The meson as a bound system in 2 D quantum chromodynamics

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . We investigate the Bethe–Salpeter-like amplitude, which involves a system comprising a quark and an anti-quark combined by gauge fields, to satisfy gauge invariance in two dimensions. We derive the equation of motion for this system and construct a series of singular integraldifferential equations. By solving these equations, we obtain an algebraic equation that determines the mass spectra for the (n + 1)th non-zero eigenvalue (where n = 0 or a positive integer). In this algebraic equation, the solution for large values of n is written in the same form as G. ’t Hooft’s W-K-B solution, and the zero mass solution arises under a boundary condition equivalent to the boundary condition that ’t Hooft derived. Thus, we show convincingly that zero mass is actually a physical mass. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .


Introduction
In 1974, Gerard 't Hooft determined the 2D mass spectra of a meson [1].He insisted that these spectra include the zero mass spectrum.Although his work produced spectra for only two dimensions, his solution demonstrated the possibility of solving the U (1) problem, at least in principle.Therefore, many papers [2][3][4][5] on this subject were published in the late 1970s.However, an explicit solution for this spectrum has not yet been found.The absence of a solution indicates that we must investigate 't Hooft's zero mass spectrum more deeply; even in 2D studies, Peccei [6] and Pak [7] arrived at different conclusions regarding the existence of the zero mass spectrum.In addition, we wish to emphasize that 't Hooft obtained the zero mass spectrum as an eigenvalue under a different boundary condition than the boundary conditions applied to other mass spectra.In this respect, we must consider the work of Hornbostel et al., who used the quantization of the 1 + 1-dimensional light-cone gauge theory [8].They concluded that a meson has zero mass for quark mass m = 0, even though the case involving the lightest state is ambiguous.Their reasoning was based on the fact that the continuum limit achieves the same form as 't Hooft's equation, which gives μ 2 = 0 for the φ = 1 solution when m = 0, where μ 2 is the eigenvalue and φ is the eigenfunction, which is the wave function in the P 0 − P 1 momentum space.However, as mentioned earlier, this φ = 1 solution violates the boundary condition required for a physical state.In addition, the φ(P) = 1 solution in momentum space is a δ-function in X space (specifically, in T − X space) and is written as δ(X ).Subsequently, the quark and anti-quark always occupy the same space-time position.This result is clearly unacceptable as a physical solution.Therefore, their assertion that zero mass can be treated as a physical mass was inconclusive.It remains crucial to consider whether 't Hooft's zero mass spectrum can be regarded as a physical mass spectrum.In this paper, we consider the equation of motion based on the Bethe-Salpeter-like amplitude and obtain an algebraic equation that determines the meson mass spectra group consistent with 't Hooft's mass spectra group.In addition, we obtain the zero mass spectrum as an eigenvalue under the same boundary conditions applied to the other mass spectra.Because our boundary condition is equivalent to 't Hooft's boundary condition, we can argue convincingly that the zero mass spectrum is actually a physical mass.

Formulation
In order to derive equations within the framework of 't Hooft's model [1], we introduce the following hadronic operator, which was proposed by Suura [8].For a confined system, Suura defined the Bethe-Salpeter-like amplitude as where the gauge-invariant bi-local operator q(1, 2) is defined as Here, α and β denote Dirac indices.The above definition is formulated in the time axial gauge (A 0 = 0) because we are interested in the Hamiltonian formalism.This description is extended to the non-Abelian gauge field by P denotes the path ordering, and the λ a 2 components are the generators of the adjoint representation of the SU(N ) color gauge group.The trace is calculated for color spin a.Because the physics for the observable color singlet should be path-independent, Suura chose a straight line for zeroth order consideration.Justification of this choice is given in Ref. [9] based upon the Tamm-Dancoff method [10,11].However, because we are presently interested in the 't Hooft model (which is 2D), choosing a straight line as the path produces an exact result.
From the definition in Eq. ( 3), the 2D version of the gauge-invariant bi-local operator is described by In two dimensions, the initial Lagrangian densities are given as follows: where a represents color indices, γ are the Dirac γ -matrices, and μ and γ are two components of space-time.Because we employ the A 0 = 0 gauge, the momentum canonically conjugate to A 1 (x) is Ȧ1 (x) which is −E 1 (x), and the momentum canonically conjugate to ψ(x) is iψ + (x).Henceforth, we drop subscript 1.
PTEP 2014, 053B01 T. Kurai Therefore, we obtain the following Hamiltonian: Here, D Ai is the covariant derivative, and i = 1 in this case.We employ the metric system and γ -matrices in the following way, which was also employed by Casher et al. [12]: The usual canonical quantization procedure leads to Both the equal time commutations between the components of A and between the components of E are zero; the equal time anti-commutations between the two ψ and the two ψ + are also zero.These Hamiltonian density and commutation relations are invariant under a time-independent SU(N ) transformation such that where U = U (x) can be any N × N unitary matrix function in which x satisfies det U = 1.Because the U (x) are time independent, the invariance group U (x) is generated by the x-dependent operators G a , which are conserved such that which subsequently produces the following: One easily obtains Consequently, G a commutes with H , and therefore Ġa = 0.By commuting H with ψ, we obtain the equation of motion for ψ.Likewise, by commuting A a , we derive the equation of motion: However, one cannot obtain the equation corresponding to Gauss's law (D • B a (x) = 0 may be obtained in the 4D case).Hence, Gauss's law must be imposed as a subsidiary condition on a physical state (namely the gauge-invariant states): Henceforth, we denote the string part with In order to obtain the equation of motion for q(1, 2), we must consider the effect of applying the gradient to the string part.By applying to the path integral the same treatment applied to the time ordering, one can write U (2, 1) as where The gradient ∂( 2) is generated by the infinitesimal displacement of x(2), while the point x( 1) is fixed.
We then obtain To obtain the third line from the second line, we used a path order argument.Because the path order is defined as the time order, a gauge field for a higher position is always situated on the left-hand side of a gauge field for a lower position.For example, we may more precisely write where Therefore, the integral from 1 to 1 + is always on the left-hand side of the integral from 0 to 1. Similarly, applying ∂(1) to U (2, 1) gives The equation of motion for quark fields is simply the usual Dirac equation with the vector gauge fields satisfying the gauge principle, which means that Here we consider the case in which the quark mass is zero.
From the definition of the gauge-invariant bi-local operator q(1, 2) in two dimensions (given by Eq. ( 4)), the time derivative of this operator is for which we introduce the notation Note that the gauge fields arising from the quark fields are canceled out by the gradient of the string part.
The term sandwiched between the physical states in this equation will vanish according to the cluster expansion theorem [13].We therefore find The integrand can be rewritten as The last line results from the fact (Appendix A) that Therefore, the last line is exact when physical states are considered.
In order to regroup the charge vertex term, we use the Fierz identity for SU(N ) [14] as follows: We thereby obtain The final form of the equation of motion for q(1, 2) is thus Placing Eq. ( 24) between the vacuum and a physical state and then taking the center-of-mass system to be After factoring out the phase part (the exponential part), Eq. (24) becomes where The derivation of the last term is given in Appendix B.
Recalling the fact that α = σ 3 and then taking the decomposition we obtain the following series of equations: Here, we set S(r ) = (iσ 3 )P r for the X 2 and X 3 cases in order to obtain a closed form for the equations; consequently, P 0 becomes W 0 , which is the mass of this system.

The 't Hooft model
Because the Schwinger model and the 't Hooft model are CP = −1 and CP = +1 models, respectively, we expect to obtain the 't Hooft model's mass spectra from Eqs. ( 29) and (30).
Changing the variables in the integral such that r − x = s and, henceforth, using the notation r = x, Eqs. ( 29) and (30) become, respectively, In order to write the equations in closed form, we take X ∓ = X 2 ∓ i X 3 and then Eqs. ( 31) and (32) become, respectively, To address the integral term, we use the definition of absolute value, which results in where ε is the step function defined as 1 for x ≥ 0, and as −1 for x < 0.
Multiplying ε(x) on both sides of the equation and considering the new function to be F + (x) = X + (x)ε(x), Eq. (33) becomes Here, for the second and fourth terms, we have used the fact that (ε(x)) 2 = 1.
Considering the x > 0 region (for which ε(x) = 1) and taking the derivative with respect to x on both sides, Eq. (35) becomes Here, to obtain the last term, we have used ∂ x ε(x − s) = 2δ(x − s).For the singular integral term, we use the slightly modified Sokhotsky formula [16]: where + (x) is the value resulting from unlimitedly approaching x from the upper hemisphere, is the value resulting from unlimitedly approaching x from the lower hemisphere, and a 1 and a 2 are non-zero arbitrary constants.Here, we have modified the original form of the Sokhotsky formula by changing + , which will be used in a later argument.Note that s and x are real values.Subsequently, Eq. ( 36) is written as In order to find the solutions for both + and (−) + , we set both sides of Eq. ( 39) to be equal to zero.Because a 1 and a 2 are non-zero constants, we can independently obtain the differential equations for (+) + and (−) + and then solve these differential equations.To address the derivative of the δ-function term, we integrate twice successively from ∞ to x.Note here that we consider the range x > 0 while Subsequently, we can see that Taking the derivative twice successively, we obtain the following differential equations: 1 We use the definition of the δ-function as (−ε z ε) 0 otherwise when ε approaches 0.
We may then conclude that This conclusion is always true (even though x approaches zero) as long as x is larger than ε when ε approaches zero.The solutions satisfying the condition that both the function and its first derivative at infinity must be zero are as follows (the derivation is given in Appendix C):
In order to determine the eigenvalue of W 0 , we must consider the boundary condition.The 't Hooft model's boundary condition was established in momentum space (specifically, in P 0 − P 1 space).Because 't Hooft divided this quantity by the maximum momentum, his boundary condition was essentially identical to the condition that, at P 1 = 0, the wave function is zero and that, at P 1 = ∞, the wave function is zero.
Because this derivation is conducted in the normal space-time, this boundary condition is equivalent to the condition that, at x = 0, the wave function is zero and that, at x = ∞, the wave function is zero (x = 0 corresponds to P 1 = ∞, and x = ∞ corresponds to P 1 = 0).
It is important to note that F + (x) in Eq. ( 38) is essentially χ + (x) = χ 2 (x) + iχ 3 (x).This component is one of the off-diagonal elements in our Bethe-Salpeter-like amplitude (the other component is χ − ).Consequently, the wave function automatically approaches zero at x = ∞ because the Bethe-Salpeter-like amplitude is represented by a linear combination of (For χ − , it is given by

− (x).)
In order to satisfy the other boundary condition, we invoke the following wave function: Here, we have taken in Eq. ( 38).These values were determined by imposing the condition that the constant term in is variable) must be canceled out in order to satisfy the boundary condition (see Appendix D).The expansion of F + (x) is then expressed by Eq. (45) when which is based on → 0, and β 2 α → 0, the fact that W κ,μ is defined by a linear combination of the Whittaker functions for M κ,μ and M κ,−μ (see Appendix D): Thus, using the fact that F + (x) = ε(x)X + (x), we find that because (as we defined earlier) ε(x) = −1 for x < 0, and +1 for x ≥ 0; thus, (ε(x))2 = 1.Therefore, our boundary condition is represented such that at x = 0, χ + (0) = 0 and, at x = ∞, χ + (∞) = 0.
As previously mentioned, χ + automatically satisfies the x = ∞ boundary condition and, therefore, we need only to consider the x = 0 boundary condition.
The condition that χ + (0) = 0 gives the following equation: We must consider two additional cases: where W 2 0 /g 2 approaches zero, and where this value is non-zero.
For the first case, we immediately obtain a zero mass solution from Eq. (45) (see Appendix D). Thus, For the second case, we use the integral representation of W κ,μ in the following manner [17,18] 2 : Therefore, after adjusting the coefficients, Eq. (47) becomes PTEP 2014, 053B01 T. Kurai From Eq. ( 48), we obtain the following equation: where x = W 2 0 g 2 .We use the following representation of the -function [17,18] 3 : where γ is the Euler-Mascheroni constant, and ζ is the Riemann ζ -function.
We can then express the -function as follows: In order to address the integral, we use the following expression: Therefore, the integral becomes Because the real part in the exponent is unchanged (while the imaginary part in the exponent differs only by a sign), Eqs. ( 51) and ( 52) can be written in the following manner: right-hand side of Eq. (51) = A(x) − i B(x), right-hand side of Eq. (52) = A(x) + i B(x). where The ratio in the integral from Eq. (49) then becomes Using the description of Eq. ( 55) for the ratio in the integral from Eq. ( 49), we obtain an algebraic equation that determines the non-zero mass spectra of the bound states to be the following: Here N = 0 or a positive integer, x = W 2 0 g 2 and A(x) and B(x) are given in Eqs. ( 53) and (54), respectively.
To compare the eigenvalue of our solution with 't Hooft's eigenvalue, we consider the case in which N is large.Because the value of arctan B(x)  A(x) is between − π 2 and π 2 , we can ignore this term and, instead, express the constant terms on the left-hand side of Eq. (56) by ρ.
Setting x/π = N + α log N + η, we can then show that To obtain the second to last line we used α = 1/π 2 and η = ρ π + 1 π 2 log π .Thus, W 2 0 = g 2 π N + 1 π 2 log N + ρ π + 1 π 2 log π is the first approximate solution of Eq. ( 56).This solution is in exactly the same form as the W-K-B solution in the 't Hooft model.
For the equation representing X − , when we replace W 0 with −W 0 , this equation is then identical to the equation representing X + , which becomes apparent when we compare Eq. (33) with Eq. (34).Consequently, the formula obtained for X − (which determines the mass spectra) is identical to Eqs. (47) and (48).Therefore, the equations obtained to determine the mass spectra of Eq. ( 56) and the zero mass spectrum are consistent in this argument.Here, we again insist that our zero mass spectrum is obtained from the same boundary condition used for the mass spectra group in Eq. ( 56); therefore, we conclude that the zero mass spectrum truly represents a physical mass.
Finally, we discuss whether our approach is equivalent to 't Hooft's method.Our approach is quite similar to 't Hooft's methodology but is not exactly identical.The reason for this difference is that the system we consider maintains gauge invariance, whereas 't Hooft's system is not gauge invariant when the quark and anti-quark are at different space-time positions.Even in momentum space, this difference results in a discrepancy; in particular, the contour of the singular integral is not closed in 't Hooft's case but is closed in our case.For this reason, 't Hooft's integral equation cannot be solved analytically, whereas our case can be solved analytically (we solved this case in x-space).For comparison, we show the equation of motion in momentum space in Appendix F.

+
/∂ x| x=∞ = 0; specifically, we are seeking solutions that satisfy the previous condition.