Confining potential under the gauge field condensation in the SU(2) Yang-Mills theory

In the study of quarkonia, QCD potential is often used. Although there are some phenomenological potentials (see, e.g., bali ), the Cornell potential $V_{CL}(r)$ cor ; cor2 is simple but workable. This potential has the Coulomb plus linear form as

$$V_{CL}(r)=-\frac{K}{r}+\sigma r,$$

(1.1)

where $K$ and $\sigma$, 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., rip ), the following Yukawa plus linear potential was obtained suz ; mts ; sst ; sst2 :

$$V_{YL}(r)=-\frac{Q^{2}}{4\pi}\frac{e^{-mr}}{r}+\left(\frac{Q^{2}m^{2}}{8\pi}\ln\frac{m^{2}+m_{\chi}^{2}}{m^{2}}\right)r,$$

(1.2)

where $Q$ is the static quark charge and $m_{\chi}$ 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. hs , based on the SU(2) gauge theory in the non-linear gauge of the Curci-Ferrari type, we also derived the potential $V_{YL}(r)$. In this case, the mass $m$ comes from the gauge field condensation $\langle A_{\mu}^{+}A_{\mu}^{-}\rangle$.

In this paper, in the framework of Refs. hs ; hs1 , we restudy the confining potential. In the next section, we briefly review Refs. hs ; hs1 , and present the potential between the static charges $Q$ and $\bar{Q}$. In Sect. 3, the equation to determine an ultraviolet cut-off $\Lambda_{c}$ is derived. In Sect. 4, using this cut-off, we show that the $Q\bar{Q}$ 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 $\sigma$, the parameters in $V_{c}(r)$ are determined in Sect. 5. In this process, we find a scale $r_{c}\approx 0.2\ \mathrm{fm}$. In the intermediate region, the scale $R_{0}\approx 0.5\ \mathrm{fm}$ has been proposed so . The meanings of the scales $r_{c}$ and $R_{0}$ for $V_{c}(r)$ are clarified in Sect. 6. We also propose a scale $\tilde{R}_{0}$ 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.

In this section, we review Refs. hs ; hs1 briefly.

In this section, we study Eq.(2.10). The free propagator $G_{\mu\nu}^{(0)}(p)$ is calculated in Appendix A as

$$G_{\mu\nu}^{(0)}(p)=\frac{1}{p^{2}+M^{2}}P^{T}_{\mu\nu}+\frac{1}{M^{2}}P^{L}_{\mu\nu},\quad P^{T}_{\mu\nu}=\delta_{\mu\nu}-\frac{p_{\mu}p_{\nu}}{p^{2}},\quad P^{L}_{\mu\nu}=\frac{p_{\mu}p_{\nu}}{p^{2}}.$$

(3.1)

Assuming that $M$ exists below a cut-off $\Lambda_{c}$, we obtain

$$G_{\mu\mu}^{(0)}(x,x)=\int_{0}^{\Lambda_{c}}\frac{d^{4}p}{(2\pi)^{4}}G_{\mu\mu}^{(0)}(p)=\frac{1}{(4\pi)^{2}}\left\{\frac{\Lambda_{c}^{4}}{2M^{2}}-3M^{2}\ln\left(1+\frac{\Lambda_{c}^{2}}{M^{2}}\right)\right\},$$

(3.2)

where the $M$-independent term $\Lambda_{c}^{2}$ is subtracted.

The VEV $v$ depends on the momentum scale $\mu$. From Eqs.(2.3) and (2.4), we find $v$ disappears above the scale $\mu_{0}=\Lambda_{\mathrm{QCD}}$. When $0\leq\mu\leq\Lambda_{\mathrm{QCD}}$, $v$ behaves as

$$v\to 0\quad(\mu\to\Lambda_{\mathrm{QCD}}),\quad v\to\Lambda_{\mathrm{QCD}}^{2}\quad(\mu\to 0).$$

Namely the maximal value of $v$ is $\Lambda_{\mathrm{QCD}}^{2}$, and the left hand side of Eq.(2.10) satisfies

$$0\leq\frac{v}{64\pi}\leq\frac{\Lambda_{\mathrm{QCD}}^{2}}{64\pi}.$$

On the other hand, the VEV $G_{\mu\mu}^{(0)}(x,x)$ only depends on the constants $M$ and $\Lambda_{c}$. Since the tachyonic mass should be removed in the overall momentum region completely, using the maximal value of $v$, we interpret Eq.(2.10) as ²²2If we consider the scale dependent $M$, Eq.(3.2) should be replaced by the integral $$\int_{0}^{\Lambda_{c}}\frac{d^{4}p}{(2\pi)^{4}}\left\{\frac{P_{\mu\nu}^{T}}{p^{2}+M(p)^{2}}+\frac{P_{\mu\nu}^{L}}{M(p)^{2}}\right\}.$$ As $M(p)$ is unknown, we cannot calculate it. But it is also $\mu$-independent.

$$\frac{\Lambda_{\mathrm{QCD}}^{2}}{64\pi}\simeq G_{\mu\mu}^{(0)}(x,x).$$

(3.3)

From Eqs.(3.2) and (3.3), we obtain

$$\frac{\pi}{2}\frac{\Lambda_{\mathrm{QCD}}^{2}}{M^{2}}\simeq\frac{\Lambda_{c}^{4}}{M^{4}}-6\ln\left(1+\frac{\Lambda_{c}^{2}}{M^{2}}\right).$$

(3.4)

We make a comment. Eq.(3.4) determines the relation of $M$ and $\Lambda_{c}$ so as to give the maximal value of $v$. Using Eq.(3.3), we find $m^{2}=2g^{2}G_{\mu\mu}^{(0)}(x,x)\simeq g^{2}\Lambda_{\mathrm{QCD}}^{2}/(32\pi)$. Although the VEV $v$ depends on the scale $\mu$, and disappears above $\Lambda_{\mathrm{QCD}}$, $G_{\mu\mu}^{(0)}(x,x)$ does not. So the mass $m$ can survive above the scale $\Lambda_{\mathrm{QCD}}$.

Usually, the potential $V$ in Eq.(2.15) is calculated as follows. Let us divide the momentum $q$ in $V_{L}$ into $q_{n}=\mbox{\boldmath$q$}\cdot\mbox{\boldmath$n$}$ and $\mbox{\boldmath$q$}_{T}$ that satisfies $\mbox{\boldmath$q$}_{T}\cdot\mbox{\boldmath$n$}=0$. Then the integral of $q_{n}$ has the infrared divergence, and the integral of $q_{T}=|\mbox{\boldmath$q$}_{T}|$ has the ultraviolet divergence. The former divergence is removed by the choice $\mbox{\boldmath$r$}\parallel\mbox{\boldmath$n$}$ sst ; hs , and the latter divergence is avoided by the cut-off $m_{\chi}$ as suz ; mts ; sst ; sst2

$$V_{L}=\frac{Q^{2}m^{2}}{(2\pi)^{2}}\int_{-\infty}^{\infty}dq_{n}\int_{0}^{m_{\chi}}dq_{T}q_{T}\frac{1-\cos q_{n}r}{(q_{n}^{2}+q_{T}^{2}+m^{2})q_{n}^{2}}.$$

(4.1)

Eq.(4.1) becomes the linear term in Eq.(1.2). The term $V_{Y}$ has the integral of $q=|\mbox{\boldmath$q$}|$ over the region of $0\leq q<\infty$, and the Yukawa potential in Eq.(1.2) is obtained.

To introduce a cut-off in a different way, we write the potential as

$$V(r)=\int dqW(\mbox{\boldmath$q$},m,r).$$

(4.2)

As we stated in Sect. 3, the mass $m$ can exist above $\Lambda_{\mathrm{QCD}}$. From Fig. 1(c), after $v$ disappears, it is expected that $m^{2}$ contributes to the correction for $M^{2}$. Since $A_{\mu}^{\pm}$ are considered to be massless above $\Lambda_{c}$, we assume that the cut-off for $m$ is $\Lambda_{c}$, and define Eq.(4.2) as

$$V(r)=\int_{0}^{\Lambda_{c}}dqW(\mbox{\boldmath$q$},m,r)+\int_{\Lambda_{c}}^{\infty}dqW(\mbox{\boldmath$q$},0,r).$$

(4.3)

Eq.(4.3) is rewritten as

$$V(r)=\int_{0}^{\infty}dqW(\mbox{\boldmath$q$},0,r)+\int_{0}^{\Lambda_{c}}dq\left\{W(\mbox{\boldmath$q$},m,r)-W(\mbox{\boldmath$q$},0,r)\right\}.$$

(4.4)

The first term becomes

$$\int_{0}^{\infty}dqW(\mbox{\boldmath$q$},0,r)=V_{1}(r)=Q^{2}\int_{D_{\infty}}\frac{d^{3}q}{(2\pi)^{3}}\frac{1-\cos\mbox{\boldmath$q$}\cdot\mbox{\boldmath$r$}}{q^{2}},\quad D_{\infty}=\{q\ |\ 0\leq q<\infty\},$$

(4.5)

and the second term leads to

	$\displaystyle\int_{0}^{\Lambda_{c}}dq\left\{W(\mbox{\boldmath$q$},m,r)-W(\mbox{\boldmath$q$},0,r)\right\}=V_{2}(r)+V_{3}(r),$
	$\displaystyle V_{2}(r)=Q^{2}\int_{D_{\Lambda_{c}}}\frac{d^{3}q}{(2\pi)^{3}}(1-\cos\mbox{\boldmath$q$}\cdot\mbox{\boldmath$r$})\frac{m^{2}}{(q^{2}+m^{2})q_{n}^{2}},$
	$\displaystyle V_{3}(r)=-Q^{2}\int_{D_{\Lambda_{c}}}\frac{d^{3}q}{(2\pi)^{3}}(1-\cos\mbox{\boldmath$q$}\cdot\mbox{\boldmath$r$})\frac{m^{2}}{(q^{2}+m^{2})q^{2}},\quad D_{\Lambda_{c}}=\{q\ |\ 0\leq q<\Lambda_{c}\},$		(4.6)

where $V_{2}(r)$ ($V_{3}(r)$) comes from $V_{L}$ ($V_{Y}(m^{2})-V_{Y}(m^{2}=0)$). Eq.(4.5) gives the usual Coulomb potential

$$V_{1}(r)=-\frac{K_{c}}{r},\quad K_{c}=\frac{Q^{2}}{4\pi}.$$

(4.7)

Now we consider $V_{2}(r)$. To satisfy $0\leq q\leq\Lambda_{c}$, the domain of integration is not $(-\infty<q_{n}<\infty,0\leq q_{T}\leq m_{\chi})$ in Eq.(4.1), but $(-\varepsilon\leq q_{n}\leq\varepsilon,0\leq q_{T}\leq\sqrt{\Lambda_{c}^{2}-\varepsilon^{2}})$ with $0<\varepsilon\ll 1$, i.e.,

$$V_{2}(r)=\frac{Q^{2}m^{2}}{(2\pi)^{2}}\int_{0}^{\sqrt{\Lambda_{c}^{2}-\varepsilon^{2}}}dq_{T}q_{T}\int_{-\varepsilon}^{\varepsilon}dq_{n}\frac{1-e^{iq_{n}r}}{q_{n}^{2}(q_{n}^{2}+q_{T}^{2}+m^{2})}.$$

Since the integrand is singular at $q_{n}=0$, we choose the anticlockwise path $\Gamma_{\varepsilon}$ in Fig. 2, and take the limit $\varepsilon\to 0$. Then we obtain

$$V_{2}(r)=\frac{Q^{2}m^{2}}{(2\pi)^{2}}\int_{0}^{\Lambda_{c}}dq_{T}q_{T}\frac{\pi r}{q_{T}^{2}+m^{2}}=\sigma_{c}r,\quad\sigma_{c}=\frac{Q^{2}m^{2}}{8\pi}\ln\left(\frac{\Lambda_{c}^{2}+m^{2}}{m^{2}}\right).$$

(4.8)

If the cut-off $\Lambda_{c}$ is replaced by $m_{\chi}$, $V_{2}(r)$ becomes the linear term in Eq.(1.2).

Finally, neglecting additive constants, we find $V_{3}(r)$ becomes

$$V_{3}(r)=\frac{Q^{2}m^{2}}{2\pi^{2}}\int_{0}^{\Lambda_{c}}dq\frac{\sin qr}{qr}\frac{1}{q^{2}+m^{2}}.$$

(4.9)

Thus the confining potential we propose is

$$V_{c}(r)=\sum_{k=1}^{3}V_{k}(r)=-\frac{K_{c}}{r}+\sigma_{c}r+\frac{Q^{2}m^{2}}{2\pi^{2}}\int_{0}^{\Lambda_{c}}dq\frac{\sin qr}{qr}\frac{1}{q^{2}+m^{2}}.$$

(4.10)

In addition to the Coulomb plus linear part, there is the term $V_{3}(r)$.

Although we presented the potential $V_{c}(r)=\sum_{k=1}^{3}V_{k}(r)$, the values of the parameters $Q^{2}$ and $m$ are unknown. To determine them, let us expand a potential $V(r)$ as

$$V(r)=V(r_{c})+V^{\prime}(r_{c})(r-r_{c})+\frac{V^{\prime\prime}(r_{c})}{2}(r-r_{c})^{2}+\frac{V^{(3)}(r_{c})}{3!}(r-r_{c})^{3}+\cdots.$$

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 $(K,\sigma)$ and the constant $V(r_{c})$ appropriately. ³³3To determine the three parameters, it is natural to choose appropriate three points $r_{k}\ (k=1,2,3)$. However, in the next step, we must determine $V_{c}$ so as to the difference between $V_{c}$ and $V_{CL}$ becomes minimum. Since $V_{3}$ in $V_{c}$ contains $r$ in the integrand, it is difficult to determine $r_{k}$. We impose the conditions

$$V_{T}^{(n)}(r_{c})=V_{CL}^{(n)}(r_{c}),\quad(n=0,1,2),$$

(5.1)

and determine $(K,\sigma)$ and $V(r_{c})$.

Next, we require that the potential $V_{c}(r)$ fits well to this $V_{CL}(r)$ at $r=r_{c}$, and use the conditions

$$V_{CL}^{(n)}(r_{c})=V_{c}^{(n)}(r_{c}),\quad(n=0,1,2,3).$$

(5.2)

We note, to determine the parameters $Q^{2}$ and $m$, the two conditions with $n=1,2$ are necessary. However, to determine $r_{c}$, the condition with $n=3$ is required.

From Eqs.(1.1) and (4.10), we have

	$\displaystyle-\frac{r^{3}}{2}V_{CL}^{\prime\prime}(r)=K,\quad\frac{1}{2r}\left(r^{2}V_{CL}^{\prime}(r)\right)^{\prime}=\sigma,$
	$\displaystyle-\frac{r^{3}}{2}V_{c}^{\prime\prime}(r)=K_{c}-\frac{r^{3}}{2}V_{3}^{\prime\prime}(r),\quad\frac{1}{2r}\left(r^{2}V_{c}^{\prime}(r)\right)^{\prime}=\sigma_{c}+\frac{1}{2r}\left(r^{2}V_{3}^{\prime}(r)\right)^{\prime}.$

Since Eq.(5.2) with $n=2$ gives $r_{c}^{3}V_{CL}^{\prime\prime}(r_{c})=r_{c}^{3}V_{c}^{\prime\prime}(r_{c})$, this condition becomes

$$K=K_{ef}(r_{c}),\quad K_{ef}(r_{c})=K_{c}-\frac{r_{c}^{3}}{2}V_{3}^{\prime\prime}(r_{c}).$$

(5.3)

In the same way, the condition $\displaystyle\left(r_{c}^{2}V_{CL}^{\prime}(r_{c})\right)^{\prime}=\left(r_{c}^{2}V_{C}^{\prime}(r_{c})\right)^{\prime}$ leads to

$$\sigma=\sigma_{ef}(r_{c}),\quad\sigma_{ef}(r_{c})=\sigma_{c}+\frac{1}{2r_{c}}\left(r_{c}^{2}V_{3}^{\prime}(r_{c})\right)^{\prime},$$

(5.4)

and $\displaystyle\left(r_{c}^{3}V_{CL}^{\prime\prime}(r_{c})\right)^{\prime}=\left(r_{c}^{3}V_{c}^{\prime\prime}(r_{c})\right)^{\prime}$ becomes

$$\left(r_{c}^{3}V_{3}^{\prime\prime}(r_{c})\right)^{\prime}=0.$$

(5.5)

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 $\sigma$. Thus we determine $V_{c}(r)$ as follows. Choosing the values of the scale parameter $\Lambda_{\mathrm{QCD}}$ and the off-diagonal gluon mass $M$, Eq.(3.4) determines the cut-off $\Lambda_{c}$. Then, substituting the values of $\Lambda_{c}$, $K$ and $\sigma$ into Eqs.(5.3)-(5.5), the quantities $Q$, $m$ and $r_{c}$ are determined numerically. As an example, we choose the values

$$\Lambda_{\mathrm{QCD}}=0.2\ \mathrm{GeV},\quad M=1.2\ \mathrm{GeV},\quad K=0.3,\quad\sigma=0.18\ \mathrm{GeV}^{2}.$$

(5.6)

We note that the off-diagonal gluon mass $M\simeq 1.2$ GeV in the region of $r\gtrsim 0.2$ fm was obtained by using SU(2) lattice QCD in the maximal Abelian gauge as . The values $K\lesssim 0.3$ and $\sigma\lesssim 0.2$ GeV² come from lattice simulations bali .

Now using the values of $\Lambda_{\mathrm{QCD}}$ and $M$ in Eq.(5.6), Eq.(3.4) gives $\Lambda_{c}\simeq 2.03$ GeV. Next, we substitute this $\Lambda_{c}$ and $(K,\ \sigma)$ in Eq.(5.6) into Eqs.(5.3)-(5.5), and solve these equations. The details are explained in Appendix B. The results are

$$r_{c}\simeq 1.145\ \mathrm{GeV}^{-1}=0.226\ \mathrm{fm},\quad a=m^{2}/\Lambda_{c}^{2}\simeq 0.263,$$

and these values lead to

$$m=1.04\ \mathrm{GeV},\quad K_{c}=\frac{Q^{2}}{4\pi}=0.285,\quad\sigma_{c}=\frac{Q^{2}m^{2}}{8\pi}\ln\left(\frac{\Lambda_{c}^{2}+m^{2}}{m^{2}}\right)=0.242\ \mathrm{GeV}^{2}.$$

(5.7)

Thus we obtain

$$V_{c}(r)=\sum_{k=1}^{3}V_{k}(r),\quad V_{1}(r)=-\frac{0.285}{r},\quad V_{2}(r)=0.242\cdot r,\quad V_{3}(r)=0.747\cdot S(r,0.263),$$

(5.8)

where

$$S(r,a)=\int_{0}^{\Lambda_{c}r}dx\frac{\sin x}{x}\frac{ar}{x^{2}+a(\Lambda_{c}r)^{2}}.$$

(5.9)

In Fig.3, 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, $V_{1}(r)+V_{3}(r)\approx 0$ for $r\gtrsim 0.35\ \mathrm{fm}$ is reasonable. Using the values of $(K,\ \sigma)$ in Eq.(5.6), $V_{CL}(r)$ becomes

$$V_{CL}(r)=-\frac{0.3}{r}+0.18\cdot r.$$

(5.10)

Eqs.(5.8) and (5.10) are plotted in Fig. 4. Since $V_{c}(r)$ is fitted to $V_{CL}(r)$ at $r_{c}\simeq 0.226\ \mathrm{fm}$, they fit very well for $0.1\ \mathrm{fm}\lesssim r\lesssim 0.4\ \mathrm{fm}$. However, when $r$ becomes large, as $V_{1}(r)+V_{3}(r)\approx 0$ and $\sigma_{c}>\sigma$, $V_{c}(r)>V_{CL}(r)$ holds for $r\gtrsim 0.4\ \mathrm{fm}$. In the same way, $K_{c}<K$ leads to $V_{c}(r)>V_{CL}(r)$ for $r<0.09\ \mathrm{fm}$.

In Sect. 5, the scale $r_{c}\simeq 0.226\ \mathrm{fm}$ appears. On the other hand, considering the force $-V^{\prime}(r)$, the intermediate scale $R_{0}$, which satisfies

$$r^{2}V^{\prime}(r)|_{r=R_{0}}=1.65,\quad R_{0}\simeq 0.5\ \mathrm{fm},$$

(6.1)

was proposed so . In successful potential models, this relation holds fairly well. For example, the $V_{CL}(r)$ with $(K=0.52,\ \sigma=0.183\ \mathrm{GeV}^{2})$ cor2 gives $R_{0}=0.49\ \mathrm{fm}$. If we substitute $V_{c}(r)$ in Eq.(5.8) into Eq.(6.1), we obtain

$$K_{ef}(R_{0})+R_{0}^{2}\sigma_{ef}(R_{0})=0.285+0.242R_{0}^{2}+0.747R_{0}^{2}\cdot H(R_{0},0.263)=1.65,$$

(6.2)

where

$$H(r,a)=\int_{0}^{\Lambda_{c}r}dx\left(\cos x-\frac{\sin x}{x}\right)\frac{a}{x^{2}+a(\Lambda_{c}r)^{2}}$$

(6.3)

comes from $V_{3}^{\prime}(r)$. Eq.(6.2) gives the solution $R_{0}=0.51\ \mathrm{fm}$. We note, $V_{CL}(r)$ with $(K=0.3,\ \sigma=0.18\ \mathrm{GeV}^{2})$ in Sect. 5 gives the larger value $R_{0}=0.54\ \mathrm{fm}$.