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There is one open parameter which appears both in position and momentum space. It is ΛM S, which sets the scale in perturbation theory, but whose exact value is hard to determine within perturbation theory. This is why we have to make use of lattice data.

Many calculations for determining the parameter have been done in the past [12, 13], which produced dierent values for ΛM S depending on the strategy and the specic data. For reasons of transparency we determine the parameter on our own, with the same underlying data that we will use later for the calculation of the bottomonium system. Our technique, which is performed in momentum space, was developed in [14]

and is used here in the same way. The major dierence is, that we make use of the continuum expression of the lattice potential which we derived in chapter 2. For the further calculation we need this lattice continuum potential in momentum space. After a continuous Fourier transform of equation (2.15) it reads

Vlat(p) = 4πα

p2 −8πσ

p4 . (3.19)

To summarize the technique, we compare the momentum space version of the perturba-tive potential (3.2) with Vlat(p) in an area in which both expressions are reliable. ΛM S

comes in with equation (3.10) and is adjusted by minimizing the distance between both potentials in the area between pmin and pmax. Figure 10 illustrates this procedure of tting the two functions.

In order not to prefer a certain setup in the calculation and to prevent possible correla-tions we perform a huge number of ts. Thereby the setup is choosen randomly. What we obtain is a mean value forΛM S which can be interpreted as a nal result. The standard deviation corresponds to the systematic error.

Concretely, we perform 20000 ts. In each t the perturbative order NNLO or NNNLO, a parameter pair (αi, σi) from the continuum list and a t intervall[pmin, pmax]with

• pmin ∈[1500,2250]MeV

• pmax∈[2250,3000] MeV

• pmax−pmin≥375MeV

is choosen randomly. There are also two dierent methods how to treat the inte-gral in equation (3.10), either inserting (3.6) and performing the integration numeri-cally (method (1)), or expanding the integrand and performing the integral analytinumeri-cally (method (2)). The latter strategy yields

ln p

ΛM S

= b1

0ln

β0αs(p) 4π

+ 2π

β0αs(p)+b2−b210

αs(p)

4π +b3−2b1b2+b310

αs(p) 4π

2

+b4−b22−2b1b3+ 3b21b2−b410

αs(p) 4π

3

(3.20) They provide compatible results. Half of the ts are done with method (1) and the remaining ones with method (2). We nally obtain

ΛM S = 302(12)MeV, (3.21)

which will be used as in input throughout the upcoming calculations. Compared to the former resultΛM S = 329(20)MeV in [14], (3.21) is≈8%smaller. This can be attributed to the continuum extrapolation, which is missing in [14].

-6e-06 -5e-06 -4e-06 -3e-06 -2e-06 -1e-06 0

1000 1500 2000 2500 3000 3500

1875 2625

V(p) [MeV-2 ]

p [MeV]

Vlat(p) Vpert(p)

Figure 10: An illustration of the method to determineΛM S. Vlat and Vpert are matched in a suitable area in momentum space in such a way that their distance is minimal. For the shown setup [NNNLO, method (2), pmin = 1875 MeV, pmax= 2625MeV], the result isΛM S = 290 MeV.

4 Combined potential

In the previous sections two expressions for theQQpotential in position space have been derived. The rst oneVpert from perturbation theory which can be used up tor Λ1 andVlatfrom lattice QCD which is valid fromr≈0.1fm. Since we aim at constructing aM S

position space potential that is applicable over the whole distance range, both expressions will be merged. The requirement, that the ranges of validity have an overlap is fullled.

In order to have the full control we prescribe the distance where the transition takes place. According to the ranges of validity of the potentials we choose the transition area between 0.08 fm ≤ r ≤ 0.20 fm. There are dierent strategies how to combine both potentialsVpert andVlat. The easiest possibility is to join the two potentials together at a given distance, for instancert= 0.15 fm, so that the full potential reads

V(r) =

( Vpert(r), r < rt Vlat(r), r≥rt

. (4.1)

Since the overall energy constant of the potential is not important, both expressions can always be shifted to achieve Vlat(rt) = Vpert(rt). The disadvantage of this solution is, that at the transition point the potential is no longer smooth because the rst derivative is not necessarily continuous. Although we found compatible results with this option, (4.1) will not be used in our nal analysis.

Another strategy is inspired by the aim to have a smooth potential. We construct a third function Vtrans(r)that interpolates between the two potentials. At the transition points r1 andr2, which we prescribe, the rst derivatives coincide. The simplest choice for this interpolating function is a quadratic function

Vtrans(r) =ar2+br+c, (4.2)

whose parameters are xed by the following boundary conditions at the transition points:

V0+Vpert(r1) = Vtrans(r1)

Solving these four linear equations yieldsa, b, c and the constant energy shift V0, which we added to the perturbative potential. In this case the full potential reads

V(r) =

Since there is one open parameterµf left in the perturbative expression, it would be also possible to merge the two potentials by xing this parameter under the condition that the derivatives of the potentials have to be continuous at only one given transition point.

Thus we would avoid the interpolation. But this is not necessarily the case or would lead to too large values for µf. During the numerical analysis we will vary the parameter in an appropriate area.

There are more parameters that inuence the exact shape of the combined potential.

When using the potential in a computation one has to do a statistical analysis by re-peating the computation many times with dierent input parameters. The following quantities are randomly choosen:

• µf ∈[3.0,7.0] ΛM S

• ΛM S from a Gaussian distribution with ΛM S = 302 MeV and∆ΛM S = 12MeV

• r1 ∈[0.08,0.12]fm

• r2 ∈[0.16,0.20]fm

• a pair(α, σ) from a 2-dimensional Gaussian distribution (characterized by (2.22)) Since we want to investigate the inuence of string breaking to the bottomonium system in chapter 5, we introduce another expression:

V(r) =









V0+Vpert(r), r < r1

Vtrans(r), r1≤r≤r2 Vlat(r), r2≤r < rsb Vlat(rsb), r ≥rsb

, (4.5)

with Vlat(rsb) =const. For the string breaking distance rsb = 1.13(10)(10) fm we refer to [15]. Figure 11 illustrates the introduced combination strategies.

-2

(a) Merging by interpolation betweenr1= 0.1 fm andr2= 0.18fm.

(b) Adding string breaking atr= 1.13fm [15].

-2

(c) Special case with only one transition point atrt= 0.15fm.

Figure 11: Dierent strategies to combine Vpert and Vlat. In the perturbative formulas µf was set toµf = 5×302MeV.

5 Bottomonium spectrum

5.1 Solving the Schrödinger equation

As an application of our combined staticQQ potential in position space which is valid across the whole distance range, we use it to reproduce the bottomonium system in the static limit. Since bottomonium is the least relativistic mesonic system and several bb states are experimentally established, it is well suited to test the potential.

The spectrum is determined by solving the non-relativistic Schrödinger equation (~= 1)

− 1

2m∆ +V(r)

ψ(r) =Eψ(r) (5.1)

with the spherically symmetric potential V(r) according to equation (4.4) or (4.5).

m = m2b is the reduced mass of the two interacting quarks with masses mb. With the introduction of polar coordinates(r, θ, φ) the Laplacian operator ∆reads

∆ = 1 We can identify the polar representation of the squared angular momentum operator

L2 =− in equation (5.2). The Schrödinger equation now reads

A separation of variables with the ansatz

ψ(r, θ, φ) =R(r)Y(θ, φ) (5.5)

splits the Schrödinger equation into an angular and a radial equation. The solutions of the angular equation are the well-known spherical harmonics Y`m(θ, φ) that are eigenstates ofL2:

L2Y`m(θ, φ) =`(`+ 1)Y`m(θ, φ), (5.6) with integer numbers`≥0 andm=−`, ..., `.

The radial equation then reads

− 1

The energyEn` will depend on `because of the eective potential (5.9). Moreover the index nindicates the quantization of the bound states.

With the substitution un`(r) =rRn`(r) and after some algebra [16] we nd

− 1 2m

d2un`

dr2 +

V(r) +`(`+ 1) 2mr2 −En`

un`(r) = 0, (5.8) which is analogous to the Schrödinger equation in one dimension with an eective po-tential

Veff(r) =V(r) +`(`+ 1)

2mr2 . (5.9)

The complete wave function

ψn`m(r, θ, φ) =Rn`(r)Y`m(θ, φ) (5.10) has to fulll the normalization condition. The spherical harmonics themselves are already normalized by denition. As a consequence we nd that un`(r) needs to be normalized

as well: Z

0

|un`(r)|2dr= 1. (5.11)

To ensure the normalization, the functionun`(r) has to vanish for large r. This can be formulated as a boundary condition:

r→ ∞: un`(r)−→0. (5.12)

To get an idea of the wavefunction's behaviour and the boundary conditions we take a look at the asymptotic behaviour. The behaviour forr →0is determined by

d2un`

dr2 ' `(`+ 1)

r2 un`(r), (5.13)

yielding the solution

r→0 : un`(r)∼r`+1. (5.14)

Looking at r → ∞ the Schrödinger equation is dominated by the linear rising part σr.

The resulting asymptotic dierential equation d2un`

dr2 ' σr

2mun`(r) (5.15)

is known as Airy equation. To guarantee the boundary condition (5.12), the solution is the Airy function of the rst kind [17]:

un`(r)'Ai

r3 r σ

2m

. (5.16)

Solving the Schrödinger equation yields the binding energy of a bound state En` and the corresponding wave function un`. Inserting ` = 0 gives the S-states of the system, respectively `= 1 the P-states.

To actually gain the masses Mn` of the bottomonium states, the quark masses have to be added to the binding energies [1]: Mn` = En`+ 2mb. Because the overall energy constant to V(r) is still arbitrary, absolute values of En` are not meaningful. We can only interpret energy dierences ∆E between the states. For a proper comparison with experimental states we xM10 ≡mexpb(1S)) = 9.399GeV and obtain any other state by adding the energy dierence between the state and the 1S result. Concretely the calibrated states are calculated by

Mn` =M10 + ∆En`, (5.17)

with∆En`≡En`−E10 and∆E10= 0.