Sequential Spin Measurements.
(a) At time t = 0 a large ensemble of spin-1/2 particles is prepared, all of them in the spin-up state (with respect to the z axis). They are not subject to any forces or torques. At time t_1 > 0 each spin is measured – some along the z direction and others along the x direction (but we aren’t told the results). At time t_2 > t_1 their spin is measured again, this time along the x direction, and those with spin up (along x) are saved as a subensemble (those with spin down are discarded). Question: Of those remaining (the subensemble), what fraction had spin up (along z or x, depending on which was measured) in the first measurement?
(b) Part (a) was easy—trivial, really, once you see it. Here’s a more pithy generalization: At time t = 0 an ensemble of spin-1/2 particles is prepared, all in the spin-up state along direction a. At time t_1 > 0 their spins are measured along direction b (but we are not told the results), and at time t_2 > t_1 their spins are measured along direction c. Those with spin up (along c) are saved as a subensemble. Of the particles in this subensemble, what fraction had spin up (along b) in the first measurement? Hint: Use Equation 4.155 to show that the probability of getting spin up (along b) in the first measurement is P_{+}=\cos ^{2}\left(\theta_{a b} / 2\right) and (by extension) the probability of getting spin up in both measurements is P_{++}=\cos ^{2}\left(\theta_{a b} / 2\right) \cos ^{2}\left(\theta_{b c} / 2\right) Find the other
three probabilities \left(P_{+\rightarrow}, P_{-+}, \text {and } P_{--}\right) Beware: If the outcome of the first measurement was spin down, the relevant angle is now the supplement of \theta_{b c} \text {. Answer: }\left[1+\tan ^{2}\left(\theta_{a b} / 2\right) \tan ^{2}\left(\theta_{b c} / 2\right)\right]^{-1} .
\chi_{+}^{(r)}=\left(\begin{array}{c} \cos (\theta / 2) \\ e^{i \phi} \sin (\theta / 2)<br /> \end{array}\right) ; \quad \chi_{-}^{(r)}=\left(\begin{array}{c} e^{-i \phi} \sin (\theta / 2) \\ -\cos (\theta / 2) \end{array}\right) (4.155).