Sum rules for parton distribution functions

Question

In Schwarz, the parton distribution functions for proton has the following sum rules, $$ \int dx f_u(x)=2\quad,\quad\int dxf_d(x)=1 $$ where $x$ is the fraction of momenta of proton carried by the parton species or Bjroken x. Then we can write $$ \sum_i\int dx f_i(x)=3 \quad \text{for any hadron includes 3 partons} $$ Then he gives the result $$ \sum_i\int xf_i(x)dx=1 \tag{1} $$ Now to prove 1 we can take the definition of x as fraction, hence we can use $\sum_i\langle x\rangle_i=1$, the average value of $x$. Then we use $$ \frac{\sum_i\int xf_i(x)dx}{\sum_i\int f_i(x)dx}=1 $$ which means $$ \sum_i\int xf_i(x)dx=3 $$ Where I am doing mistake.

References : Schwarz pp 676

Answer 1

The problem with this approach is that we are treating $f_i(x)$ purely as probability distribution function. As such, a different normalisation condition would have been imposed over it. Refering to "Fundamentals of Particle Physics" by Pascal Paganini

They represent the probability densities (strictly speaking, they rather represent the number densities as they are normalised to the number of partons) to find a parton of type $i$ carrying a momentum fraction $x$ of the proton. In other words, $f_i(x) dx$ is the number of partons of type $i$ within the proton carrying a momentum fraction between $x$ and $x + dx$.

The correct approach to understanding PDF would be following. The average momentum carried by $n$ partons (number of parton - number of anti-parton = $n$) of species $i$ is given by: $$\sum_{n}\langle x_i\rangle_{n}=\sum_n\int_0^1 x f_{in}(x)dx$$ The average momentum carried by any particular $n-$th Parton belonging to species $i$ is: \begin{align} \langle x_i\rangle_{n}&= \frac{\sum_n \langle x_i \rangle_n}{n}\\ &=\frac{\sum_{n}\int_0^1 xf_{in}(x)dx}{\sum_{n}\int_0^1 f_{in}(x)dx}\\ &= \int_0^1 xf_{in}(x)dx \frac{\sum_n 1}{\sum_{n}\int_0^1 f_{in}(x)dx}\\ &=\int_0^1 xf_{in}(x)dx \end{align} The sum rule can be seen as: $$\sum_{n}\underbrace{\int_{0}^{1}f_{in}(x)dx}_{=1} = \sum_{n}1= n$$ Instead of having to rewrite the $`\sum_n'$ over and over again, we redefine the PDF by suppressing the index $n$ and the associated sum. Under this redefinition, we use the sum rule as normalisation condition over the PDFs. Returning to the problem in hand, \begin{align} \sum_i\sum_n \langle x_i\rangle_n &=\sum_{i}\sum_{n}\frac{\sum_{n}\int_0^1xf_{in}(x)dx}{\sum_{n}\int_0^1 f_{in}(x)dx}\\ &=\sum_{i}\underbrace{\sum_{n}\int_0^1xf_{in}(x)dx}_{\int_0^1 xf_{i}(x)dx}\frac{\sum_{n}1}{\sum_{n}\int_0^1 f_{in}(x)dx}\\ &= \sum_i \int_0^1 xf_{i}(x)dx=1 \end{align} We see that there is no contradiction here.