# I Higher order terms in perturbation theory (QFT)

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1. Apr 29, 2017

### Frank Castle

I'm fairly new to QFT and I'm currently trying to understand perturbation theory on this context.
As I understand it, when one does a perturbative expansion of the S-matrix and subsequently calculates the transition amplitude between two asymptotic states, each order in the perturbative calculation simply provides corrections to the leading order contribution, taking into account all the possible interactions that can take place in between the initial and final states. Provided that, as the order increases, the contributions become more and more suppressed relative the the leading order term, them we can trust our perturbative calculation and furthermore the first few terms in the expansion provide a good approximation to the true value of the transition amplitude.

First of all, is what I wrote in the previous paragraph the correct intuition for what's "going on"?!

Secondly, do the higher order contributions correspond to anything physical other than describing all the possible intermediate interactions that can take place between an initial and final state? Do they correspond to more energetic interactions or is it simply analogous to perturbation theory in QM in that the higher order terms just serve as corrections to the leading order term due to the quantum nature of the system?

Apologies if this is a stupid question, just want to get a correct intuition for what's going on

2. Apr 29, 2017

### Staff: Mentor

Most of the time, yes.

If you look at differential cross sections, there are regions of the phase space that cannot happen at leading order, for example a net transverse momentum of the final state particles considered. The leading order does not provide a good approximation there, it will be completely wrong.
But most of the time leading order gives you some first approximation, next to leading order is more accurate, and so on.
How could it get more physical than that?

It is exactly perturbation theory.

3. Apr 30, 2017

### Frank Castle

Sorry, what I meant was that do higher order terms correspond to more energetic interactions or is the physical interpretation simply that they describe all possible intermediate interactions that can occur (exchange of gauge bosons, virtual particle-antiparticle loops, etc.) inbetween the initial and final states, and their inclusion simple serves to provide a more accurate description of the physical process?!

Can one interpret the leading order term as the "classical" interaction (that may or may not be physically possible classically due to conservation of energy and momentum), and then higher order terms correspond to quantum corrections (or "radiative" corrections) which take into account interactions (that wouldn't occur classically) that can occur due to the quantum mechanical nature of the system?

Last edited: Apr 30, 2017
4. Apr 30, 2017

### Staff: Mentor

Some higher orders (e. g. involving heavy particles) only become relevant at higher energies, but in general they are not related to the energy scale of the process.
Sometimes.

5. Apr 30, 2017

### Frank Castle

Ah ok, so they simply correspond to corrections to lower order terms, and physically to the possible intermediate interactions that can take place in between the initial and final states. (Sorry to go on about this point, but I just want to check that I've understood this correctly).

Is the philosophy that we wish to mathematically model an interaction by assuming that it results from a small perturbation of the "free" (i.e. no interactions) case. Given this, one then formally expands in terms of some coupling constant, with the assumption that as the order increases the contribution becomes more and more suppressed, such that one can reasonably approximate the interaction by calculating the first few terms in the expansion?!

When is this interpretation reasonable?

6. Apr 30, 2017

### Staff: Mentor

Right.
I'm not sure if we "wish" that. It is the most accurate method we have to make calculations (if the coupling constant is small).
Yes. That only works if the coupling constant is small, of course.
The Rydberg energies in atoms are the result of the 1/r potential, higher orders lead to the fine structure, for example.

7. Apr 30, 2017

### Frank Castle

Ok, I think I understand a bit better now.

So, in the case where the coupling isn't small, the perturbation isn't small since higher order terms aren't sufficiently suppressed with respect to leading order terms and hence the approximation breaks down. Is this what is meant by an interaction being non-perturbative, i.e. one cannot approximate by a perturbative expansion of the S-matrix?! Are there are reasons why a theory can be non-perturbative other than the coupling constant being large?

8. Apr 30, 2017

### Staff: Mentor

Right.
Not to my knowledge.

Without practical relevance, but interesting in terms of mathematics: the pertubative series in terms of Feynman diagrams is not convergent. After many orders, the number of diagrams grows too fast, and increasing orders lead to larger contributions instead of smaller ones. But we are far away from the point where this would become relevant.

9. Apr 30, 2017

### Frank Castle

Ok cool, thanks for the insights!

On a slight tangent (perhaps I should put this in a separate question), what is the reasoning behind requiring that free-field Lagrangians are at most quadratic in the fields? Is it simply because this ensures that the EOM for the field are linear and hence the solutions satisfy the superposition principle implying (at least in the classical) sense, that wavepackets do not interfere with one another as they propagate past one another, i.e. they are free-fields?!

Furthermore, what is the motivation for including the term $\frac{1}{2}m^{2}\phi^{2}$ in the free-field case? I get that the parameter $m$ is attributed to the mass of the field a posteriori, but is the reason for the inclusion of such a term in the first place? Is it simply because a priori there is no reason not to - one should include all possible terms up to quadratic order?! Or is there also some physical motivation as well, in that from quantum mechanics, the wave function of a relativistic particle (of mass $m$) should satisfy the Klein-Gordon equation?!

10. Apr 30, 2017