# How to find eigenvectors/eigenspinors

1. May 26, 2014

### mark.laidlaw19

1. The problem statement, all variables and given/known data

Hi all, I have a net spin operator that the problem has asked me to find:$$S=\frac{\hbar}{2} \left(\begin{array}{cc}\cos\alpha&\sin\alpha\\\sin\alpha&-\cos\alpha\end{array}\right)$$ and I need to write out the matrix representation with respect to the $S_z$ spinor basis.

2. Relevant equations

I know that to do this, I need to find the eigenvalues and eigenvectors (or eigenspinors) of this matrix, and then rewrite them in terms of the $S_z$ basis. From the characteristic equation of this matrix, I have found the eigenvalues to be $\pm\frac{\hbar}{2}$

3. The attempt at a solution

I am pretty confident with most of this problem, however, when I try to find the eigenvectors of the spin operator above, I end up with this:

$$\left(\begin{array}{cc}\cos\alpha&\sin\alpha\\\sin\alpha&-\cos\alpha\end{array}\right) \left(\begin{array}{cc}a\\b\end{array}\right) = \left(\begin{array}{cc}a\\b\end{array}\right)$$
I can't seem to solve these equations for a or b in order to find the eigenspinors, as once I substitute one equation into the other, the a and the b cancel out, and I'm wondering if there is a simple step that I am missing.

Many thanks in advance

Last edited: May 26, 2014
2. May 26, 2014

### unscientific

From your eigenvector equation you obtain:

$$a cos(\alpha) + b sin(\alpha) = a$$
$$\frac{a}{b} = \frac{sin (\alpha)}{1 - cos(\alpha)}$$

Can you use the double angle identity to simplify this? $\sin2x = 2sinx cos x$ and $sin^2x = \frac{1- cos2x}{2}$

3. May 26, 2014

### dauto

Off course it does. That's always going to be true for an eigenvalue equation. You already found the eigenvalues and subbed them into the eigenvalue equation. That means the two equations are not independent anymore. If you use both of them you just get an identity such as 1=1 (duh...) What you have to do is use only one of the equations to find a/b like unscientific did and than impose some normalization condition on the resultant eigenvector.

4. May 26, 2014

### unscientific

Exactly, because both equations are exactly the same, simply rearranged.

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