# MUBs and Symmetry Breaking

One can obtain a Hilbert space from a wide variety of physical situations. Suppose we are to use a Hilbert space to model elementary particles, for instance. One the one hand, we’d like to have our model cover as many particles as possible. But to do this, we may have to assume a symmetry that is not exactly correct. For example, to get a model that includes isospin, we have a natural inclination to assume that isospin is a perfect symmetry when, in reality, it cannot be. Assuming perfect symmetry as an approximation leaves us with the need to later correct for this by including a symmetry breaking.

Recently, we’ve been discussing using Mutual Unbiased Bases (MUBs) to model the elementary particles. If we want to include very disparate objects as states in different MUB bases, we need to also include a symmetry breaking mechanism.

When elementary particle theory is built from symmetry alone, one finds that there are a nearly infinite number of ways of breaking symmetry. The freedom to do this makes it difficult to find our way. In the case of MUBs, however, we can write down some very natural ways to break symmetry if we treat the Hilbert space as the vector space of a Clifford algebra like the Dirac algebra. In this post we will show how the natural symmetry breaking in the Dirac algebra can lead to structures that are reminiscent of the structure of the elementary particles.

As discussed a couple posts ago, we can write down a MUB for the Dirac algebra (which has 16 elements, but only 15 if you ignore the scalar 1) by distributing the Dirac bilinears into 5 groups of 3. Each group of 3 commutes, and any two of the three multiply to give the other (or its negative). This is a good way to illustrate the natural symmetry breaking that arises when a MUB is built from a Clifford algebra that is associated with external symmetries. (Yeah, I know that I’m violating Coleman-Mandula. So shut up and just give me a ticket, I can pay the fine.)

How Many Dirac MUBs are there???

The symmetry breaking will depend on how the Dirac bilinears are split into MUBs. So first, let’s figure out just how many possible splittings there are. We’ll enumerate them. For this post, I will not actually write out the MUB but instead just write out the various divisions of the Dirac bilinears into commuting groups. Consequently, I will not have to make a choice of signature.

The Dirac bilinears, written in my short hand notation that replaces $\gamma^1 = x, \gamma^2 = y$, etc., consist of the following 15 terms (leaving off the scalar 1):

{x, y, z, t, xy, xz, yz, xt, yt, zt, xyz, xyt, xzt, yzt, xyzt}

We can repackage these 15 elements into 3×5 form of five sets that commute in more than one way. For example, the set of 3 that includes xyz can include any of these bilinears: x, y, z, xy, xz, yz. However, if xyz and x are used in the same basis generating set, then that set also includes their product, yz. So these six choices of partners with xyz come down to the following 3 choices for the group that contains xyz:

{x, yz, xyz}, {y, xz, xyz}, {z, xy, xyz}.

These three choices are symmetric or cyclic in {x,y,z}. They amount to choosing a preferred axis in 3-space similar to how the Pauli algebra chooses a preferred axis when it is the z-axis that is diagonalized in the Pauli spin matrices. In writing down all the possible choices for the Dirac MUB, we may as well make the third choices, and put xyz along with z and xy. We can obtain the other three cases by cyclic substitution.

So far we’ve split the Dirac bilinears as follows:
{z, xy, xyz}, {x, y, t, xz, yz, xt, yt, zt, xyt, xzt, yzt, xyzt}

For the next choice, let’s look at the things that x commutes with: {yz, yt, zt, xyt, xzt}. We can’t choose yz because that would imply xyz and we’ve already used that degree of freedom. So there are two choices {x,yt,xyt} and {x,zt,xzt}:

{z, xy, xyz}, {x,yt,xyt}, {y, t, xz, yz, xt, zt, xzt, yzt, xyzt}
{z, xy, xyz}, {x,zt,xzt}, {y, t, xz, yz, xt, yt, xyt, yzt, xyzt}

Since y commutes with xz, xt, zt, xyt, and yzt, but we can’t use xz for the previous reason, we have two possible groups for y: {y, xt, xyt} and {y, zt, yzt}. This would split the above two MUB choices into four except that xyt is already used in the first line and zt is already used in the second line. Consequently, we still have 2 possibilities after we assign y:

{z, xy, xyz}, {x,yt,xyt}, {y, zt, yzt}, {t, xz, yz, xt, xzt, xyzt}
{z, xy, xyz}, {x,zt,xzt}, {y, xt, xyt}, {t, xz, yz, yt, yzt, xyzt}

Now t commutes with xz, yz, xzt, and yzt. This gives the groups {t, xz, xzt} and {t, yz, yzt}. As before, each of these intersects with something already chosen so for our final result, we still have just two choices for the complete MUB generating sets:

{z, xy, xyz}, {x,yt,xyt}, {y, zt, yzt}, {t, xz, xzt}, {yz, xt, xyzt}
{z, xy, xyz}, {x,zt,xzt}, {y, xt, xyt}, {t, yz, yzt}, {xz, yt, xyzt}

In addition to these, there are 3x as many under the cyclic permutation of {x,y,z}, so the total number of choices for the Dirac MUB is only six, a manageable number:

{z, xy, xyz}, {x,yt,xyt}, {y, zt, yzt}, {t, xz, xzt}, {yz, xt, xyzt}
{x, yz, xyz}, {y,zt,yzt}, {z, xt, xzt}, {t, xy, xyt}, {xz, yt, xyzt}
{y, xz, xyz}, {x,zt,xzt}, {x, yt, xyt}, {t, yz, yzt}, {xy, zt, xyzt}
{z, xy, xyz}, {x,zt,xzt}, {y, xt, xyt}, {t, yz, yzt}, {xz, yt, xyzt}
{x, yz, xyz}, {y,xt,xyt}, {z, yt, yzt}, {t, xz, xzt}, {xy, zt, xyzt}
{y, xz, xyz}, {z,yt,yzt}, {x, zt, xzt}, {t, xy, xyt}, {yz, xt, xyzt}

Dirac Bilinears as Operators

To break the symmetry, we need to look at the bilinears as operators. It is possible that the reader is unfamiliar with this relation as most textbooks pretty much ignore qubit theory in favor of the more general, and more difficult, theory that includes momentum. If so, see the excellent wikipedia article for a start. The operator for charge is t, and the operator for spin in the x, y, and z directions are respectively yz, -xz, and xy. This gives us interpretations for four of the basis vectors.

The Dirac bilinears xt, yt, and zt correspond to velocity (not momentum since we are in qubits) in the x, y, and z directions as can be shown by substitution. Let $\psi(x,t)$ be a wave moving in the +z direction at speed c=1 so that $\psi(x,t) = \psi(z-t)$. To arrange for this to satisfy the massless Dirac equation (where I have added hats to the Dirac operators to distinguish them from the coordiantes):
$(\hat{x}\partial_x + \hat{y}\partial_y + \hat{z}\partial_z + \hat{t}\partial_t))\psi(z-t) = 0$.
The above simplifies since psi is a function of z and t only. In addition, we can multiply on the left by the operator t to get (uh, using the -+++ signature convention):
$(-\hat{z}\hat{t}\partial_z -\partial_t)\psi(z-t) = 0.$
In the above, the first minus sign comes from the anticommutation rules and the second comes from the signature choice. Working out the partial derivatives we have:
$(-\hat{z}\hat{t} +1) \dot{\psi}(z-t) = 0.$
The above is satisfied if $\dot{\psi}$ is a multiple of the ideal (1+zt)/2. Consequently, zt is the operator for velocity c in the z direction.

Since zt is the operator for velocity in the z direction, and xy is the operator for spin in the z direction, their product, xyzt, is the helicity operator. The helicity operator is frequently written $\gamma^5$, a source of continuing confusion in string theory and other theories that add one or more hidden dimensions. That this is the correct operator for helicity is verified by its use in picking out the chiral portions of wave functions. The reader is probably familiar with the projection operator $(1+\gamma^5)/2$, which implies that xyzt is the helicity operator.

The remaining bilinears are determined by the above assignments. For example, x = -xt t, so x, y, and z are operators for “charge x velocity.” (In deriving the Koide formula for the lepton masses, the assumption was made that the 1 operator gives mass, but that gets us too far afield.) The resulting 15 assignments are:

t: charge
xyzt: helicity
xyz: charge x helicity (reversed helicity for anti-particles)
xy, -xz, yz: spin
xt, yt, zt: velocity
xyt, -xzt, yzt: charge x spin (reversed spin for anti-particles)
x, y, z: charge x velocity (reversed velocity for anti-particles).

Dirac Bases

Of course it is well known that one has several different choices of quantum numbers for the electron and positron. The most common choice is to split them into spin up or spin down, and particle or anti-particle. Because of the importance of helicity in the standard model, the usual choice for elementary particles is right or left helicity, and particle or anti-particle. If we were to diagonalize the Dirac algebra according to these two different choices of quantum numbers, we would end up with two different bases. We will call these bases the UD and LR basis, standing for up-down and left-right. This is a small step towards MUB, and it’s useful to see how these two choices show up in our splitting of the Dirac algebra into MUBs.

I guess it should go without saying that if we wish the quantum states picked out by our basis choices to be eigenstates of an operator, we need for that operator to be included in the three commuting operator defining that basis. From the density operator point of view this is quite obvious.

If it’s not, get out a piece of paper and pencil. Choose your favorite representation of the Dirac algebra, that is pick some Dirac matrices. Compute (1+iyz)(1+xt)/4 and verify that this is a pure density matrix. That is, verify that it has trace 1 and squares to itself. It is a 4×4 matrix. Take one of its columns (don’t pick one that is zero), and treat it as a spinor (i.e. a ket). Normalize the ket. Compute the bra. Multiply the ket by the bra and verify that you get back the pure density matrix. Now verify that the spinor is an eigenstate of yz, xt, and xyzt.

That was quite a pain in the rear, wasn’t it. Now you know why I am writing a book on density operator theory; quantum mechanics is much easier on the other side.

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The obvious Dirac operator that picks out charge (i.e. the difference between particle and anti-particle) is t. So both the UD and LR bases need to include t or something equivalent to it. The UD basis also includes spin in the z-direction, xy, so a natural choice for a UD basis is simply {t, xy, xyt}. This choice is included in two of the six possible MUB sets:

The Up / Down bases:
{x, yz, xyz}, {y,zt,yzt}, {z, xt, xzt}, {t, xy, xyt}, {xz, yt, xyzt}
{y, xz, xyz}, {z,yt,yzt}, {x, zt, xzt}, {t, xy, xyt}, {yz, xt, xyzt}

The LR basis needs to distinguish states by helicity and charge. The operators that include helicity and charge in their definitions are t = charge, xyzt = helicity, but these two operators do not commute, and consequently one cannot choose a diagonal basis for them. More generally, the Dirac bilinears that are “neutral” in orientation (i.e. some sort of scalar or time oriented operator) are 1, t, xyz, and xyzt. All the other bilinears can be written as products of these four with x, y, or z. But t, xyz, and xyzt each anticommute with the other so they cannot be used to define a basis. So in writing down a basis that diagonalizes helicity, we have to make an orientation choice to go along with it. The usual choices involve z, and the two relevant MUBs are:

{y, xz, xyz}, {x,zt,xzt}, {x, yt, xyt}, {t, yz, yzt}, {xy, zt, xyzt}
{x, yz, xyz}, {y,xt,xyt}, {z, yt, yzt}, {t, xz, xzt}, {xy, zt, xyzt}

In both cases, the LR basis includes spin in the z-direction. Note that the above two choices are converted into each other by swapping x and y.

MUB quantum numbers

Suppose we are to interpret all the MUBs as defining elementary particles. There are 5 bases and each of them defines 4 particles. So there will be a total of 20 states. Each group of four will take quantum numbers with respect to their bilinear generators of +-1. In the example of the electron Dirac theory, the charge will be either +1 or -1, and spin in the z-direction will be +1/2 or -1/2 (or in the helicity basis, helicity will be either +1 or -1). Doubling the spin 1/2 quantum numbers so that they look like the other quantum numbers and take values of +1 or -1, the quantum numbers of the four states of a basis are:
(1,1), (1,-1), (-1,1), (-1,-1).

These four states are not eigenstates of any of the other Dirac bilinears (with the exception of the 1 operator which has all states as trivial eigenstates with an eigenvalue of 1). Now each basis set includes 3 commuting operators, so actually the quantum numbers can be written as 3-vectors instead of the 2-vectors drawn above, but since two basis operators multiply to give the third operator, the third quantum number is determined by the other two, it’s their product.

So instead of 15 non trivial operators for a MUB, we instead have only 10. The remaining 5 are products of the first 10. Each operator is nonzero for only one of the five particle sets. The structure of the quantum numbers looks as follows.

Taking the example of the first of the MUBs:
{z, xy, xyz}, {x,yt,xyt}, {y, zt, yzt}, {t, xz, xzt}, {yz, xt, xyzt}
We will take the first two operators from each group of three to give the 10 operators {z, xy, x, yt, y, zt, t, xz, yz, xt}. The 20 elementary particles will carry quantum numbers as follows:

 .z xy .x yt .y zt .t xz yz xt -1 -1 .0 .0 .0 .0 .0 .0 .0 .0 -1 +1 .0 .0 .0 .0 .0 .0 .0 .0 +1 -1 .0 .0 .0 .0 .0 .0 .0 .0 +1 +1 .0 .0 .0 .0 .0 .0 .0 .0 .0 .0 -1 -1 .0 .0 .0 .0 .0 .0 .0 .0 -1 +1 .0 .0 .0 .0 .0 .0 .0 .0 +1 -1 .0 .0 .0 .0 .0 .0 .0 .0 +1 +1 .0 .0 .0 .0 .0 .0 .0 .0 .0 .0 -1 -1 .0 .0 .0 .0 .0 .0 .0 .0 -1 +1 .0 .0 .0 .0 .0 .0 .0 .0 +1 -1 .0 .0 .0 .0 .0 .0 .0 .0 +1 +1 .0 .0 .0 .0 .0 .0 .0 .0 .0 .0 -1 -1 .0 .0 .0 .0 .0 .0 .0 .0 -1 +1 .0 .0 .0 .0 .0 .0 .0 .0 +1 -1 .0 .0 .0 .0 .0 .0 .0 .0 +1 +1 .0 .0 .0 .0 .0 .0 .0 .0 .0 .0 -1 -1 .0 .0 .0 .0 .0 .0 .0 .0 -1 +1 .0 .0 .0 .0 .0 .0 .0 .0 +1 -1 .0 .0 .0 .0 .0 .0 .0 .0 +1 +1

This is generic for the Dirac MUBs. Where the different MUBs differ is in the assignment of the operators to the particles.

Symmetry Breaking

Consider t, the Dirac bilinear for charge. The particles that are eigenvalues of charge will show up with charges of +1 or -1. The particles that are not eigenstates of charge will show up with charge 0. In the example of the leptons, the charged particles are the electron and positron which carry charges of -1 and +1. The neutral particles are the neutrinos which take charges of 0. In this example, the charged particles have much heavier masses than the neutral particles. So this is an example of symmetry breaking.

Suppose that we write these elementary particle masses as a quadratic equation in the Dirac bilinears. To arrange for the charged leptons to weigh much more than the neutral leptons, we simply assign a larger contribution to the charge quantum number. There will be some other quantum number that the neutrinos possess and the charged leptons do not, but there is no reason for us to automatically assume that all the Dirac bilinears should carry the same weight. In our usual use of the Dirac bilinears the operators are not at all equivalent so there is no reason for us to suspect that they should contribute equally to particle properties.

This sort of assumption will break the symmetry of the MUBs. As an example, suppose we assign a weight of 1 to Dirac scalar (which is present in all states), a weight of the Dirac vectors x, y, z, and t of 3, and give 9 to the Dirac bilinears xy, xz, yz, xt, yt, and zt, a weight of 27 to the Dirac psuedovectors xyz, xyt, xzt, yzt, and a weight of 81 to the Dirac psuedoscalar xyzt. Let’s take absolute values so that we only have to calculate 5 weights. Then the 5 weights for the two choices of MUB work out as follows:

{z, xy, xyz}, {x,yt,xyt}, {y, zt, yzt}, {t, xz, xzt}, {yz, xt, xyzt} =>
1+3+9+27, 1+3+9+27, 1+3+9+27, 1+3+9+27, 1+9+9+81 = 40, 40, 40, 40, 100

{z, xy, xyz}, {x,zt,xzt}, {y, xt, xyt}, {t, yz, yzt}, {xz, yt, xyzt}
1+3+9+27, 1+3+9+27, 1+3+9+27, 1+3+9+27, 1+9+9+81 = 40, 40, 40, 40, 100

In other words, under this simple assumption, no matter which set of MUBs we choose, the pattern in the particle multiplets will be the same. There will be 4×4 = 16 light particles, and 4×1 = 4 heavy particles.