# I Derivation of splitting function from cross section

1. Nov 9, 2016

### CAF123

Consider the real emission correction to the tree level process $e^+ e^- \rightarrow q \bar q$ involving a final state gluon emitted from either the outgoing quark or antiquark line. The differential cross section for producing a quark with fractional energy $x_1$ in the $q \bar q g$ final state is $$\frac{1}{\sigma_0} \frac{d \sigma}{d x_1} = \frac{2 \alpha_s}{3 \pi} \int dx_2 \frac{x_1^2 + x_2^2} {(1-x_1)(1-x_2)},$$ where $\sigma_0$ is the tree level cross section for $e^+ e^- \rightarrow q \bar q$. The singularities present for $x_i \rightarrow 1$ are associated with soft and collinear divergences which are removed upon consideration of virtual corrections to the tree level process for $q \bar q$ production. Thus, $$\int_0^1 dx_1 \frac{d \sigma_R^{(1)}}{d x_1} + \sigma_V^{(1)} = \frac{\alpha_s}{\pi} \sigma_0 = \text{finite}$$ This may be rewritten as $$\int_0^1 dx_1 \left( \frac{d \sigma_R^{(1)}}{d x_1} + \left( \sigma_V^{(1)} -\frac{\alpha_s}{\pi} \sigma_0 \right) \delta(1-x) \right) = \text{finite}$$ Now, define $$F(x)_+ = \text{lim}_{\beta \rightarrow 0} \left( F(x) \theta(1-x-\beta) - \delta(1-x-\beta) \int_0^{1-\beta} F(x')dx' \right),$$ where $\beta$ acts as a regulator. We can then write $$\frac{1}{\sigma_0} \frac{d \sigma^{(1)}}{dx_1} = \frac{1}{\sigma_0} \left(\frac{d \sigma^{(1)}}{dx_1}\right)_+ +\alpha_s R \delta(1-x)\,\,\,\,\,(1)$$ where $R = \sigma_0(1+\alpha_s/\pi)\,\,\,\,(2)$ and $$\int_0^1 dx_1 \left(\frac{d \sigma^{(1)}}{dx_1}\right)_+ = 0\,\,\,\,\,(3)$$ and so $$\left(\frac{d \sigma^{(1)}}{dx_1}\right)_+ = \frac{\alpha_2}{2\pi} P_{q \rightarrow qg}(x_1) \cdot L + \alpha_s f(L),\,\,\,\,\,(4)$$ with $L$ the logarithmically divergent piece and $f(L)$ a left over function depending on how the z integration was regularised.

Most of this was cited from https://www.ippp.dur.ac.uk/~krauss/Lectures/QuarksLeptons/QCD/DGLAP_2.html. I'm just wondering if someone could explain where equations (1)-(4) come from? I think (3) is clear from the definition of the plus distribution but I am not sure how (4) comes from the earlier equations and how (1) and (2) are initially obtained.

Thanks!

2. Nov 10, 2016

### RGevo

Firstly, I think eq. (1) is wrong, since the second part on the RHS (which is meant depict the born+virtual alpha_s correction) would actually
lead to two powers of alpha_s here (which can't be, since its a single gluon emission). In addition, the dimensions don't seem to match up for each term.

The splitting functions are functions which can be derived from squaring matrix elements (and only some diverge when integrating over them from 0-1). The divergences in the real emission appear when performing the phase space integration (integrating over x, z), which at the end of the day cancel against those present in the virtual corrections (this is the KLN theorem).

It can be more clear to perform independently the virtual correction (with its divergence isolated in some way) and combine this the real emission piece integrated entirely over the phase space (with its divergence isolated). Then you see specifically that they cancel. I can suggest trying to do the virtual calculation your self. I think there are more clear examples - I like the discussions in https://cds.cern.ch/record/454171/files/p53.pdf.

If you do the calculation differentially, then you have to perform some tricks to combine the two (regulate each piece separately with a beta-regulator in this language). This would allow you to do:
1) Evaluate the contributions from Leading Order + Virtual (+divergent part ) + the real emission part in the divergent limit (which depends on beta). (the divergent parts cancel here)
2) You combine this with the real emission part not in the divergent limit (which would also depend on beta)
In total you get an answer which is not divergent, and not dependent on beta.

Confusing right?

3. Nov 11, 2016

### CAF123

Hi RGevo, thanks for reply, I hope we can continue this discussion.
As far as I understand, the KLN theorem states that provided we make an inclusive summation over all (degenerate) initial and final states our end result will be free of divergences (it will be 'infrared safe'). In the case of $e^+ e^- \rightarrow q \bar q$, the order $\alpha_s$ corrections amounts to considering final state gluons as well as the interference of the born level result with the self energy gluonic corrections and QCD vertex correction (the former we set to zero assuming we have a consistent UV regularisation and renormalisation scheme in place). Their sum is supposed to be finite as is expressed in the equation $$\int_0^1 dx_1 \frac{d \sigma_R^{(1)}}{dx_1} + \sigma_V^{(1)} = \text{finite}$$ I am interested in doing a proper analysis to see this is true but why should this be finite? The KLN theorem states that we have to sum over all degenerate initial and final states. So I have two questions:
1) We did not sum over all degenerate initial states so why is the KLN theorem not violated? - that is, why is the result finite even though we didn't consider initial state degeneracy (e.g degeneracy associated with an electron state and an electron emitting soft/collinear photon(s)) ?
2) What does degenerate mean exactly? I understand that a quark together with a cloud of collinear gluons is degenerate with a quark but in the computation of the $e^+ e^- \rightarrow q \bar q g$ cross section, we make a phase space integral over all momentum of the gluon so wouldn't this result in summing over configurations where the gluon contribution is no longer degenerate with just a quark (i.e contributions where the emitted gluon is no longer collinear with the quark)?

Thanks!

Last edited: Nov 11, 2016
4. Nov 11, 2016

### RGevo

Hi CAF,

These are important questions.

1) For the first. The KLN theorem in this case is telling us that we should not consider the real photon corrections without the corresponding virtual corrections.

In this case, one would consider photon emissions from the electron lines, and also the final state quark lines. This real emission contribution (as you say) has infrared divergences associated to the photon becoming soft and/or collinear with either electron or quark lines.

One would then have to also consider the virtual photon corrections to both the initial and final state (a photon correction to the eeGamma/Z-vertex or the qqbarGamma/Z-vertex). In addition, one should also include the photon correction between the initial electron and final state quark lines (a box-diagram in this case).

Assuming we had done the UV renormalisation/regularisation first. Then this would only be finite after I included all real emission + virtual corrections to this fixed-order in perturbation theory. Which would be:
alpha^3

The case in the previous example (the QCD correction) was:
alpha^2 alpha_s

So practically, the KLN theorem is telling you to simultaneously compute real emission diagrams and virtual diagrams to a particular order in perturbation theory.

2) If I understand correctly, the contribution which is not degenerate corresponds to the contribution when the q qbar and gluon partons are all distinguishable. Experimentally, this would be when you reconstruct three distinct jets. When the partons are not soft/collinear, there are no infrared divergences. So this phase space contribution gives you the cross section for the three-jet rate (while the soft/collinear pieces + the virtual are together giving you the two-jet rate).

Does this make more sense?

5. Nov 12, 2016

### CAF123

Ok, thanks, so if I understood correctly, if I just consider the emission of a photon off an electron line but do not also consider the virtual correction to the eeGamma/Z vertex then I would end up with a divergence in my result?

In $e^+e^- \rightarrow q \bar q$ annihilation at $\mathcal O(\alpha^2, \alpha_s)$, we don't consider any photon emission off the electron line/ vertex correction - is that because at this order we simply don't have such contributions? So at higher orders in perturbation theory, i.e at $\mathcal O(\alpha^3, \alpha_s)$ we would need to consider these photon emissions off the quark and electron lines and the respective vertex corrections to get finite result? I think that is basically what you wrote but I just wanted to rewrite it to see if I understood you correctly.

If I have two soft emissions off the electron line, what is it that cancels the divergence in this case?

In the phase space integral over the momentum k of the emitted gluon, the regions of k around zero (ie those below the factorisation scale or below the experimental resolution parameter etc) are those which give the soft/collinear divergences which show up as poles in epsilon for example in dim reg. These poles then cancel those in the virtual eeGamma/Z g vertex. The finite pieces of this integral (k above factorisation scale) contribute to the 3 jet event? I’m just wondering why we even perform a full phase space integral over k in the first place when all we want to describe is a degenerate gluon+quark state and quark state? Because even performing the full phase space integral the finite pieces are not just neglected in the $e^+ e^- \rightarrow q \bar q g$ correction so we seem to be including a sector of the integral in the correction which does not contribute to a degenerate state.

Thanks again!

6. Nov 13, 2016

### RGevo

Hi again,

It is correct that you should consider all diagrams which occurs at a given fixed-order in the expansion. So, as you say at alpha^3 alpha_s you would consider:

real-real (so a QED and QCD emission of a photon and gluon respectively)
virtual-real (either a virtual QED + real QCD or vice versa)
virtual-virtual (so both QCD and QED)

Typically the order is defined at the level of the squared amplitude. So I would have to consider in the virtual-virtual for example:

1-loop virtual QCD amplitude * 1-loop virtual QED amplitude
also
2-loop virtual (QCD*QED amplitude) * born level amplitude
...etc.

Regarding the phase space integration, let's consider an observable relevant for the ee>qqg final state. We do not observe partons (but jets), and so we must also do the same for our fixed-order prediction. Let us now define the observable delta_phi between the two leading final state jets (say leading in transverse momentum).

For the simple ee>qq prediction, all configuartions are back-to-back in phi (since only two final state partons). If we compute the cross section as:
dsigma/d deltaphi, then the cross section all falls in the back-to-back bin. Now we consider our QCD correction, it is true that a large part of the QCD correction (the virtual, the soft+collinear+also some finite contribution) ends up in the first bin - corresponding to two back-to-back jets. However, there is also a contribution in the other bins corresponding to configurations of three well seperated jets and dphi of the two leading jets != pi.

To get the inclusive cross section, you must also consider these configurations. These contribute to the total cross section, and also give you more differential information about the process - which you can measure to test the underlying theoretical predictions. If I wanted to do this as a small project, I would to the following:

1) Compute the born, real, virtual amplitudes
2) Renormalise the virtual amplitudes
3) Squared up my renormalised amplitudes to order alpha^2 alpha_s
4) Perform a regularisation (so typical approaches are dipole subtraction or the old school phase-space slicing method) procedure
5) Perform a numerical integration of my squared matrix elements with a numerical algorithm such as Vegas, applying whatever
cuts I was interested in. This would allow me to have a fully differential description of ee>qqbar to NLO in QCD.

I could instead at step (3) perform the phase space integrals of the 2-body (qqbar) and 3-body (qqg) in d-dimensions. This would allow me
to analytically isolate the infrared divergences. The infrared divergences from the three-body integration (corresponding to the soft and
collinear divergences) are cancelled with equivalent 'soft and collinear' divergences in the one-loop integrals. The one-loop integrals
are also typically performed in d-dimensions and the (UV and infrared) poles isolated as 1/eps poles. Different one-loop integrals, (depending
on whether the internal fermion lines of the integral are massless for example) have different infrared/UV structure.

Again, a lot of information to digest/understand for your self.

7. Nov 25, 2016

### CAF123

Thanks it makes sense to me. Upon looking back at the total cross section for $e^+ e^- \rightarrow q \bar q$ annihilation I came up with another question -

the total cross section for this process comes about from considering all QCD permissible final states so e.g. $q \bar q g, q \bar q q \bar q, q \bar q gg$ and so on. We have more particles in the final state other than just a $q \bar q$ to reflect the fact that soft or collinear emissions of these real particles would be undetectable in the experiment with the result of not being distinguishable from a virtual correction. This mathematically shows up as having divergences in the results. (I think what I wrote here is correct).

It is commonly said in the literature and books that the $\mathcal O (\alpha_s)$ correction receives contributions from the square of the two single real emission diagrams and the interference of a virtual correction with the born level result. This makes sense. The divergences arising in each of these contributions are then seen to cancel. What I was wondering was why don't we also consider the interference between a graph with two gluon real emissions and the born level result. This would also be $\mathcal O(\alpha_s)$.

Thanks again!

8. Nov 25, 2016

### RGevo

Hi Caf,

Glad I was of help. In this case, the `double real' (since two emissions) would contribute to the matrix element at level $\mathcal{O}(\alpha_s)$ as you say. However, it won't contribute to the cross section until $\mathcal{O}(\alpha_s^2)$,
since the specific distinct final state:
e-e+ > q qbar parton_i parton_j only interferes with other diagrams with this specific initia/final state configuration.

In other words, the born and the virtual interfere since they are both ee > qqbar. Same for the 2> 3 (it interferes with its self and so on).

9. Nov 26, 2016

### CAF123

Thanks! So if I let A be the diagram where two gluons are emitted from a quark line and B the diagram where they are emitted from the antiquark line then |A+B|2 will contribute at $\mathcal O (\alpha_s^2)$ to the cross section and I suppose the divergences present in this result will cancel against those present in the interference of the born level result and a diagram where we have a double gluon vertex correction?

Does the fact that only diagrams with the same initial/final state interfere come from elementary quantum mechanics? Like for example in the double slit experiment, we put light in a state |i> through a diffraction grating and observe the position |f> of the light quanta hitting the detector on the screen but not through which slit it went. So we have a quantum interference between the two possible paths from |i> to |f>.

Or e.g interference between 2 -> 2 t and u channel processes where we have same initial and final state but different attachments of final state external legs to the vertices.

Similarly here we have quantum interference at each order in the strong coupling between what happens between the incoming e^-/e^+ pair and the outgoing q qbar pair.

Thanks!

Last edited: Nov 26, 2016
10. Feb 13, 2017

### CAF123

Hi RGevo,
I see that in going from the second to the third equation in my OP (that is also written in the link I posted at the bottom of the OP), there is an input of a delta function. It seems to me that the replacement $\int_0^1 dz\, \delta(1-z) =1$ is made. I'm not sure why this step is justified because the zero of the delta here occurs at the end point of the integration limit so is not well defined. Yet I have seen this step made in many physics books/resources in the discussions of virtual and real cancellations.
Do you have any comments regarding this?
Thanks!