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I am having trouble understanding why it is consistent to include "Breamsstrahlung" diagrams in computations of scattering amplitudes.

For example, consider the scattering of two electrons to two electrons. According to Peskin and Schroder (I'm looking at chapter 5), it seems that in addition to including one-loop diagrams, you should also include diagrams with radiated photons in the final state. I would think however that these diagrams correspond to an entirely different process, namely a process in which two electrons scatter to two electrons and a photon. Why should these diagrams be included in electron-electron scattering in which no photon is produced? P&S almost seem to argue that they need to be included for the simple reason that, if they didn't include them, they would get a nonsensical (i.e. infinite) result.

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P&S almost seem to argue that they need to be included for the simple reason that, if they didn't include them, they would get a nonsensical (i.e. infinite) result.

Well, I am confident that Peskin and Schroeder not only "seem" to argue in this way but they explicitly and comprehensibly enough write this fact because it is both true and important. It's the most important insight one may say about infrared divergences in the simplest process in which infrared divergences may be discussed.

It's the whole point of this discussion about Bremsstrahlung that one has to include the cross sections of different processes – with different final states – in order to get a finite result for the cross section. Of course, because the final states are different, there is no interference between the different processes and it would make no sense to add the complex amplitudes. If one were adding the amplitudes for different processes, he would be mixing apples and oranges, indeed.

However, one may add and one must add the cross sections of processes without photons as well with those with an arbitrary number of photons (especially the very low-energetic, soft ones) in order to get a finite answer. This is not mixing apples and oranges; this is just about the addition of ordinary probabilities for possible outcomes that are similar, needed to get the "overall probability".

The reason is that in QED, the probability that two charged particles get fully elastically reflected from each other is really zero exactly. As far as the exact result goes, it is 100% certain that some photons are emitted during the process. In fact, the average number of photons that are emitted (the expectation value of the number of photons in the final state) is infinite. You may heuristically say that it's because electromagnetism is a long-range force that never ceases to act. In other words, Bremsstrahlung is always emitted because accelerated charged objects always emit electromagnetic radiation (that's why we have to build collider rings such as the LHC that are as large – and straight – as possible).

When you compute the cross sections perturbatively, what happens? You know that the final result for the pure elastic (photonless) process' cross section is zero. But you obviously don't get zero at the tree level: you get a finite contribution corresponding to the tree-level diagram. The loop diagrams' job is to conspire and drive the nonzero finite initial approximate answer towards zero.

This would suggest that the perturbative expansion breaks down completely: if the leading tree-level term is finite and if the perturbative corrections are relatively smaller and they should be smaller, they can't eliminate the finite term and bring you to zero.

However, the perturbative expansion for the total cross section is actually OK but one must correctly organize it. The ability of the "seemingly small" loop corrections to eliminate the "seemingly large" tree-level term is manifested by infrared divergences of the loop diagrams. Already at one loop, you will see that there is a correction to the amplitude and the mixed term, the tree-vs-one-loop product in the cross section, gives you an infinite negative term to the cross section which drives you towards zero. It's important for you to try to calculate the one-loop diagrams and see the infrared-divergent term – i.e. term whose divergence arises from very small values of the loop momenta.

However, this infrared divergence doesn't signal any problem with the theory.

Instead, all infrared divergences always just tell you that you have asked a wrong question.

You shouldn't have computed the cross section for the process without photons at all. The total cross section is really zero and the tree-level approximation is a bad one. Instead, you should have calculated an inclusive cross section with the same initial state but with an adjustable final state that can contain, aside from the charged particles, an arbitrary number of photons with energy $E<E_{\rm min}$.

Note that if you were imagining that photons with really tiny energies are unlikely to be produced, the limit of the inclusive cross section for $E_{\rm min}\to 0$ that I just described would be the same thing as the cross section for the photonless subprocess only. That would be because it would be infinitely unlikely for the energies of the photons to be infinitely small.

However, in the real world, it is not infinitely unlikely. It's actually guaranteed that there are an arbitrary number (infinite number in total) of photons with arbitrarily low energies, the so-called soft photons. So if you take a small $E_{\rm min}$ and calculate the inclusive cross section – one that allows the soft photons below $E_{\rm min}$ in the final state as well – you get a divergent correction to the cross section from the new processes that exactly cancels the negative divergent correction from the one-loop diagram (interfering with the tree-level one). The divergent terms go away at this level and this success gets repeated at any loop level.

To isolate the actual finite part that is left – a finite term in the cross section that obviously depends on $E_{\rm min}$ (or other parameters specifying what soft photon final states you have still included) but just "mildly" – requires some work but this finite contribution from the loop diagram with $k$ loops gets suppressed by the factor of $e^{2k}$ just like you expect, so multi-loop diagrams get really unimportant very quickly (and they're not amplified by any divergent coefficients anymore because the divergences get subtracted).

So the infrared, long-distance divergence of the photonless process' cross section isn't an inconsistency of the theory. It is just a sign you should have asked a better question. Because the presence of very soft photons with $E<E_{\rm min}$ in the final state can't be experimentally detected if $E_{\rm min}$ is really tiny, the final states with such soft photons are operationally indistinguishable from the photonless final states and the theory actually forces you to acknowledge this fact and compute the total (inclusive) cross section only.

The expansion for "exclusive cross sections" that can't really be measured in the real world – e.g. the cross section for the scattering in which arbitrarily soft photons are strictly forbidden in the final state (which can't be guaranteed by any apparatus) – may lead to a power expansion in $e$ that contains infrared-divergent terms. However, if you compute any cross section that can actually be measured by an apparatus – e.g. one that allows the "undetectable" soft photons with very low energies – QED provides you with a perfectly fine perturbative expansion in $e$ whose all terms are finite and drop geometrically as $e^{2k}$ as expected from loop corrections.

It's one of the "positivist hallmarks" of modern physics that physical theories don't have to provide us with answers to questions that actually can't be measured in the real world. QED – like other theories – gives us a perfectly consistent expansion for the measurable quantities that take all the real-world "in principle" limitations into account (including our inability to detect some very soft photons) and in science, it is not obliged to predict anything else. More generally, this positivist theme is present everywhere in modern physics. For example, quantum mechanics never tells us "what is really going on" before a measurement. It only discusses facts or their probabilities – what can actually be measured – and discourages us from thinking that there are any "hard facts" before the measurement.

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Do you know any other example in physics where on the one hand two processes are in principle different so that there is no interference, but, however, on the other hand considering each of them individually doesn't make sense so that the probabilities of both processes must be compulsorily added up? –  drake Jun 19 '13 at 23:00
    
Dear @drake, I don't know which examples you would agree to classify as "other" example. Whenever it is (almost) guaranteed that some additional objects are created or additional events occur during a process, you must calculate the inclusive probability to get a meaningful number because the probability that the process happens without them (like 2-electron repulsion without any soft photons whatsoever) is simply zero. It doesn't have to be a photon, it can be graviton or another massless particle if one exists. –  Luboš Motl Jun 20 '13 at 4:57
    
In the real Universe around us, everything may be boiled down to processes between particles etc. and the need to include the creation of extra soft massless particles - of any kind that exist (probably only photons and gravitons) - is therefore the only example. In models of different universes than ours (but related), there are other massless particle species (e.g. moduli) and therefore other examples. –  Luboš Motl Jun 20 '13 at 4:58
    
In non-fundamental physics, you will find lots of examples of this "broader principle". For example, when two planets collide, you want to calculate the final velocities or their probabilities not only for the elastic collision which is in principle distinguishable from the inelastic ones but also for inelastic ones in which small pieces of rocks are shot around - because the latter is what will almost definitely happen. The simple reason why we don't spent time with the "clean process" is that its correctly calculated probability is zero. –  Luboš Motl Jun 20 '13 at 5:00
    
First, thank you for your kindness. What somehow bothers me is the following: On the one hand, the output of the "two" processes (with and without soft photons) are experimentally indistinguishable and therefore the probabilities must be added up. On the other hand, the amplitudes are not added up (no interference) because the two processes are "in principle" distinguishable. What does "in principle distinguishable but experimentally indistinguishable" mean in a positivistic theory like QFT? –  drake Jun 20 '13 at 18:49

This is closely related to What's the difference between inclusive and exclusive decays?

To take this from a experimental point of view, you have to ask if the instrument can actually deliver a sample of "pure" events or if it must deliver a mixed sample. In the case of these corrections there is a continuum between a "pure" and "corrected" events, so any instrument will deliver a mixture.

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