Let $|\uparrow\rangle$ and $|\downarrow\rangle$ denote the eigenstates of $S_{z}$ for a particle of spin $\frac{1}{2}$. Show that

$$|\uparrow \theta\rangle=\cos \frac{\theta}{2}|\uparrow\rangle+\sin \frac{\theta}{2}|\downarrow\rangle, \quad|\downarrow \theta\rangle=-\sin \frac{\theta}{2}|\uparrow\rangle+\cos \frac{\theta}{2}|\downarrow\rangle$$

are eigenstates of $S_{z} \cos \theta+S_{x} \sin \theta$ for any $\theta$. Show also that the composite state

$$|\chi\rangle=\frac{1}{\sqrt{2}}(|\uparrow\rangle|\downarrow\rangle-|\downarrow\rangle|\uparrow\rangle),$$

for two spin- $\frac{1}{2}$ particles, is unchanged under a transformation

$$|\uparrow\rangle \mapsto|\uparrow \theta\rangle, \quad|\downarrow\rangle \mapsto|\downarrow \theta\rangle$$

applied to all one-particle states. Hence, by considering the action of certain components of the spin operator for the composite system, show that $|\chi\rangle$ is a state of total spin zero.

Two spin- $\frac{1}{2}$ particles A and B have combined spin zero (as in the state $|\chi\rangle$ above) but are widely separated in space. A magnetic field is applied to particle B in such a way that its spin states are transformed according to $(*)$, for a certain value of $\theta$, while the spin states of particle A are unaffected. Once this has been done, a measurement is made of $S_{z}$ for particle A, followed by a measurement of $S_{z}$ for particle B. List the possible results for this pair of measurements and find the total probability, in terms of $\theta$, for each pair of outcomes to occur. For which outcomes is the two-particle system left in an eigenstate of the combined total spin operator, $S^{2}$, and what is the eigenvalue for each such outcome?

$$\left[\sigma_{x}=\left(\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}\right), \quad \sigma_{y}=\left(\begin{array}{cc} 0 & -i \\ i & 0 \end{array}\right), \quad \sigma_{z}=\left(\begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array}\right) \cdot\right]$$