# The Generalized Pauli Group

I’ve been watching Perimeter Institute lectures again, and one that dips into the stuff I’m working came up. In addition to the lecture, there is an acrobat file that contains the slides and photos of the blackboard. This is 124 pages long and it is these pages that I will reference.

Applications of the generalized Pauli group in quantum information
Speaker(s): Thomas Durt – Vrije Universiteit Brussel
Abstract: It is known that finite fields with d elements exist only when d is a prime or a prime power. When the dimension d of a finite dimensional Hilbert space is a prime power, we can associate to each basis state of the Hilbert space an element of a finite or Galois field, and construct a finite group of unitary transformations, the generalised Pauli group or discrete Heisenberg-Weyl group. Its elements can be expressed, in terms of the elements of a Galois field. This group presents numerous applications in Quantum Information Science e.g. tomography, dense coding, teleportation, error correction and so on. The aim of our talk is to give a general survey of these properties and to present recently obtained results in connection with three problems: -the so-called ”Mean King’s problem” in prime power dimension, -discrete Wigner distributions, -and quantum tomography . Finally we shall discuss a limitation of the possible dimensions in which the so-called epistemic interpretation can be consistently formulated, in relation with the existence of finite affine planes, Euler’s conjecture and the 36 officers problem.
Date: 10/10/2007 – 4:00 pm

This has been discussed a little over at Kea’s blog and at Matti’s blog as well, and I figure I should write something up here as well, since I am also applying “Mutually Unbiased Bases” (MUBs); what I call snuarks are the Pauli algebra example of a set of MUBs.

I’ve added an explicit calculation in density matrix formalism for a maximal MUB for the Dirac algebra, and I’ve converted the solution into spinor form. This is a wonderful illustration of how much more powerful the density matrix formalism is than the spinor formalism.

I’d have talked about this here before but when various people (including Kea and Matti) brought up the subject of finite fields (i.e. fields that have only a finite number of elements) I hadn’t paid much attention. I work with Clifford algebra, and matrices whose elements are taken from a Clifford algebra. These things most assuredly have an infinite number of elements, so I didn’t see why I should risk my brain stretching it to accomadate an understanding of finite fields as well. So I’m going to rewrite the applicable parts of his lecture into my own notation.

Page 4 Example d=2

What is going on here is not that the Hilbert space is finite (as I’d assumed at first glance), but instead that the Hilbert space has a finite dimension, and therefore a basis set for the Hilbert space has a finite number of elements. In the case of qubits, or the Pauli algebra, the Hilbert space has dimension d=2. The usual choice of basis set is spin up = |+z> and spin down = |-z>. Durt first considers the permutations on this set of 2 elements. There are 2! elements in the permutation group, the identity “I” and the “negation”, which I will call “J”, which swaps the two basis elements. We can write these in matrix form as follows:

An interesting fact is that the negation operator, which, with an abusive overloading of the notation I will call $J_z$, is the same as $\sigma_x$, the matrix for spin in the x direction. Yet another interesting fact is that the identity operator for the z basis, which I will naturally call $I_z$ is twice the density matrix for an unpolarized qubit.

Page 6: Definition of MUBs

One can repeat the process for the x and y directions, and one naturally finds that the x, y, and z related cyclically. Of course the unpolarized density matrix is unique, so $I_x=I_y=I_z$. We have:

Now most quantum mechanics uses only a single basis for a Hilbert space. However, there is an interesting attribute of the above three bases: the transition probabilities between states taken from one basis to any of the four states taken from the other two bases, are all equal to 1/2. More generally the definition of a set of MUBs for a Hilbert space of dimension d is that the transition probabilities between states from different bases be 1/d.

The reason one wants a transition probability of 1/d is that we want transition probabilities to be unbiased. In the example of a Hilbert space with 3 elements, we want the transition probabilities to all be 1/3 because we need the three transition probabilities to add to unity.

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An aside: Democratic Quantum States MUBs are related to the “democratic density matrix” which is important in defining representatons of Clifford algebras in matrices. Suppose that one has chosen a basis set for a Clifford algebra and one wishes to produce a matrix representation of the Clifford algebra that has this basis set diagonalized. For specificity, let us suppose that the associated Hilbert space has dimension d and that we are looking for a representation of the Clifford algebra in dxd matrices. For example, with the Dirac algebra, d=4.

There are only a finite number of ways of choosing the diagonal matrices that will represent the chosen basis set for the Hilbert space; one can permute them around so the number of ways of doing this is d!. But even after one has chosen the diagonal matrices, there are still an infinite number of ways of choosing the representation.

For example, in the case of the Pauli algebra, this can be seen by observing that one can transform the matrices by a rotation that leaves the z-axis unchanged. This will change the matrix representations of $\sigma_x$ and $\latex \sigma_y$, but will leave $\sigma_z$ unchanged.

But if, in addition to the basis set that we wish to diagonalize, we also choose what I have called a “democratic state”, that is, a state that has equal transition probabilities to all of the states of the basis set, then one can complete and uniquely define the matrix representation for the full Clifford algebra (not just the diagonal elements) by assigning the “democratic density matrix” to the democratic state. The “democratic density matrix” is the matrix with all elements equal to 1/d.

This should be fairly obvious. If the reader wishes to see a proof, see Chapter 4, section 3 of my book on density matrices.

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Page 8-9 Estimating Density Matrices

Suppose one has an ensemble of identical particles prepared in some particular density matrix state. In order to estimate the values of that density matrix, it is enough to know the transition probabilities from that density matrix to the states of a MUB. In addition, the process is optimal in that there is no redundancy shared between the bases; no information is wasted.

Durt’s lecture continues with the case of general Hilbert spaces. My own efforts have been with Clifford algebras, and collections of Feynman diagrams that correspond to bound states. It’s kind of cute to write this in the language of MUBs. The Clifford algebra defines the fundamental fermions of the theory. The Pauli principle would require that we consider the states only with respect to some particular basis. Rather than slavishly following the Pauli principle, which has only been verified for the observed state of matter and consequently we can imagine does not apply to preons, I allow these states to be taken from MUBs instead of just a single basis. One can then compute the transition probabilities and Berry phases between the states, which we can attribute to a preon gauge boson that is similar to a gluon. A typical interaction looks like:

Note the product of two pure density matrices in the top right corner of the above: $(1+x)/2(1+z)/2$ multiplied by a coupling constant. This is my short hand for the product of two density matrices taken from MUBs: $(1+\sigma_x)/2(1+\sigma_z)/2$, and can also be written as a complex multiple of $|+x\rangle\langle+z|$.

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As an aside, I prefer writing the products of MUBs in pure density matrix form, rather than in state vector form: $|+x\rangle\langle+z|$ because products such as this are quite sloppy; its value depends on the choice of phase for the spinors. One can solve this problem by choosing state vectors for each of the MUBs, but this is rather inelegant. In fact, if one wishes to choose state vector phases for all the states in a non trivial Hilbert space, one will find that one cannot do so in a continuous manner. Since the MUBs are discrete the failure of continuity is not a problem, but by contrast, density matrices do not have arbitrary phases and their products are well defined without any of these issues (except that every now and then their products are zero).

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Page 46
As Hilbert spaces, the Clifford algebra density matrices have dimension $2^n$. According to this page of Durt’s lecture, these objects have MUBs with n+1 basis sets. We’ve already seen the solution in the Pauli algebra where n=2. As my aside above noted, the MUBs of a Clifford algebra are very closely related to the democratic density matrix states. To see how to write down the MUBs for a (complex) Clifford algebra (which also solves the MUB problem for 2^n), one needs to understand how one chooses a democratic density matrix, because what is going on here is that one is picking two democratic density matrices that are democratic with respect to each other.

The method is probably best explained with an example, and is a little too involved to write into a blog, so I will only outline the procedure. For more details, read chapter 4 of my density matrix book.

A Clifford algebra is generated by a set of “canonical basis vectors” . For the Dirac algebra, we can efficiently write these as x, y, z, and t. In the usual notation these are $\gamma_1, \gamma_2, \gamma_3, \gamma_0$, but I’m not going to write out all that tex when the letters will do as well. These four canonical basis vectors generate the Dirac algebra, which ends up with dimension 16. The 16 objects that define the basis for the Clifford algebra are called the “canonical basis elements”. For the Dirac algebra, they are 1, x, y, z, t, xy, xz, yz, xt, yt, zt, xyz, xyt, xzt, yzt, and xyzt, and you may know them as the “Dirac bilinears.”

To define a basis set for the Hilbert space, one first chooses a maximal set of commuting square roots of unity. For the Dirac algebra, such a “maximal set” has 2 elements. We can take them to be ixy and zt because ixy and zt commute, and square to unity: (ixy)(ixy) = (zt)(zt) = 1. [Note: I am using the -+++ signature convention.] These define our basis set for the Hilbert space (written in density matrix form, which is far more elegant than spinor form of course) as: the four pure density matrix states: $(1\pm ixy)(1\pm zt)/4$. Of course if you just have to have your ugly spinors, you can convert these pure density matrices into 4×1 spinor form by writing them out in 4×4 matrix form, and choosing any non zero column of the matrix as your spinor. Ugly, but that is the way most of you do it.

Now the secret to the procedure is to realize that we can trivially compute the average value of an operator of the Dirac algebra with respect to a quantum state chosen from our basis set $(1\pm ixy)(1\pm zt)/4$ as follows: First, write the operator as a sum over canonical basis elements (i.e. as a sum over bilinears). You can ignore any canonical basis element that anticommutes with ixy or anticommutes with zt as these elements, when they are commuted past the state, will permute the basis elements. Since two different basis elements annihilate, the product of a state and a permutation of it will be zero.

An example will assist in this. One computes the average of an operator in density matrices by taking the trace of the products of the states. We wish to compute $tr(x \rho)$. We first note that $\rho^2 = \rho$ and the trace allows us to move stuff around so that we write the computation as $tr(x\rho^2) = tr(\rho x \rho)$. Writing $\rho$ out in terms of the square roots of unity we have:

On the other hand, if a canonical basis element commutes with both ixy and zt, then, because the set {ixy, zt} is a maximal set of commuting roots of unity, that canonical basis element has to be either 1, ixy, zt, or ixyzt (or an imaginary multiple of one of these). But the basis quantum states are eigenstates of these four operators with eigenvalues of $\pm 1$ according as the quantum state has a +1 or -1 in its corresponding position.

From the above, it should be clear that for our new states to be “unbiased” they must be built out of those canonical basis elements that take average values of zero, along with unity, which all pure density matrices need. (Actually, I’m not sure of this, but it is true for the Clifford algebras I care about, that is, the Pauli algebra, the Dirac algebra, and C(4,1).) So to define a basis that is unbiased with respect to our original basis of {ixy, zt}, we need only choose a basis that avoids ixy, zt, and ixyzt.

So for our second basis, we can choose pretty much anything we like, for example {x, iyz} so our two basis sets are {ixy, zt} and {x, iyz}. To get the third basis, we need to avoid commuting with either of these, so this eliminates ixy, zt, ixyzt, x, iyz, and ixyz. What is left is y, z, t, ixz, xt, yt, ixyt, ixzt, and iyzt. We need to choose two of these that commute. The first two that commute is {y, xt} and a set of 3 MUBs for the Dirac algebra is {ixy, zt}, {x, iyz}, and {y,xt}.

Can we add a fourth basis set? You bet. Adding {y,xt} to our MUB left the remaining degrees of freedom as z, it, ixz, yt, xzt, and yzt. We can choose {z,yt}, which leaves it, ixz, and xzt. The fifth basis set is therefore {it,ixz}. This gives the five MUBs as:

{ixy, zt}, {x, iyz}, {y,xt}, {z,yt}, {it,ixz}.

Note that each of these uses three degrees of freedom, for example, {ixy,zt} uses up ixy, zt, and ixyzt. There are 16 total degrees of freedom in the Dirac bilinears, all the MUBs share unity. With 5 basis sets and unity, we have 5×3 +1 = 16 and we’ve used up all the degrees of freedom of the Dirac bilinears.

When I get around to it, I’ll convert the above into 5 sets of four 4×1 spinors that are give the same dot product. For now, simply note that when you compute the expectation values of one state with respect to a state from another basis set, everything will be annihilated except the 1/4, so the measurement will give 1/4. Thus the transition probabilities are all equal to 1/4.

Okay, the algebra wasn’t too difficult. With a little help from Java, I substituted in the chiral representation for the Dirac algebra into the above 5 sets of basis sets. This gives 5 sets of spinors that are all a distant 1/4 from each other sure enough. The 5 sets are as follows, in the order given above.

The basis set for {ixy, zt} (which is diagonalized in the chiral representation):

The other 4 basis sets are not diagonalized and consequently are more complicated. I will give twice the normalized spinors, and will choose phase so that the top element of the spinor is positive real. Then the four sets are, in order:

Now, wasn’t the density matrix version of all that a lot more elegant, less arbitrary, more understandable, and more useful than the ugly, arbitrary, confusing, and useless spinor method? Density matrices are your friend. Learn them and use them.

Mass

Now what does all this have to do with mass? I can write this in MUB notation, perhaps it will make it easier to understand. My neutrino mass formula was built on the assumption that the elementary fermions are composites built from three preons each, and these three preons are taken from three different bases that are unbiased. There is only one Feynman diagram, and it corresponds to basis permutation. When analyzing bound states, one writes all the very long lived bound state Feynman diagrams and divides them into classes according to how they permute the basis elements. Then one sums over the Feynman diagrams in each class. (The sums are finite and exist because the calculation is done in qubits, see this post for the mathematical details and literature reference.)

Since the particle is bound, it’s state after a long time can be assumed to be identical to its initial state. Furthermore, this process, the long time propagation of a bound state, has to be idempotent. To get the propagator for a doubly long time, one uses the usual methods of QFT in momentum space. This turns the problem into a matrix multiplication problem. One looks for solutions to the matrix square problem, $\rho^2 = \rho$.

An important detail is that when you convert the problem from products of Feynman diagrams into matrix multiplication, you have to take into account the quantum or Berry phase. This introduces $\pi/12$ into the complex phases, which is how it comes to be that my calculation for the neutrino massees has this factor. It was the resulting formula that got my references in the peer reviewed literature despite my never “publishing” anything.