Arcadian Functor

occasional meanderings in physics' brave new world

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Location: New Zealand

Marni D. Sheppeard

Wednesday, December 31, 2008

Happy New Year

Happy New Year! Tonight I will be washing dishes in a hotel, having successfully found some temporary employment.

Tuesday, December 30, 2008

Still Standing Still

It turns out that I will not be talking at Imperial College on Jan 7 after all, because unknown representatives of the UK government are holding my passport while my visa application is being processed. I was informed that the application had been deemed non standard (I cannot imagine how, having answered all the questions and, as far as I can tell, obtained the maximum possible score on the points system) which sounds ominously like 'you'll be waiting a while to hear from us'.

Tuesday, December 16, 2008

Standing Still

After a few pleasant days in Wanaka I found myself in Christchurch once again, so I went to see the movie The day the Earth stood still. I especially enjoyed the bit where the alien crosses out the Dark Force term in the equations of relativity.

Wednesday, December 10, 2008

Moving North V

OK, I think it is finally time to put Carl out of his misery.

At the end of the month I will be flying to the UK for my first ever postdoc, in quantum information theory at the University of Oxford. My new boss will be Bob Coecke. Bob is also one of the organisers of the January workshop on Categories, Logic and Physics at Imperial College.

Monday, December 08, 2008

M Theory Lesson 246

The inverses of $C^{n}$ (that is, powers of the circulant $C = (0,1,1)$) also behave very nicely. A little algebra shows that for $C^{n} = (x,y,y)$, $C^{-n} = (a,b,b)$, where

$a = \frac{x+y}{x^2 - 2y^2 + xy}$
$b = \frac{1 - ax}{2y}$

In the special cases of interest, $(x,y) = (n,n+1)$ or $(n+1,n)$, we find (respectively) that

$C^{-n} = (\frac{-(2n+1)}{3n+2}, \frac{n+1}{3n+2}, \frac{n+1}{3n+2})$ or
$C^{-n} = (\frac{2n+1}{3n+1}, \frac{-n}{3n+1}, \frac{-n}{3n+1})$

and $|a| + |b| = 1$ in all cases. For example, the inverse of $(2,3,3)$ is the circulant $(1/8) (-5,3,3)$. These forms for the inverse of a positive circulant hold even when $n$ is not an ordinal. For the more general case of a positive circulant of the form $(n,n+d,n+d)$, the sum $|a| + |b| = 1/d$.

M Theory Lesson 245

The $3 \times 3$ code circulant $C = (0,1,1) = (231) + (312)$ appears in a standard generating method for a three dimensional representation of the braid group $B_3$, which we may discuss later. For now, observe that powers of $C$ define circulants $C^{n} = (x,y,y)$ where the sequences of $x$ and $y$ are given by

$x_n \in 0,2,6,10,22,42, \cdots$
$y_n \in 1,3,5,11,21,43, \cdots$

After the initial terms, $y_n$ settles down to $y_n = y_{n-1} + x_{n - 1}$, which is always equal to $x_n \pm 1$. The sequence $x_n$ is number A078008 in the database. That is, $3 x_n \equiv A_n$ gives the chromatic polynomial for 3 colours of the cyclic polygon graph on $n$ sides. In terms of the Tutte polynomial $T(t)$, $A_n (t)$ may be expressed as

$A_{n} (t) = (-1)^{n-1} t T(1-t,0)$

Similarly, for the $4 \times 4$ case $C = (0,1,1,1)$, the sequence $x_n$ is given by A054878. This sequence is associated with paths on a square, whereas the $3 \times 3$ sequence is associated with paths on a triangle. Note that $C$ is the adjacency matrix for the triangle.

In general, calculating the chromatic polynomial of a graph is an NP-complete problem. By splitting the circulant matrix elements into 2 sequences, it is easy to define $x_n$ in terms of the simple recursion $x_n = y_n + (-1)^{n-1}$, where $y_n$ itself is built from the Fibonacci type rule above.

Sunday, December 07, 2008

Gravity Probe Update

According to the last status report for Gravity Probe B, they will soon post a final analysis of the data for both GR tests.

Friday, December 05, 2008

Darkness Rising

A recent PI talk by the theoretician Itay Yavin discusses a gauge theoretic WIMP explanation of dark matter evidence, based on work with various authors, including Neal Weiner.

He was not afraid to question the validity of some of the experimental results, in which he was helped along by an audience member, who sounded suspiciously like John Moffat (I cannot view the video). However, Yavin pointed out that it is difficult to collectively dismiss results from DAMA, PAMELA, ATIC, CDF (multi muons), the WMAP haze and Integral.

The model in question looks at a breaking of a dark sector $SU(2) \times U(1)$ symmetry, resulting in a $4 \times 4$ mass matrix (for 4 bosonic states) whose parameters can be selected to fit all results mentioned above, although not via the same decay processes. To everyone's amusement, an audience member asked whether or not there were more parameters in the model than data points. Anyway, the CDF lepton jets are supposed to originate from dark photon decays. One suspects that we will see many more talks along these lines in the near future.

Thursday, December 04, 2008

M Theory Lesson 244

The list of $d+1$ operators, whose columns form a set of MUBs in any prime dimension $d$, is most easily described by the procedure outlined in the paper by Monique Combescure. For $d = 3$, up to factors of $\sqrt{3}$, this operator set is The standard basis is read off the identity matrix, and M theorists will recognise the Fourier operator, which defines the second basis. In general, the third operator is defined by the 1-circulant

$M_3 = (1, \omega^{-1}, \omega^{-3}, \cdots, \omega^{-k(k+1)/2},1)$

and the remaining bases are specified by circulant powers of $M_3$. For $d = 3$ there only remains $M_4 = M_{3}^{2}$. The operator $M_{3}$ diagonalises $VU$, for the two Weyl generators $U$ and $V$. For $d=3$, $V = (231)$ and $U$ is the diagonal $(1, \omega, \omega^{2})$.

Combescure extends this result to all odd dimensions, in which case $j+1$ MUBs are constructed, where $j>2$ is the smallest divisor of $d$, and $M_{3}^{j-1}$ is the highest non trivial power of $M_3$. In even dimensions, there are only three operators which provide MUBs, and $M_3$ is defined differently. In particular, one requires the root $\sqrt{\omega}$, forcing factors of $i$ into the Pauli MUB algebra.

Tuesday, December 02, 2008

M Theory Lesson 243

Recall that the two dimensional Fourier type operator, which diagonalises a $2 \times 2$ circulant, is combined with a three dimensional Fourier operator to obtain the tribimaximal neutrino mixing matrix. Using the more conventional Fourier operators, with entries $F_{ij} = \omega^{ij}$ and $0 \leq i,j \leq d - 1$, the mixing matrix is expressed Note that the zero sum of the last column of the $2 \times 2$ Fourier operator, in combination with the top row of ones on the $3 \times 3$ operator, is entirely responsible for the zero entry of the MNS matrix. A product of standard Fourier operators always has this property, since the final column cycles through all the $d$-th roots of unity, which sum to zero.

In two dimensions, instead of cycling all three Pauli operators, like our usual choice of operator, the standard Fourier operator generates a 2-cycle of the form The standard basis for $\mathbb{C}^{2}$ forms the eigenvector set for $\sigma_{Z}$, and $\sigma_{Z}$ along with $\sigma_{X}$ form the generators for the noncommutative Weyl algebra, out of which further MUBs are constructed, in this case simply $\sigma_{Y} = -i \sigma_{Z} \sigma_{X}$.

The wikipedia article links to the original MUB paper by Julian Schwinger, who already knew about the quantum Fourier transform.

Time Essayed

The deadline for the fqxi essay competition on The Nature of Time is at hand. Some entries of interest to AF readers include those by:
Louise Riofrio
Carl Brannen
Matti Pitkanen
David Finkelstein
Lawrence Crowell
Steven Weinstein
Julian Barbour
Christine Dantas
David Hestenes
Philip Gibbs
Tony Smith