MUBs (for Mutually Unbiased Bases) are quite popular at the moment. Kea is running a mini-series Mutual Unbias as is Carl Brannen. Further, the Perimeter Institute has a good website for its seminars where they offer streaming video (I like their MacromediaFlash format giving video and slides/blackboard shots simultaneously, in distinct windows) including a talk on MUBs (as well as an old talk by Wootters).

So what are MUBs to mathematicians? Recall that a d-state quantum system is just the vectorspace equipped with the usual Hermitian inproduct . An observable is a choice of orthonormal basis consisting of eigenvectors of the self-adjoint matrix . together with another observable (with orthonormal basis ) are said to be mutally unbiased if the norms of all inproducts are equal to . This definition extends to a collection of pairwise mutually unbiased observables. In a d-state quantum system there can be at most d+1 mutually unbiased bases and such a collection of observables is then called a MUB of the system. Using properties of finite fields one has shown that MUBs exists whenever d is a prime-power. On the other hand, existence of a MUB for d=6 still seems to be open…

The King’s Problem¹ is the following : A physicist is trapped on an island ruled by a mean king who promises to set her free if she can give him the answer to the following puzzle. The physicist is asked to prepare a d−state quantum system in any state of her choosing and give it to the king, who measures one of several mutually unbiased observables on it. Following this, the physicist is allowed to make a control measurement on the system, as well as any other systems it may have been coupled to in the preparation phase. The king then reveals which observable he measured and the physicist is required to predict correctly all the eigenvalues he found.

The Solution to the King’s problem in prime power dimension by P. K. Aravind, say for , consists in taking a system of k object qupits (when one qupit is a spin l particle) which she will give to the King together with k ancilla qupits that she retains in her possession. These 2k qupits are diligently entangled and prepared is a well chosen state. The final step in finding a suitable state is the solution to a pure combinatorial problem :

She must use the numbers 1 to d to form ordered sets of d+1 numbers each, with repetitions of numbers within a set allowed, such that any two sets have exactly one identical number in the same place in both. Here’s an example of 16 such strings for d=4 :

11432, 12341, 13214, 14123, 21324, 22413, 23142, 24231, 31243, 32134, 33421, 34312, 41111, 42222, 43333, 44444

Here again, finite fields are used in the solution. When , identify the elements of with the numbers from 1 to d in some fixed way. Then, the of number-strings are found as follows : let and take as the first 2 numbers the ones corresponding to these field-elements. The remaning d-2 numbers in the string are those corresponding to the field element (with ) determined from by the equation

where is the field-element corresponding to the integer i ( corresponds to the zero element). It is easy to see that these strings satisfy the conditions of the combinatorial problem. Indeed, any two of its digits determine (and hence the whole string) as it follows from and that .

In the special case when d=3 (that is, one spin 1 particle is given to the King), we recover the tetracode : the nine codewords

0000, 0+++, 0—, +0+-, ++-0, +-0+, -0-+, -+0-, –+0

encode the strings (with +=1,-=2,0=3)

3333, 3111, 3222, 1312, 1123, 1231, 2321, 2132, 2213

1. actually a misnomer, it’s more the poor physicists’ problem… [↩]

Time to wrap up this series on the Bost-Connes algebra. Here’s what we have learned so far : the convolution product on double cosets

is a noncommutative algebra, the Bost-Connes Hecke algebra , which is a bi-chrystalline graded algebra (somewhat weaker than ’strongly graded’) with part of degree one the group-algebra . Further, has a natural one-parameter family of algebra automorphisms defined by and .

For any algebra together with a one-parameter family of automorphisms one is interested in KMS-states or Kubo-Martin-Schwinger states with parameter , (this parameter is often called the ‘invers temperature’ of the system) as these are suitable equilibria states. Recall that a state is a special linear functional on (in particular it must have norm one) and it belongs to if the following commutation relation holds for all elements

Let us work out the special case when is the matrix-algebra . To begin, all algebra-automorphisms are inner in this case, so any one-parameter family of automorphisms is of the form

where is the matrix-exponential of the nxn matrix . For any parameter we claim that the linear functional

is a KMS-state.Indeed, we have for all matrices that

(the next to last equality follows from cyclic-invariance of the trace map). These states are usually called Gibbs states and the normalization factor (needed because a state must have norm one) is called the partition function of the system. Gibbs states have arisen from the study of ideal gases and the place to read up on all of this are the first two chapters of the second volume of “Operator algebras and quantum statistical mechanics” by Ola Bratelli and Derek Robinson.

This gives us a method to construct KMS-states for an arbitrary algebra with one-parameter automorphisms : take a simple n-dimensional representation , find the matrix determining the image of the automorphisms and take the Gibbs states as defined before.

Let us return now to the Bost-Connes algebra . We don’t know any finite dimensional simple representations of but, sure enough, have plenty of graded simple representations. By the usual strongly-graded-yoga they should correspond to simple finite dimensional representations of the part of degree one (all of them being one-dimensional and corresponding to characters of ).

Hence, for any ( details) we have a graded simple -representation with action defined by

Here, is computed using the ‘chinese-remainder-identification’ ( details).

Even when the representations are not finite dimensional, we can mimic the above strategy : we should find a linear operator determining the images of the automorphisms . We claim that the operator is defined by for all . That is, we claim that for elements we have

So let us compute the action of both sides on when . The left hand side gives whereas the right-hand side becomes

proving the claim. For any parameter this then gives us a KMS-state for the Bost-Connes algebra by

Finally, let us calculate the normalization factor (or partition function) . Because we have for that the trace

is equal to the Riemann zeta-value (at least when ).

Summarizing, we started with the definition of the Bost-Connes algebra , found a canonical one-parameter subgroup of algebra automorphisms and computed that the natural equilibria states with respect to this ‘time evolution’ have as their partition function the Riemann zeta-function. Voila!