• Keine Ergebnisse gefunden

Special case: identical masses in the final state of the decay

10.2 Rescattering and Khuri–Treiman equations

11.1.2 Special case: identical masses in the final state of the decay

Consider now the special case withπ‘š1≑ 𝑀andπ‘š2= π‘š3 = π‘š4≕ π‘šas well as𝑠thrπ’₯ = 4π‘š2. That is, the aim is to develop an understanding of the singularities in the decay of one particle into three particles of equal mass, where the intermediate two-particle states consist of particles of the same species as the ones in the decay product. In this case, Eq. (11.15) and Eq. (11.16) simplify to

0 =π‘₯3[𝑧⋆2βˆ’ 1] + 2π‘₯2[(𝑀2βˆ’ 5π‘š2) βˆ’ 𝑧⋆2𝑅]

+ π‘₯ [𝑧⋆2𝑅2βˆ’ (𝑀2βˆ’ 5π‘š2)2] βˆ’ 4𝑧⋆2π‘š2(𝑀2βˆ’ π‘š2)2. (11.19) Solving for𝑧⋆2yields

𝑧sing⋆2 = π‘₯ (5π‘š2βˆ’ 𝑀2+ π‘₯)2

(π‘₯ βˆ’ 4π‘š2) [π‘₯2βˆ’ 2π‘₯ (𝑀2+ π‘š2) + (𝑀2βˆ’ π‘š2)2]. (11.20) The original path of integration in the𝑧⋆plane extends fromβˆ’1to 1, i.e. it covers the interval[0, 1]

on the real axis in the𝑧⋆2plane. Solving Eq. (11.19) forπ‘₯with𝑧sing⋆2 ∈ {0, 1}fixed results in1 𝑧sing⋆2 = 0 ⇔ π‘₯sing ∈ {0, 𝑀2βˆ’ 5π‘š2} ,

𝑧sing⋆2 = 1 ⇔ π‘₯sing = 𝑀2βˆ’ π‘š2

2 . (11.21)

Moreover,

π‘₯β†’βˆžlim 𝑧sing⋆2 = 1. (11.22)

Again, the original path of integration in theπ‘₯plane extends along[4π‘š2, ∞). As long as𝑀2< 9π‘š2, that is as long as the decay is kinematically forbidden, the singularities Eq. (11.21) stay below4π‘š2, and thus they do not interfere with the original path of integration. However, as soon as𝑀2β‰₯ 9π‘š2, both 𝑀2βˆ’ 5π‘š2 β‰₯ 4π‘š2 and(𝑀2βˆ’ π‘š2)/2 β‰₯ 4π‘š2. That is, the original path of integration fails to work inside the decay region. Another special point at which significant simplifications arise is 𝑀2 = 9π‘š2, that is the boundary of the decay region (in𝑀2). In this case Eq. (11.19) boils down to

0 = (π‘₯ βˆ’ 4π‘š2)2[π‘₯ (𝑧⋆2βˆ’ 1) βˆ’ 16π‘š2𝑧⋆2] , (11.23) which is solved by

π‘₯sing ∈ {4π‘š2, βˆ’16π‘š2 𝑧⋆2

1 βˆ’ 𝑧⋆2} . (11.24)

The second solution is non-positive for 𝑧⋆2 ∈ [0, 1), but the first one interferes directly with the lower limit of integration inπ‘₯, pointing again to the fact that the original contours of integration fail to work as soon as the decay is kinematically allowed.

The correct way to circumvent this problem is to analytically continue Eq. (11.9) as a function of𝑀2 via a deformation of the contours of integration. There are different ways to perform this continuation, since there are two integrals, both of which can be deformed. Here, the choice is made to keep the contour of integration in 𝑧⋆, i.e. 𝑧⋆ ∈ [βˆ’1, 1], while distorting the contour of

1Note that in both cases Eq. (11.19) boils down to a quadratic equation inπ‘₯.

integration inπ‘₯. Hence, in the following the singularities in theπ‘₯plane as a function of𝑧⋆2 ∈ [0, 1]

are studied.

The endpoints π‘₯sing(𝑧⋆2 = 0) and π‘₯sing(𝑧⋆2 = 1) are given by Eq. (11.21). In addition, the singularities π‘₯sing(𝑧⋆2) for 𝑧⋆2 ∈ (0, 1) are obtained by solving the in this case cubic Eq. (11.19).

This can be done, e.g., using Cardano’s formula.

To be precise, denote an𝑛-th root of unity by πœ‰π‘› ≔exp[2πœ‹π‘–

𝑛 ] , πœ‰π‘›π‘› = 1. (11.25)

Furthermore, for arbitrary𝜁 ∈ β„‚, define the principal value of the𝑛-th root as 𝜁1/𝑛≔exp[1

𝑛ln(𝜁)] . (11.26)

Here ln(𝜁 ) = ln(|𝜁 |) + 𝑖arg(𝜁 ) denotes the principal branch of the logarithm, with a cut along (βˆ’βˆž, 0]. The value of the𝑛-th root on its other𝑛 βˆ’ 1Riemann sheets is accordingly given as

exp[1

𝑛(ln(𝜁) + π‘Ÿ2πœ‹π‘–)] =exp[1

𝑛ln(𝜁)]exp[π‘Ÿ2πœ‹π‘–

𝑛 ] = 𝜁1/π‘›πœ‰π‘›π‘Ÿ, (11.27) withπ‘Ÿ ∈ 1, … , 𝑛 βˆ’ 1. In the special case of a cube root,

πœ‰3= βˆ’1 + π‘–βˆš3

2 , πœ‰32 = πœ‰3βˆ— (11.28)

hold.

Now consider the equation

3

βˆ‘

𝑖=0

𝑐𝑖π‘₯𝑖 = 0, (11.29)

where the coefficients 𝑐𝑖 in the case of interest are fixed via comparision with Eq. (11.19). The solutions of Eq. (11.29) read [224, 225]

π‘₯sing,π‘˜ = βˆ’ 1

3𝑐3[𝑐2+ πœ‰3π‘˜Ξ”Β±+ Ξ”0

πœ‰3π‘˜Ξ”Β±] , π‘˜ ∈ {0, 1, 2} . (11.30)

Here Ξ”0≔ 𝑐22βˆ’ 3𝑐3𝑐1,

Ξ”1≔ 2𝑐23βˆ’ 9𝑐3𝑐2𝑐1+ 27𝑐32𝑐0, Δ± ≔ (Ξ”1Β± βˆšΞ”21βˆ’ 4Ξ”30

2 )

1/3

,

(11.31)

where the choice of the sign inΔ±is arbitraryβ€”according to Eq. (11.27) it merely corresponds to a change of the sheet of the square rootβ€”as long asΔ± β‰  0. However, if it equals 0 for one choice of the sign, the other one is to be picked. Note that due to Eq. (11.27) the different solutions correspond to different choices of the Riemann sheet of the cube root inΔ±.

In the following, consider arbitrary 𝑀2, π‘š ∈ ℝ and𝑧⋆2 ∈ (0, 1). Thus, in the case of interest 𝑐𝑖, Ξ”0, Ξ”1 ∈ ℝ. Moreover,πœ‰3Δ± is real for 𝑀2 > 9π‘š2. Hence,π‘₯sing,1 ∈ ℝ, moreover,π‘₯sing,0 = π‘₯sing,2βˆ— due to Eq. (11.28):

Δ± = πœ‰33Δ± = πœ‰32πœ‰3Δ± = πœ‰3βˆ—πœ‰3Δ± = (πœ‰3πœ‰3Δ±)βˆ—. (11.32)

The singularities π‘₯sing,π‘˜ for π‘˜ = 0, 1, 2are plotted in the π‘₯ plane for all values of𝑧⋆2 ∈ [0, 1]

and for different values of 𝑀2 in Fig. 11.1. Independently of the value of 𝑀2, the negative real axis is always singular. Moreover, it is the only singular region as long as𝑀2 is non-positive. As soon as𝑀2 > 0, a structure symmetric around the real axis starts to form, which in the beginning is connected to the negative real axis. A further increase in 𝑀2 results in a detachment of this structure from the negative real axis, it moves steadily towards the branch point at 4π‘š2. To be precise, as soon as it detaches from the negative real axis at𝑀2= 5π‘š2the structure is given as

π•Š ≔ {π‘₯sing,π‘˜ ∢ π‘˜ ∈ {0, 2}and𝑧⋆2∈ (0, 1)} βˆͺ {𝑀2βˆ’ 5π‘š2,𝑀2βˆ’ π‘š2

2 } . (11.33)

Here the two elements given at the end correspond to Eq. (11.21). When 𝑀2 reaches the decay threshold, i.e.𝑀2 = 9π‘š2,π•Šcollapses to a single point at exactly the branch point4π‘š2. For𝑀2 >

9π‘š2,π•Šemerges again; it moves along the real axis to the right and increases in size with increasing 𝑀2. Moreover,

max{Re(𝑦) ∢ 𝑦 ∈ π•Š} = {

𝑀2βˆ’π‘š2

2 , 𝑀2< 9π‘š2, 𝑀2βˆ’ 5π‘š2, 𝑀2> 9π‘š2, min{Re(𝑦) ∢ 𝑦 ∈ π•Š} = {𝑀2βˆ’ 5π‘š2, 𝑀2< 9π‘š2,

𝑀2βˆ’π‘š2

2 , 𝑀2> 9π‘š2

(11.34)

holds.

The problem is not only that for𝑀2 > 9π‘š2the singularities interfere with the original path of integration, but also that at𝑀2 = 9π‘š2 they hit the branch point, which marks the beginning of the path of integration. Since the endpoints of an integration contour are fixed, this is a serious obstacle for any analytic continuation.

The way out is to equip 𝑀2 with a small, positive imaginary part, that is, to perform the re-placement𝑀2↦ 𝑀2+π‘–πœ–withπœ– > 0, and consider the limitπœ– β†˜ 0in the end. This does not alter the singularities depicted in Fig. 11.1 in any significant2way except for the location ofπ•Šat𝑀2= 9π‘š2: instead of π•Š|𝑀2=9π‘š2 = {4π‘š2}nowπ•Š|𝑀2=9π‘š2+π‘–πœ– = {4π‘š2+ π‘–πœ–}holds. That is, the singularities do not hit the branch point.

To understand how the contour of integration in theπ‘₯plane needs to be deformed, start with Eq. (11.9) for𝑀2< 9π‘š2, where the original contour[4π‘š2, ∞)is fine. Now increase𝑀2continuously, while sticking to the+π‘–πœ–prescription. At𝑀2 = 9π‘š2the singularity structureπ•Špasses by the start of the integration contour above the real axis, for larger values of 𝑀2 the structure π•Š grows in size. Hence, to avoid any collision of the contour with π•Š, the contour needs to be deformed into the lower half plane. This is illustrated in Fig. 11.2 for𝑀2 = 12π‘š2. Note that the depicted choice of the deformed path, denotedC, is just one possible choice out of many. Any path that starts at 4π‘š2, extends towards infinity, and avoids the singularities via bending into the lower half plane is legitimate.

2For𝑀2β‰  9π‘š2there is no change inπ•Š, since the limitπœ– β†˜ 0can be performed without running into any problems.

However, forπœ– β‰  0the parts ofπ•Šthat are traced out byπ‘₯sing,0andπ‘₯sing,2are partly interchanged, due to the branch cuts of the roots appearing in Eq. (11.31).

βˆ’2 0 2

Im(π‘₯)/π‘š2 𝑀2= βˆ’3π‘š2 π‘₯sing,0

π‘₯sing,1 π‘₯sing,2

βˆ’2 0 2

Im(π‘₯)/π‘š2 𝑀2= 0π‘š2

βˆ’2 0 2

Im(π‘₯)/π‘š2 𝑀2= 3π‘š2

βˆ’2 0 2

Im(π‘₯)/π‘š2 𝑀2= 6π‘š2

βˆ’2 0 2

Im(π‘₯)/π‘š2 𝑀2= 9π‘š2

βˆ’2 0 2

Im(π‘₯)/π‘š2 𝑀2= 12π‘š2

βˆ’4 βˆ’2 0 2 4 6 8 10

Re(π‘₯)/π‘š2

βˆ’2 0 2

Im(π‘₯)/π‘š2 𝑀2= 15π‘š2

Figure 11.1: The singularities for different values of𝑀2. The gray blob denotes the branch point at4π‘š2, while the original path of integration is represented by the dotted gray line. At𝑀2 = 9π‘š2 the singularitiesπ‘₯sing,0andπ‘₯sing,2shrink down to a point located exactly at the branch point4π‘š2.

βˆ’4 βˆ’2 0 2

Re(π‘₯)/π‘š2 4 6 8 10

βˆ’1.5

βˆ’1.0

βˆ’0.5 0.0 0.5 1.0 1.5

Im(π‘₯)/π‘š2

C π‘₯sing,0 π‘₯sing,1 π‘₯sing,2

Figure 11.2: The singularities for 𝑀2 = 12π‘š2as well as the deformed contour of integration, de-noted byC. As before the gray blob represents the branch point at4π‘š2, while the original path of integration is depicted as a dotted gray line.