Primitive Idempotents and Generations

In the previous post, we looked at long term bound snuark states, came up with three coupled quadratic equations in complex numbers I, J, and K:
Quadratic equations for I, J, and K as functions of I, J, and K

noted that these equations are equivalent to a matrix idempotency equation:
Cyclic 3x3 matrix idempotency equation

wrote down the eight solutions:
Eight solutions to the coupled quadratic equations

and promised to interpret them in the next (this) post. Okay.

First, let’s note that the object we are looking at is a representation of an effective virtual propagator for a long term snuark bound state. As such, it has to have the structure of any virtual propagator — it must be a density matrix. Density matrices are idempotent. For example, in the usual spinor notation (and using the example of a spin-1/2 state oriented in the +x direction) we have |+x><+x| , and this squares to itself. Put in matrix form with the Pauli matrices this amounts to:
Idempotency equation for Pauli density matrix +x

All the Pauli density matrices are idempotent, but not all idempotent 2×2 matrices are Pauli density matrices. Three counter examples illustrate the three ways that a 2×2 matrix can be idempotent but not a Pauli density matrix:
2x2 idempotent but not Pauli densiy matrix examples

These examples are the zero matrix, the unit matrix, and a non Hermitian idempotent matrix. None of our solutions produce a 3×3 cyclic matrix that is non Hermitian so we will not discuss this further (though the topic is very attractive). The zero matrix would have corresponded to a zero spinor and no information is contained. The corresponding 3×3 cyclic matrix is our first solution I = J = K = 0. As an operator, the zero matrix maps all spinors to zero.

The unit matrix is not a density matrix because its trace is 2 instead of the required 1. The corresponding cyclic matrix is our last solution, I = 1, J = K = 0. As an operator, the unit matrix is trivial.

If we trust the principle that a density matrix needs to have a trace of 1, then 3 of our 8 solutions are valid density matrices, namely the ones with I=1/3 because the trace of the cyclic 3×3 matrix is just three times I. This is what we will decide, but it is worthwhile explaining the choice another way.

Density matrices that are not pure can be used to represent statistical mixtures. These matrices still have trace 1 and so do not include the unit matrix, but there is a statistical density matrix that is a multiple of (i.e. half) the unit matrix. This is the mixture one gets by taking 50% spin up and 50% spin down (or 50% spin +x and 50% spin -x, etc.). Writing this matrix as a sum of 0.5 spin up and 0.5 spin down we have:
Statistical mixture density matrix as sum of spin up and down

Schwinger’s Measurement Algebra

There is a not very widely known version of quantum mechanics invented by Julian Schwinger that uses operators to represent quantum states. In this, “measurement algebra”, a quantum state is represented by the projection operator that picks out that particular state. In this theory, the zero and unit operators have an interpretation.

The zero operator is the measurement that does not allow any particles to pass. In the language of particle physics experiments, it is a beam stop. The one operator is a null measurement in that it allows any particle to pass. This is a free beam.

In the measurement algebra, two measurements are incompatible if they multiply to zero. From an experimental point of view, this would happen if one installed a spin-up filter after a spin-down filter. Any particle with spin up will be incompatible with spin down, so the multiplicative product of a spin up filter and a spin down filter is zero. Put into Pauli matrix form, the product is:
Measurement product of spin up and down is zero

In this interpretation, you can always take two incompatible measurements (which correspond to a beam with only those quantum states present) and add them together to make a composite measurement (which corresponds to a beam with two composite states present). An experimenter could do this with a beam by combining two beams into one.

These interpretations are about a beam with a lot of particles in it. Our theory needs to be about particles one at a time. But the interpretation of quantum mechanics is that the identical particles in a beam are characterized identically. Looking at that, a natural interpretation for the unit measurement (Pauli) algebra element is that it is a quantum state with two particles present, one spin up the other spin down. (Or one spin +x, the other spin -x, etc.)

To apply this measurement algebra interpretation to our 8 quadratic equation solutions, requires a multiplication. It is by multiplication that we can determine that two states are incompatible (they multiply to zero). Now we could take “multiplication” to automatically mean matrix multiplication apply this to the cyclic matrix form of our solutions and in fact we would get the right answer. But a more instructive method of defining multiplication of solutions is to take two of our solutions, and convolve them with the laws of QFT.

To convolve two solutions, we can use one solution to represent the Feynman diagrams for a time period of length T, and the other solution to represent the Feynman diagrams for another time period of length T, and then compute the resulting Feynman diagrams for the period of length 2T formed by taking one than the other with the usual rules for multiplying and adding Feynman diagrams.

There are 8 solutions and so 8 possible choices for the solution to be applied to both the first and second time periods. This is a total of 8×8 = 64 multiplications. Each is a fairly simple calculation, but that’s a lot of work. Instead, let’s do one representative product and leave the rest as an exercise for the reader. We will work out the product of the 2nd and 7th solutions, that is, we will convolve the following two solutions:
https://carlbrannen.files.wordpress.com/2007/09/twotoconv.png2nd and 7th quadratic equation solutions

In convolving these two equations, we will need to make 3×3 = 9 complex multiplications, and we will add these up into three results. The convolution is of the form (I,J,K) x (I,J,K) -> (I,J,K). Using, as before, a= \exp(\pm 2i\pi/3) , b = -1/2 \pm i\sqrt{3}/2 , for brevity, here are the calculations:
Details of QFT product of 2nd and 7th solutions to coupled quadratic equations
In short, the convolution product of these two solutions to the quadratic equation is just the first solution, I = J = K = 1/3.

Similarly, in the Pauli algebra, the product of any arbitrary density matrix with the unit matrix is just the density matrix. In the measurement interpretation, that’s simply because that arbitrary density matrix is already present as a component of the unit matrix (which is composite). In our multiplication, our result means that the 2nd solution is a component of the 7th solution which therefore is composite. Our interest is in the non composite solutions, which we call “primitive”. These are the primitive idempotents of the long term bound states of the QFT.

One can apply this method to compute the other 63 convolution products. I guess if you look at the above multiplication, and imagine replacing one of the solutions with zero, it will be clear that the resulting product is zero. Similarly, replacing one of the solutions with the unit solution, that is, the solution I=1, J=K=0, the product will be the other solution. In fact, these act just like matrix multiplication. The remaining products are manageable.

With all this careful analysis, we can now interpret the 8 solutions. The first is just the zero solution. It corresponds to no particle at all. The next three, the solutions with I=1/3 (which have trace 1) are the single particle solutions that we are interested in. The next three solutions have I=2/3 (with trace 2) and are solutions with two particles present. The last is just the unit solution. It corresponds to the presence of all three possible particles, the full beam.

Our interest is in the single particle solutions. As you can see, there are three. These correspond to the three particle generations. Which means we are yet another small step closer to deriving the Koide mass relations.

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2 Comments

Filed under physics

2 responses to “Primitive Idempotents and Generations

  1. Great stuff! I hope you get more recognition for this, and will refer to it in a blog post.

  2. Pingback: A Fictitious Snuark Vacuum State « Mass

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