Rooting out the Answer

I have a new paper out today, with Jacob Bourjaily, Andrew McLeod, Matthias Wilhelm, Cristian Vergu and Matthias Volk.

There’s a story I’ve told before on this blog, about a kind of “alphabet” for particle physics predictions. When we try to make a prediction in particle physics, we need to do complicated integrals. Sometimes, these integrals simplify dramatically, in unexpected ways. It turns out we can understand these simplifications by writing the integrals in a sort of “alphabet”, breaking complicated mathematical “periods” into familiar logarithms. If we want to simplify an integral, we can use relations between logarithms like these:

\log(a b)=\log(a)+\log(b),\quad \log(a^n)=n\log(a)

to factor our “alphabet” into pieces as simple as possible.

The simpler the alphabet, the more progress you can make. And in the nice toy model theory we’re working with, the alphabets so far have been simple in one key way. Expressed in the right variables, they’re rational. For example, they contain no square roots.

Would that keep going? Would we keep finding rational alphabets? Or might the alphabets, instead, have square roots?

After some searching, we found a clean test case. There was a calculation we could do with just two Feynman diagrams. All we had to do was subtract one from the other. If they still had square roots in their alphabet, we’d have proven that the nice, rational alphabets eventually had to stop.


So we calculated these diagrams, doing the complicated integrals. And we found they did indeed have square roots in their alphabet, in fact many more than expected. They even had square roots of square roots!

You’d think that would be the end of the story. But square roots are trickier than you’d expect.

Remember that to simplify these integrals, we break them up into an alphabet, and factor the alphabet. What happens when we try to do that with an alphabet that has square roots?

Suppose we have letters in our alphabet with \sqrt{-5}. Suppose another letter is the number 9. You might want to factor it like this:

9=3\times 3

Simple, right? But what if instead you did this:

9=(2+ \sqrt{-5} )\times(2- \sqrt{-5} )

Once you allow \sqrt{-5} in the game, you can factor 9 in two different ways. The central assumption, that you can always just factor your alphabet, breaks down. In mathematical terms, you no longer have a unique factorization domain.

Instead, we had to get a lot more mathematically sophisticated, factoring into something called prime ideals. We got that working and started crunching through the square roots in our alphabet. Things simplified beautifully: we started with a result that was ten million terms long, and reduced it to just five thousand. And at the end of the day, after subtracting one integral from the other…

We found no square roots!

After all of our simplifications, all the letters we found were rational. Our nice test case turned out much, much simpler than we expected.

It’s been a long road on this calculation, with a lot of false starts. We were kind of hoping to be the first to find square root letters in these alphabets; instead it looks like another group will beat us to the punch. But we developed a lot of interesting tricks along the way, and we thought it would be good to publish our “null result”. As always in our field, sometimes surprising simplifications are just around the corner.

10 thoughts on “Rooting out the Answer

  1. Alf Johanson

    Extra dimensional, to me, is any dimension beyond 3+1, Even the “+1”, time dimension (i^2 t^2) could be considered abstractly orthogonal to 3-space, and therefore extra-dimensional, so where should the line be drawn?
    You mentioned in a post from sometime earlier this year that Nima had found a small negative component to the amplituhedron. I am wondering if the negative roots you have found to cancel out might, in fact, be the same objects that appeared in Nima’s research.
    The idea of extra dimensions enters in as my understanding of the amplituhedron is that it is positive real, so a negative component would seem to be off shell, or extra dimensional.
    Since your mathematical formulation has reduced the integrals to real components, I am still wondering if the negative roots might appear as some other additively canceling objects that might have an extra dimensional basis.


    1. 4gravitons Post author

      I don’t think it’s all that helpful to view complex numbers as extradimensional. You can represent them with extra dimensions, and sometimes that’s helpful, but physically they don’t have to correspond to an extra dimension. In particular, it doesn’t correspond to off-shell vs. on-shell: there are some amplitudes (for example, for just three particles) where writing them on-shell means you actually need to use complex numbers.

      I don’t remember what the negative part of the amplituhedron mentioned is a reference to. There was another space, not the same as the amplituhedron, where he noticed a surprisingly small area that was negative. But that’s a very different context, and if I remember correctly was a totally different theory.

      Regardless, the amplituhedron being “positive” in Nima’s sense doesn’t actually mean the amplitude is always positive. It means the amplitude is positive if the particle momenta are “positive”, i.e. if they satisfy certain conditions. The momenta won’t always satisfy those conditions, even without any extra dimensions, so you can still have a negative or even complex-valued amplitude.

      Liked by 1 person

  2. Andrew Oh-Willeke

    “Things simplified beautifully: we started with a result that was ten million terms long, and reduced it to just five thousand. And at the end of the day, after subtracting one integral from the other… We found no square roots!”

    If that isn’t a cry of absolute triumph, I don’t know what is! 🙂


  3. Jan Reimers

    Those two diagrams you show don’t actually look like Feynman diagrams to me. Are they a modern enhancement of Feynman diagrams or are they actually something else, … Penrose diagrams?



    1. 4gravitons Post author

      Indeed, they’re not exactly Feynman diagrams, well spotted!

      This is a notation that has to do with generalized unitarity. The diagrams function like normal Feynman diagrams, but instead of specific particles the lines are sort of generic “any particle” lines (in case you know some QFT: scalar propagators, where we add in some specific numerators given by the wavy lines). It represents a particular Feynman integral, but not one corresponding to an individual actual Feynman diagram.


  4. Jan Reimers

    I should have said “are those Twistor diagrams”, I realize that Penrose Diagrams are something completely unrelated. They guy just came up with too many types of diagrams!!

    Liked by 1 person


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