Eigenfunction of a Jones Vector (System)

In summary, the conversation is about solving for the eigenfunction in a system using the eigenvalue equation. The recommended approach is to first find the eigenvalue using the determinant of the matrix and then substituting it into the equation Av = λv. This will result in two equations linking the components of the vector, and a normalization procedure can be used to fix the constant. Additionally, it is important to note that the matrix is an operator while the vector is the eigenfunction itself.
  • #1
KasraMohammad
20
0
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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  • #3
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.
 
  • #4
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?
 
  • #5
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.
 

1. What is an eigenfunction of a Jones Vector?

An eigenfunction of a Jones Vector is a vector that represents a specific state or mode of polarization in a light beam. It is a mathematical representation of the polarization state and can be used to analyze and manipulate light in various optical systems.

2. How is an eigenfunction of a Jones Vector calculated?

The eigenfunction of a Jones Vector is calculated using a matrix representation of the polarizing elements in the optical system. The matrix is multiplied with the Jones Vector to obtain the eigenfunction, which is a complex vector with two components representing the amplitude and phase of the polarization state.

3. What is the significance of eigenfunctions in optics?

Eigenfunctions are essential for understanding and manipulating light in optical systems. They allow for the analysis of complex polarization states and can be used to design and optimize optical devices such as polarizers, waveplates, and retarders.

4. Can eigenfunctions be used to describe non-linear polarization effects?

Yes, eigenfunctions can be used to describe non-linear polarization effects in optical systems. This is because the eigenfunction represents the state of polarization at a specific point in the optical system, and this state can change due to non-linear effects such as birefringence or Kerr effect.

5. Are eigenfunctions unique for each optical system?

Yes, each optical system will have its unique set of eigenfunctions. This is because the eigenfunctions are dependent on the specific polarizing elements and their arrangement in the system. Therefore, different optical systems will have different eigenfunctions, even if they have similar components.

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