Eigenfunction of a Jones Vector (System)


by KasraMohammad
Tags: eigenfunction, jones, vector
KasraMohammad
KasraMohammad is offline
#1
Nov15-12, 02:35 PM
P: 20
I am trying to find out just how to solve for the eigenfunction given a system, namely the parameters of an optical system (say a polarizer) in the form of a 2 by 2 Jones Vector. I know how to derive the eigenvalue, using the the constituent det(λI -A) = 0, 'A' being the system at hand and 'λ' the eigenvalue. How do you go about solving for the eigenfunction?
Phys.Org News Partner Physics news on Phys.org
Sensitive detection method may help impede illicit nuclear trafficking
CERN: World-record current in a superconductor
Beam on target: CEBAF accelerator achieves 12 GeV commissioning milestone
mathman
mathman is offline
#2
Nov15-12, 03:14 PM
Sci Advisor
P: 5,935
This is a pure math question. Try the following:

http://mathworld.wolfram.com/Eigenvalue.html
http://www.math.hmc.edu/calculus/tutorials/eigenstuff/
http://www.sosmath.com/matrix/eigen2/eigen2.html
Philip Wood
Philip Wood is offline
#3
Nov16-12, 05:15 PM
P: 860
Once you've found λ, you can substitute its value into Av = λv. If you then multiply out the left hand side and equate components, v1 and v2, of v on either side, you'll get two equivalent equations linking v1 and v2. Eiter will give you the ratio v1/v2. This is fine: the eigenvalue equation is consistent with any multiplied constant in the eigenvector. There will be a normalisation procedure for fixing the constant.

KasraMohammad
KasraMohammad is offline
#4
Nov16-12, 07:39 PM
P: 20

Eigenfunction of a Jones Vector (System)


so I got λ = 1. 'A' I assume is the system matrix or my Jones Vector, which is given as a 2 by 2 matrix. So that makes Av=v, thus A must be 1?? The 'v' values must be the same, but isn't 'v' the eigenfunction itself? The equation Av=v eliminates the 'v' value. What am I doing wrong here?
Philip Wood
Philip Wood is offline
#5
Nov17-12, 02:39 AM
P: 860
v is the vector and A is the matrix. The matrix isn't a vector, but is an operator which operates on the vector.

Try it with a matrix A representing a linear polariser at 45 to the base vectors. This matrix has all four elements equal to 1/2. This gives eigenvalue of 1, and on substituting as I explained above, shows the two components, v1 and v2, of the vector to be equal, which is just what you'd expect.


Register to reply

Related Discussions
What does it mean that vector is independent of coordinate system Introductory Physics Homework 5
show that this system is not a vector space Calculus & Beyond Homework 3
Jones vector of circularly polarized light Classical Physics 1
Jones gives quantum algorithm for Jones knot polynomial Beyond the Standard Model 0