# Qutrit Mutually Unbiased Bases (MUBs)

Mutually Unbiased Bases are sets of bases for a Hilbert space that are “unbiased:” the transition probabilities between any two states from different bases are equal. For a Hilbert space of dimension 3 (i.e. qutrits), the transition probability is 1/3. The operator space of a Hilbert space of dimension n is n^2, in this case the operator space has 9 dimensions. Each base consists of 3 quantum states. It turns out that a base uses up 3-1 = 2 degrees of freedom of the operator space, and the scalar part of the operator space is shared by all. So for a 3-dimensional Hilbert space, there are at most four mutually unbiased bases. In this post I will derive a set of four such bases.

In yesterday’s post I derived a complete set of mutually unbiased bases for the Hilbert space with dimension 4. I did this by adding a geometric assumption to that Hilbert space, the Dirac algebra. That is, the Dirac algebra has 4×1 spinors so the vectors are a 4-dimensional Hilbert space.

The Dirac Algebra (4-dimensional Hilbert space) Method

The method was to write the 16-dimensional operator space of the Dirac algebra (and therefore the 16-dimensional operator space of the 4-dimensional Hilbert space) as degrees of freedom of the 16 Dirac bilinears. A basis for the Dirac spinors is generated by two commuting square roots of unity. The Dirac bilinears all square to +1 or -1; to make them all into square roots of unity, we multiply the ones that square to -1 by i. For example, the bilinear $\gamma^1\gamma^2$ squares to -1 regardless of one’s choice of signature. To turn this into a square root of unity, one replaces it with $i\gamma^1\gamma^2$.

If I take a pair of Dirac bilinears, they will either commute or anticommute. To define a basis for the Dirac spinors, one can take two Dirac bilnears that commute, say A and B. Then $(1\pm A)(1\pm B)/4$ is a pure density matrix for any of the four choices of signs. These four pure density matrices sum to unity and annihilate each other, and therefore define four states that make up a basis set for the Dirac algebra. To actually get the basis set, one converts the pure density matrices into spinors by writing them out in matrix form and taking any nonzero column as the spinor.

If I consider a Dirac bilinear as an operator, it will take a nonzero average value with respect to these four quantum basis states if and only if that Dirac bilinear commutes with all of the square roots of unity that were used to generate the basis. To arrange for another basis to be mutually unbiased with respect to this first basis, it follows that I need only require that I build the next basis from Dirac bilinears that do not commute with all the Dirac bilinears I’ve used so far. What is going on here is that the 15 non scalar Dirac bilinears are being divided into five groups of 3. Each group of 3 commutes (and if one multiplies two of them together, you will get the third). On the other hand, no two groups fully commute with each other. One can show that under these assumptions, the 5 groups of Dirac bilinears generate 5 basis sets that are mutually unbiased. For yesterday’s post, the 16 Dirac bilinears were split as follows: In the more effiicent notation I use for my own work, the above sets are written as {ixy, zt}, {x, iyz}, {y,xt}, {z,yt}, {it,ixz}. This is in the -+++ signature and i’s convert the bilinears to square roots of unity. It’s interesting that in the mathematical Clifford algebra literature, what I’ve been calling “square roots of unity” are instead called “roots of unity”. To solve the MUB problem for qutrits, I will indeed use cube roots of unity.

The Qutrit Operator Algebra

To do the same thing for 3-dimensional Hilbert spaces, I need to build a structure that is similar to the “commuting square roots of unity” structure that works so nicely in the Dirac algebra. The first problem is that 3 is not a power of 2, so I can’t write the basis sets as $(1\pm A)/2$. But I still want to write the problem in terms of primitive idempotents, and therefore in terms of pure density matrices.

Now the Dirac bilinears are of the form $A^aB^bC^cD^d$ where A, B, C, and D are matrices (i.e. $\gamma^1, \gamma^2, \gamma^3, \gamma^0$), and a, b, c, and d are integers. By the rules of the Clifford algebra, we need only consider a, b, c, and d to be 0 or 1. Any other values can be cancelled out by using the anticommutation relations and the signature.

But the qutrit operator algebra has 9 dimensions. To write these dimensions in the form Dirac form, I will have to find two matrices, which I will call J and M, and write the qutrit operator algebra degrees of freedom as $J^jM^m$. To get the right dimension 9 = 3×3, I have to have j and m values as 0, 1, and 2. Therefore, I need $J^3 = M^3 = 1$.

Now for this to work, I have to be able to use these things to write a basis for qutrits. In the case of the Dirac algebra, this feature was provided by the fact that a Dirac bilinear B, when a square root of unity, can be turned into an idempotent by (1+B)/2. This follows from B^2 = 1, but now I have J^3 = 1.

Now the thing they drill into you in Galois theory is that in these sorts of things one must look at polynomials. In this case, a polynomial that is made from J and is idempotent is: (1 + J + JJ)/3. If I square this polynomial, I get ((1+J+JJ) + (J+JJ+JJJ)+(JJ+JJJ+JJJJ) )/9 = (1+J+JJ)/3, as desired. The two related polynomials are (1 + wJ +wwJJ)/3 and (1 + wwJ + wJJ)/3, where $w = \exp(2i\pi/3)$. Now these things are idempotent, but to make sure that they are primitive, I need for them to have traces of 1. In a 3-dimensional Hilbert operator space, the trace of 1 is 3, so I will have primitive idempotents if the trace of J is zero.

There is one other thing I need. In the case of the Dirac algebra, the reason we were able to divide the degrees of freedom up into the basis sets was because different Dirac bilinears either commute (in which case they would have a non zero quantum average) or anticommute (in which case their quantum average would be zero). The zero results because when one commutes the operator around a basis state, it changes the basis state to a new basis state, and the new basis state annihilates the old one. See the previous post if you need more detailed explanations.

In this case anticommutation wouldn’t work. To get J and M to work, I need to have a commutation relation that leaves the product multiplied by w or its square:
JM = w MJ or JM = ww MJ.
This has the addition benefit that it will let the cross terms be made into cubed roots of unity (and can therefore be built up into another basos for qutrits). If it is unclear how this will help, perhaps a computation will help. We will compute the average value for the operator M over the quantum state (1+wwJ + wJJ)/3 under the assumption that JM = w MJ: In the above, the product in the final line is zero because the two states, (1+wwJ+wJJ)/3 and (1+J+JJ)/3, are different basis states generated by J. Therefore they annihilate each other.

J and M matrix choice

It remains to show that matrices J and M exist. From our work in the quark and lepton bound states, we already know one answer: It should be clear that J^3 = 1 as required. And (1+J+JJ)/3, (1+wJ+wwJJ)/3, and (1+wwJ+wJJ)/3 are, in fact, the three circulant primitive idempotents (qutrit pure density matrices): The other obvious qutrit basis is the diagonal basis. To write it as a function of a traceless matrix M, we can choose M as: It is easy to verify that JM = wMJ, as required.

The 9 degrees of freedom of the qutrit operator space, as written in terms of the qutrit “bilinears” are: {1, J, JJ, M, MM, JM, JJMM, JJM, JMM}. The “1” degree of freedom is shared by all the basis generators. The remaining 8 bilinears divide into four groups (each of which commutes amongst itself) as follows: {J,JJ}, {M,MM}, {JM,JJMM}, {JMM,JJM}. And these basis sets generate a complete MUB for qutrits. Each of these pairs generate a set of 3 pure density matrix quantum states in the form (1+A+AA)/3, (1+wA+wwAA)/3, and (1+wwA+wAA)/3.

To get from here to a spinor basis set, one first computes the 3×4 = 12 pure density matrices in 3×3 matrix form (using the J and M matrices above), and then converts these pure density matrices into spinors by taking any nonzero column from the matrix and normalizing it. One ends up with four groups of 3 orthogonal spinors each. Ah, what the heck, let’s write them out, with the common factor of sqrt(1/3) factored out of the last three sets: Now the spinor form is a very compact way of writing these things down, but the calculations are much easier in density matrix form. Part of the problem with spinors is that the phase is arbitrary. That creates confusion that makes searches for these things harder than it has to be. I’m now working on the MUB problem for the 6 dimensional case, of course without success so far.

Part of the problem with the 6 case is that you can’t just factor it into the 2 and 3 problem. The reason is that if you do this, you end up with states like (1+A)(1+B+BB)/6, and to be unbiased with respect to this state you have to find a state that does an anticommutation thingy with both A and B. That means you can’t do a tensor product of the 2 and 3 dimension parts.  Hmmmmmmm.

Applications to Mass

In the context of my preon theory, the most obvious interpretation is to take the three basis states generated by {J, JJ} as three generations of leptons. This is related to the extension of Koide’s mass formula for the charged leptons to the neutrinos, see this paper.

The diagonal states are the individual (free) preons which have never been seen as free particles. This suggests that the remaining two basis sets might be something in between, maybe the quarks. However, there are too many such states. Hmmmmmmmm. A more natural fit might be to suppose that the four MUBs correspond to a lepton / quark pair, with J generating the lepton (charged or neutrino), and the M, MJ, MJJ generating the three colors of a quark (down or up).

However, rather than do something as simplistic as this, I think one needs to replace the complex elements of the matrices with Pauli matrices (which represent the SU(2) part of the U(1)xSU(2)xSU(3) symmetry of the standard model). This changes the problem from being one of a 3-dimensional Hilbert space into a 6-dimensional Hilbert space.

It is known that the 6-Hilbert space has a MUB set with 3 elements, but it is not known if this is complete, though it is a subject of current research. Analysis of the dimensions of certain sets of 1-parameter submanifolds of basis solutions suggests that there should be at least 4 elements. But this is all work in progress. Perhaps we can make some sort of progress by working entirely in pure density matrices.

There are two problems with applying all this to the elementary particles. The first is that we haven’t included any symmetry breaking. I’m planning to write that up tomorrow or the next day. The other is that since these are unbiased, all the transition probabilities are equal so they don’t do a very good job of describing particles whose mixing angles depend on generation. To correct these things, as with mass in the case of the leptons, one must take into account Berry or quantum phase, another great subject for a post.

1. emmanuelwjy