# Cubic Matrix Models, Quarks and Leptons

Lee Smolin recently put up an arXiv article, Matrix universality of gauge and gravitational dynamics, 0803.2926. The first sentence of the abstract is “A simple cubic matrix model is presented,which has truncations that, it is argued, lead at the classical level to a variety of theories of gauge fields and gravity.” A cubic matrix model gives nonlinear field equations. From the Smolin paper:

The dynamics cannot be linear because we want its solutions to reproduce those of non-linear field equations. The simplest non-linear dynamics are quadratic equations, which arise from a cubic action. The simplest possible non-linear action for matrices is
$S = Tr(M^3).$

.

This raises the issue of what happens when one attempts to represent the quantum numbers of the elementary particles as the solutions to quadratic equations. In this post we will show that a very simple set of quadratic equations do give the quantum numbers for the fermions.

Nonlinearity and Pure Density Matrices

The central assumption of Smolin’s paper is that one needs nonlinearity in order to get a realistic quantum theory and this implies that a reasonable place to look for an underlying order is in cubic models. My efforts at putting together pure density matrix models also are based on a desire for nonlinearity, but I believe that density matrices are a much more natural way of obtaining nonlinearity in a quantum theory. This needs some explanation.

When one has linear equations of motion, one can take two solutions, and add them together to obtain another solution. Nonlinear equations of motion do not allow this. Linearity is very attractive from a mathematical point of view for this reason; one can do perturbation theory by adding together small contributions.

Since a Hilbert space is a vector space, it is fundamentally linear. Two vectors add to produce another vector. And it is natural to keep that linearity when writing equations of motion (and therefore work with quadratic theories). However, one obtains probabilities from a Hilbert space by taking the squared magnitude of the vectors. The squaring itself is a nonlinear function, and it this nonlinearity that produces the nonlinearity of pure density matrices.

To get a pure density matrix $\rho$, from a state vector $\psi$, one takes two copies of the state vector:
$\rho(x,x') = \psi^*(x')\psi(x)$
or in the bra ket notation, $\rho = |\psi\rangle \langle\psi|$. And therefore density matrices are essentially nonlinear, as compared to state vectors.

If one begins with a normalized state vector (as one should), the pure density matrix one obtains satisfies a nonlinear equation: $\rho^2 = \rho$, the idempotency equation. This is the simplest nonlinear equation one can write. And this is the quadratic equation that we will analyze. If $\rho$ were scalar, there would only be two solutions, 0 and 1. To get more complicated solutions, we need to require that $\rho$ be more complicated. We can do this by assuming that $\rho$ is made up of several elements but we need to define their multiplication law. A natural way of doing this is to use a finite group to define how to multiply the elements of $\rho$.

Permutations on 3 Elements

The permutation group on 3 elements is a group with 6 elements. Thinking of the colors red, green, and blue, let’s label the 3 elements as R, G, and B. The 6 group elements are I, which leaves everything alone, J and K which are the even permutations, and the three elements that each leave one of R, G, or B fixed. These last we can label by the element they leave fixed. This gives us the 6 group elements as {I, J, K, R, G, B} and their operations on {R, G, B} are as follows:

The reason for talking about the permutation group on three elements is by analogy with the baryons. The baryons are composed of three quarks of three different colors forming a bound state. The bound state is maintained by the exchange of gluons. The action of a gluon on a quark is to change its color. This is a 3-body interaction; incoming quark, outgoing quark, and gluon. If we label the quarks according to their color, we can think of the action of the gluons as permuting the colors. For this to work, we have to ignore quark bubbles where a gluon decays into a quark / anti-quark pair, but over the long term, so long as the baryon eventually returns to a condition where it is composed of only three quarks (and no antiquarks), we will have a situation which we can think of as a permutation of the quarks.

Now each of these permutations can be thought of as a collection of Feynman diagrams. And Feynman diagrams define amplitudes, that is, complex numbers. And there is a natural multiplication of Feynman diagrams of permutations according to which one takes the product of the amplitudes as representing the amplitude of a diagram one obtains by pasting together the two diagrams. This is natural in the momentum space and follows from the fact that the Fourier transform of the convolution is the product of the Fourier transforms. Thus the product of two permutations becomes a complex multiplication.

So let us muddy our notation even further, and let I, J, K, R, G, and B be complex numbers; the amplitudes of the corresponding sets of Feynman diagrams. As is usual in quantum mechanics, we have to include all possible things that can happen. So after some long time, we can describe the bound state by these 6 complex numbers. To get probabilities from these numbers we will be taking squared magnitudes so the phases of these numbers may have some arbitrariness to them which we will get back to further on.

The idempotency requirement of a density matrix is that $\rho^2 = \rho$. To translate this into a multiplication on our complex numbers we use the permutation group product rule. Given two sets of 6 complex numbers, say $(I_u,J_u,K_u,R_u,G_u,B_u)$ and $(I_v,J_v,K_v,R_v,G_v,B_v)$, we need to define their product $(I_{uv},J_{uv},K_{uv},R_{uv},G_{uv},B_{uv})$.

All the ways we can get I from two of {I,J,K,R,G,B} are I = II = JK = KJ = RR = GG = BB. Therefore, when we convolute two of our 6-complex numbers, the rule for the I value of the convolution is:
$I_{uv} = I_uI_v + J_uK_v + K_uJ_v + R_uR_v + G_uG_v + B_uB_v.$
The rules for the other 5 convolution products are:
$J_{uv} = I_uJ_v + J_uI_v + K_uK_v + R_uG_v + G_uB_v + B_uR_v,$
$K_{uv} = I_uK_v + K_uI_v + J_uJ_v + R_uB_v + G_uR_v + B_uG_v,$
$R_{uv} = I_uR_v + J_uG_v + K_uB_v + R_uI_v + G_uK_v + B_uJ_v,$
$G_{uv} = I_uG_v + J_uB_v + K_uR_v + R_uJ_v + G_uI_v + B_uK_v,$
$B_{uv} = I_uB_v + J_uR_v + K_uG_v + R_uK_v + G_uJ_v + B_uI_v.$
This defines a multiplication [uv] = [u][v] where [x] is a vector of six complex numbers.

The requirement that a pure density matrix be idempotent becomes a rule that a bound state [b], in order to be self consistent with respect to permutations, need to satisfy [b] = [b][b]. Another way of saying this is that bound states don’t change with time, so their propagators need to convolute with themselves to give themselves. But since we can’t distinguish phases, we need to allow some latitude in our complex phases. We can do this by treating (I,J,K,R,G,B) the same as a complex multiple of the same vector. This generalizes our equation to be something like [b] = k[b][b] where k is an arbitrary complex phase. For this post, let’s restrict k to be real, that is, +1 or -1.

Weak hypercharge

Assuming k=+1, we substitute I, J, K, R, G, B into the above equations to obtain a set of 6 coupled quadratic equations:

If I negate all six numbers I, J, K, R, G, and B, the right hand side will be unchanged, but the left hand side will change sign. Therefore, the solutions for k=-1 are just the same as the solutions for k=+1 but with I, J, K, R, G, and B negated. So I only really need to solve one set of the equations, for k=+1, to get the solutions for both cases.

And what are the solutions to these coupled quadratic equations? The symbolic algebra package I use, Maxima, was unable to figure it out. Perhaps your package is better. But by cranking on them by hand, I eventually found the solutions. There are ten solutions that are “discrete” that is, that consist of single points in the 6-dimensional space. And in addition, there are four solutions that are 2-manifolds. The 10 discrete solutions are:

where w is one third the cubed root of unity (which the LaTeX engine here seems to refuse to let me parse). The four continuous solutions are:

where $\alpha, \beta$ are arbitrary complex numbers, and $\gamma = \sqrt{\alpha^2 + \beta^2 +\alpha\beta-1/12}.$

So can we associate any of the above numbers with the quantum numbers of the elementary fermions? Yes, the values given for I, that is taking k=+1 and k=-1 together: {-1, -2/3, -1/2, -1/3, -1/6, 0, +1/6, +1/3, +1/2, +2/3, +1}, match the weak hypercharge quantum numbers of the elementary fermions exactly, no more, no less. Also, one finds that the weak isospin quantum numbers appear as the sum of the R, G, and B quantum numbers. What’s more, these two sets of quantum numbers match. That is, the elementary particle with a given value of weak hypercharge ends up with the correct value for weak isospin.

This is only part of the story. When one takes into account other ways that the arbitrary phases can be assigned, one ends up with the generation structure. A previous post discusses this: Quarks, Leptons, and Generations.

Can this be simplified? Maybe. Let I be weak hypercharge, and define 3R=3G=3B to be weak isospin, call it T. Then we can reduce the 6 coupled quadratic equations to 4 coupled quadratic equations in I, J, K, and T:

where k=+1 or -1. Beyond that I’ve not simplified it, but I haven’t put much effort on it. I think the relationship with the permutation group on 3 elements is more significant.

The above 4 equations can be thought of as a requirement on a circulant matrix with elements I, J, K, and a democratic matrix with entries T:

From the above it is clear that if we put T=0, we can solve for {I,J,K} by looking for the idempotent circulant matrices. There are 8 such solutions and these are included in the 14 solutions given earlier. The three of them with I=1/3 form a basis for the Hilbert space of 3 dimensions. That is, those three matrices are the pure density matrices one obtains from a basis set for 3-vectors. The remaining matrix, the democratic matrix, is also a pure density matrix (when T=1/3), and it is from a basis for the Hilbert space that is unbiased with respect to the circulant basis. This is another place where Mutually Unbiased Bases (MUBs) shows up in the elementary particles. The explicit solutions to the above equations are discrete. For k=+1, the solutions are as follows:

where w is the cubed root of unity. The solutions for k=-1 are the same, but with all terms negated. If one ignores the J and K values, the results for I and T are exactly the values needed for the weak hypercharge and weak isospin quantum numbers. The particle content is as follows:

To get the particle content for the negative solutions, convert particle to antiparticle by swapping bar for no bar, and switching left L for right R.

There should be a way of eliminating J and K and getting rid of the unwanted multiplicity in those entries. Alternatively, one might look for a set of quadratic equations that gave the weak hypercharge and weak isospin and some color quantum numbers.

The general form of the I, J, K, T equations suggests that we should take a quick look at the simplest set of coupled idempotency equations:
A = AA + BB,
B = 2AB.
There are exactly four solutions to these equations, (A,B) is in {(0,0), (1/2,-1/2), (1/2,+1/2), (1,0)}. In terms of weak hypercharge and weak isospin, these are the quantum numbers of the leptons.

A single quadratic equation typically has two solutions. The above two coupled quadratic equations have four solutions. If we want to model color, we need to have the quarks appear with three colors. This gives a total of 16 particles and 16 antiparticles for a total of 32. So maybe we should look at systems of five coupled quadratic equations.

To add color with 5 quantum numbers, assuming we still have weak hypercharge and weak isospin, the next three quantum numbers are going to be R, G, and B. Each lepton will have R=G=B, while the quarks will have at least one different. The natural values to assign here would be either {+1,-1} or {0,1}.