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Eigenfunction of a Jones Vector (System)

by KasraMohammad
Tags: eigenfunction, jones, vector
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KasraMohammad
#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?
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mathman
#2
Nov15-12, 03:14 PM
Sci Advisor
P: 6,038
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
#3
Nov16-12, 05:15 PM
PF Gold
P: 939
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
#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
#5
Nov17-12, 02:39 AM
PF Gold
P: 939
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.


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