Polarization Controller Stanford Lefevre

AN EXAMINATION OF THE MANUAL FIBER POLARIZATION CONTROLLER PEADAR COYLE

Abstract. The problem statement for my own exposition in the MPC

1. Introduction A Manual fiber polarization controller can be modelled as two circles at right angles to each other, joined by a ’connecting helix’. The ’connecting helix’ is what we are trying to find. The Manual Fiber polarization controller is believed to be controlled by stress induced birefringence and geometric birefringence. An important aspect of the geometric phase, is the global nature of it, this global nature is believed to be useful in the Fault Tolerance of quantum computing networks. 2. The manual fiber polarization controller Quantum Computing, is an exciting and fast growing field. There is a need to control the polarization of quibits, and one wat to do this is to use waveplates. However waveplates can be expensive and not as easily adaptable as ’bat ear’ or to give it a more understandable name ’the manual fiber polarization controller’. The Manual Fiber Polarization controller was invented circa 1980 by Lefevre, in Stanford (?). n 1980, H. C. Lefevre devised an all fiber polarization controller based on the same tri-waveplate concept (see Figure 1). In this device, the three wave plates were replaced with optic fiber coils. Coiling the fiber induces stress on the fiber, which in turn produces birefringence via the photoelastic effect. The amount of birefringence is inversely proportional to the diameter squared of the coil. By adjusting the diameter of the coil and number of turns in the coil, any desired fiber wave plate can be created. Because the wave plates are made of a fiber, it is not necessary to bring the light out of the fiber, thus eliminating the time consuming process of collimation, alignment, and refocusing. In addition, material cost is greatly reduced because these fiber wave plates are much less expensive than the bulk wave plates. Although numerous names, such as the Micky Mouse Ears, the Dog Ears, the Butterfly, the Three Panels, and the Fiber Coils, have been given to this clever idea, the Lefevre technique represents a major step forward from the triwaveplate controller because it utilizes the intrinsic properties of an optical fiber. In this short paper one proposes to see how much of the actual change in polarization can be explained by the geometric phase. It shall use the rather well understood optical observation of the Berry Phase in coiled optical fibers, and see how much of this can be explained by photoelastic effects and how much by geometric effects. Date: March 15, 2009. 1

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3. Proposed Model Using the well understood mathematics of differential geometry, the model of the Manual Fiber Polarization controller (MPC) will be using the Frenet Frame of the Cylindrical Helix. We shall model the MPC (illustrated below) as two circles at right angles to each other. Of course the MPC won’t always have the two wires at right angles (due to the range of motion of the paddles), however this is the first simple case. The aim of the following draft is to find the connecting helix of these two circles. One way we could model these two circles is using the Frenet Frame of the Helix. The Frenet Frame is a 3 times continuously differential frame, which is easily mapped to the electromagnetic field. Let one circle in the x-y plane, be translated by +1 from the origin of the cartesian axis. Let the torsion be zero, and let the circle be a space curve representable by Radius and Curvature. We know that a helix is defined as a a mapping from <− > <3 3.1. Problem Definition. For the purposes of this problem definition, we define the circles to be two helices defined with zero torsion, translated from the y-axis by 1. Let the parameterization be based on the Serret-Frenet frame, with β being the angle around the circle. (3.1)

γxy = [Cosβ + 1, Sinβ + 1, 0]

(3.2)

γyz = (0, Cosβ + 1, Sinβ + 1)

(3.3) L. et us before moving on remind ourselves of some differential geometry formulas. p (3.4) τ= 2 R + p2 (3.5)

κ=

R2

R + p2

Where κ and τ are curvature and torsion respectively. Case 1. Notice in γ case κ=1/R which is a well known result for the circle. i.e the radius of a circle is equal to the inverse of its curvature This is not technically a theorem but i’ll try to treat it as thus 4. Key Formulae

(4.1)

γ 0 = TA = (−Sinβ, Cosβ, 0);

(4.2)

NA = TA0 = (−Cosβ, −Sinβ, 0) ;

(4.3)

BA = NA0 = (Sinβ, −Cosβ, 0)

The other circles formula are listed below: (4.4)

γ 0 = TB = (0, −Sinβ, Cosβ);

(4.5)

NB = TB0 = (0, −Cosβ, −Sinβ) ;

(4.6)

BB = NB0 = (0, Sinβ, −Cosβ)

BAT EARS

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5. Geometric Comparison of the ends of the connecting helix Since we are essentially trying to find where the circle mends to the helix, I shall provide a list of formula I’ve derived. Is the ’mending’/ blending idea coherent. I eagerly request feedback on this issue. 6. Let us define a helix Firstly I’ll give a bit of revision of the differential geometry formulae. /kappa and /tau are both scalars and subsequently invariant. (6.1)

τ = p/[R∧ 2 + p∧ 2];

(6.2)

κ = R/[R∧ 2 + p∧ 2]

(6.3)

q = (RCos(t), RSin(t), pt)

(6.4)

T = (−RSin(t), RCos(t), p)

(6.5)

N = (−RCos(t), −RSin(t), 0)

(6.6) where R is the radius of the circle, and p is the pitch length along the z-axis divided by 2π. Some more equations comparing a parameter along the connecting helix. We compare the values of T and N using dotproducts, using 0 inputted as parameters. (6.7)

R2 Sint T (0).N (t) = p p2 + R2

(6.8)

p2 + CosR2 t T (0).T (t) = p p2 + R2

(6.9)

CosR2 t N (0).N (t) = p p2 + R 2 7. Cases for the Photon

In the literature it has been both experimentally and theoretically examined that the photon, the massless spin-1 boson that carries light, can exhibit manifestions of the Berry Phase. The Berry Phase is a non-integrable phase factor that arises from the adiabatic transport of a system around a closed path in parameter space. According to Simon this can be viewed as parallel transport. q = RCost, RSint, bt (7.1)

Where R is the radius (which is inverse to the curvature κ and b is the pitch of the helix. The pitch is defined as ...... Since the torsion is zero therefore the helix (circle) is defined as .... There are alternative methods to solving this problem, and one major problem will be the problem of discontinuities. For instance does the curvature turn to zero on the connecting helix, what happens if it does? A method proposed by Bishop in 1975 was to use Relatively Parallel Frames, the method of parallel transport (which is defined in the review article as being from a result in differential geometry) around the frame. Derivations are shown below....

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8. The Definition of the Problem The following are the Frenet Parameters at the end of the helix. The object is to find a connecting helix. (8.1)

q = {Cosβ + 1, Sinβ + 1, 0}

(8.2) (8.3)

TA = q 0 = {−Sinβ, Cosβ, 0}

(8.4)

NA = q” = {Cosβ, −Sinβ, 0}

(8.5) (8.6)

BA = q”’ = {−Sinβ, Cosβ, 0}] qA = {0, Cosβ − 1, Sinβ − 1}

(8.7) (8.8)

TA = q 0 = {0, −Sinβ, Cosβ}

(8.9) (8.10)

NA = q” = {0, Cosβ, −Sinβ}

(8.11)

BA = q”’ = {0, −Sinβ, Cosβ}]

Geometrically compare the two ends of the helix, (8.12)

TA .TB = −SinβCosβ

(8.13)

NA .TB (β) = (Sin)∧ 2β

(8.14)

NB .TA (β) = (Cos)∧ 2β

1 (Sin2β) 2 T. he other important thing to do is to compare a parameter at the ends of the helix. R2 Sint (8.16) T (0) · N (t) = − p p2 + R2 (8.15)

NA .NB (β) = −SinβCosβ =

The fraction in front of the R2 Sint is the normalisation coefficient. p2 + CostR2 (8.17) T (0) · T (t) = p2 + R2 . . His other statement, which I took from an email from him was, that we cut the circle at some point (call this α). 9. His final part of the problem statement There are going to be a few possible results when solving this equation. No solution A continuous solution A discrete set of results

BAT EARS

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10. Derivation of the Berry Phase in this system The following interpretation is the Z p (10.1) θ = τ (Cos[t] + 1)2 + (Sin[t] + 1)2 dt Where θ is the Berry Phase and τ is the scalar invariant classifying the deviation of planarity of the curve, or the torsion. Z p (10.2) (1 + Cos[t])2 + (1 + Sin[t])2 dt I appreciate feedback about the integral above. Z q

(10.3)

sin2 (t) + 2 sin(t) + cos2 (t) + 2 cos(t) + 2 dt

The result is:  √

      √  3  π π 1 1 1 1 3 F1 − ; − , − ; ; 2 + √ 4 sin t + +3 2 , √ −2 2 cos t + 4 2 2 2 2 4 2 2       √ 1 π 1 1 1 3 +3 2 + 2(sin(2t) − 1) F1 ; , ; ; 4 sin t + 4 cos(t) − sin(t) 2 2 2 2       √  √  3  3 π π √ √ 2+ +3 2 , +3 2 4 sin t + − 2 4 sin t + 4 4 2 2 r r   √ π π (6 sin(t) + 6 cos(t) + 9) + 2 − sin t + − 1 1 − sin t + 4 4 r r    π π p / − sin t + − 1 1 − sin t + 4 sin(t) + 4 cos(t) + 6 4 4 As one can see this is a non-analytic solution, which means I’m not exactly sure where to proceed.