Magnetic Resonance — Question 3

Question 3

A resonant pulse area fixes the spin-flip probability

A resonantly driven spin begins in its lower Zeeman state. A co-rotating circular magnetic field of magnitude $B_1$ gives $P_e(t)=\sin^2(\Omega t/2)$ with $\Omega=|\gamma|B_1$. Use $|\gamma|/(2\pi)=28.0\,\mathrm{GHz/T}$ and $B_1=10.0\,\mu\mathrm T$.

1. Find the first complete-inversion pulse duration.

2. Find the first equal-population pulse duration and its excitation probability.

Question 3 - Solution

A resonant pulse area fixes the spin-flip probability

(a) Method and calculation

$$\Omega=2\pi(28.0\times10^9)(10^{-5})=2\pi(2.80\times10^5)\,\mathrm{s^{-1}}.$$$$\Omega t_\pi=\pi\Rightarrow\boxed{t_\pi=1.786\,\mu\mathrm s}.$$

(b) Interpretation and further calculation

$$\Omega t_{\pi/2}=\pi/2\Rightarrow\boxed{t_{\pi/2}=.8929\,\mu\mathrm s}.$$$$\boxed{P_e(t_{\pi/2})=\sin^2(\pi/4)=1/2}.$$

Physical check

Here B1 is the amplitude of the co-rotating circular field. For a linearly oscillating transverse field, its resonant rotating component has half the peak amplitude in the rotating-wave approximation.