Preparing a state $|\chi\rangle=\frac{1}{\sqrt{2}}|\frac{1}{2},+\frac{1}{2}\rangle+\frac{1}{\sqrt{2}}|\frac{1}{2},-\frac{1}{2}\rangle$ 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 $|\frac{1}{2},-\frac{1}{2}\rangle$). 


*

*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.

*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.

*How does one prepare a state with $\phi_1=\phi_2=0$ so that $$|\chi\rangle=\frac{1}{\sqrt{2}}\left|\frac{1}{2},+\frac{1}{2}\right\rangle+\frac{1}{\sqrt{2}}\left|\frac{1}{2},-\frac{1}{2}\right\rangle\tag{2}?$$
 A: Set up a Stern-Gerlach experiment, filter the beam about $\hat z$ with a magnetic gradient along this axis, measure the $\hat x$ component of spin:   your state is a combination of the $\pm$ eigenstates of $\sigma_x$. 
A: *

*If you measure $S_z$ with equal probabilities of finding $=1/2$ or $-1/2$ it means that the spin system was prepared in a direction perpendicular to $z$. Without loss of generality let us call it the $x$ direction. The state you wrote is just the general state of spin in the $x$ direction, which is a linear combination
$$|\chi\rangle=a|+1/2\rangle+b|-1/2\rangle,$$
constrained to a normalization condition,
$$|\langle\chi|\chi\rangle|^2=1,$$
and to the probabilities of finding $S_z=\pm 1/2$ conditions,
$$|\langle+1/2|\chi\rangle|^2=1/2, \quad |\langle-1/2|\chi\rangle|^2=1/2.$$
The solution for the coeficientes is $a=e^{i\phi_1}/\sqrt 2$ and $b=e^{i\phi_2}/\sqrt 2$.

*The state
$$|+x\rangle=\frac{1}{\sqrt 2}\left(|+1/2\rangle+|-1/2\rangle\right),$$
is such that when measuring $S_x$ it gives $+1/2$ always. Now let us say you start with an ensamble of spins prepared with $S_z=+1/2$. If you want now to prepare the system to the state $S_x=+1/2$, just rotate the apparatus (a Stern-Gerlach device for instance) into the $x$ direction and measure the spin. Half of the measurements will show $S_x=+1/2$ and those spins are all prepared into $|+x\rangle$. The other half gives $S_x=-1/2$ and are prepared into $|-x\rangle$.
A: Based on the information you've laid out,

[If we measure $S_z$, then] 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 $|\frac{1}{2},-\frac{1}{2}\rangle$)

there is not enough information to tell whether the system is in a pure or a mixed state. As two concrete examples, both


*

*the completely mixed state $\rho=\frac12\big(|\frac{1}{2},-\frac{1}{2}\rangle\langle\frac{1}{2},-\frac{1}{2}|+|\frac{1}{2},\frac{1}{2}\rangle\langle\frac{1}{2},\frac{1}{2}|\big)$, and

*the pure state $|\psi\rangle=\frac{1}{\sqrt{2}}\big(|\frac{1}{2},-\frac{1}{2}\rangle+|\frac{1}{2},\frac{1}{2}\rangle\big)$,


are compatible with those measurement statistics. The most general state that will produce those measurement outcomes has a density matrix of the form
$$
\rho = \begin{pmatrix}\frac12 & \rho_{12}^* \\ \rho_{12} & \frac12\end{pmatrix},
$$
in the $\{|\frac{1}{2},+\frac{1}{2}\rangle,|\frac{1}{2},-\frac{1}{2}\rangle\}$ basis, where $\rho_{12}$ is the coherence between the two basis states; this can be zero (giving the completely mixed state) or any complex number with a modulus up to $1/2$ (in which case the state is a pure state, with the phase of $\rho_{12}$ indicating the relative phase between the basis states in the superposition), or somewhere in between (indicating partial coherence between the two states).

If you explicitly know that you don't have any more information about the state, then you default to the completely mixed state. This is because, as you note, to be able to talk about a pure state with those statistics, it needs to be an even-weights superposition state of the form 
$$|\chi_\phi\rangle=\frac{1}{\sqrt{2}}\left|\frac{1}{2},+\frac{1}{2}\right\rangle+\frac{e^{i\phi}}{\sqrt{2}}\left|\frac{1}{2},-\frac{1}{2}\right\rangle,$$
and you need to specify a relative phase $\phi$ between the two (the global phase, on the other hand, is irrelevant), and if you don't have any more information then you can't know what to put in for $\phi$. One way to do this is to do an averaging where all the different phases are equally likely, and this gives you...
\begin{align}
\rho
& = \int_0^{2\pi} |\chi_\phi\rangle \langle \chi_\phi| \:\mathrm d\phi
\\& = \int_0^{2\pi}\frac12 \begin{pmatrix}1& e^{-i\phi} \\ e^{i\phi} & 1\end{pmatrix} \mathrm d\phi
\\& = \frac12 \begin{pmatrix}1& 0 \\ 0 & 1\end{pmatrix}
\end{align}
the completely mixed state.
On the other hand, if you do have more information about the state (like measurement statistics on $S_x$ and $S_y$) then you can say more about the phase, but in such a generalized situation, there's very little more you can say.

And finally, as to how you prepare the positive-phase superposition,
$$|+\rangle=\frac{1}{\sqrt{2}}\left|\frac{1}{2},+\frac{1}{2}\right\rangle+\frac{1}{\sqrt{2}}\left|\frac{1}{2},-\frac{1}{2}\right\rangle,$$
the most convenient way is to measure $S_x$ and wait for an outcome of $+1/2$, but really, there's an infinite number of ways to arrive at that outcome, so it's not a particularly answerable question.
A: *

*Yes, the spin state is not uniquely specified. You can actually omit the phase $e^{i\phi_1}$ and then $|{\chi}> = \frac{1}{\sqrt2}|\frac{1}{2},+\frac{1}{2}> + \frac{e^{i\phi}}{\sqrt2}|\frac{1}{2},-\frac{1}{2}>$, since it doesn't matter if the system is isolated.


*It's just like, when you know the direction of $\hat z$, can you point out the direction of $\hat x$ ?  No, you can only determine x-y plane and have to define $\hat x$ if you want. So you have to define the direction of $|\chi+> = \frac{1}{\sqrt2}|\frac{1}{2},+\frac{1}{2}> + \frac{1}{\sqrt2}|\frac{1}{2},-\frac{1}{2}>$ first.
Let's say the direction of $|\frac{1}{2},+\frac{1}{2}>$:  $\hat z$ , and the direction of your wanted $|\chi+>$:  $\hat x$ . You can then measure an ensemble of state in direction $\hat x$, and pick up those states that collapse into $|\chi+>$
