses by adding up the quark energies.We take the proton as an example, and the formalism can be easily generalized to other hadrons.The proton wave function is given by superposing the quark states where û and d are the quark field operators, without writing down the spin-flavor and color indices explicitly, and |p B (0) represents the proton state from the bag model, centered in the coordinate.The experimental proton charge radius of 0.840 fm [19] corresponds to R = 5.85 GeV −1 in the bag model, leading to the proton and ∆ masses as which are close to the experimental value, However, the neutron charge radius is essentially zero in the bag model, which is inconsistent with the experiments.
There are several mass corrections, which can be summarized as follows [3]: • The energy of the bag is proportional to its volume, given by E V = 4πBR 3 /3 with B the bag energy density.
• The zero-point energy is found to be negative empirically, given by E 0 = −Z 0 /R.
• The quark-gluon interaction introduces the strong coupling constant of α s .
Here, B, Z 0 and α s are treated as free parameters in the model.On the other hand, the bag radii are determined by minimizing the hadron masses, given as [3] ∂M where M H and R H are the mass and bag radius of H, respectively, resulting in different hadron bag radii.Typically, the baryon and meson radii are found to be around 5 and 3 GeV −1 , respectively.Except for u and d, the quark masses are taken to be free parameters in the model, fitted to be [9] M s = 0.279 GeV , M c = 1.641GeV , M b = 5.093 GeV .
which can be viewed as the effective masses when the quarks swim inside the bags.
Although the mass corrections from the bag and the zero-point energies are intuitively satisfactory, they are problematic under the spacetime symmetry.For instance, particles at rest are invariant under the space translations.How can their masses related to a finite volume?How do the zero-point energies transform under the Lorentz boost?These problems partly come from that the description of the bag model is a semi-classical one.Quarks are quantum objects as they satisfy the Dirac equation, whereas the bag itself, having a definite position as well as a concrete boundary, is a classical object.These are the essential problems that have haunted the bag model ever since it was proposed, closely related to the CMM.
As a pioneer of quark models, it provides an excellent framework to understand the hadrons in relativistic systems.However, it can hardly be applied to the particle decay processes.
Another issue after considering the mass corrections is that we lost the ability to track the t-dependencies of the individual quarks.The best we can get is the t-dependency of the proton wave function, read as Here, we can no longer take t 1,2,3 to be different as done in Eq. ( 6) for 2E k u + E k d = M p after including the bag and zero-point energies.Due to this reason, we can not apply a Lorentz boost on Eq. (11) , which would mix up the spacetime coordinates.The problem can be traced back to that the t = 0 plane is not invariant under a Lorentz boost, which we will discuss more details in Sec.IV, in which the wave functions are spell out in terms of the creation operators.

III. THE WAVE PACKET AND ELLIPSOIDAL BAG APPROACHES
We are interested in the applications of the bag model in the decay system, in which momentum eigenstates are required.In the literature, there are two methods concerning decays with the bag model.These methods have their own advantages but fail to achieve a consistent framework with the Poincaré symmetry.Nonetheless, they shed light on the CMM problem and we discuss them briefly.

A bag as a wave packet
The most naïve solution to explain the CMM is the wave packet approach, which treats the bag wave function as a localized wave packet.We extract the momentum states based on the Fourier analysis, read as where the left hand side is a proton state with p the 3-momentum, leading to the quark state decomposition as One of the advantages of this method is that the t-dependencies of the quark states are not required, so it can easily cooperate with the bag and zero-point energies 1 .

The ellipsoidal bag approach
The contradiction occurred in the wave packet approach is attributed to that we can not have a wave packet with a definite energy.To solve the problem, the Lorentz boost for the bag state is needed [20][21][22][23].Boosting Eq. ( 6), the wave function is read as with v the velocity and where S v and Λ v are the Lorentz boost matrices toward ẑ direction for Dirac spinors and spacetime coordinates, respectively.In this work v is always chosen in the ẑ direction, In contrast to Eq. ( 6), the spherical bag deforms to an ellipsoid due to the Lorentz contraction, and the bag itself is moving.
Clearly, the t-dependencies (E q ) of the quark states are needed by Eq. ( 15).A reasonable range for the up and down quark energies is given as or numerically where we allocate E 0 and E V evenly among the quarks.Notice that Eq. ( 14) admits implicitly that the energies of the quarks are independent to each others, which is not true if there have interactions among them.As a consequence, we have E u,d > M p /3, and Eq. ( 17) serves as a major uncertainty in the evaluation.Nonetheless, for the processes where the initial and final baryons have the same velocity, the Lorentz boost is not required and therefore the calculation does not suffer from the uncertainty.
If we take Eq. ( 14) as a momentum eigenstate, there are couple requirements to be added by hand before the calculations : • To have the same bag volume, the initial and final hadrons must be opposite in velocities.
• As the bag is moving, a specific timing for the computation has to be chosen.
The studies of the ellipsoidal bag approach are carried out in the decays with the b → c transition [24], where the overlapping between the heavy quarks is treated by the heavy quark symmetry.On the other hand, the overlapping of the light quarks would lead to the v-dependency of the Isgur-Wise function, read as with where l is the spectator quark in the heavy hadron transition, and D v l ( x ∆ ) is a function of both velocity and position, where the dependence on x ∆ is defined for the latter convenience.
In Ref. [24], it was found that B(B 0 → D − l + ν l ) = 2.12%, which is close to (2.31 ± 0.10)% given by the experiments [25].Remarkably, Eq. ( 18) can be understood intuitively, in which γ −1 comes from the Lorentz contraction of the bag volume, while the exponential in the integral causes the damping, which is a punishment for not being at the same velocity.However, it is unarguable that both the Lorentz and translational symmetries are broken in the ellipsoidal bag approach, since a specific initial frame and timing are demanded in the computation.Furthermore, Eq (18) tell us that the inner products between |l ±v are nonzero, violating the energy momentum conservation.

IV. THE HOMOGENEOUS BAG MODEL
The homogeneous bag model, first proposed in Ref. [26], has been widely applied in various decay systems [27][28][29][30].It is meant to reconcile the Poincaré symmetry and the bag model, without necessarily introducing a new parameter.However, in Ref. [26] only the scalar operators were considered.In this work, we would like to generalize the formalism to the vector and tensor operators as well.
In the following discussion, we will construct hadron states by combining the features of the wave packet and ellipsoidal bag approaches.Remind that the wave packet approach violates the Lorentz symmetry as masses depend on velocities, whereas the ellipsodial bag one breaks the translational symmetry as the wave functions are localized.We will show that the inconsistencies with the Poincaré symmetry are resolved in our framework.We also take the neutron β decay for an instance of the computation, as it is independent of free parameters.

Hadron wave functions
We start with a hadron state at rest.As a zero momentum state has to distribute homogeneously over the space, we linearly superpose infinite bags at different locations.A proton state at rest is constructed as where N H is the normalization constant for the hadron H. Notice that the formula is identical to Eq. ( 12) when p = 0, but the idea is totally different.Eq. ( 12) is meant to extract the specific momentum component from the wave packet, whereas we are trying to build up a momentum eigenstate from infinite static bags here.
For completeness, we would write down the baryon wave functions with color and spinor indices.It can be accomplished by adopting the quark creation operators, given by along with where [d 3 x] = d 3 x 1 d 3 x 2 d 3 x 3 , the Greek (Latin) letters stand for the color (spinor) indices, the arrows represent the spin directions of the quarks, and the Fermi statistic is guaranteed by the anti-commutation relation For the sake of compactness, the quarks operators are evaluated at t = 0 if not stated otherwise.
To obtain a nonzero momentum, we have to apply the Lorentz boost U v on Eq. (21).
Recall that the transformation rule of the quark operators is given as It states that even if we start with the operators at an equal time, we inevitably have to deal with the quark operators with timelike distances as the quarks have different positions in the baryon at rest.The problem can be traced back to that the plane t = 0 is not invariant under the Lorentz boost, which leads to a unequal time among the operators.The unequal time commutators require the knowledge of the dynamical details, which can not be perturbatively calculated.To overcome the problem, we utilize that which stem from that the quarks are energy eigenstates in the bag model at least for the first order approximation.
The wave functions after the Lorentz boost can be obtained by the following trick.Without lost of generality, we write the wave functions after boosting as where Ψ v is a function to be determined.Note that as J z commutes with U v , the proton remains as an eigenstate of J z .Applying the annihilation operators, we arrive at with x v = (x, y, γz).In the third line of Eq. ( 27), we have used that the vacuum is invariant under Lorentz boosts, and the fourth and fifth ones can be obtained by Eqs. ( 24) and (25), respectively.By comparing the first and fifth lines of Eq. ( 27), we deduce that To obtain N p , we calculate the overlaping which can be derived from Eqs. ( 23), ( 26) and (28).The spin indices have not been written down explicitly as they are irrelevant as long as the inital and final quarks are in the same directions and we have taken S † v ′ = S v by anticipating that v = v ′ or else the integral vanishes.To simplify the integral, we adopt the following variables: Now, the integral is read as where (q 1 , q 2 , q 3 ) = (u, d, u), 1/γ 3 comes from the Jacobian in Eq. ( 30), and D 0 q ( x ∆ ) is defined in Eq. ( 19) with v = 0. To get the correct δ function, we have to demand that for the integral of x A , which is the main source of errors in our model.However, the range in Eq. ( 16) shall cover the reasonable values.The δ function can be interpreted as that the overlapping of D 0 q ( x ∆ ) for the bags with a distance x ∆ occurs infinite times in the integral, which is essentially a result of the translational symmetry.
By taking the normalization for a momentum state as where u p and λ p are the Dirac spinor and spin of the proton, we find 1 It is important that N p must be independent of the velocity because the Lorentz boosts are unitary for physical states.Here, our result shows that it is indeed the case in contrast to Eq. ( 12) .The wave functions and normalization constants of other baryons can be obtained straightforwardly with trivial modifications.In Appendix A, we give the baryon wave functions at rest that are used in this work, and the evaluation of N p can be found in Appendix B.

Transition matrix elements
After the hadron wave functions are constructed, the calculations of the transition matrix elements are straightforward.Here, we choose the neutron-proton transition as an example with d → u at quark level.For the calculation, we adopt the Briet frame, where n and p have opposite velocities, and without lost of generality, we take v ẑ.By sandwiching the quark transition operators with the hadron states, we arrive at along with where Γ is an arbitrary Dirac matrix and x ± = x ± x ∆ /2.Here, λ u,d ∈ (↑, ↓) are the spins of the annihilated up and down quarks, and N λpλn λuλ d represents the overlapping with specific λ p,n,u,d , which can be computed by matching the LHS and RHS of Eq. ( 35).As the values of N λpλn λuλ d is independent of the velocity, it can be obtained by taking v = 0 for convenient.From the angular momentum conservation, we have that It states that if the baryon spin is (un)flipped by the operator, then the spin of the quark shall also be (un)flipped.On the other hand, by the Wiger-Eckart theorem, we have Consequently, there are only two independent numbers in N λpλn λu,λ d , given as In the n → p beta decays, we have (N nonflip , N flip ) = (1, 5/3).
Each term in Eq (35) has a concrete physical meaning, which can be summarized as follows: • D v q ( x ∆ ) are the overlapping coefficients of the spectator quarks (u and d ) in the initial and final states as found in the ellipsoidal bag approach.Note that the centers of the quark wave functions are separated at a distance of x ∆ ; • Γ λpλn ud ( x ∆ ) is the overlapping coefficient of d → u at quark level.Again, the centers of the bags are separated at a distance of x ∆ .
Here, we have found that the overlapping integrals of the spectator quarks with different velocities (D v q ) do not vanish, which is a feature inherited from the ellipsoidal bag approach.However, we have the energy-momentum conservation when we consider the whole wave functions [26].It can be viewed as the spectator quarks are kicked by the bag, which are in turn kicked by the quark transition operators.
The main ambiguity of the homogeneous bag model comes from E q in the exponential as shown in Eq. ( 16).However, the deviations are insensitive at low velocities as E q are always followed by v, which make the calculation for the neutron-proton transition unaffected.
For the neutron β decay, the dimensionless form factors F V,A 1 are defined by [31] where q corresponds to the 4-momentum difference of the neutron and proton, and v n and v p are the velocities of the neutron and proton, respectively.F V 1 and F A 1 can be extracted straightforwardly after computing the transition matrix elements with Γ = 1 and Γ = γ 0 γ 1 γ 5 , respectively.In the model, u and d are taken as masslesss, and thus there is only one free parameter R, having the length dimension.The twist is that F V,A 1 do not rely on the bag radius since the length dimension can not be canceled.At the v → 0 limit, we find that which are close to the experimental value of 27 .The details of numerical evaluation can be found in Appendix B. Compared to the result of 09, given previously in the bag model [3], the ratio improves significantly after considering the correction from the CMM.
Notice that to compute the matrix elements we have taken the Briet frame, where the initial and final hadrons have opposite velocities.However, in principle, it can be calculated in other Lorentz frames also.For an illustration, with Γ = γ 0 γ µ , we have FIG. 1: The definitions of the angles θ, φ and φ, where the right figure is the adopted cylindrical coordinate in evaluating D v q ( x ∆ ) .
where J 0 is the zeroth Bessel function, j ± (0,1)q ≡ ω q(+,−) j (0,1) (p q r ± ) , (B4) with r ± = | x ± |.Here, we have absorbed N to the overall normalization constant N n,p for convenience.Due to the finite bag radius, the integrals are bounded as For the sake of compactness, the regions of the integrations for ρ and z ′ are not written down explicitly as long as there is no confusion.We have dropped χ • σ]χ as v is the only specified direction.It vanishes since we always choose the spin directions at ±v.In the last line of Eq. (B3), we have used with the integrand being an even function of z ′ .Notice that Eq. (B3) is consistent with Eq. (B1) due to that J 0 (a) = J 0 (−a), which implies that J 0 (2E q v sin θρ) = J 0 (2E q v √ 1 − cos 2 θρ).To conclude, the number of integrals in Eq. ( 19) is reduced as two, which greatly shorten the evaluating time.
To compute the normalization constant, we take v = 0 in Eq. (B3) and arrive at where we employ the isospin symmetry D v u = D v d .Now we turn our attention to Γ λpλn ud ( x ∆ ).From Eq. (B3), we find that D v q ( x ∆ ) is an even function of x ∆ .Thus, we can drop the terms that are odd regarding to x ∆ in Γ λpλn ud ( x ∆ ).In this work, we use Γ = 1 and Γ = γ 0 γ 1 γ 5 as examples.To evaluate Γ = 1, we take λ n = λ p =↑ in Eqs. ( 35) and (36), resulting in where E di = E u + E d is the energy of the spectator quarks.Finally, we obtain In the neutron β decay, we can safely set v → 0 and neglect the contributions from F V,A 2,3 .Comparing the right hand sides of Eqs. ( 40) and (B9), we find F V 1 = 1.Now we turn our attention to Γ = γ 0 γ 1 γ 5 .The trick of Eq. (B1) can not be applied as Γ provides an extra direction.In the cylindrical coordinate described in Eq. (B2), we have where φ is the azimuthal angle between the v ⊗ x ′ and v ⊗ x ∆ planes.In addition, we have ŷ′ = cos φ sin φ − sin φ cos θ cos φ ρ + sin φ cos θ sin φ + cos φ cos φ + sin θ sin φẑ ′ .
Here, x′ and ŷ′ point toward the x and y directions, respectively, when we choose v ẑ with Cartesian coordinate system (see FIG. 1).We define To calculate Eq. ( 36), we choose (λ n , λ p ) = (↑, ↓), resulting in that where we have used Eqs.( 37) and (39) along with S v γ 0 γ 1 γ 5 = γ 0 γ 1 γ 5 S −v .The integrand I can be further simplified by noting where χ ↑ and χ ↓ stand for the quark spins pointing toward the v and −v directions, respectively, and the first line of Eq. (B15) is due to that we only consider the even part of the integrand regarding to x ∆ .For the sake of compactness, we have defined In the cylindrical coordinates described in Eq. (B2), J nm depends on ρ , z ′ and r ∆ only, with the following property Accordingly, we find that J 00 and J 11 are even functions of z ′ , whereas J 01 /r + ± J 10 /r − are even and odd, respectively.
On the other hand, from Eq. (B15), we see that I 3 depends also on the azimuthal angle φ.
To compute we interchange the order of the integrals of dφ and d 3 x.In addition, we make use of that D v q are independent of φ, leading to B10) x′ = − sin φ sin φ − cos φ cos θ cos φ ρ + cos φ cos θ sin φ − sin φ cos φ + sin θ cos φẑ ′ ,