Stability of asymmetric tetraquarks in the minimal-path linear potential

The linear potential binding a quark and an antiquark in mesons is generalized to baryons and multiquark configurations as the minimal length of flux tubes neutralizing the color, in units of the string tension. For tetraquark systems, i.e., two quarks and two antiquarks, this involves the two possible quark--antiquark pairings, and the Steiner tree linking the quarks to the antiquarks. A novel inequality for this potential demonstrates rigorously that within this model the tetraquark is stable in the limit of large quark-to-antiquark mass ratio.

( v 1 s 1 + v 2 s 1 + s 1 s 2 + s 2 v 3 + s 2 v 4 ) . (2) The first two terms of U describe the two possible quark-antiquark links, and their minimum is sometimes referred to as the "flip-flop" model, schematically pictured in Fig. 1(b). It was introduced by Lenz et al. [11], who used, however, a quadratic instead of linear rise of the potential as a function of the distance. The last term, V 4 , is represented in Fig. 1(c) and corresponds to a connected flux tube. It is given by a Steiner tree, i.e., it is minimized by varying the location of the Steiner points s 1 and s 2 . The choice of this potential is inspired by Refs. [3,12,13,14], and has been discussed in the context of lattice QCD [15,16]. The four-body problem in quantum mechanics is notoriously difficult. For instance, Wheeler proposed in 1945 the existence of a positronium molecule (e + , e + , e − , e − ) which is stable in the limit where internal annihilation is neglected, i.e., lies below its threshold for dissociation into two positronium atoms. In 1946, Ore published a four-body calculation of this system [17] and concluded that his investigation "counsels against the assumption that clusters of this (or even of higher) complexity can be formed". However, in 1947, Hylleraas and the same Ore published an elegant analytic proof that this molecule is stable [18]. It has been discovered recently [19].
Similarly, the above model (2), in its linear version, was considered by Carlson and Pandharipande, who entitled their paper [20] "Absence of exotics in the flux tube model", i.e., did not find stable tetraquarks 1 . However, Vijande et al. [21] used a more systematic variational expansion of the wave function and in their numerical solution of the four-body problem found a stable tetraquark ground state. Moreover, unlike [20], they considered the possibility of unequal masses, and found that stability improves if the quarks are heavier (or lighter) than the antiquarks, in agreement with previous investigations (see, e.g., [21] for Refs.).
It is thus desirable to check whether this minimal-path model supports or not bound states. The present attempt is based on an upper bound on the potential, which leads to an exactly solvable four-body Hamiltonian.
With the Jacobi vector coordinates and their conjugate momenta, the relative motion is described by the Hamiltonian where µ, given by µ −1 = m −1 + M −1 , is the quark-antiquark reduced mass. Using the scaling properties of H, one can set m = 1 without loss of generality. The simplest bound on the potential U is as the tree with optimized Steiner points s 1 and s 2 is shorter than if the junctions are set at the middles of the quark separation v 1 v 2 and antiquark separation v 3 v 4 . This leads to a separable upper bound for the Hamiltonian Now, the ground state e 0 of p 2 x + x corresponds to the radial equation −u (r) + ru(r) = e 0 u(r) with u(0) = u(∞) = 0 and is the negative of the first zero of the Airy function, e 0 = 2.3381 . . . By scaling, the ground state of αp 2 x + β x , with α > 0 and β > 0 is α 1/3 β 2/3 e 0 . Thus the lowest eigenvalue of H is with µ = M/(1 + M ). By comparison, the threshold of (QQqq) is made of two identical (Qq) mesons, each governed by the Hamiltonian h = p 2 /(2µ) + r , where p is conjugate to the quark-antiquark separation r. Thus the threshold energy is and it is easily seen that E > E th for any value of the quark-to-antiquark mass ratio M , i.e., the bound (5) cannot demonstrate binding.
A better bound will be proved below. If there is a genuine Steiner tree 2 linking the quarks to the antiquarks, then But if V 4 is not associated to a genuine Steiner tree, this inequality is often violated. Consider for instance a rectangular configuration with ||v 1 v 2 || = ||v 3 v 4 || ||v 1 v 3 || = ||v 2 v 4 || (in this case the mathematical Steiner tree problem would require a Steiner point linking v 1 and v 3 , another Steiner point linking v 2 and v 4 , but the corresponding fluxes are not permitted by the color coupling in QCD), then z ∼ 0 and V 4 ∼ x + y , so (9) does not hold. However, it will be shown that for any configuration of the quarks and antiquarks, i.e., for any x, y and z. Then the ground state of H is bounded as As shown in Fig. 2, this bound E significantly improves the previous one, E . It is easily seen than E becomes smaller than E th for very large values of the mass ratio, more precisely for M > 6402, and thus that the tetraquark is bound at least in this range of M . The numerical estimate of [21] actually indicates stability for all values of M , even M = 1. To summarize, we obtained an analytic upper bound on the ground state energy of tetraquarks systems with two units of open flavor, (QQqq), using a model of linear confinement inspired by the strong-coupling regime of QCD. The key is an inequality on the length of a Steiner tree with four terminals. The bound confirms a recent numerical investigation, in which this potential was shown to bind these tetraquarks below the threshold for dissociation into two mesons. It remains to investigate whether this stability survives refinements in the dynamics, such as short range corrections, spin-dependent forces, etc.
It is our intention to extend this investigation to the case of the pentaquark (one antiquark and four quarks) and hexaquark configurations (six quarks), which have been much debated in recent years.
Three terminals The three-point problem is very much documented in textbooks [22,23,24,25,26]. Let v 1 v 2 v 3 be the triangle, with side lengths a 1 = ||v 2 v 3 ||, . . . and angles α 1 = ∠v 1 v 2 v 3 , etc. The problem of finding a path of minimal length sv 1 + sv 2 + sv 3 linking the three vertices has been solved by Fermat and Torricelli. See, e.g., [22]. The result is the following: if one of the angles, say α 1 , is larger than 120 • , s coincides with v 1 , otherwise each side of the triangle is seen from s with an angle of 120 • . The Steiner point s is thus at the intersection of three arcs of circles, see Fig. 3(a).
The three-terminal problem is also linked to Napoleon's theorem, which states that if one draws external equilateral triangles on each side, v 1 v 2 w 3 , v 1 v 3 w 1 and v 1 v 1 w 2 , the centers of these triangles form an equilateral triangle (dashed lines in Fig. 3(b)), a nice example of symmetry restoration. The junction s is just the intersection of v 1 w 1 , v 1 w 2 and v 3 . Note that sv 1 + sv 2 = sw 3 , and similar relations, and thus the potential is simply The point w 3 and its symmetric with respect to v 1 v 2 , t 3 form the toroidal domain associated to the subset {v 1 , v 2 }. The length of the minimal Steiner tree is the maximal distance between v 3 and the domain {w 3 , t 3 }.
FIG. 3: Left: the junction lies on each arc from which a side is seen under 120 • . Right: the three-terminal Steiner problem as a side product of Napoleon's theorem From the above properties, one can estimate the string potential in a closed form. If α 1 ≥ 120 • , then V 3 = a 2 + a 3 , and similarly for large , and by summation Now, 1 2 being four times the area of the sv 1 v 2 triangle, the second term in the above equation is four times the whole area of v 1 v 2 v 3 , which is given by the Henon theorem. Altogether, in the case of a genuine Steiner tree [7] which can be computed quickly.
The planar tetraquark problem. For the four-point problem, there are many special cases, which can be treated by inspection. If, for instance the quark v 2 is on the back of v 1 , as in Fig. 4(a), the problem reduces to the Steiner problem for {v 1 , v 3 , v 4 }. Another special case is shown in Fig. 4(b), where the quarks are close to the antiquarks. For the standard Steiner problem of geometry, the solution would correspond to the Steiner tree shown as a dotted line, with a Steiner point s 3 linked to v 1 and v 3 and another one, s 4 , linked to v 2 and v 4 . This is not allowed by the different color properties of quarks and antiquarks, hence our best tree, shown as a solid line, has only one junction. But in estimating the potential U of Eq. (2) for this configuration, the minimum is the flip-flop term d 13 + d 24 .
Let us turn to the case of a genuine Steiner tree (v 1 v 2 )s 1 s 2 (v 3 v 4 ) as in Fig. 5. The string of Fig. 1(c) is minimized with respect to s 1 and s 2 . Hence for fixed s 2 , it assumes the Fermat-Torricelli minimization of v 1 v 2 s 1 , a well-known iteration property of Steiner trees. Hence ∠v 1 s 1 v 2 = 120 • and v 1 v 2 is the bissector of ∠v 1 s 1 v 2 and passes through the point w 12 which completes an equilateral triangle v 1 v 2 w 12 in the quark sector. Similarly, it also passes through w 34 which makes v 3 v 4 w 34 equilateral in the antiquark sector.
The junction points s 1 and s 2 are just the other intersections of the straight line w 12 w 34 with the circumcircles of v 1 v 2 w 12 and v 3 v 4 w 34 , as shown in Fig. 5. There is a possible ambiguity about on which side s 1 or s 2 should be, but this is easily solved by the requirement that the total length of the string is minimum. Crucial is the observation that V = w 12 w 34 , so that the determination of the Steiner points s 1 and s 2 is not required to compute V 4 .
A variant is that is t 12 is the symmetric of w 12 with respect to v 1 v 2 , the set {w 12 t 12 } is the toroidal domain associated to the quarks, and similarly {w 34 t 34 } for the antiquarks, the length of the Steiner trees is the maximal distance between these two sets. This construction, which is a special case of the Melzak's algorithm [27], leads to a very easy computation. If each vector v i is identified with its affix (complex number) v i , etc., then those of w 12 and w 34 are easily deduced, for instance w 12 = −j 2 v 1 −jv 2 or −jv 1 − j 2 v 2 (depending on which side is w 12 ), if one uses the familiar root of unity j = exp(2iπ/3). Once w 12 and w 34 are determined, V = w 12 w 34 . If one wishes to locate the Steiner points, it is sufficient to remark that w 12 s 2 .w 12 w 34 = w 12 c 34 2 − r 2 34 and w 34 s 1 .w 34 w 12 = w 34 c 12 2 − r 2 12 , where c 12 is the center of the circle v 1 v 2 w 12 and r 12 = d 12 √ 3/2 its radius and c 34 and r 34 are defined similarly in the antiquark sector.
The spatial tetraquark problem In general, the four constituents do not belong to the same plane. The minimum is achieved for v 1 v 2 s 1 s 2 coplanar, and v 3 v 4 s 1 s 2 also coplanar, but in a different plane. The toroidal domain to which the point w 12 belongs is the Melzak circle, of axis v 1 v 2 and radius r 12 = v 1 v 2 √ 3/2, and similarly for w 34 in the antiquark sector. The straight line w 12 w 34 has to intersect these two circles as well as the lines v 1 v 2 and v 3 v 4 . The problem consists of constructing such a straight line. The reasoning can be made on Fig. 5, if one imagines that v 3 v 4 s 2 is not coplanar to v 1 v 2 s 1 . As stressed in [28], the key is to determine p and q, the intersections of s 1 s 2 with v 1 v 2 and v 3 v 4 , respectively. In this paper, the following coupled equations are derived for the abscissa x p of p along v 1 v 2 and x q of q along v 3 v 4 . These abscissas are from the common perpendicular uv to v 1 v 2 and v 3 v 4 (u ∈ v 1 v 2 and v ∈ v 3 v 4 ), with uv = h, uh = m and vk = n, where h is the middle of v 1 v 2 and k that of v 3 v 4 . The equations (14) can be solved by iterations, with remarkably fast convergence. Once x p and x q , i.e., p and q, are determined, the Steiner points are determined by imposing they are on the circles v 1 v 2 w 12 and v 3 v 4 w 34 , respectively. For instance, if s 1 = p + t(q − p), t obeys a second order equation 3 .
If one is interested only in the length of the Steiner tree and not in the position of the Steiner points, an alternative formalism consists of locating p through p = h + x (v 2 − h) and q = k + y (v 4 − k). With this notation, the length of the tree is simply which is easily minimized by varying x and y. The minimisation is equivalent to solving the coupled equations which expresses that w 12 , p, s 1 , s 2 , q and w 34 are collinear. These equations are easily solved by iteration or any other means. We believe that, besides checking the particular cases with large angles or a single Steiner point, the fastest computation of the connected four-quark potential consists of minimising (15) or solving (16). We expect a dramatic improvement in computing time from the above algorithm.
However, it is aesthetically appealing to attempt a further reduction of the number of variables to be determined numerically, and to provide an almost analytic estimate of the interaction as a function of the coordinates of the quarks and antiquarks. Finding V 4 = w 12 w 34 , the maximal distance between the Melzak circles C 12 and C 34 , is very similar to the problem of the minimal distance beween two circles in space, as addressed e.g., in [29,30]. Neff [29] has shown that with the help of Lagrange multipliers and Gröbner type of elimination performed by computer-algebra sofware, the squared stationary distance V 2 4 obeys an eighth-order polynomial equation whose coefficients are rational functions of the coordinates of v 1 , v 2 , v 3 and v 4 .
Eberly [30] showed that if m is associated to an angle θ along C 12 , and n to φ along C 34 , then imposing mn 2 to be stationary, results in two equations of the type where α i , β i and γ i contain constants and terms linear in cos φ and sin φ. Solving (17) as two linear equations, as if cos θ and sin θ were independent, and then imposing cos θ 2 + sin θ 2 = 1 gives an equation for cos φ and sin φ, which is transformed into an 8 th order equation in cos φ. It is slightly faster to rewrite (17) using t = tan(θ/2) and u = tan(φ/2) as where the coefficients are quadratic in u. The compatibility of two such equations is simply and is directly a polynomial in u, of order 8. The confining potential V4 for the tetraquark system (v1v2v3v4) is the minimal length of the tree v1s1 + v2s1 + s1s2 + s2v3 + s2v4 when s1 and s2 are varied. It is also the maximal distances between the circles C12 and C34, i.e., the distance w12w34. The Melzak circle C12 is centered at the middle of v1 and v2, has v1v2 as axis and a radius v1v2 √ 3/2, and C34 has analogous properties in the antiquark sector.
Proof of the inequality (10) If we have a positively oriented edge from s 1 to s 2 , i.e., the Steiner tree is non degenerate, then we have is valid, regardless of whether V 4 is a degenerate or non degenerate Steiner tree. We follow the variational method introduced in [31]. The problem is formulated as a global optimisation problem as follows; Define L as the length of the formal Steiner tree spanned by the four vertices. This length is obtained from the distance between the farthest points on the two Melzak circles. In terms of the usual Steiner tree components, L = v 1 s 1 + v 2 s 1 ± s 1 s 2 + v 3 s 2 + v 4 s 2 ). We get the positive sign for s 1 s 2 if there is a real Steiner tree. On the other hand, if the Steiner vertices have interchanged position, so that on the line between the two farthest Melzak points, s 2 is closer to the Melzak point for v 1 , v 2 than s 1 , then we have the negative sign for s 1 s 2 . So we can construct a formal tree on the six vertices v 1 , v 2 , v 3 , v 4 , s 1 , s 2 where the edge joining the two Steiner vertices is 'negatively oriented'. Now it is easy to see that L ≤ ( x + y ) √ 3/2 + z . So if V = L then the desired inequality follows trivially. So we only need to consider the situation where L < V , i.e the Steiner tree is formal rather than a real Steiner tree. Now by the inequality above, if either of d 13 + d 24 , d 14 + d 23 is not larger than L, then clearly the required inequality follows. So we only need to consider the case when d 13 + d 24 > L and d 14 + d 23 > L.
We can parametrise the points v 1 , v 2 , v 3 , v 4 by the numbers v 1 s 1 , v 2 s 1 , ± s 1 s 2 , v 3 s 2 , v 4 s 2 . (It is easy to see that these four points are determined up to rotation, translation by five parameters.) By rescaling, we can assume that the sum of these five numbers is 1, without loss of generality for the inequality. It is easy to see that all the numbers are then bounded so the domain becomes compact. So we seek a maximum of the ratio of R = min{d 13 + d 24 , d 14 + d 23 } and ( x + y ) √ 3/2 + z = B over this domain. Now suppose that we rotate the triangles v 1 v 2 v 3 and v 1 v 2 v 4 around an axis line through v 1 v 2 . Clearly we can think of one triangle as being fixed and the other as moving relative to the first one. The quantity R does not change by this rotation, but obviously B does. Hence a maximum of the ratio R/B corresponds to a minimum for B under such a rotation. Now an elementary argument shows that such a minimum for B occurs for the configuration being planar, i.e when the vertex v 4 moves into the plane of v 1 , v 2 , v 3 . Now assume that some initial configuration satisfies R/B > 1 and the Steiner tree is formal rather than real. As the triangle v 1 v 2 v 4 rotates around an axis line through v 1 v 2 , it is easy to see that the two Melzak circles move apart. At some intermediate point, if they cross, then we find that the Steiner tree changes from being formal to being real. At this intermediate point, it is trivial to see that R/B < 1. But this is impossible, since we have initially R/B > 1 and R/B is increasing, since B is decreasing and R is fixed.
On the other hand, if the Melzak circles never intersect, then this must be true for the planar configuration. So we would have such a configuration for which the Steiner tree is still formal but R/B > 1. It is elementary to prove that this is impossible. So this completes the argument.