How to find the direction for which the single spin is 1? I have the state $\left|\psi\right.\rangle = \alpha\left|\right.0\rangle + \beta\left|\right.1\rangle$.
I want to show that there is a direction $\vec{n}$ for which the spin is $+1$, so $\langle\vec{n}\cdot \vec{\sigma}\rangle = 1$.
I started by calculating the expectation value of the general matrix $$A=\begin{pmatrix} 
a &b \\ 
 c&d 
\end{pmatrix}$$
So $\langle\psi|A|\psi\rangle = 1$ got me to the result $$A = \begin{pmatrix} 
1 &0 \\ 
 0&1 
\end{pmatrix}$$
After which I get kinda stuck. Is my result so far correct and how can I continue to get $\vec{n}$?
Edit: I got to that result by using $\alpha^2 + \beta^2 = 1$.
 A: Hopefully your homework deadline has expired, so I'm not doing your homework for you.
Recall that "direction" is a unit vector, so, in spherical coordinates,
$\hat n= (\sin \theta \cos\phi, \sin\theta\sin\phi,\cos\theta)^T$. Consequently,
$$
\hat n\cdot \vec \sigma= \begin{pmatrix} \cos\theta & e^{-i\phi}~ \sin\theta \\  e^{i\phi}~ \sin\theta & -\cos\theta\end{pmatrix} ~,
$$
with the celebrated eigenvector of eigenvalue 1,
$$
| \psi_+\rangle =  \begin{pmatrix} 1+\cos\theta   \\  e^{i\phi}~ \sin\theta  \end{pmatrix} {1\over \sqrt{2(1+\cos\theta)}}~,
$$
up to an over-all irrelevant phase.
Your given vector then, (recalling $|\alpha|^2+|\beta|^2=1$), must be this eigenvector above, up to an over-all   phase Φ which cancels with its conjugate in the expectation value; and so is tuned to make the upper component real by cancelling the phase of α,
$$ e^{i\Phi}\begin{pmatrix}  \alpha \\  \beta\end{pmatrix}  = 
 \begin{pmatrix} 1+\cos\theta   \\  e^{i\phi}~ \sin\theta  \end{pmatrix} {1\over \sqrt{2(1+\cos\theta)}}~~.
$$
Thus,
$$
\cos\theta= 2 e^{2i\Phi} ~ \alpha^2-1, \qquad e^{i\phi}= \frac{  \beta }  {\sqrt{ e^{-2i\Phi} -\alpha^2}}~~.
$$
For example, if α were already real, then Φ=0, and
$$
\cos\theta= 2   \alpha^2-1, \qquad e^{i\phi}= \frac{  \beta}{\sqrt{1-\alpha^2}}
 ~~.
$$
Of course, for $\phi=0$, you need $\beta=\sqrt{1-\alpha^2}$, and $\cos\theta = \alpha^2-\beta^2$.
