Hi guys, I've been wondering about this for a while, but can't seem to find any satisfactory answers yet. Example: For nonradiative Bhabha scattering (\(e^+e^- \to e^+e^-\), no gammas), at tree level you get contributions from two diagrams: a photon exchange (scattering) diagram, and a pair annihilation-reconstitution diagram. In exercise 5.2 of Peskin & Schroeder, using Mandelstam variables they write the net differential cross-section as \(\frac{d\sigma}{dcos\theta}=\frac{\pi\alpha^2}{s} \left[ u^2 \left(\frac{1}{s}+\frac{1}{t}\right)^2+ \left(\frac{t}{s}\right)^2+\left(\frac{s}{t}\right)^2\right]\) (also written this way on Wikipedia) The tree-level exchange diagram contributes a divergence to the differential cross-section (t=0) when the virtual photon being exchanged has zero energy and momentum and thus an infinite propagation amplitude, i.e. when the incoming and outgoing particles have their momenta unchanged. This means that, at tree level, you get a divergence in both the differential and integrated cross-sections for low-angled scattering. Same problem happens in Moller scattering when a virtual photon with no energy or momentum is exchanged. In practice, if this cross-section were valid at low angles, it ought to mean you have virtually zero probably of seeing any electrons and positrons scatter at appreciable angles. The divergence is similar to the case of Rutherford scattering at low angles, but in this case you're not dealing with impact parameters and definite outcomes, just hypothetical infinite plane waves at sharply-defined momenta (or wave packets with narrow momentum spread), so the simple workaround for divergence in Rutherford's case won't hold here. All I've read so far is that physicists don't usually look at the ultra low-angle stuff scattering off a beam of colliding electrons and positrons, so it doesn't affect them. But I still have a problem with it, because if it were completely accurate down to angle 0, that means electrons and positrons would never scatter at anything but 0 angle or at least within an extremely narrow angle range, and colliders would need to shoot an infinite number of them around to see interesting stuff. Now I first learned the details of Bhabha scattering 2 years ago and I've had this question in my head for a long time, but I thought at some point the answer would come when I had learned more. My supervisor, for instance, insisted that by adding up the infinite number of soft photon contributions to this process, I could cancel out the zero angle divergence. Yet in all the books I've looked through, it appears soft-photon contributions are only used to cancel infrared divergences arising from certain loop corrections, not tree-level stuff. What I've learned about field strength renormalization doesn't seem to help much either. Can anyone help me out here? Any more info I can provide besides linking to the Wikipedia article? Any insights others could provide would be greatly appreciated, as Bhabha scattering is an important part of my studies and I don't feel like I understand the fine details at this moment.
I think I might have found the answer to my question, but I'd appreciate it if someone could verify it. Basically, now that I recall after looking into it again, I believe that higher order loop corrections involving the photon's vacuum polarization will lead to an effective propagator for a photon with a small mass. That seems to be what the textbooks say, but I'd appreciate it if someone could correct me in the event I'm wrong. So with these vacuum polarization corrections to the photon propagator, you'd have a high differential cross-section for low-angle Moller and Bhabha scattering, but it wouldn't diverge to infinity as you approach zero angle. Thus QED renormalization solves the problem in a way I hadn't originally considered.
Well I did some more reading, and it looks like vacuum polarization is a necessary correction, but it doesn't get rid of the low-angle divergence. For starters, I found out it's referred to as "collinear" divergence in the literature. Secondly, it looks like this divergence is cancelled out by adding diagrams for low-angle photon emission up to a certain cutoff detection angle, much in the way soft Brehmsstrahlung cancels the UV divergences in certain loop corrections. I'm curious if it's possible to compute the integrated cross-section all the way to angle zero, i.e. if this cross-section remains bounded as the experimental photon cutoff angle approaches zero. If you have some QED process with an unbounded cross-section, doesn't that imply any process with a finite cross-section has 0% probability to occur?
I can't offer much in the way of specific comments, I'm too rusty at QED to remember the details of this stuff, but hopefully qualitative comments might be useful. The thing which immediately strikes me about the issue of taking the photon's energy and momentum to zero is that you make the two Feynman diagrams degenerate, they both essentially become the same 4 point interaction. This is the same sort of phenomena you get if you use Fermi's effective theory of weak interactions, in that if you work at much much lower energies than the weak scale then the exchange of say a Z boson is done over such a short scale that it seems like it doesn't exist. In full electroweak theory anything which is mediated by a photon can also be mediated by a Z boson, since they have all the same quantum numbers via electroweak symmetry breaking etc. However, you only see a process which looks like the Feynman diagrams in your Wiki link, but with a Z, when you near 90GeV. As a result Fermi's electroweak approximation has a clear range of applicability in terms of energy. But what happens if you do the same with a photon, ie dialling down the energy? It has no rest mass so in principle you cannot form the same sort of valid effective theory for photon processes as you can for Z. In the Z case its clearly the mass of the Z which is responsible for the effective theory being valid in some energy scale, which you can see by the fact an effective theory typically involves Lagrangian expansions in parameters like \(\frac{Q^{2}}{M^{2}}\) where M is the the energy scale cut off of your theory, which the electroweak scale provides for the Z boson. No such M in QED, the only way you'd get a valid low photon energy limit is if the photon has a rest mass. The renormalisation of the propagator does pretty much the job, though in a subtle way, in that doing the naive \(p^{\mu}(\gamma) \to 0\) limit is not going to agree with the result you get if you give the photon a mass, take the limit and then turn off the mass term. That is certainly a different issue than the usual IR or UV divergences. Turning off \(\gamma\) gives you an effective 4 lepton interaction which relates to the perturbative expansion of your Lagrangian in \(\frac{Q^{2}}{M^{2}}\) except M=0 for the photon, until you do the propagator renormalisation. That make much sense?
I think I understand what you've said so far, especially relating to the Z-boson contribution being negligible at energies \(\ll 90GeV\). For the studies I'm doing, we don't need to worry about these kinds of effects too much, although they're included with the Monte Carlo generator I use. As far as photon mass renormalization goes, I should just advise everyone that what I wrote about it a few weeks ago is crap. I rememeber how you sum up all the possible combinations of 1PI diagram that can be stuffed between the vertices of a propagator, get a geometric sum and ultimately a new renormalized propagator, but I believe I was wrong to assert this propagator is identical to what you'd get for a photon with a rest mass (and aren't there identities like Ward-Takahashi or what have you, that prove it must always be treated as massless?). I wasn't actually thinking about photon mass regularization, because you still let \(m_{photon}\to 0\) at the end with that technique. I think the concept of treating low-angle photons the same way as soft Brehmsstrahlung was the concept I was missing. I've been reading some papers on the subject from arXiv today, plus some Peskin & Schroeder review, and it seems there's a famous paper back from the 1960's written by Yennie, Frautschi and Suura where they introduce and develop the YFS exponentiation scheme in detail, so that might be next on my list. Is it not true that the total cross-section for something to happen in QED must still be finite? I don't pretend to remember all the exact details on how you go from calculating transition amplitudes to scattering cross-sections and decay rates, so I could be terribly wrong about this.
Ok, time for another update now that I've looked into the problem in far more thorough detail. The difficulty I see can be generalized to a much larger variety of QED processes. In general, I think there's a divergence in any scattering process involving a diagram where two otherwise separate processes are connected by a single photon. If you want to find the total integrated cross-section, you have to include all the outcomes where a single zero-momentum photon gets exchanged, and that's why you get zero-angle divergences in tree-level amplitudes for Bhabha, Moller etc. YFS exponentiation with soft Brehmsstrahlung and virtual photon loops will not fix this particular divergence, because those are all higher \(\mathcal{O}(\alpha)\) corrections, and the divergences should separately cancel at every order of \(\alpha\). Same deal for semi-hard Brehmsstrahlung with cutoff angles. The general problem I see is illustrated, but not described or noted, in section 6.2 of Peskin and Schroeder. For instance, you have equation 6.28 for the matrix amplitude of an electron scattering off a very heavy charged fermion, and only exchanging a single photon with that heavy fermion in the process. The rest is about corrections from the electron's own self-interactions and radiation after it changes momenta, which is a separate issue and what renormalization deals with (along with vacuum polarization corrections to the photon which also fail to kill the zero-exchange-momentum divergence). Equation 6.28 reads as follows: \(i\mathcal{M}=ie^2\left(\bar{u}(p')\Gamma^\mu(p',p)u(p)\right)\frac{1}{q^2}\left(\bar{u}(k')\gamma_ \mu u(k)\right)\) \(p\) and \(p'\) refer to the initial and final 4-momenta of the scattered electron, \(k\) and \(k'\) are the counterparts for the heavy fermion. \(\Gamma^\mu\) is a miscellaneous 4x4 matrix absorbing all the possible corrections one could make to the electron scattering vertex, and \(q\) is the 4-momentum for the single photon exchange. I could scan a picture of the Feynman diagram schematics, but I don't want to plagiarize Peskin and Schroeder's book, plus most QFT people probably have it anyhow. Can do it by PM if requested, though. So my issue is, what kills the divergence when \(q^2\to 0\)? When you want to find the total integrated cross-section, that singularity has to be included, which is why you get a \(\frac{1}{sin^4\theta}\) divergence in the tree-level Bhabha cross-section as you go to zero angle. I've looked through several sources and haven't found much, but I did find this one paper which discusses the problem in a slightly different context, and suggests the issue might be with the way cross-sections and asymptotic states are defined in these particular cases. I was wondering about that myself, because there are lots of simplifying assumptions and approximations used in defining cross-sections in the first place, perhaps some of these assumptions break down when you're dealing with the case of a null interaction. Anyway, here's the link on ArXiv. Also on a final note, what I mused in an earlier post about Ward-Takahashi was junk. Ward-Takahashi is the QED version of classical charge conservation, whereas the requirement of a massless photon comes from gauge invariance.
