Spin magnetic resonance

Consider a spin one-half system (e.g., a bound electron) placed in a uniform $z$-directed magnetic field, and then subjected to a small time-dependent magnetic field rotating in the $x$-$y$ plane. Thus,

$${\bf B} = B_0 \hat{\bf z} + B_1(\cos\omega t \hat{\bf x} + \sin\omega t \hat{\bf y}),$$ (756)

where $B_0$ and $B_1$ are constants, with $B_1\ll B_0$. The rotating magnetic field usually represents the magnetic component of an electromagnetic wave propagating along the $z$-axis. In this system, the electric component of the wave has no effect. The Hamiltonian is written

$$H = - \mbox{\boldmath$\mu$}\! \cdot \! {\bf B} = H_0 + H_1,$$ (757)

where

$$H_0 = \frac{e B_0}{m_e} S_z,$$ (758)

and

$$H_1 = \frac{e B_1}{m_e} \left(\cos\omega t S_x + \sin\omega t S_y\right).$$ (759)

The eigenstates of the unperturbed Hamiltonian are the `spin up' and `spin down' states, denoted $\vert+\rangle$ and $\vert-\rangle$, respectively. Thus,

$$H_0 \vert\pm \rangle = \pm \frac{e \hbar B_0}{2 m_e} \vert\pm \rangle.$$ (760)

The time-dependent Hamiltonian can be written

$$H_1 = \frac{e B_1}{2 m_e} \left[\exp( {\rm i} \omega t) S^- + \exp(-{\rm i} \omega t) S^+\right],$$ (761)

where $S^+$ and $S^-$ are the conventional raising and lowering operators for the spin angular momentum. It follows that

$$\langle + \vert H_1 \vert+\rangle = \langle - \vert H_1\vert-\rangle = 0,$$ (762)

and

$$\langle - \vert H_1\vert + \rangle = \langle + \vert H_1\ver... ... = \frac{e \hbar B_1}{2 m_e} \exp( {\rm i} \omega t).$$ (763)

It can be seen that this system is exactly the same as the two-state system discussed in the previous section, provided that we make the identifications

$$\vert 1\rangle \rightarrow \vert-\rangle,$$ (764)

$$\vert 2\rangle \rightarrow \vert+\rangle,$$ (765)

$$\omega_{21} \rightarrow \frac{e B_0}{m_e},$$ (766)

$$\gamma \rightarrow \frac{e \hbar B_1}{2 m_e}.$$ (767)

The resonant frequency, $\omega_{21}$, is simply the spin precession frequency for an electron in a uniform magnetic field of strength $B_0$. In the absence of the perturbation, the expectation values of $S_x$ and $S_y$ oscillate because of the spin precession, but the expectation value of $S_z$ remains invariant. If we now apply a magnetic perturbation rotating at the resonant frequency then, according to the analysis of the previous section, the system undergoes a succession of spin-flops, $\vert+\rangle \rightleftharpoons \vert-\rangle$, in addition to the spin precession. We also know that if the oscillation frequency of the applied field is very different from the resonant frequency then there is virtually zero probability of the field triggering a spin-flop. The width of the resonance (in frequency) is determined by the strength of the oscillating magnetic perturbation. Experimentalist are able to measure the magnetic moments of electrons, and other spin one-half particles, to a high degree of accuracy by placing the particles in a magnetic field, and subjecting them to an oscillating magnetic field whose frequency is gradually scanned. By determining the resonant frequency (i.e., the frequency at which the particles absorb energy from the oscillating field), it is possible to calculate the magnetic moment.