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In the literature on parton evolution one often reads that the DGLAP splitting kernels $P_{ij}$ represent the probabilities for a parton of type $j$ to split into a parton of type $i$ and another parton.

At leading order the kernels are: \begin{align} P_{qq}(z)&=C_F\left(\frac{1+z^2}{[1-z]_+}+\frac{3}{2}\delta(1-z)\right),\\ P_{qg}(z)&=T_F\left(z^2+(1-z)^2\right),\\ P_{gq}(z)&=C_F\frac{1+(1-z)^2}{z},\\ P_{gg}(z)&=2C_A\left(\frac{z}{[1-z]_+}+\frac{1-z}{z}+z(1-z)\right)+\frac{\beta_0}{2}\delta(1-z), \end{align} where $\beta_0=\frac{11}{3}C_A-\frac{4}{3}T_FN$ (see pp.681) in M.D. Schwartz: Quantum Field Theory and the Standard Model.

If the splitting kernels are to be interpreted as the probability densities for a parton of type $i$ to branch into a parton of type $j$, we should have $\int^1_0 dz \sum_j P_{ji}(z)=1-P(\mathrm{no branching})$.

But integrating the above expressions gives

\begin{align} \int^1_0dzP_{qq}(z)&=0,\\ \int^1_0dzP_{qg}(z)&=\frac{2}{3}T_F,\\ \int^1_0dzP_{gq}(z)&=C_F\left(2\int^1_0\frac{dz}{z}-\frac{3}{2}\right),\\ \int^1_0dzP_{gg}(z)&=C_A\left(2\int^1_0\frac{dz}{z}-\frac{11}{6}\right)-\frac{2}{3}NT_F, \end{align}

i.e. two of the kernels are clearly not integrable over $[0,1]$, which is a prerequisite for any function to be a probability density.

How can the kernels still be interpreted as probability densities? Or are they maybe no probability densities to begin with? Maybe they can only be interpreted as such, if they are defined on a domain $[x,1]$ with $0<x<1$?

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  • $\begingroup$ I don't understand why the integrals should sum up to one, could you elaborate on why you think so? This is just the probability for particles to radiate particles and the divergence tells us that it's way more probable to have low energy gluons... $\endgroup$ – Ismasou Apr 30 at 14:10
  • $\begingroup$ @Ismasou If I have a set of events $A,B,C$, that have the possibilities $P(A), P(B), P(C)$ to happen, and there are no other events that could happen, then we should have $1 = P(A)+P(B)+P(C)$, right? What troubles me is that the sum over the integrated splitting kernels seems to diverge. $1/z$ is not integrable over $[0,1]$ and thus it cannot be a probability density. If I understand correctly, then $P_{gq}(z)dz$ is the probability for a quark to branch into a gluon with momentum fraction $z+dz$ and something else, right? Then the integral over $z\in[0,1]$ should be finite but it is infinite. $\endgroup$ – Thomas Wening Apr 30 at 14:25
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    $\begingroup$ a) The events don't have to happen, a quark doesn't have to radiate, this is just the probability that it will radiate if it does, (1-P(A)-P(B)-P(C)) will be the probability that the quark doesn't radiate. This is how I understand it. b) I think the divergence has to be cut with a scale, usually I use these in medium and there you use the debye scale as a lower cut, I don't know about vacuum. c) if you use this in a kinetic equation you find that at some point these particles thermalize and the merging processes will cancel the splitting, this makes more sense to me. $\endgroup$ – Ismasou May 1 at 10:04
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    $\begingroup$ I believe the the correct statements are $\int_0^1 dz z ( P_{qq}(z) + P_{gq}(z) ) = 0$ and $\int_0^1 dz z (2 N_f P_{qg}(z) + P_{gg}(z) ) = 0$ which follow from the sum of probabilities of the possible quark and gluon branchings and no branchings. $\endgroup$ – CAF May 1 at 18:19
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    $\begingroup$ As CAF says, the splitting functions conserve momentum (and perhaps can be thought of as probabilities in that sense), but they're really better understood as related to differential particle number operators $\endgroup$ – WAH May 3 at 19:08
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The splitting functions tell you the probability for a given radiance to occur, and some exhibit manifest divergences (e.g. as $z \rightarrow 0$) signifying a growth in that particular branching in a certain part of the phase space. The correct statements are $$\int_0^1 z \, \text{d}z \, (P_{q \leftarrow q}(z) + P_{g \leftarrow q}(z)) = 0\,\,\,\,\,\,\,;\,\,\,\,\,\,\,\,\int_0^1 z\, \text{d}z\, (2n_f P_{q \leftarrow g}(z) + P_{g \leftarrow g}(z)) = 0 $$ They may be derived by totalling the sum of all fractional constituent nucleon momentum to unity, means $$ \int_0^1 \text{d}x \,x \, \left[ \sum_{i} \left( q_i(x) + \bar{q}_i(x) \right) + g(x) \right] = 1 $$ and taking the derivative w.r.t $\mu^2 \,d/d \mu^2$ in each of the terms as follows $$\sum_i \mu^2 \frac{d}{d \mu^2} ( q_i(x) + \bar{q}_i(x) ) = \frac{\alpha_s}{2 \pi} \sum_i \left( q_i \otimes P_{q \leftarrow q} + 2 g \otimes P_{q \leftarrow g} \right),$$ $$ \mu^2 \frac{d}{d \mu^2} g(x) = \frac{\alpha_s}{2 \pi} \sum_j \left( q_j \otimes P_{g \leftarrow q} + g \otimes P_{g \leftarrow g} \right),$$ where $P_{\bar q \leftarrow q} = 0$ at leading order and $P_{\bar q \leftarrow g} = P_{q \leftarrow g}$ with $$(X_i \otimes P_{i \leftarrow j})(x) \equiv \int_{x}^1 \frac{d y}{y} X_i(y) P_{i \leftarrow j} \left(\frac{x}{y} \right).$$ Their addition gives $$\sum_i \mu^2 \frac{d}{ d \mu^2} \left( q_i(x) + \bar{q}_i(x) + g(x) \right) = \frac{\alpha_s}{2 \pi} \left( \sum_i q_i \otimes (P_{q \leftarrow q} + P_{g \leftarrow q}) + g \otimes ( P_{g \leftarrow g} + 2 n_f P_{q \leftarrow g} ) \right)$$ and so $$ \frac{\alpha_s}{2 \pi} \int_0^1 \text{d} x \, x \, \left( \sum_i q_i \otimes (P_{q \leftarrow q} + P_{g \leftarrow q}) + g \otimes ( P_{g \leftarrow g} + 2 n_f P_{q \leftarrow g}) \right) = 0.$$ Now, noting that $$ \int_0^1 \text{d}x \int_x^1 \text{d}y = \int_0^1 \text{d}y \int_0^y \text{d}x,$$ and making a change of variables $x = yz$ decouples the $x$ and $y$ integrations allowing for the relations given above for all $q_i$ and $g$. These relations are valid only at leading order, at the next order additional flavour effects come into play which makes the DGLAP evolution matrix structurally more involved.

Additional comments: The factor of $z$ included means it is the total momentum fraction that is conserved, not the number density. In this sense, it is $zP_{ij}$ that are the probability densities over $z$. Note the sum of probabilities of all possible splittings plus probability for no splitting is what sums to unity. You can trade the non integrable singularity at $z=1$ in the splitting functions for something integrable plus explicit IR divergences centred at $z=1$, this is done practically via the plus prescription. Schematically, this gives the following

$$\int_0^1 \text{d}z\, z \,( P^0_{g \leftarrow g} + P_{q \leftarrow g} + A_0 \delta (1-z) + \delta(1-z) ) = 1 \Rightarrow \int_0^1 \text{d}z\, z \,(P_{g \leftarrow g} + P_{q \leftarrow g} ) = 0, $$ where $P_{g \leftarrow g}$ is the plus-prescribed (regularised) splitting function and $A_0$ is of course tuned to cancel the divergence at $z=1$.

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  • $\begingroup$ I don't understand how the first two equations show that the kernels are probability distributions. I would expect that if they were probabilities, that $\int^1_0dz\sum_iP_{i\leftarrow q}=1-P(\mathrm{no branching})$, instead of zero. Also, why is there an additional $z$ in the integrand? Does that mean that the kernels themselves are no probability distributions but only the kernels times $z$? $\endgroup$ – Thomas Wening May 5 at 11:09
  • $\begingroup$ Basically what you did is show the momentum sum rule for the splitting kernels, right? But that the sum rule rather shows that zP(z) describe the momentum flow of the branchings. I however want to know how the P themselves can be probability densities, i.e. non-negative (clear), integrable (not clear at all to me), normalised functions in z. $\endgroup$ – Thomas Wening May 5 at 12:12
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    $\begingroup$ @ThomasWening The factor of z included means it is the total momentum fraction that is conserved, not the number density. Note the sum of probabilities of all possible splittings plus probability for no splitting is what sums to unity. You can trade the non integrable singularity at z=1 in the splitting functions for something integrable plus explicit IR divergences centred at z=1, this is done practically via the plus prescription. Putting this all together gives you the equations I derived. Hope it’s a bit clearer! $\endgroup$ – CAF May 5 at 18:05
  • $\begingroup$ I still don't understand how they can sum to unity, if there is a non-integrable singularity at $z=0$. $\endgroup$ – Thomas Wening May 5 at 18:17
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    $\begingroup$ IMO yes, this is why it’s common to see on plots of Parton distribution functions vs. x, the y axis labelled as $x f_i$. In my work, I am almost always using $ xg(x)$ for the gluon and never just $g(x)$... $\endgroup$ – CAF May 5 at 18:34

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