Where Am I Going Wrong in Sakurai's Quantum Spin Eigenvalue Problem?

[Theage]

Homework Statement

I am currently working on a seemingly straightforward eigenvalue problem appearing as problem 1.8 in Sakurai's Modern QM. He asks us to find an eigenket $\vert\vec S\cdot\hat n;+\rangle$ with $$\vec S\cdot\hat n\vert\vec S\cdot\hat n;+\rangle = \frac\hbar 2\vert\vec S\cdot\hat n;+\rangle$$ where the unit vector n is defined by polar angle alpha and azimuthal angle beta.

Homework Equations

The definitions of the Pauli sigma matrices, along with the formula $$\hat n = (\sin\alpha\cos\beta,\sin\alpha\sin\beta,\cos\alpha)$$ for conversion to Cartesian coordinates.

The Attempt at a Solution

$$\vec S\cdot\hat n = \frac\hbar 2\sin\alpha\cos\beta\begin{pmatrix}0&1\\1&0\end{pmatrix}+\sin\alpha\sin\beta\begin{pmatrix}0&-i\\i&0\end{pmatrix}+\cos\alpha\begin{pmatrix}1&0\\0&-1\end{pmatrix} = \frac\hbar 2\begin{pmatrix}\cos\alpha&\sin\alpha e^{-i\beta}\\\sin\alpha e^{i\beta}&-\cos\alpha\end{pmatrix}.$$ Thus we have the $\lambda=1$ eigenvalue problem for the matrix $$\begin{pmatrix}\cos\alpha&e^{-i\beta}\sin\alpha\\e^{i\beta}\sin\alpha&-\cos\alpha\end{pmatrix}.$$ This becomes the system of equations $$\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}x\cos\alpha+ye^{-i\beta}\sin\alpha-x\\-y\cos\alpha+xe^{i\beta}\sin\alpha-y\end{pmatrix}.$$ Thus beta appears to be completely arbitrary, and alpha = n*pi, however this appears to have no solutions as we get cosines equaling 2 which is nonsense. The characteristic polynomial predicts a lambda=1 eigenvalue - where am I going wrong?

 

[TSny]

Thus we have the $\lambda=1$ eigenvalue problem for the matrix $$\begin{pmatrix}\cos\alpha&e^{-i\beta}\sin\alpha\\e^{i\beta}\sin\alpha&-\cos\alpha\end{pmatrix}.$$

This looks good to me.

This becomes the system of equations $$\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}x\cos\alpha+ye^{-i\beta}\sin\alpha-x\\-y\cos\alpha+xe^{i\beta}\sin\alpha-y\end{pmatrix}.$$

I don't understand your matrix on the right hand side. How did you get the $-x$ and the $-y$ terms on the far right of the matrix?

$\alpha$ and $\beta$ are fixed by the choice of $\hat{n}$. You need to find $x$ and $y$.