Confining potential under the gauge field condensation

SU(2) gauge theory in the nonlinear gauge of the Curci-Ferrari type has a gluon condensation $\langle A_{\mu}^+A_{\mu}^-\rangle$ in low-energy region. Instead of the magnetic monopole condensation, this condensation makes classical gluons massive, and can yield a linear potential. We show this potential consists of the Coulomb plus linear part and an additional part. Comparing with the Cornell potential, we study this confining potential in detail, and find that the potential has two implicit scales $r_c$ and $\tilde{R}_0$. The meanings of these scales are clarified. We also show that the Cornell potential that fits well to this confining potential is obtained by taking these scales into account.


Introduction
In the study of quarkonia, QCD potential is often used.Although there are some phenomenological potentials (see, e.g., [1]), the Cornell potential V CL (r) [2,3] is simple but workable.This potential has the Coulomb plus linear form as where K and σ, that is called the string tension, are constants.The Coulomb part is expected from the perturbative one-gluon exchange, and the linear part represents the confinement.
Is it possible to derive V CL (r) from QCD? Using the dual Ginzburg-Landau model (see, e.g., [4]), the following Yukawa plus linear potential was obtained [5][6][7][8]: where Q is the static quark charge and m χ is the momentum cut-off.In this model, the mass m is related to the vacuum expectation value (VEV) of the monopole field.In Ref. [9], based on the SU(2) gauge theory in the non-linear gauge of the Curci-Ferrari type, we also derived the potential V Y L (r).In this case, the mass m comes from the gauge field condensation A + µ A − µ .In this paper, in the framework of Refs.[9,10], we restudy the confining potential.In the next section, we briefly review Refs.[9,10], and present the potential between the static charges Q and Q.In Sect.3, the equation to determine the ultraviolet cut-off Λ c is derived.
In Sect.4, using this cut-off, we show that the Q Q potential becomes the confining potential V c (r) = V CL (r) + V 3 (r), where V 3 (r) is the additional potential.The potential V c (r) has several parameters.Comparing V c (r) with V CL (r), and choosing the appropriate values of K and σ, the parameters in V c (r) are determined in Sect. 5.In this process, we find a scale r c ≈ 0.2 fm.In the intermediate region, the scale R 0 ≈ 0.5 fm has been proposed [11].The meanings of the scales r c and R 0 for V c (r) are clarified in Sect.6.We also propose a scale R0 that is related to R 0 .Based on this analysis, we obtain V CL (r) that fits well to V c (r). Section 7 is devoted to a summary and comments.In Appendix A, the propagator for the off-diagonal gluons is presented.The equations in Sect.5, that determine the values of the parameters in V c (r), are solved in Appendix B.

Condensate A
First, we review Refs.[9,10] briefly.We consider the SU(2) gauge theory in Euclidean space.The Lagrangian in the nonlinear gauge of the Curci-Ferrari type [12] is where B = −B + igc × c, α 1 and α 2 are gauge parameters, and w is a constant.Introducing the auxiliary field ϕ, which represents α 2 B, L N L is rewritten as [13] In Ref. [14], by integrating out c and c with momentum µ ≤ p ≤ Λ, we have shown that gϕ A acquires the VEV gϕ 0 δ A3 under an energy scale µ 0 .In the presence of the VEV v = gϕ 0 , ghost loops give tachyonic masses to gluons.The scale µ 0 , the value of the VEV v = gϕ 0 and the tachyonic gluon mass terms are summarized as (2.3) We note, when Λ ≫ µ 0 ≫ µ, the relation holds [14], where Λ QCD is the QCD scale parameter.
Although the tachyonic mass terms appear, the interaction (gA µ × A ν ) 2 produces the VEV and the tachyonic mass terms are removed.We divide the A = 3 component as A 3 µ = a µ + b µ , where a µ (b µ ) is the quantum (classical) part.Then the effective Lagrangian after integrating out c and c becomes where where G µν is the propagator for the massive fields A ± µ .Now we consider the confining potential.As the classical field b µ , we choose the dual electric potential B µ in Ref. [9].The field B µ describes the monopole solution of the dual gauge field.The color electric current j µ is incorporated by the replacement where n • ∂ = n µ ∂ µ , and n µ = (0, n) with n • n = 1.We note this is the Zwanziger's dual field strength [15].Thus the classical part of L in Eq.(2.6) becomes The equation of motion for B µ is and B µ is solved as (2.9) If we use Eq.(2.9),Eq.(2.8) gives To derive the static potential between the color electric charges Q and −Q, the static current Then it leads to the potential where r = a − b, q = |q| and q n = q • n.Although the term V Y (V L ) becomes the Yukawa (linear) potential in Eq.(1.2) usually, we restudy Eq.(2.12) in Sect. 4.
3 Cut-off Λ c and the mass M In this section, we solve Eq.(2.7).The propagator G µν is calculated in Appendix A as , this is the propagator in the R ξ gauge.Substituting Eq.(3.1) into Eq.(2.7), we obtain Since this integral diverges, we introduce the ultraviolet cut-off Λ c and replace M 2 as The second term, which is independent of M, is left for convenience.As we explain in Appendix A, the existence of the condensate v implies α ′ 2 → ∞.This limit is necessary to maintain the BRS invariance in the case of M = 0. Thus we choose α ′ 2 as where we set α ′ 2 = −α 1 in the second term to make the propagator transverse.Substituting Eq.(3.4) into Eq.(2.7), we obtain where the constant d 4 p p 2 , that comes from the terms independent of M and vanishes in the dimensional regularization, has been neglected.We rewrite Eq.(3.5) as and determine Λ c in Sect. 5.
We make a comment.If we choose α ′ 2 = −α 1 , and take the limit M → 0 in Eq.(3.2),Eq.(2.7) becomes The right hand side is a constant, and vanishes if we use the dimensional regularization.So M = 0 implies v = 0.However, as we cannot take the limit M → 0 in Eq. (3.4).Since v = 0 implies α ′ 2 → ∞, we cannot make M → 0 after taking the limit α ′ 2 → ∞.However if we take the limit Λ c → 0 in Eq.(3.4), the part with M disappears, and it leads to Eq.(3.7).As m 2 = g 2 v/(32π), even if the limit M → 0 is unable to take, we can consider the case of m = 0 by taking the limit Λ c → 0.

Confining potential
Usually, the potential V in Eq.(2.12) is calculated as follows.Let us divide the momentum q in V L into q n = q • n and q T that satisfies q T • n = 0. Then the integral of q n has the infrared divergence, and the integral of q T = |q T | has the ultraviolet divergence.The former divergence is removed by the choice r n [7,9], and the latter divergence is avoided by the cut-off m χ as [5-8] Eq.(4.1) becomes the linear term in Eq.(1.2).The term V Y has the integral of q = |q| over the region of 0 ≤ q < ∞, and the Yukawa potential in Eq.(1.2) is obtained.
To introduce a cut-off in a different way, we write the potential as In Sect.3, we already have the cut-off Λ c , and m disappears in the limit Λ c → 0. So we define Eq.(4.2) as When Λ c → 0, the first term disappears, and the Coulomb potential is realized.Eq.(4.3) is rewritten as Fig. 1 The integration path Γ ε for q n .
The first term becomes and the second term leads to where Now we consider V 2 (r).To satisfy 0 ≤ q ≤ Λ c , the domain of integration is not .
Since the integrand is singular at q n = 0, we choose the anticlockwise path Γ ε in Fig. 1, and take the limit ε → 0. Then we obtain If the cut-off Λ c is replaced by m χ , V 2 (r) becomes the linear term in Eq.(1.2).
Finally, neglecting additive constants, we find V 3 (r) becomes (4.9) Thus the confining potential we propose is Λc 0 dq sin qr qr (4.10) In addition to the Coulomb plus linear part, there is the term V 3 (r).

Determination of parameters
Although we presented the potential V c (r) = 3 k=1 V k (r), the values of the parameters Q 2 and m are unknown.To determine them, let us expand a potential V (r) as We assume there is a true confining potential V T (r).Then we require the Cornell potential V CL (r) fits well to V T (r) at a point r = r c .This is achieved by choosing (K, σ) appropriately.We impose the two conditions1 and determine (K, σ).
Next, we require that the potential V c (r) fits well to this V CL (r) at r = r c , and use the conditions We note, to determine the parameters Q 2 and m, two conditions are necessary.However, to determine r c , the third condition is required.
From Eq.(5.2) with n = 2, we have r 3 c V ′′ CL (r c ) = r 3 c V ′′ c (r c ).Using Eqs.(1.1) and (4.10), this condition becomes In the same way, the condition and Of course, the true potential V T (r) is unknown.So, instead of the first step proposed in Eq.( 5.1), we choose appropriate values of K and σ.Thus we determine V c (r) as follows.
Choosing the values of the scale parameter Λ QCD and the off-diagonal gluon mass M, Eq.(3.6) determines the cut-off Λ c .Then, substituting the values of Λ c , K and σ into Eqs.(5.3)-(5.5), the quantities Q, m and r c are determined numerically.As an example, we choose the values We note that the off-diagonal gluon mass M ≃ 1.2 Gev in the region of r 0.2 fm was obtained by using SU(2) lattice QCD in the maximal Abelian gauge [16].The values K 0.
Thus we obtain Fig. 2 The potentials V 1 , V 2 , V 3 and The unit of r is fm, and the unit of the potentials is GeV.
If the unit of r is changed over from GeV −1 to fm, the potential becomes In Fig. 2, the potentials V k (r), (k = 1, 2, 3) and V c (r) are plotted.Since V 1 (r) + V 3 (r) is a substitute for the Yukawa potential, V 1 (r) + V 3 (r) ≈ 0 for r 0.35 fm is reasonable.Using the unit fm for r, V CL (r) with (K = 0.3, σ = 0.18) becomes (5.11) Eqs.(5.10) and (5.11) are plotted in Fig. 3. Since V c (r) is fitted to V CL (r) at r c ≃ 0.226 fm, they fit very well for 0.1 r 0.4.However, when r becomes large, as V 1 (r) + V 3 (r) ≈ 0 and σ c > σ, V c (r) > V CL (r) holds for r 0.4 fm.In the same way, K c < K leads to V c (r) > V CL (r) for r < 0.09 fm.6 The scales R 0 and R0 In Sect.5, the scale r c ≃ 0.226 fm appears.On the other hand, considering the force −V ′ (r), the intermediate scale R 0 , which satisfies Fig. 3 The potentials V c and V CL with (K = 0.3, σ = 0.18 GeV 2 ).Additive Constants are chosen to become was proposed [11].In successful potential models, this relation holds fairly well.For example, the V CL (r) with (K = 0.52, σ = 0.183) [3] gives R 0 = 0.49.If we substitute V c (r) in Eq.(5.8) into Eq.(6.1), we obtain where comes from V ′ 3 (r).Eq.( 6.2) gives the solution R 0 = 0.51 fm.We note, V CL (r) with (K = 0.3, σ = 0.18) in Sect. 5 gives the larger value R 0 = 0.54 fm.
To see the meanings of these scales, the effective Coulomb coupling 3) and the effective string tension σ ef (r) = σ c + V ′ 3 (r) + r 2 V ′′ 3 (r) in Eq.(5.4) are plotted in Fig. 4. We find K ef (r c ) = 0.3 is the maximal value and σ ef (r c ) = 0.18 is the minimal value.Namely r c is the position where K ef (r) is maximum and σ ef (r) is minimum.
In Fig. 5, r 2 V ′ c (r) = K ef (r) + r 2 σ ef (r) and r 2 σ ef (r) are plotted.We find that r 2 V ′ c (r) satisfies r 2 V ′ c (r) ≈ K ef (r) for r < 0.2 fm, and r 2 V ′ c (r) ≈ r 2 σ ef (r) for r 0.5 fm.Namely, the force-related quantity r 2 V ′ c (r) is almost saturated with the string part r 2 σ ef (r) above Fig. 4 The effective Coulomb coupling The unit of r is fm.
Before closing this section, based on the above analysis, we present the potential V CL (r) that fits to V c (r) better in the region of r > 0.6 fm.When r is small, the Coulomb part K ef (r) dominates.So keeping the condition Eq.(5.3) intact, we set K = 0.3.When r becomes large, the string part dominates.To determine the value of σ, it is reasonable to set the condition r 2 V ′ (r)| r= R0 = 2.13, R0 ≃ 0.58 fm.(6.4)We find V CL with (K = 0.3, σ = 2.12 GeV 2 ) satisfies Eq.(6.4).We note this V CL satisfies Eq.(6.1) as well.
Changing the unit of r to fm, this potentila becomes The potential V c (r) and V CL (r) in Eq.(6.5) are plotted in Fig. 6.As we explained in Sect.5, the behavior V c (r) > V CL (r) comes about for large r.However Eq.(6.5) fits fairly well for r < 1.2 fm.Fig. 6 The potentials V c and V CL with (K = 0.3, σ = 0.212 GeV 2 ).

7 Summary and comments
In this paper, we considered the SU(2) gauge theory, and studied the Q Q potential Eq.(2.12).This potential is derived under the gauge field condensation.In Refs.[5][6][7][8], the dual Ginzburg-Landau model, which describes the monopole condensation, leads to the potential.
In our approach [9,10], the ghost condensation v = 0 induces the VEV A + µ A − µ = v/(64π).If we divide the diagonal gluon as A 3 µ = a µ + b µ , the classical part b µ acquires the mass m = g v/(32π), whereas the quantum part a µ is massless, The off-diagonal gluons A ± µ acquire the mass M through Eq.(2.7).As the classical solution b µ , we choose the monopole solution of the dual gauge field B µ [10].Then the propagator of B µ leads to the Q Q potential Eq.(2.12).
In calculating Eqs.(2.7) and (2.12), ultraviolet cut-off is necessary.Since the masses M and m are related, we introduced a single cut-off Λ c so as to m vanishes and the potential V (r) becomes the Coulomb potential in the limit Λ c → 0. Then Eq.(2.7) and Eq.(2.12) become Eq.(3.5) and Eq.(4.10), respectively.The confining potential under the VEV A + µ A − µ is V c (r) in Eq.(4.10).It has the linear potential V 2 (r) and, instead of the Yukawa potential, the Coulomb potential V 1 (r) and the additional term V 3 (r).
Although we derived V c (r), there are unknown parameters.To determine them, using Eq.(3.6) and the values of v and M in Eq.(5.6), the cut-off Λ c = 2.03 GeV was obtained.Next, assuming that the Cornell potential V CL (r) with (K = 0.3, σ = 0.18) describes a true potential well at some point r c , we required V c (r) ≃ V CL (r) near r = r c .To realize this requirement, we imposed the conditions in Eq.(5.2).By solving these conditions, r c ≃ 0.226 fm, the values of m and Q 2 in Eq.(5.7), and V c (r) in Eq.(5.8) were obtained.
There are two implicit scales r c and R 0 (or R0 ) in V c (r).To understand them, the effective Coulomb coupling K ef (r) in Eq.(5.3) and the effective string tension σ ef (r) in Eq.(5.4) were studied.Since V 3 (r) contributes to them, they depend on r.At r = r c , K ef (r) becomes maximum and σ ef (r) becomes minimum.For r > R 0 ≃ 0.5 fm, 0.23 < σ ef (r) < 0.25 holds, and K ef vanishes at R0 ≃ 0.58 fm.
Namely the main force between Q and Q is the effective Coulomb force −K ef /r 2 for r < r 0 , and the effective string force −σ ef (r) for r > R 0 .Although V c (r) was determined to fit to V CL (r) with (K = 0.3, σ = 0.18) at r = r c , it becomes larger than V CL (r) for r > 0.4 fm.The Cornell potential is often used to fit lattice Using these values, Eqs.(B1) and (B2) give In the same way, using S(r, a) in Eq.(5.9),Eq.(4.9) becomes
). Namely, although the quantum part a µ is massless, the classical part b µ and the off-diagonal components A ± µ are massive.The mass M, that is introduced as the source term of the local composite operator A +