Question 4.36: An electron is at rest in an oscillating magnetic field B = ...

(a)

H =-\gamma B \cdot S =-\gamma B_{0} \cos \omega t S _{z}=-\frac{\gamma B_{0} \hbar}{2} \cos \omega t\left(\begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array}\right) .

(b)

\chi(t)=\left(\begin{array}{l} \alpha(t) \\ \beta(t) \end{array}\right), \text { with } \alpha(0)=\beta(0)=\frac{1}{\sqrt{2}} .

i \hbar \frac{\partial \chi}{\partial t}=i \hbar\left(\begin{array}{c} \dot{\alpha} \\ \dot{\beta} \end{array}\right)= H \chi=-\frac{\gamma B_{0} \hbar}{2} \cos \omega t\left(\begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array}\right)\left(\begin{array}{l} \alpha \\ \beta \end{array}\right)=-\frac{\gamma B_{0} \hbar}{2} \cos \omega t\left(\begin{array}{c} \alpha \\ -\beta<br /> \end{array}\right) .

\dot{\alpha}=i\left(\frac{\gamma B_{0}}{2}\right) \cos \omega t \alpha \Rightarrow \frac{d \alpha}{\alpha}=i\left(\frac{\gamma B_{0}}{2}\right) \cos \omega t d t \Rightarrow \ln \alpha=\frac{i \gamma B_{0}}{2} \frac{\sin \omega t}{\omega}+\text { constant } .

\alpha(t)=A e^{i\left(\gamma B_{0} / 2 \omega\right) \sin \omega t} ; \quad \alpha(0)=A=\frac{1}{\sqrt{2}}, \quad \text { so } \alpha(t)=\frac{1}{\sqrt{2}} e^{i\left(\gamma B_{0} / 2 \omega\right) \sin \omega t} .

(c)

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

=i \sin \left[\frac{\gamma B_{0}}{2 \omega} \sin \omega t\right] . \quad P_{-}^{(x)}(t)=\left|c_{-}^{(x)}\right|^{2}=\sin ^{2}\left[\frac{\gamma B_{0}}{2 \omega} \sin \omega t\right] .

(d) The argument of \sin^2 must reach \pi / 2(\text { so } P=1) \Rightarrow \frac{\gamma B_{0}}{2 \omega}=\frac{\pi}{2}, \text { or } B_{0}=\frac{\pi \omega}{\gamma} .