If a Hilbert space is d-dimensional, we expect that the number of elementary particles we can describe with it is d. For example, a Pauli spinor is 2-dimensional and electrons come in two states, spin-up and spin-down. The Dirac spinors are 4-dimensional and the related theory describes electrons plus their antiparticles, and therefore 4 states.

So long as we think of elementary particles as the things that can only be represented by spinors (i.e. the 2×1 or 4×1 vectors of the Pauli or Dirac theory, respectively), each Hilbert space can only provide a home for a number of particles given by the dimension of the vectors of that space. There are d complex degrees of freedom for each particle and therefore, since quantum mechanics is more elegant with complex wave functions, there is room for only d particles represented by that spinor. That’s all there is, and that’s how elementary particles has been done for many decades.

In the usual theory, the Clifford algebra acts on the states. When we discuss a “basis” here it will be in the context of the d-dimensional Hilbert space, that is, we will mean a basis for the states that the Clifford algebra acts upon. A base defines the quantum states that we can consider to be different particles, or different aspects of the same particle. For the Dirac algebra, one basis has the four particles: {spin up electron, spin down electron, spin up positron, spin down positron}. But this is not the only possible basis for the Dirac algebra quantum states. We could instead pick {right handed electron, left handed electron, right handed positron, left handed positron}.

From a quantum information point of view, the splitting of the Hilbert space into a particular basis is a somewhat inadequate description of the information contained in a quantum state on the space. Suppose we have a large number of identical states and we wish to determine what state it is. If we make measurements with respect to just one basis, we will get the right answer if the state is entirely within that basis, but this won’t necessarily be all the information about the state.

For example, suppose the state is a spin-1/2 state. If it is polarized in the y-direction, then anytime we measure it in the z-direction we will get the value + or – with equal probabilities. This will not allow us to distinguish between, an +y oriented state and a -y oriented state. And vice versa. To determine what state we really have we need to measure it with respect to more than just the usual single direction. And this is where MUBs come in. It turns out that if you measure an arbitrary quantum state (even a mixed density matrix state) with respect to a “complete set of MUBs”, you will get just barely enough information to completely determine the quantum state.

The idea we will discuss here is to use MUBs to describe elementary particles, a subject of current research. We will briefly discuss the motivation, the history, and some of the advantages of this idea, and finally show how it can be used to derive some of the features of Garrett Lisi’s E8 model

**Geometry and the Zoo**

It would be wonderful to be able to derive the elementary particles as the result of the geometry of spacetime. The Dirac algebra is a step in the right direction in that it is intimately connected to the geometry of special relativity and the geometry of spacetime. However, the Dirac algebra is woefully inadequate in that it provides for only 4 orthogonal states. This is not enough.

To get an idea of how large the zoo has become, Garrett Lisi recently used an E8 240-plet to organize the elementary particles. Almost all of those 240 states were known particles. To fit that many particles into a single finite Hilbert space we’d need it to be 240 dimensional or more.

If we wanted to generalize the Pauli / Dirac algebra to a higher dimension we could eventually get one big enough to hold 240 states. The generalization of these algebras are the Clifford algebras. Clifford algebras are generated by a set of vectors. For the Pauli and Dirac algebras, there are one vector per dimension, the Pauli is built from the 3 dimensions of space, while the Dirac algebra is built from the 4 dimensions of spacetime. (Actually, this is true if the Pauli algebra is treated as a real Clifford algebra while the Dirac algebra is treated as a complex Clifford algebra so we are comparing apples to oranges here but we’ll ignore that and deal only with complex Clifford algebras.)

How many dimensions does it require to make a Clifford algebra big enough for the nearly 240 elementary particles? Well 240 < 256 = 2^8. It turns out that to get this, we need 16 dimensions. The appropriate Clifford algebra would be Cl(16). (For the cognoscenti, I’m ignoring signature because I’m assuming a complex Clifford algebra here.) This would be appropriate for a spacetime with 12 curled up dimensions, along with the 3 spatial and one time dimension we observe.

This is similar to what string theory has done and the result is likely to be that we will have the same difficulty that the string theorists find themselves in: we don’t know how to treat all those extra dimensions. There are too many ways to compactify them. But that’s not all. We don’t know how to define their interactions. We want a unified field theory to be simple and elegant.

**Packing In Extra Particles**

A Clifford algebra is not an algebra of quantum states, at least in the usual sense of state vectors in quantum mechanics, but instead can be thought of as operators on those states. Frequently they’re written in matrix form. Since a matrix can be written as a collection of vectors, another idea for getting lots of particles into a small dimension Clifford alegbra is to put the particles into the vectors, or equivalently, into the elements of the Clifford algebra matrices rather than into the vectors that these matrices act upon. This idea was pursued a few years ago by Greg Trayling in hep-th/9912231, by Trayling and W. E. Baylis in hep-th/0103137, by Garrett Lisi in 0511120, by F. Nesti in hep-th 0706.3304 and probably by others whose papers I’ve forgotten or didn’t see.

In putting the elementary particles into a Clifford algebra matrix form instead of a spinor form, the dimensionality of the Clifford algebra is greatly reduced. For 256 states, one needs a 16×16 matrix. To get to a Clifford algebra of this size requires only 8 dimensions. (The rule is that if you have 2n dimensions, you will end up with 2^n x 2^n matrices, or a total of 4^n matrix elements. Since 4^4 = 256, one requires 4×2 = 8 dimensions.) This is getting closer to the observed 4, but is still too many by a factor of two.

The idea of using MUBs to define the particles is an idea kind of similar to the idea of packing them into matrix form in that either you way you end up with a lot more particles for a low dimension Clifford algebra, but the MUB technique has the advantage that it automatically defines the particle interactions for you.

A set of MUBs is “complete” if measurements with respect to them are sufficient to completely define the quantum states. It turns out that for a d-dimensional Hilbert space, this requires d+1 dimensions. Since each of these bases will have d states, this gives us a total of d(d+1) particles for a d-dimensional MUB theory. This is approximately equal to the d^2 one gets with the above matrix theory papers.

**Particle Interactions**

In a MUB model, the particle interactions are automatically defined by the usual spinor rule of inner product. Two particles in the same basis do not interact as they are orthogonal. Any other pair of particles interact with equal probability. This is quite elegant and simple, and it follows naturally the rules of quantum mechancis, but it also has two problems with modeling the elementary particles: (a) all the probabilities are equal while the known particles have coupling constants that are very different; (b) for a d-dimensional theory, a given particle interacts with d^2 other particles which is far too large for the standard model.

A solution to both these problems is to assume a symmetry breaking. That is, we assume that the symmetry used to create the Clifford algebra is only approximate and that some of the coupling constants are deeply suppressed. This gets rid of the excess couplings and has the added advantage that it can be used to make the coupling constants depend on the pair of particles involved.

To get an idea of how many coupling constants have to be eliminated we can look at Lisi’s E8 paper. In his model, three elementary particles interact if their quantum numbers sum to zero. Out of the 240 x 239 x 238 / 3! = 2275280 triples of particles, there are a total of 74410 that are allowed, (at least according to my java applet). This is about one in 31.

There is another problem here. The elementary particles are modeled with 3-body (or more) interactions, but the natural particle interactions of a MUB are only 2-body interactions. Fortunately, there is a history in elementary particles of the number of legs in interactions decreasing as time has gone on. For example, the (lowest order contribution to the) weak decay of the neutron is drawn nowadays as two 3-body interactions:

but in earlier theory, before the W particle was discovered, the interaction was a 4-body one:

Now it would seem impossible to get much use out of 2-body interactions, but there are situations where this might be suitable in a field theory. For instance, one can think of the mass contributions for a field as a 2-body interaction between the left and right handed particles. And if it is a good approximation to ignore one of the three particles (for instance it could be a graviton or even a neutrino), then it may be possible to get a good result with a 3-body interaction approximated by a 2-body one. This would be particularly useful if the 3rd body had neglible energy and momentum, but nevertheless its emission had a radical effect on the other two particles.

An example of a near 2-body interaction seen in physics is a very low energy inverse beta decay of a heavy atom. In this sort of decay, an electron is taken from a low orbital and turns a proton into a neutron. The atom changes to a new element, but if the decay has very little energy, the neutrino can be ignored and at the atomic level the decay appears to be a 2-body decay.

To write 3-body interactions out of 2-body ones, we have to assume that at least some of the elementary particles are composites made from preons. An interaction might look like this:

or it could be a lot simpler.

Yet another natural reason for using 2-body interactions to model what is actually a 3-body interaction is if the elementary particles are made from preons that interact with an extremely heavy boson. That heavy boson restricts the interaction to such a small neighborhood that to our energy scales we can approximate it as a point interaction. In such a case a MUB theory would give particle interactions not for the known elementary particles, but instead for the preons assumed to compose them.

**Bound States of Preons**

If we suppose that the standard model particles are composites made from states modeled by MUB preons, we have two issues to deal with: (a) In order to have the correct number of interactions, and to allow those interactions to have disparate levels, we must include symmetry breaking. (b) We have to explain why the preons themselves cannot be observed.

If the MUBs are generated from a Clifford algebra, there is a natural way to break symmetry in a MUB. We assign different weights to the various blades. This was discussed in MUBs and Symmetry breaking, a recent post. The advantage of this is that it is purely geometric and fairly restrictive. And it gives a natural solution to the problem of defining the standard model particles in terms of bound states of MUBs; we simply assume that the same symmetry breaking that would make some interactions extremely high energy will also make the bare preons, or the unwanted combinations of preons, have extremely high energy.

As an example, let’s use the Dirac algebra as the Clifford algebra for the MUBs. There are 16 degrees of freedom. If we distinguish between space and time vectors, we can divide the bilinears of the algebra into the following classes:

{1}

{x,y,z}

{t}

{xy,yz,xz}

{xt, yt, zx}

{xyz}

{xyt, xzt, yzt}

{xyzt}.

Thought of as symmetry elements of a vibration defined by a differential equation, the vectors of a Clifford algebra correspond to mirror planes that are symmetries of the vibration. This idea is best explained in David Hestenes’ paper on the application of Clifford algebra to the classification of the crystal symmetries:

Point Groups and Space Groups

in Geometric Algebra. So the natural choice of symmetry breaking is to require that the longer bilinears sum to zero.

If we make the cutoff at the vector level, we will have five allowed nonzero numbers for the observed bound states. Now a given observed bound state could have various values for these 5 quantum numbers. So we would expect that in this case, the elementary particles could be described by 5 quantum numbers. The remaining 11 bilinears would correspond to quantum numbers that cannot be observed.

The above sort of definition needs to be sharpened a bit because if A+B+C happens to be zero for higher blades, then so will 2A+2B+2C. We need to recognize the first item as a valid bound state but reject the second. In other words, we require a Pauli principle: in a single bound state, each MUB can appear only once or not at all.

Also, if there is more than one way of putting preons together to get the same sum, then it is possible that we could require extra quantum numbers to distinguish between them. In particular, the fermions come in three generations, each with the same quantum numbers, but differ only because of mass. If the mass comes from the how those 11 hidden quantum numbers are cancelling, then in modeling the system we would have to add extra quantum numbers appropriately. To distinguish between 3 generations, we need one or two extra quantum numbers, depending on how they’re used.

Lisi’s model of E8 uses 8 quantum numbers for each particle, so his quantum numbers are about the right size for fitting a MUB preon theory with blade symmetry breaking based on a 5-d or 6-d Clifford algebra. Or we could get 8 quantum numbers out of the Dirac algebra itself by assuming that any quantum numbers with must sum to zero. Of the 16 bilinears, there are just 8 that satisfy that requirement so we would turn the Dirac algebra MUBs into an 8-dimensional set of quantum numbers immediately.

**Particle Interactions**

Let’s make the assumption that the bound states of a MUB set are as described above, the set of collections of MUB states that happen to sum to zero except for the scalar and vector portion. Furthermore, let’s make the assumption that the reason we can write these interactions as 2-body interactions is because the bosons that change the states are all exchanged at a single point. Then it’s natural to assume that the total quantum numbers are always preserved; what a boson takes from an initial state, it delivers to a final state.

Let’s also make the assumption that the non hidden quantum numbers fully describe a state. That is, given those quantum numbers we could, if we knew the theory, write down the basis states that make it up. Under these assumptions, can we derive a relationship that will tell us that it is possible for three bound states to interact?

A first requirement is that there be no net change in quantum number. If we choose our quantum numbers conveniently, this can be rearranged to be that they three vectors of quantum numbers sum to zero. This is good because Garrett used a similar rule for particle interactions in his E8 model.

Next, let’s suppose that I have three bound states that happen to have quantum numbers that add to the quantum numbers of a third state. By the above assumptions, this means that I know what the internal states and it’s fairly natural to expect that this will be a valid particle interaction for the bound states. (Though I really don’t think I’ve proved it here.) In any case, this is just the kind of behavior we’d expect for a theory of this sort.

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Ooops. Handedness does not commute with charge so there is no such thing as a left handed fermion antiparticle.