Question 4.35: In Example 4.3: (a) If you measured the component of spin an...

(a) Using Eqs. 4.151 and 4.163:

\chi_{+}^{(x)}=\left(\begin{array}{c} 1 / \sqrt{2} \\ 1 / \sqrt{2} \end{array}\right),\left(\text { eigenvalue }+\frac{\hbar}{2}\right) ; \chi_{-}^{(x)}=\left(\begin{array}{c} 1 / \sqrt{2} \\ -1 / \sqrt{2} \end{array}\right),\left(\text { eigenvalue }-\frac{\hbar}{2}\right)          (4.151).

\chi(t)=\left(\begin{array}{c} \cos (\alpha / 2) e^{i \gamma B_{0} t / 2} \\ \sin (\alpha / 2) e^{-i \gamma B_{0} t / 2} \end{array}\right)         (4.163).

c_{+}^{(x)}=\chi_{+}^{(x) \dagger} \chi=\frac{1}{\sqrt{2}}\left(\begin{array}{ll} 1 & 1 \end{array}\right)\left(\begin{array}{c} \cos \frac{\alpha}{2} e^{i \gamma B_{0} t / 2} \\ \sin \frac{\alpha}{2} e^{-i \gamma B_{0} t / 2} \end{array}\right)=\frac{1}{\sqrt{2}}\left[\cos \frac{\alpha}{2} e^{i \gamma B_{0} t / 2}+\sin \frac{\alpha}{2} e^{-i \gamma B_{0} t / 2}\right] .

P_{+}^{(x)}(t)=\left|c_{+}^{(x)}\right|^{2}=\frac{1}{2}\left[\cos \frac{\alpha}{2} e^{-i \gamma B_{0} t / 2}+\sin \frac{\alpha}{2} e^{i \gamma B_{0} t / 2}\right]\left[\cos \frac{\alpha}{2} e^{i \gamma B_{0} t / 2}+\sin \frac{\alpha}{2} e^{-i \gamma B_{0} t / 2}\right] .

=\frac{1}{2}\left[\cos ^{2} \frac{\alpha}{2}+\sin ^{2} \frac{\alpha}{2}+\sin \frac{\alpha}{2} \cos \frac{\alpha}{2}\left(e^{i \gamma B_{0} t}+e^{-i \gamma B_{0} t}\right)\right] .

=\frac{1}{2}\left[1+2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2} \cos \left(\gamma B_{0} t\right)\right]=\frac{1}{2}\left[1+\sin \alpha \cos \left(\gamma B_{0} t\right)\right] .

(b) From Problem 4.32(a):  \chi_{+}^{(y)}=\frac{1}{\sqrt{2}}\left(\begin{array}{l} 1 \\ i \end{array}\right) .

c_{+}^{(y)}=\chi_{+}^{(y) \dagger} \chi=\frac{1} {\sqrt{2}}\left(\begin{array}{ll} 1 & -i \end{array}\right)\left(\begin{array}{c} \cos \frac{\alpha}{2} e^{i \gamma B_{0} t / 2} \\ \sin \frac{\alpha}{2} e^{-i \gamma B_{0} t / 2} \end{array}\right)=\frac{1}{\sqrt{2}}\left[\cos \frac{\alpha}{2} e^{i \gamma B_{0} t / 2}-i \sin \frac{\alpha}{2} e^{-i \gamma B_{0} t / 2}\right] .

P_{+}^{(y)}(t)=\left|c_{+}^{(y)}\right|^{2}=\frac{1}{2}\left[\cos \frac{\alpha}{2} e^{-i \gamma B_{0} t / 2}+i \sin \frac{\alpha}{2} e^{i \gamma B_{0} t / 2}\right]\left[\cos \frac{\alpha}{2} e^{i \gamma B_{0} t / 2}-i \sin \frac{\alpha}{2} e^{-i \gamma B_{0} t / 2}\right] .

=\frac{1}{2}\left[\cos ^{2} \frac{\alpha}{2}+\sin ^{2} \frac{\alpha}{2}+i \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}\left(e^{i \gamma B_{0} t}-e^{-i \gamma B_{0} t}\right)\right] .

=\frac{1}{2}\left[1-2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2} \sin \left(\gamma B_{0} t\right)\right]=\frac{1}{2}\left[1-\sin \alpha \sin \left(\gamma B_{0} t\right)\right] .

(c)

\chi_{+}^{(z)}=\left(\begin{array}{l} 1 \\ 0 \end{array}\right) ; \quad c_{+}^{(z)}=\left(\begin{array}{ll} 1 & 0 \end{array}\right)\left(\begin{array}{c} \cos \frac{\alpha}{2} e^{i \gamma B_{0} t / 2} \\ \sin \frac{\alpha}{2} e^{-i \gamma B_{0} t / 2} \end{array}\right)=\cos \frac{\alpha}{2} e^{i \gamma B_{0} t / 2} ; \quad P_{+}^{(z)}(t)=\left|c_{+}^{(z)}\right|^{2}=\cos ^{2} \frac{\alpha}{2} .