Precession of a spin in a magnetic field

We have $\hat H=-\gamma\,\mathbf B\cdot\hat{\mathbf S}=-\gamma BS_x=\omega_0S_x=\frac{\hbar\omega_0}{2}\sigma_x$. Since the eigenvalues of $\sigma_x$ are $\pm1$, those of $\hat H$ are $\pm\hbar\omega_0/2$, with eigenvectors the states $\ket{x\pm}=(\ket+\pm\ket-)/\sqrt2$ of spin $\pm\hbar/2$ along $x$. The separation between the two levels is $\hbar\omega_0$. The initial state $\ket+$ is not an eigenstate: it is an equal-weight superposition of the two levels.

The time-evolution operator is $U(t)=e^{-i\hat Ht/\hbar}=e^{-i(\omega_0t/2)\sigma_x}$. From the formula given above with $\mathbf n=\mathbf e_x$ and $\alpha=\omega_0t/2$,

$$U(t)=\cos\frac{\omega_0t}{2}\,\mathbf{1}-i\sin\frac{\omega_0t}{2}\,\sigma_x .$$

Since $\sigma_x\ket+=\ket-$, we obtain

$$\ket{\psi(t)}=\cos\frac{\omega_0t}{2}\,\ket+-i\sin\frac{\omega_0t}{2}\,\ket- .$$

The probabilities are

$$P_+(t)=\cos^2\frac{\omega_0t}{2},\qquad P_-(t)=\sin^2\frac{\omega_0t}{2}.$$

The spin periodically changes from the state $\ket+$ to the state $\ket-$, at the angular frequency $\omega_0$, equal to the Bohr frequency of the system.

Let $c_+=\cos(\omega_0t/2)$ and $c_-=-i\sin(\omega_0t/2)$ denote the components of the state. For a state $c_+\ket++c_-\ket-$, we have

$$\langle\sigma_x\rangle=2\operatorname{Re}(c_+^*c_-),\qquad \langle\sigma_y\rangle=2\operatorname{Im}(c_+^*c_-),\qquad \langle\sigma_z\rangle=|c_+|^2-|c_-|^2 .$$

Here $c_+^*c_-=-i\cos\frac{\omega_0t}{2}\sin\frac{\omega_0t}{2}=-\frac i2\sin(\omega_0t)$, and therefore

$$\langle S_x\rangle=0,\qquad \langle S_y\rangle=-\frac\hbar2\sin(\omega_0t),\qquad \langle S_z\rangle=\frac\hbar2\cos(\omega_0t).$$

The vector $\langle\mathbf S\rangle$, whose norm remains constant at $\hbar/2$, rotates in the $(y,z)$ plane, perpendicular to the field, at the angular frequency $\omega_0$: on the Bloch sphere, the state traces the great circle perpendicular to the $x$ axis. This is the Larmor precession of the spin about the magnetic field.

With $\mathbf B=B\mathbf e_x$, we have $\langle\mathbf S\rangle\times\mathbf B=B\bigl(0,\ \langle S_z\rangle,\ -\langle S_y\rangle\bigr)$. The precession equation therefore reads, with $\gamma B=-\omega_0$,

$$\frac{d\langle S_x\rangle}{dt}=0,\qquad \frac{d\langle S_y\rangle}{dt}=-\omega_0\langle S_z\rangle,\qquad \frac{d\langle S_z\rangle}{dt}=\omega_0\langle S_y\rangle .$$

Let us verify this using our results: $\frac{d}{dt}\bigl(-\frac\hbar2\sin\omega_0t\bigr)=-\omega_0\frac\hbar2\cos\omega_0t=-\omega_0\langle S_z\rangle$, and $\frac{d}{dt}\bigl(\frac\hbar2\cos\omega_0t\bigr)=-\omega_0\frac\hbar2\sin\omega_0t=\omega_0\langle S_y\rangle$. All three equations are satisfied. We thus recover exactly the classical equation for the precession of a magnetic moment, as a consequence of Ehrenfest's theorem and the angular-momentum commutation relations.

The spin is in the state $\ket-$ with certainty when $P_-(t)=\sin^2(\omega_0t/2)=1$, that is, when $\omega_0t=\pi$: $t_\pi=\pi/\omega_0$. During this time, the Bloch vector has completed a half-turn, from the north pole to the south pole; this is called a $\pi$ pulse. Numerically, $\omega_0=eB/m_e\simeq1{,}76\times10^{11}\times10^{-3}\simeq1{,}76\times10^{8}$ rad s$^{-1}$, and

$$t_\pi=\frac{\pi}{\omega_0}\simeq1{,}8\times10^{-8}\ \mathrm{s}\simeq18\ \mathrm{ns}.$$