Les Houches Parametrisation

We follow the Les Houches parametrisation as presented in Ref. [A+05].

Free parameters in the Les Houches Parametrisation

As described in Ref. [A+05], the Les Houches parametrisation assumes that the total sea, \(\Sigma=u+\bar{u}+d+\bar{d}+s+\bar{s}\), is constrained to be made 40% by up and anti-up, 40% by down and anti-down, and 20% by strange and anti-strange, which means that we can write:

\[\begin{split}u+\bar{u}=0.4\Sigma,\notag\\ d+\bar{d}=0.4\Sigma,\\ s+\bar{s}=0.2\Sigma.\notag\end{split}\]

It is also assumed that there is no difference between \(\bar{u}\) and \(\bar{d}\), so we are only left with four active flavours, namely \(g, u_{v}, d_{v}\) and \(\Sigma\). Furthermore, \(\epsilon_g\), \(\gamma_g\), \(\epsilon_\Sigma\) and \(\gamma_\Sigma\) are all set to zero. We are therefore left with the set of equations:

(1)\[\begin{split}xf_g(x,Q_0) &= A_g\,x^{\alpha_g}\,(1 - x)^{\beta_g}, \notag \\ xf_{u_v}(x,Q_0) &= A_{u_v}\,x^{\alpha_{u_v}}\,(1 - x)^{\beta_{u_v}}\, (1+\epsilon_{u_v}\sqrt{x}+\gamma_{u_v} x), \\ xf_{d_v}(x,Q_0) &= A_{d_v}\,x^{\alpha_{d_v}}\,(1 - x)^{\beta_{d_v}}\, (1+\epsilon_{d_v}\sqrt{x}+\gamma_{d_v} x), \notag \\ xf_\Sigma(x,Q_0) &= A_\Sigma\,x^{\alpha_\Sigma}\,(1 - x)^{\beta_\Sigma}.\notag\end{split}\]

This amounts to 16 parameters. Moreover, not all parameters are independent. \(A_g\) is related to \(A_\Sigma\) by the momentum sum rules:

(2)\[A_g\int_0^1 x^{\alpha_g}\,(1 - x)^{\beta_g} dx + A_\Sigma \int_0^1 x^{\alpha_\Sigma}\,(1 - x)^{\beta_\Sigma}\, dx = 1,\]

and the \(A_{u_v}\) and \(A_{d_v}\) parameters are determined by the valence sum rules:

(3)\[\begin{split}& A_{u_v}\,\int\, x^{\alpha_{u_v}-1}\,(1 - x)^{\beta_{u_v}}\, (1+\epsilon_{u_v}\sqrt{x}+\gamma_{u_v} x) dx = 2, \notag \\ & A_{d_v}\,\int\,x^{\alpha_{d_v}-1}\,(1 - x)^{\beta_{d_v}}\, (1+\epsilon_{d_v}\sqrt{x}+\gamma_{d_v} x) dx = 1,\end{split}\]

leaving 13 free parameters [1] .

[1]

In ref. [A+05], \(\epsilon_{u_v}\) is fixed to its best-fit value, \(\epsilon_{u_v} = -1.56\), in order to avoid instability due to a very high correlation between \(u_v\) parameters. They therefore left only 12 parameters free to vary. We decide to leave \(\epsilon_{u_v}\) free because we don’t believe we will encounter this problem.

Normalisations

In order to be able to perform a fit, we would like to write all PDFs explicitly in terms of free parameters, x and \(Q_0\).

We therefore write the expressions for \(A_g\), \(A_{u_v}\) and \(A_{d_v}\) explicitly by solving the integral spelled out in the sum rules, Equations (2) and (3), which are of the form of Euler beta functions, given by:

(4)\[\int_0^1 dt \, t^{\alpha -1} (1-t)^{\beta -1} = \frac{\Gamma(\alpha) \Gamma(\beta)}{\Gamma(\alpha + \beta)},\]

where, for positive integer \(n\), \(\Gamma(n)\) is defined as:

\[\Gamma(n) = (n-1)!.\]

We find that:

\[A_g = \frac{\Gamma(\alpha_g + \beta_g + 2)}{\Gamma(\alpha_g+1)\Gamma(\beta_g+1)}\left[ 1 - A_{\Sigma} \frac{\Gamma(\alpha_\Sigma + 1) \Gamma(\beta_\Sigma + 1)}{\Gamma(\alpha_\Sigma + \beta_\Sigma +2)} \right],\]

\[A_{u_v} = \frac{2}{\Gamma(\beta_{u_v}+1)}\left[ \frac{\Gamma(\alpha_{u_v})}{\Gamma(\alpha_{u_v} + \beta_{u_v} + 1)} + \epsilon_{u_v} \frac{\Gamma(\alpha_{u_v} + 1 / 2)}{\Gamma(\alpha_{u_v} + \beta_{u_v} + 3 / 2)} + \gamma_{u_v} \frac{\Gamma(\alpha_{u_v} + 1)}{\Gamma(\alpha_{u_v} + \beta_{u_v} + 2)} \right]^{-1},\]

\[A_{d_v} = \frac{1}{\Gamma(\beta_{d_v}+1)}\left[ \frac{\Gamma(\alpha_{d_v})}{\Gamma(\alpha_{d_v} + \beta_{d_v} + 1)} + \epsilon_{d_v} \frac{\Gamma(\alpha_{d_v} + 1 / 2)}{\Gamma(\alpha_{d_v} + \beta_{d_v} + 3 / 2)} + \gamma_{d_v} \frac{\Gamma(\alpha_{d_v} + 1)}{\Gamma(\alpha_{d_v} + \beta_{d_v} + 2)} \right]^{-1}.\]

The Les Houches Parametrisation in the evolution basis

Colibri works in the evolution basis, whose elements can be written as a linear combination of the elements of the flavour basis.

We start by writting the elements of the evolution basis in terms of quark flavours, which is as follows:

(5)\[\begin{split}\Sigma &= u+\bar{u}+d+\bar{d}+s+\bar{s}, \notag \\ T_3 &= (u + \bar{u}) - (d + \bar{d}), \notag \\ T_8 &= (u+\bar{u} + d + \bar{d}) - 2(s+\bar{s}), \\ V &= (u-\bar{u}) + (d-\bar{d}) + (s-\bar{s}), \notag \\ V_3 &= (u - \bar{u}) - (d - \bar{d}), \notag \\ V_8 &= (u-\bar{u} + d-\bar{d}) - 2(s-\bar{s}). \notag\end{split}\]

Noting that \(u_v = u - \bar{u}\), \(d_v = d - \bar{d}\) and that, since there are no valence strange quarks, \(s_v = s - \bar{s} = 0\), and applying the assumptions stated above, we find:

(6)\[\begin{split}T_3 &= (u-\bar{d})-(d-\bar{u}) = u_v - d_v = V_3, \notag \\ T_8 &= \Sigma - 3(s+\bar{s}) = 0.4\Sigma, \\ V_8 &= u_v + d_v - 2 \cdot 0 = V. \notag\end{split}\]

Therefore, we are again left with only four active flavours; \(\Sigma\), \(V\), \(V_3\) and the gluon.

We already have an explicit parametrisation for \(f_\Sigma\) and \(f_g\), as stated in Eq. (1). We have the ingredients to write analogous expressions for \(f_V\) and \(f_{V_3}\), which are given by:

(7)\[\begin{split}x f_V &= x f_{u_v} + x f_{d_v} \\ &= A_{u_v}\,x^{\alpha_{u_v}}\,(1 - x)^{\beta_{u_v}}\, (1+\epsilon_{u_v}\sqrt{x}+\gamma_{u_v} x) + A_{d_v}\,x^{\alpha_{d_v}}\,(1 - x)^{\beta_{d_v}}(1+\epsilon_{d_v}\sqrt{x}+\gamma_{d_v} x), \notag\end{split}\]

(8)\[\begin{split}x f_{V_3} &= x f_{u_v} - x f_{d_v} \\ &= A_{u_v}\,x^{\alpha_{u_v}}\,(1 - x)^{\beta_{u_v}}\, (1+\epsilon_{u_v}\sqrt{x}+\gamma_{u_v} x) - A_{d_v}\,x^{\alpha_{d_v}}\,(1 - x)^{\beta_{d_v}}(1+\epsilon_{d_v}\sqrt{x}+\gamma_{d_v} x). \notag\end{split}\]