 Solution

The $+x$ eigenstate is $|+x⟩=(|\uparrow ⟩+|\downarrow ⟩)/\sqrt{2}$, since ${\sigma }_x(1,1{)}^T=(1,1{)}^T$. Use a resonant quarter-turn pulse, ${\tau }_{\pi /2}=\pi /(2\Omega )$, and choose its phase so thatThe laboratory-frame relative amplitude then is $i{e}^{-i\pi /2}=1$, givingIn rotating axes the equivalent target is obtained with $\phi =\pi /2$; the extra laboratory phase above compensates Larmor precession during the pulse. After switching the RF field off, the spin continues to precess in the static field, so this statement specifies the preparation instant rather than a state fixed along laboratory $x$ forever.