Two methods of solving QCD evolution equation Aleksander Kusina, Magdalena Sławińska

2 Multiple gluon emission from a parton participating in a hard scattering process. The parton with hadron’s momentum fraction x 0 emits gluons. After each emission its momentum decreases: x 0 >x 1 >... > x n-1 > x n The evolution is described by momentum distribution function of partons D(x, t). t denotes a scale of a process. t = lnQ

3 Evolution presented in a (t, x) diagram. The change of momentum distribution function D(x, t) is presented on a diagram by lines incoming and outgoing from a (t, x) cell.

4 Evolution equation for gluons From many possible processes we consider only those involving one type of partons (gluons). The evolution equation is then one-dimentional: where z denotes gluon fractional momenta kernel P(z, t) stands for branching probability density

5 We use regularised kernel: where P   represents outflow of momentum and P  – inflow of momentum. We discuss simplified case of stationary P. Proper normalisation of D, namely: requires: leading to:

6 Monte Carlo Method t0t0 t1t1... t n -1 tntn x0x0 x1x1 x n -1 xnxn t max We generate values of momenta and ”time” according to proper probability distribution for each point in the diagram. (x 0, t 0 )->(x 1, t 1 )->...->(x n-1, t n-1 ) We obtain an evolution of a single gluon. Each dot represents a single gluon emission. Repeating the process many times we obtain a distribution of the momentum x.

7 Monte Carlo Method Iterative solution We introduce the following formfactor:

8 By using substitution we transform the evolution equation to the integral form: and obtain the iterative solution:

9 to obtain the markovian form of the iterative equation we define transition probability: Which is properly normalized to unity Applying this probability to the iterative solution we obtain the markovian form: Markovianisation of the equation

10 Now we introduce the exact form of the kernel so that we can explicitly write the probability of markovian steps The transmission probability factorizes into two parts where

11 Once more the final form of the evolution equation The Monte Carlo algorithm: 1. Generate pairs (t i, z i ) from distributions p(t) and p(z) 2. Calculate T i = t 1 + t t i,x i = z 1 z 2... z i 3. In each step check if T i > t max (t max – evolution time) 4. If T i > t max, take the pair (T i -1, x i -1 ) as a point of distribution function D(x, t max ) and EXIT 5. Repeat the procedure: GO TO POINT 1 MC algorithm

12 Results Starting with delta – distribution, now we demonstrate, how the gluon momenta distribution changes during evolution t=2t=5

13 t=10t=15 t=50t=25

14 From the histograms we see the character of the evolution – momenta of gluons are softening and the distribution resembles delta function at x=0. Now we investigate how the evolution depends on coupling constant  s :

15  s =0.3  s =1

16 Semi- analytical Method The model Problems: ● How to interpret probability P(z) ? ● Discrete calculations Solutions: ● Many particles in the system  their distribution according to P(z) distribution ● Calculations performed on a grid ● evolution steps of size  t ● momenta fractions  N bins of width  x ● k th bin represents momentum fraction (k + ½)  x

17 Since time steps and fractional momenta are descreet, so must be the equation The interpretation of P(z) within this model: In each evolution step particles move - from k to k-1, k-2,..., 0 - from N-1, N – 2,..., k + 1 to k where

18  s =0.3

19 This is to emphasise that both calculation methods and computational algorithms differ very much. In MC the history of a single particle is generated according to probability distributions and its final momentum is remembered. These operations are repeated for 10 8 events (histories) so that a full momenta distribution is obtained. In semi- analytical approach, a momenta distribution function is calculated by considering all 10 4 emiter particles. At each scale a number of particles changing position from (t, i) to (t+1, k) is calculated. All particles are then redistributed and a new momenta distribution is obtained. To compare the methods we divided corresponding histograms. Comparison of the methods

20 As we can see from division of final distribution functions, both methods give the same distribution within 2%! T = 4T = 10 T = 18

21 References: [1] R. Ellis, W. Stirling and B. Webber, QCD and Collider Physics (Cambridge University Press, 1996)