Let $f_i(x)$ be a parton distribution function as known from QCD factorisation theorems. Is $f_i(x)$ non-negative for $0<x<1$? If so, how can this be seen from the definition of PDFs in terms of operator products?
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2$\begingroup$ I assume that you are referring to a probability density function. Is this correct? $\endgroup$– NotMeCommented Apr 6, 2020 at 16:41
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$\begingroup$ No, I do mean the parton distribution functions from the QCD-improved parton model. $\endgroup$– Thomas WeningCommented Apr 6, 2020 at 16:43
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1$\begingroup$ Probability distributions are non-negative by axiomatic conditioning. When they are not, violating Kolmogorov's axioms, they are called quasi-probability distributions, a very different beast, not relevant here. $\endgroup$– Cosmas ZachosCommented Apr 6, 2020 at 16:43
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$\begingroup$ @CosmasZachos That is true but the question is not about probability distributions but parton distribution functions instead. $\endgroup$– Thomas WeningCommented Apr 6, 2020 at 16:45
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$\begingroup$ Parton distribution functions are probability distribution functions, of course, so positive semi-definite, as already stated. GPDs in phase space, by contrast, are epically not probability distributions, but quasi-probability distributions , instead, named so for a reason: they may get negative, by violating probability rules. Is that what you are asking about? $\endgroup$– Cosmas ZachosCommented Apr 6, 2020 at 16:58
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3 Answers
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By definition parton distribution function is a re-scaled probability density : $$ f_i(x) = K\, |\phi |^{2} $$ So YES, it must be non-negative, however it can be greater than $1$.
Also take a look into this research :
where authors claim that :
The VFN scheme is valid only at asymptotically large values of Q and cannot be routinely extrapolated to the low-Q region.
So it can't take negatives values for sure.
answered Apr 6, 2020 at 16:58
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$\begingroup$ How are they rescaled and what is the underlying probability density $|\Phi|^2$? $\endgroup$ Commented Apr 6, 2020 at 17:08
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$\begingroup$ From the wikipedia: Parton_distribution_functions : "A parton distribution function (PDF) within so called collinear factorization is defined as the probability density for finding a particle with a certain longitudinal momentum fraction x at resolution scale $Q^2$." $\endgroup$ Commented Apr 6, 2020 at 17:17
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OK, as the question eventually morphed, the answer is
- For plain (non phase space GPDs!) pdfs, they really are scaled probability distribution functions, always positive semidefinite, and their integrals over x are positive numbers.
They are an assumed input (not an output) of parton estimates, fitted to data.
Typically, as, e.g., in Schwartz 32.1.4, they count gluons and constituent and sea quarks. You know there is a constituent d in the proton. If there were no sea, you'd have $$ \int dx ~~f_d(x) =1, $$ and the function would be a bona-fide probability distribution: it would map what fraction of the proton's momentum that quark is likely to carry.
But there is a sea, so also an extra contribution to $f_d(x)$, which actually blows up at small xs. However, that extra contribution is exactly matched by $f_{\bar d} (x)$, so that $\int dx ~ f_d(x) =1+ \int dx ~f_{\bar d}(x)$, $$ \int dx ~ ( f_d(x) -f_{\bar d}(x))=1, $$ implying that the actual normalization of the probability distribution $f_d(x)$ is a positive number, left undisclosed, and dependent on other features of the problem (Q). (If you had it, you could normalize your $f_d(x)$ with it, assuming that mattered to you.)
Likewise, now, since there are 2 valence u quarks, it makes sense to scale their probability distribution normalization by a factor of 2, so you count them in all, $\int dx ~f_u(x)=2$, but, given the sea, again, $$ \int dx ~ ( f_u(x) -f_{\bar u}(x))=2, $$ etc; so that, at the end of the day, $$ \int dx ~ ( f_u(x) -f_{\bar u}(x) + f_d(x) -f_{\bar d}(x) +f_s(x) -f_{\bar s}(x)+ f_c(x) -f_{\bar c}(x)+ ... ) =3, $$ for 3 constituent quarks in all.
- Each f is a positive probability, suitably normalized, by assumption, and obeys the rules of probability.
[By contrast, GPDs in phase space are not, and violate probability axioms with aplomb, but since you did not ask, let's leave them out for now...]
answered Apr 6, 2020 at 18:58
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$\begingroup$ Hi Cosmas How does your statement 'Each $f$ is a positive probability, suitably normalized, by assumption, and obeys the rules of probability' depend on the order of perturbative QCD that one is working at? Beyond LO, I know that pQCD radiative corrections upset the probabilisitic interpretation of parton distrubutions. See e.g. penultimate slide of www2.physics.ox.ac.uk/sites/default/files/2014-03-31/…. I have never understood why, especially, as when defined rigorously as matrix elements of number density operators in the light cone gauge $\endgroup$– CAFCommented Feb 16, 2021 at 14:31
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$\begingroup$ @CAF There are subtleties which merit a separate question, but "basically" you may keep a probabilistic picture, sum rules, normalizations, etc.. to all orders in perturbation theory; there is also fuss and muss outranging this picture. The cautionary statement you cite does not imply that "anything goes"! Good reviews step in where angels fear to tread... $\endgroup$ Commented Feb 16, 2021 at 14:40
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$\begingroup$ Is it possible for the quark distribution of a hadron to be unnormalizable, due to its sea component? If the same divergence appears in both the quark distribution (sea component) and anti-quark distribution, they would cancel and could satisfy the relation $\int (f_q(x)-f_{\bar{q}}(x))dx=1$. That would suggest though, an infinite total number of quarks and antiquarks in the hadron, which doesn't seem possible if the quarks are massive, but what if they were massless? $\endgroup$ Commented Dec 28, 2021 at 21:58
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$\begingroup$ Could you tell me the source where I could verify the part that probability distribution functions are scaled? $\endgroup$ Commented Dec 5, 2023 at 13:51
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In light cone gauge parton distribution functions are matrix elements of number operators inside a hadron, so they appear to be manifestly positive, as expected from the naive parton model. There is a subtlety, because these matrix elements are divergent, and require counter terms. As a result, I do not think that there is a formal proof of positivity.
Of course, parton distributions determine cross sections, and cross sections must be positive.
