Jtdylktuy

Hi everyone I am trying to understand the geometric phase along an optical
optical fiber. I am a geometer and I am almost ok with the geometric part
but I have some diffuculties to extract physical computation. So I need your
help hoping to get my answer. Let me talk about some background on the
subject.

I am considering a geometric phase along the optical fiber as I told. To do
that I consider a curve $x$ and suppose that the optical fiber is determined
by this curve. According to Frenet frame it is known that
begin{eqnarray*}
mathbf{t}_{s} &=&kappa mathbf{n}, \
mathbf{n}_{s} &=&-kappa mathbf{t}+tau mathbf{b}, \
mathbf{b}_{s} &=&-tau mathbf{n},
end{eqnarray*}%
where $kappa ,tau ~$are the curvature and torsion of the given curve.

I suppose that $mathbf{E}$ be the complex three component electric field
vector at the point $x(s)$. The magnitude $mathbf{E}$ does vary along the
ray while its direction evolves via Fermi-Walker derivative. Then we can
normalized $E$ as $mathbf{A}=frac{mathbf{E}}{left( mathbf{E}^{ast }%
mathbf{E}right) frac{1}{2}},, $where $mathbf{E}^{ast }~$is the
conhugate of the $mathbf{E}.$ From now on, I choose to follow the
following method since it is much more close to geometry, which can also be
found in the article written by Mukunda and Simon 1993. According to this
approach, thinking of each $mathbf{t},mathbf{n},~and~mathbf{b}$ as three
component real column vectors we can write
begin{equation}
ifrac{d}{ds}Amathbf{=}HA,
end{equation}%
$H=ikappa (nt^{T}-tn^{T}). $In addition to the evolution eq. 1 there is a
constrain $t^{T}A=0. $To solve the eq. 1 let us take
$$
A=alpha frac{mathbf{n}+imathbf{b}}{sqrt{2}}+beta frac{mathbf{n}-i%
mathbf{b}}{sqrt{2}}.
$$
For the the left hand side of the Eq. 1 if I use the Frenet frame I got
$$
ifrac{d}{ds}A=frac{i}{sqrt{2}}(kappa (-alpha -beta )mathbf{t+(}alpha
^{^{prime }}-alpha itau +beta ^{^{prime }}+beta itau )mathbf{n+(}%
alpha ^{^{prime }}i+alpha tau -beta ^{^{prime }}i+beta tau )mathbf{%
b.}$$
Now if I plug this into the Eq. 1 I got
begin{equation}
frac{i}{sqrt{2}}(kappa (-alpha -beta )mathbf{t+(}alpha ^{^{prime
}}-alpha itau +beta ^{^{prime }}+beta itau )mathbf{n+(}alpha
^{^{prime }}i+alpha tau -beta ^{^{prime }}i+beta tau )mathbf{b=}%
HA=ikappa (nt^{T}-tn^{T})A.
end{equation}%
Now I have no idea how to deal with the right hand side of the equation. I
do not know what kind of operation do I need to reach to the following
results.
begin{eqnarray*}
alpha ^{^{prime }} &=&itau alpha , \
beta ^{prime } &=&-itau beta .
end{eqnarray*}%
I have an idea but I could not convince myself. If for some reason I assume
the right hand side of the eq. 2 is zero then I have
begin{eqnarray*}
(-alpha -beta ) &=&0, \
mathbf{(}alpha ^{^{prime }}-alpha itau +beta ^{^{prime }}+beta itau
) &=&0, \
mathbf{(}alpha ^{^{prime }}i+alpha tau -beta ^{^{prime }}i+beta tau
) &=&0.
end{eqnarray*}%
From the last two equation I actually get $alpha ^{^{prime }}=itau alpha
,~and, beta ^{prime }=-itau beta .$ However, I made the assumption
with no sense and also I do not know waht to do the first equation $(-alpha
-beta )=0$ even my assumption is correct coincidentally.