Consider the spin state of a particle of spin $s=\frac{1}{2}$. As far as the spin degrees of freedom are concerned, the operators $\textbf{S}^2$ and $S\_z$ form the complete set of commuting observables, i.e., $[\textbf{S}^2,S\_z]=0$. Suppose we measure both $\textbf{S}^2$ and $S\_z$. Measurement of $S\_z$ will either yield the values $m\_s=+\frac{1}{2}$ or $m\_s=-\frac{1}{2}$. Also, assume that 50% of the time we find the particle to be in the $+\frac{1}{2}$ state (denoted by $|\frac{1}{2},+\frac{1}{2}\rangle$) and the other 50% of the time in the $m\_s=-\frac{1}{2}$ state (denoted by $|\frac{1}{2},-\frac{1}{2}\rangle$). 

1. Based on this information, what does the above ensemble represent? Is it a pure or a mixed ensemble? My guess is that it is a mixed ensemble.
2. If pure (which I doubt), the normalized spin state $|\chi\rangle$ of the particle can be written as $$|\chi\rangle=\frac{e^{i\phi\_1}}{\sqrt{2}}\left|\frac{1}{2},+\frac{1}{2}\right\rangle+\frac{e^{i\phi\_2}}{\sqrt{2}}\left|\frac{1}{2},-\frac{1}{2}\right\rangle\tag{1}$$ where $\phi\_1,\phi\_2$ are arbitrary phases. Hence, the spin state of the system is not uniquely specified because the relative phase can not be fixed.
3. How does one prepare a state with $\phi\_1=\phi\_2=0$ so that $$|\chi\rangle=\frac{1}{\sqrt{2}}\left|\frac{1}{2},+\frac{1}{2}\right\rangle+\frac{1}{\sqrt{2}}\left|\frac{1}{2},-\frac{1}{2}\right\rangle\tag{2}?$$