Eigenvectors of exponential matrix (pauli matrix)

AI Thread Summary
The discussion focuses on finding the eigenvectors and eigenvalues of the matrix exp(iπσx/2), where σx is the x Pauli matrix. The matrix exponential is expressed as a series, allowing the separation into sums for even and odd powers of σx. The participant notes that there are two linearly independent eigenvectors for each sum but struggles with the multiplication of the imaginary unit. It is clarified that eigenvectors of σx are (1,1) and (1,-1), which must satisfy the eigenvalue equation exp(iπσx/2)v = Av. The conversation emphasizes the connection between the eigenvectors of the exponential matrix and the original Pauli matrix.
Punctualchappo
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Homework Statement


Find the eigenvectors and eigenvalues of exp(iπσx/2) where σx is the x pauli matrix:
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Homework Equations


I know that σxn = σx for odd n
I also know that σxn is for even n:
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I also know that the exponential of a matrix is defined as Σ(1/n!)xn where the sum runs from n=0 to infinity

The Attempt at a Solution



Using knowledge of the matrix exponential, I can say that exp(iπσx/2) = Σ(1/n!)(iπσx/2)n. I can then split this into two sums, one for even n and one for odd n. This allows me to take the power off the matrices for easier summing. It's then that I get stuck. I've attached a file showing the two sums because it's easier and clearer to show you this way.[/B]

I'd really appreciate any help that can be given.
 

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I've realized that when written as the two sums, it's clear that there are only two linearly independent eigenvectors for each sum. I'm not sure how it works when the i's get multiplied in though. Aren't there infinite eigenvalues to reflect the infinite nature of the sums?
 
Hi chap, welcome to PF :),

What would an eigenvector of exp(iπσx/2) look like ? What conditions would it have to satisfy ?

And what about eigenvectors of σx itself ?
 
So the eigenvectors would have to satisfy exp(iπσx/2)v = Av where A is the eigenvalue. I know that eigenvectors of σx are (1,1) and (1,-1)
Thanks for your help.
 
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