# I What order should we take when calculating QFT amplitudes?

1. Oct 13, 2017

### davidge

I have been reading about QFT amplitudes. It seems that difficulty increases as we consider more and more terms in the Dyson's expansion for the Scattering operator, and we need to normalize each of them if we want to get a sensible result.

My question is, nature usually uses what order? I mean, what type of Feynman diagrams "is most used by nature"? I imagine that she randomly uses any order for a same type of interaction, at each time. So my question would be better posed as, what type of Feynman diagram occurs with more frequency?

2. Oct 13, 2017

### thephystudent

Nature does not throw a dice and then chouses a diagram to follow. Feynman diagrams are terms in a perturbative expansion, meaning that for every interaction, all diagrams are used by nature simultaneously.

If you're lucky, each additional vertex factor makes the contribution of that diagram smaller, so that the series converges and you can take some cutoff and neglect the most exotic diagrams.

3. Oct 13, 2017

### davidge

So should we use expansion in all orders in any and every case we are eventually considering?

4. Oct 14, 2017

### sandy stone

5. Oct 14, 2017

### davidge

Thanks, sandy.

I didn't want to post a message in that thread, because they are masters and I'm a noob.

From what I read, it seems that those calculations are computer-made. The calculation goes for, say, 430 loops. Is it correct?

6. Oct 14, 2017

### sandy stone

I doubt I know any more about the subject than you do. I just thought you might find the info you were looking for in that thread.

7. Oct 14, 2017

### davidge

Ah, ok

8. Oct 14, 2017

### Staff: Mentor

No, because there are an infinite number of orders, and in most cases you can't find a closed-form formula that sums up all of the infinite number of contributions. The best you can do is, as @thephystudent said in post #2, to hope that the terms get smaller as you go to higher and higher orders, so you can cut off the sum at some finite (and reasonably small) order and still get an answer that's a good approximation to the true one. Computer calculations help you to be able to extend this out to more orders before the computation becomes intractable.

9. Oct 14, 2017

### davidge

Thanks

10. Oct 15, 2017

### king vitamin

Many (most?) perturbation expansions in physics do not converge. They are asymptotic series, meaning that taking finitely many terms gives you an increasingly better answer, but then the series begins to diverge. This is true for most Feynman diagram expansions in QFT as well. This can sometimes be handled using resumption methods.

11. Oct 15, 2017

### Staff: Mentor

The 430 discussed there was a purely theoretical exercise, we have no way of calculating things like that.
5 loops have been done, but only for one particular process where extreme precision is necessary.

Typically calculations are NLO (next to leading order) or NNLO. A few processes have been calculated with NNNLO precision, while some Monte Carlo generators might still work with LO.
As rule of thumb for measurements done at the LHC: LO gives you a rough idea how frequent the process is, NLO gives you a good estimate, and NNLO adds a good estimate about the uncertainty of the calculation.

12. Oct 15, 2017

Staff Emeritus
That's not what an asymptotic series is, nor is it the typical behavior in a Dyson series.

13. Oct 15, 2017

### king vitamin

Perhaps you are misinterpreting my post? I meant that an asymptotic series gives a good estimate after truncating a particular number of terms, and begins to become a worse approximation after that. Both QED and phi^4 theory have this property (as well as perturbing the quantum 1D simple harmonic oscillator with an x^4 term in QM). What I am saying is not anything different than the first sentence in Wikipedia's definition of an asymptotic series, or what Urs Schreiber says here: https://www.physicsforums.com/threa...rimental-relevance.924921/page-2#post-5840472