Eigenvalues and eigenvectors, pauli matrices

[ma18]

Homework Statement

Look at the matrix:

A = sin t sin p s_x + sin t sin p s_y +cos t s_z

where s_i are the pauli matrices

a) Find the eigenvalues and normalized eigenvectors (are they orthogonal)?

b) Write the eigenvector of s_x with positive eigenvalue as a linear combination of the eigenvalues of A

Homework Equations

The Attempt at a Solution

I got an answer but I am not confident as to it's veracity.

The matrix A is :

(cos t , sin t cos p - i sin t sin p
sin t cos p + i sin t sin p, - cos t)

or

(cos t, sin t (e^-i p)
sin t (e^ip), - cos t)

Finding the eigenvalues

(cos t - lambda, sin t (e^i p)
sin t (e^i p), - cos t - lambda)

You get

lambda^2 - 1 = 0

So the eigenvalues are +-1

The problem is finding the eigenvectors;

For +1

(cos t - 1, sin t e^(-i p )
sin t e^i p, - cos t - 1)

Using the cofactor method you get the equations

(cos t - 1) x +(sin e^i p) y = 0

(sin t e^i p) x + (-cos t - 1) y = 0

I get

v1 = C_1 (-e^-i p tan (t/2), 1)
v2 = C_2 (-e^i p tan (t/2), 1)

Then normalizing

C_1 = 1/sqrt(e^-2 i p tan (t/2)^2 + 1)

C_2 = 1/sqrt(e^-2 i p cot (t/2)^2 + 1)

I don't know how to do b

Any help/checking would be helpful

 

[vela]

The problem is finding the eigenvectors;

For +1

(cos t - 1, sin t e^(-i p )
sin t e^i p, - cos t - 1)

Using the cofactor method you get the equations

(cos t - 1) x +(sin e^i p) y = 0

(sin t e^i p) x + (-cos t - 1) y = 0

Looks good so far. Try using the trig identities $\sin^2 \frac \theta 2 = \frac{1 - \cos \theta}{2}$ and $\sin \theta = 2 \sin\frac \theta 2 \cos \frac \theta 2$. That'll make the algebra a bit easier to deal with.

 

[ma18]

Thanks for all your help guys!