Infrared Correction to Mass I

Earlier we postulated the snuark mass interaction for a transition from a left to a right handed state as a simple iq vertex stuck between two propagators:
Snuark interaction to first order

We postulated that the mass interaction that converts a left handed lepton to a right handed lepton involved three of these transitions happening simultaneously. We put the left handed snuarks as +x, +y, and +z, and the right handed snuarks as -x, -y, and -z. In doing this, because the snuark interaction forbids transitions between incompatible quantum states (for example from +z to -z), we found that there were only two ways this could happen. We labeled these two complex interactions as J and K. These two interactions amounted to the even permutations on three objects (not including the identity). They were (x,y,z) goes to (y,z,x) or (z,x,y). To distinguish the right and left handed states, we wrote these as (+x,+y,+z) goes to (-y,-z,-x) or (-z,-x,-y). All other permutations on the three objects (x,y,z) were forbidden because they included a forbidden interaction such as +x goes to -x.

Our analysis was correct in that we did find all the final states (two of them), but it was incorrect in that we did not include all the possible intermediate states. In this post and the next, we will sum over these more complicated interactions and compute an effective mass interaction. The calculation is quite similar to a similar calculation in standard QFT: the treatment of soft photons as a vertex correction to the interaction between an electron and a hard photon. See section 6.5 of Peskin and Schroeder.

In our argument for why there are only two possible final states in the transitions from +x, +y, and +z to -x, -y, and -z, we split the transitions into two groups J and K:
J and K sets of mass interactions

Before, we noted that if we took one interaction from the J group and one interaction from the K group, then even though these two interactions were not forbidden, the third interaction would be forbidden. For instance, if +x goes to -y and +y goes to -x (from the J and K sets respectively), then this would imply that +z goes to -z which is forbidden.

Let’s start a complex interaction with +x goes to -y and +y goes to -x. Obviously this is not going to do. But we can fix things by now arranging for -x to go back to +y and then have +y go to -z and then +z goes to -x. This is a total of 5 snuark interactions, so this diagram is of order 5 rather than our previous mass interaction which is of order 3. In Feynman diagrams, this complex interaction looks like this:
Complex mass interaction example 1

The above interaction can also be written as a modification to the free snuark propagators, that is, as a modification to the propagators rather than the vertex. Regrouping the above Feynman diagram, we have:
Regrouped complicated mass interaction to show propagator radiation

Grouped as above, the new interactions interrupt the +y propagator. Therefore, they form a radiative correction to the +y propagator, not a correction to the mass interaction.

When you find a new way of writing quantum mechanics, you use it to do the “easy” calculations first. In our case, that is the Koide mass relationships. The mass calculations are easy because all the pre gravitons are just soft bosons that fly off to infinity and are not picked up in experiments. This means that we only have to keep track of one complicated bound state going into the interaction and one coming out. The fine structure constant, by contrast, will involve three complicated bound states interacting. We will do those calculations after we’ve collected all the low hanging fruit. Or maybe the right phrase is “after we’ve banged up the low hanging fruit with a stick and left them bruised for the efforts of the next person.”

Next to Leading Order Calculation

We look for another correction diagram, again starting with the +y to -x interaction. From here, we need to choose another quantum state U and then transition from that quantum state to -z. We are looking for -x to U to -z. We can’t use -x or -z because these are the end points. We can’t use +x or +z because these are annihilated by the -x and +z propagators. The only possible states we can use for U are +y and -y. We just noted that using +y makes the correction into a radiative correction of the propagator. This leaves only -y:
Complex mass interaction that is not a propagator radiative correction

In our usual diagrams for left to right handed transitions, the left handed particles all have + signs in their direction vectors, while the right handed particles have – signs. The above correction has two -y propagators. The top one is just the usual right handed propagator. But the lower -y propagator, the one in the complicated interaction, is instead left handed.

Our next step is to convert the +y to -x to -y to -z interaction into an effective +y to -z interaction. The usual +y to -z interaction is first order, it takes a scalar vertex of +iq. Our correction to it will be 3rd order since there are now three vertices in the spin chain. Let’s call the effective scalar amplitude k. As before, we will compute the value of k by using Pauli matrices. I don’t actually use Pauli matrices in my calculations, I use geometric methods that are easier. I’ll cover those in a later post.

We will write the constant k as the product of two factors, k = k_p k_q . The first is the contribution due to propagator arithmetic, the second is the contribution due to the various factors of +- iq.

In order to make our notation shorter, sexier, and more natural from the point of view of the measurement algebra, let’s make up a notation to designate a left or right handed snuark traveling in a given direction. For a left handed snuark going in +y, we will use Y_L . For a right handed snuark traveling in the -y direction, we will use \bar{Y}_R . This will allow us to keep track of which snuarks are left and right handed more easily. With this notation, the calculation for the propagator contribution to k is:
YXYZ as k YZ

Translating this into Pauli projection operators (density matrices), and factoring out the factors of 0.5 used in the X and Y propagators, we have:
YXYZ as k YZ in Pauli matrices

Multiply this out and you get:
YXYZ as YZ answer
Evidently, k_p = i/2 .

Now let’s compute k_q , the contribution due to the factors of +/- iq. We assigned different signs to the vertex according to whether the interaction was a part of the J or K interaction types. The rule was x to y, y to z and z to x took +iq, while the opposite transitions took -iq. Examining the Feynman diagram for this term, we have one K type transformation (i.e. +y to -x) and two J type transformations (i.e. -x to -y and -y to -z). This gives k_q = (-iq)(iq)(iq) = iq\;q^2 . The first part, iq, is just the leading order term.

Putting all this together, we have k = iq(iq^2/2) , other than a possible symmetry factor for the identical bosons.

The Other 3rd order contributions

The calculation above was just the correction for a path that went from +y to -x to -y to -z. We need to find the other methods of getting from +y to -z in three 90 degree steps. From the +y, there are four possible steps, namely everything but +/-y. As with our first example complicated mass interaction, some of these paths are going to be radiative corrections to the propagator and we won’t deal with them right now.

It turns out that there are eight ways to get from +y to -z in exactly three right angle steps. They are:
All eight 3rd order corrections to mass vertex

Of the eight paths, four are vertex corrections and four are propagator corrections. We ignore the four propagator corrections but are interested in the vertex corrections. The vertex corrections are marked as “Corrects Vertex” in the above figure.

A Short Respite
If I were giving a lecture in a classroom, here is where I would stop talking about physics or math, and begin talking about problem solving and student homework. One good reason for doing this is so that you don’t burn out the brains of your grad students.

Back when I was a physics grad student, I found QFT calculations difficult. I guess everyone does, and most admit it. A lot of the time we are assigned a standard calculation, one where you knew the answer in advance. Somehow, it seemed that I could never get the answer exactly. I was always off by a factor of 2, or 2 pi, or something like that. I slaved over those problems, but I could never get them to work out exactly.

So I was very surprised when my homework kept getting “A” grades. Eventually I asked the instructor why he was giving me As when I was always wrong. He told me that when these calculations were originally published, they frequently had errors and had to be corrected in later papers. He said he was giving me an A because despite my errors, my homework was the best in the class.

Now back to the problem at hand. It turns out that when I calculate the above four vertex corrections two of them end up with +iq(iq^2/2) and the other two end up with -iq(iq^2/2) . So I’m going to end up with a zero result. Oops. So let’s just assume I made an arithmetic error, or I’ve missed some subtle point about what the boson scalar vertices should look like, and the four contributions are all the same as the above calculation, +iq(iq^2/2) . This gives the total NLO calculation as +iq(2 iq^2)

As to why I get minus signs on half the calculations, I don’t know, but I don’t think it’s worth holding this calculation back until I’ve got every bug worked out of it. It’s clear that there has to be an infrared correction to the mass interaction. What’s not clear is what the correct calculation is. I’m pretty sure that my spin calculations are correct. I don’t do these with Pauli matrices, I cheat; I use a much easier method based on measurement algebra theory. What I’m not sure about is the proper scalar vertex factors for the bare interaction. If you fiddle with these you get the “right” answer but I’m not sure how to justify it.

Symmetry Factors

When one converts a Feynman diagram into an amplitude, one must be careful to account for symmetry. This is a familiar detail to students of QED. These show up because you are going to integrate over all possible momenta and energy (or position and time). For qubits, there is no momenta or position information, so the only integrations are over energy (or time).

One could avoid including symmetry factors by arranging for ones integrals to only add in a given situation once. For example, if we have two identical bosons that can take energies from 0 to 1, then the natural way to write the integral over their energies would look like:
Integral of f(e1,e2) over the square 0 to 1, 0 to 1

But physically, the situation f(0.1,0.3) is identical to the situation f(0.3,0.1) so we are adding it into the summation twice. To fix that, we could fix our order of integration to eliminate that sort of thing. If we want to keep only the contributions where e_1 < e_2 , then we can use:
Integral of f(e1,e2) over the triangle

but since f(e_1,e_2) = f(e_2,e_1) , the above can be rearranged to get:
Integral of f(e1,e2) over square but divided by 2!

which is a simpler integral. More generally, if we are integrating over n energies, we can use a symmetry factor of 1/n! since there are n! ways of reordering n objects and we only want to keep one of them.

The symmetry factors only apply to identical bosons. In our case, we have two pre gravitons emitted in left to right transitions and one emitted in a right to left transition. As we will see when we compute the gravitational force, the left to right pre gravitons will stimulate other particles to have left to right transitions. That is, the left to right pre gravitons have interactions only with left handed particles. And the right to left pre gravitons will influence only right handed particles. Thus the two types of pre gravitons are distinct and there is no symmetry factor between them.

But there are two left to right bosons in the final state, so we have to multiply by a factor of 1/2! = 1/2. The resulting next to leading order contribution is k_3 = iq\; (iq^2) .

That should be enough for the first half of the calculation. In the next half, we do the same thing for the next to next to leading order. Then we’ll make a guess for the higher orders, and sum them up as an exponential. Hopefully, we’ll have time to then compare the resulting exponential to the Koide mass formula, and thereby determine the value for the coupling constant, q.

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2 Comments

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2 responses to “Infrared Correction to Mass I

  1. Kea

    Oooh, goody! Of course I’m assuming you already have the argument for that damned number worked out. And maybe this time its tantalising enough to warrant interest from more people – so the bugs can soon be ironed out. Personally, I’m just too stupid.

  2. carlbrannen

    Kea, that damned number is supposed to show up as an infinite sum that looks like an exponential. This is similar to the usual QFT calculation for soft bosons.

    In addition to my inadequacies as a calculational machine and QFT expert, the truth is that we really aren’t sure how to treat tripled Pauli statistics differently from regular Pauli statistics.

    And if my ideas on tripled Pauli statistics are completely wrong, then the soft boson calculation could give an exponential factor with the usual QFT methods, and that would give an explanation for that damned number.

    And I’ve got a decent argument (but no mathematics) for the ratio of the electron and neutrino masses. I think I mentioned that the way I was doing them, they were coming out reversed.

    The new explanation is that the neutrino snuarks have ixyzt charge +2 while the electron snuarks have charge 0. Thus the neutrinos take a pi/12 while the charged leptons take zero as a geometric angle.

    And the calculations I’m doing here apply to the neutrinos but do not apply to the charged leptons. Instead, the charged leptons have a zero coupling except at higher order (persumably where you split their snuarks into PIs).

    Because the left and right halves of the electron couple in a more complicated fashion, the two halves are unable to exactly cancel each other’s fields as well as the neutrino does. The higher uncancelled fields leads to a higher energy and therefore mass. (Treating mass as a vector ala Michael Rios.)

    Suppose you have two oppositely charged spheres separated by a distance R. The total energy of the field created by them depends on R. The smaller R is, the less the total field energy. (Is that true?)

    A weaker coupling between left and right makes for a bigger object, hence a higher total field energy, hence a higher mass.

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