15
$\begingroup$

In a particle physics lecture I had today it was stated that the magnetic moment, $g$, is not quite equal to 2, and the difference is accounted for by QED.

Later it was stated that we can see this because as well as first-order decays, there are higher order ones, but using Feynman diagrams and the appropriate rules we can also see that these occur much less often, but this effect is big enough to be measurable rather easily with modern technology.

What primarily confused me was the justification for why these higher order decays aren't as important.

Apparently, for a decay such as $\pi^-p \to \mu^+\mu^-$, there is 1 way to decay for order 1, 72 for order 3, and 12672 for order 4, but with each extra order the chance of a particular decay happening drops by a factor proportional to $1/\alpha^2$. (Sorry, I don't have a source for this besides my lecture notes, but my University doesn't often get things like this wrong.)

It seems to me the number of ways you can make a decay happen increases with some function of the factorial of the order (I have no idea how to calculate this exactly, but factorial seems right), but the chance of a decay happening falls off like a polynomial. So, why doesn't contribution of higher orders tend to infinity? What am I missing? What do I have wrong?

$\endgroup$
  • $\begingroup$ Note that the factor is polynomial in $\alpha$, but exponential in the order, which is what you are worried about. $\endgroup$ – fqq Feb 11 at 18:30
16
$\begingroup$

Congratulations. You have uncovered the dirty little secret of particle physics.

It is generally believed that expanding in a perturbation series (of Feynman diagrams), including successively higher powers of $\alpha$, or whatever the coupling constant is, gives a result which converges, getting closer and closer to the true value. You obtain a power series in $\alpha$, the coefficent of each term being harder and harder to calculate and needing powerful computers and/or clever theorists, but with no limit in principle to the accuracy.

The truth is that this is an asymptotic series. It does not converge. It improves to begin with as higher order corrections are included, but there comes a point beyond which the increased number of diagrams at the next order more than outweighs the extra power of $\alpha$ and the result diverges. So calculating successive diagrams gives you better accuracy, but only up to a point; beyond that things get worse.

Theorists know this but don't talk about it, perhaps because it doesn't do their image any favours. Experimentalist generally are not aware of this limit and continue to believe that their theorist colleagues are (given enough time and computers) capable of producing perfect predictions.

Having said all that, the calculations for $g-2$ are well within the region of continuous improvement at the order to which they are currently calculated.

$\endgroup$
  • $\begingroup$ Thanks for the answer, could you point me towards somewhere i could read more about this? E.g, a specific textbook that covers this well? $\endgroup$ – T. Smith Feb 11 at 19:38
  • 3
    $\begingroup$ Here is Dyson’s classic 1952 paper on this: journals.aps.org/pr/abstract/10.1103/PhysRev.85.631 $\endgroup$ – G. Smith Feb 11 at 20:43
  • $\begingroup$ Ben Carl Bender also talks about asymptotic series and lots of other thing, in case any one is interested in youtube.com/… $\endgroup$ – onurcanbektas Feb 16 at 6:11
6
$\begingroup$

A picture is worth a thousand words. To echo @RodgerJBarlow's point, see below for an illustration (see this paper) of tendency of blowing up in an asymptotic series: enter image description here

Mathematically speaking, one can resort to techniques along the lines of Borel transformation to convert a divergent series into a convergent one.

$\endgroup$
  • $\begingroup$ Question would be though why Borel summation should lead to the physical correct value. $\endgroup$ – lalala Feb 12 at 17:34
2
$\begingroup$

Not all Feynman diagrams for Standard Model processes become more accurate as quickly as in QED.

It takes more loops in QCD to get the same level of precision as one would in QED, and each new loop in QCD is harder to calculate than the corresponding one in QED.

One of the reasons for the difference between the two is that photons don't have interactions with other photons, while gluons do have interactions with other gluons, profoundly increasing the number of possibilities at each loop, which means that there are more diagrams at each loop, each of which can make a contribution to the total.

Also, loops in the series are grouped by powers of the relevant coupling constant. So the first loop terms are a function of the coupling constant, the next loop terms are a function of the coupling constant squared, etc. This is so because the coupling constant related to the probability of one step in a Feynman diagram occurring.

For example, in a two loop diagram the coupling constant related to the probability of the first loop occurring, and also to the probability of the second loop occurring, so the square of the coupling constant is related to the probability of both loops in that diagram occurring, since both probabilities must occur for that two loop diagram to take place.

By analogy if you want to know the probability of rolling a 12 with two dice, you have to have first a six and then another six. So, the probability of that happening is (1/6)*(1/6)=1/36. The coupling constant is analogous to the probability of a particular outcome occurring in one roll of a die.

The coupling constant of QED is further from 1 than the QCD coupling constant is from 1. For example, at the Z boson mass energy scale, the QED coupling constant is approximately 1/127 while the QCD coupling constant is 1/9. (These numbers are fair to compare because both coupling constants are dimensionless numbers and are used in analogous series of Feynman diagrams.)

(1/127)^2 is much smaller than (1/9)^2; (1/127)^3 is much smaller than (1/9)^3, and so on. Therefore, the successive terms in QED, which are multiplied by the relevant power of the coupling constant at each loop, gets much less weight at each new loop relative to the previous terms, than it does in QCD.

For many physics applications, three significant digits is sufficient, which can be reached in QED which just a couple of terms. At four or five loops in QED (which still has far fewer terms than a four or five loop QCD computation which is much less precise because later terms are weighted so much more heavily and have so many more diagrams at each loop), the number of significant digits gets great indeed.

It is also worth noting that the QED coupling constant gets smaller at lower energies (to about 1/137 at the theoretical limit of zero energy), and larger at higher energies. So, higher order Feynman diagrams contribute less relatively speaking in lower energy interactions in QED than they do in higher energy interactions in QED. In orders words, at higher energies in QED (for example, in interactions with the LHC or a future linear collider) you need to consider more Feynman diagrams to get the same level of precision than you would for interactions in a laboratory at room temperature.

As @RodgerJBarlow notes in his answer:

It improves to begin with as higher order corrections are included, but there comes a point beyond which the increased number of diagrams at the next order more than outweighs the extra power of 𝛼 and the result diverges.

But, physicists don't worry about this much because the number of diagrams where this starts to happen exceeds the number of diagrams necessary to consider to produce theoretical predictions more precise than other any theoretical predictions in physics and predictions more precise than we can measure experimentally.

Ultimately, renormalization is a trick by which one can truncate infinities in calculations such as those associated with this divergent series, in a consistent manner, so as to prevent the theoretical possibility of a divergent series from mucking up the amount of precision that can be obtained by calculating out to the first few loops of the series.

For a long time it wasn't definitely known if this trick was mathematically sound, even though it seemed to work just fine whenever we needed it to. But, the mathematical propriety of this truncation of a divergent infinite series in renormallization was established with mathematical rigor sometime within the last half dozen years or so.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.