Monday, 21 January 2013

Physics at Different Scales

Back in November I wrote two posts about the problem of infinities in quantum field theories.  In the first of those, I introduced the two types of infinities that arise: infrared infinities, that happen when a kinematic variable goes to zero; and ultraviolet infinities, that come from the need to integrate (sum) amplitudes over an infinite volume.  In the second post, I explained how infrared divergences are resolved: the infinity occurs in something that is unmeasurable.  For example, it is impossible to distinguish between the production of an electron-positron pair, and the production of the same two particles plus a photon, if the energy of the photon is very small.  When making predictions, we need to include not only the things we do observe, but also the things we cannot; only then can we get well-defined answers.

Today, I want to return to this mini-series and discuss ultraviolet infinities.  I hope to outline the solution to this problem, which is a bit trickier.  This will lead on to the point I really wanted to get to from the beginning, which I'll have to leave for a final post by the end of the month.

Let's review the situation.  In a quantum field theory, we can represent the perturbation series using Feynman diagrams.  Some of these diagrams will have loops; for example,
Sample loop process; the loop is marked in green, and the momentum in the loop may take any value.
These loops involve an integral over all possible four-momenta in the loop.  Such an integral can be unbound, or loosely speaking infinite.  This would seem to suggest that the entire perturbation series is meaningless; yet, we often get pretty good predictions based on the diagrams without loops, the tree-level contribution.

Let me make a small detour to consider the nature of infinite integrals.  It's not unreasonable to think that any integral over an infinite volume is necessarily infinite; but this turns out not to be the case.  As a simple example, consider the integral
$I = \int_1^L \frac{d x}{x^n}$
Where L is a large number we will ultimately take to be infinity.  Using the standard rules for differentiation of simple polynomials, we quickly derive the result
$I = \frac{1}{n} - \frac{1}{n L^{n-1}} $
unless n = 1.  The second term above decreases as L increases for n > 1; in such cases, we can send L to infinity and get a finite result for I.  If n < 1, then we can not do this; the integral for L infinite is not finite.  For n = 1, we need to use logarithms to see what happens; it turns out that the integral is divergent in this case too.

This simple example is a useful guide.  It suggests that, for an integral over an infinite volume to be finite, you need more powers of the integration variable in the denominator than the numerator (counting the differential itself in the numerator).  Strictly this is neither a necessary nor a sufficient condition.  It will meet our needs, however, as the functions generated from Feynman diagrams are generally simple and well-behaved.

We can return to the example above to see how this matches up.  We have three different lines marked in green in out diagram above; one for the muon, one for the anti-muon and one for the photon.  Each of these will give a contribution to the amplitude that depends on the loop momentum, which we label l.  Roughly speaking, the photon line is proportional to one over l squared, and the two fermion lines to one over l.  We also have the integration measure, which, since it is a four-dimensional integral, is roughly l to the fourth.  Neglecting everything else,
$A \sim \int d^4 l \, \frac{1}{l} \cdot \frac{1}{l} \cdot \frac{1}{l^2}$
Counting the powers of l in the numerator and denominator, we get the same value in each case: 4.  Since the denominator does not have a larger value, we expect this integral to be divergent; and since the two powers are equal, the divergence is logarithmic.

Now that we have a clear statement of the problem, what's the answer?  To get there, we'll need to go through a few steps along the way.  First of all, we'll need to do a purely mathematical step to turn our expression for the quantum amplitude into a finite number.  This is known as regularisation, and can be done in multiple ways.  For example, we could simply impose an upper limit on our loop momentum L.  Our amplitude would then have the rough form
$A \sim \mathrm{log} L + \ldots$
The omitted terms are all finite when we take L to infinity.  The hope is that at the end of our calculation, terms proportional to log L or to positive powers of L will all cancel.  Something similar happens with infrared divergences, where it is common to treat the photon as having a non-zero mass and only set it to zero at the end of the calculation.

Unfortunately, in this case there are no extra diagrams we have forgotten to include that can cancel our unwanted terms.  What we have overlooked is more subtle, and also requires a little more detail on Feynman diagrams than I have offered.  One of the objects in our theory is a photon-muon vertex:
Whenever this appears in a Feynman diagram, the Feyman rules tell us to include a factor of the coupling constant e (and a few other things, omitted for clarity).  What is e?  Naively, it seems to be the electric charge; after all, classically the coupling between a charged object and the electromagnetic field is pretty much the definition of charge.  A simplistic derivation based on the Lagrangian would lead to the same conclusion.

But if quantum field theories are the correct description of reality, then they must also describe how we would measure the electric charge.  Essentially, the charge of the muon is given by the sum of all possible diagrams with one external photon, one external muon and one external anti-muon:
The true muon charge is related to the coupling e through the sum of diagrams hinted on the lower line.
In short, the true muon charge that we measure in an experiment is related to the Feynman diagram coupling e by terms proportional to log L.  As long as the true charge is small, we can invert the function implied by the above series and get an expression
$e = e(\mathrm{true})  - e^3(\mathrm{true})  \mathrm{log} L + \ldots$
Substituting this back into our starting expression, corresponding to the first diagram in this post, we find that the unwanted terms indeed cancel.  We can set L to infinity and get a finite result: the ultraviolet divergence has cancelled.

It's somewhat reasonable that the infinities cancel in this casew, because the diagrams are very similar.  Compare the last diagram in the last figure with our initial one, and the resemblance should be clear.  But what about other cases?  Are these infinities under control to all orders in the perturbation series, and for all processes?  A proof like that is pretty hard, and I'm not going to repeat it here.  (I don't even know the proof off-hand.)  It has been proved, before my parents were born in some cases, though the results are interesting; they restrict the types of theories you can write down in a non-trivial way.  I will discuss that, and some of the other implications of this work, in my final post on this subject.

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