### Spin dynamics (Video)

A magnetic moment and its dynamics

A magnetic moment $\bm{\mu}$ in an external field $\mathbf{B}$ has an energy

$$E = - \mathbf{\mu}\cdot\mathbf{B}.$$

Since $|\bm{\mu}|=\mu$ is fixed, the only parameter in the equation is the angle $\psi$ between $\bm{\mu}$ and $\mathbf{B}$,

$$E = -\mathbf{\mu} B \cos\psi.$$

Changes of the energy due to the angle $\psi$ generate a torque

$$\tau = -\frac{d E}{d \psi} = \mathbf{\mu} B\, \sin\psi,$$

in vector form

$$\mathbf{\tau} = \bm{\mu}\times \mathbf{B}.$$

Equation of motion

For the angular momentum $\bm{L}$ we have $d\bm{L}/dt=\bm{\tau}$, while the magnetic moment is related to the angular momentum, $\bm{\mu}=\gamma\bm{L}$ where $\gamma$ is the gyromagnetic ratio.

We obtain for the equation of motion of the magnetic moment

$$\frac{d\bm{\mu}}{dt} = \gamma\,\bm{\mu}\times\bm{B} = \gamma_0 \bm{\mu}\times\bm{H}$$

where $\bm{B}=\mu_0\bm{H},\;\gamma_0 = \gamma\mu_0.$

Consider the constant magnetic field $\bm{H}=H\bm{\hat{e}}_z=(0,0,H)$. The equations for the components $\bm{\mu}=(\mu_x,\mu_y,\mu_z)$ are

$$\begin{cases} \dot{\mu}_x & = \gamma_0 H\,\mu_y \\ \dot{\mu}_y & = -\gamma_0 H\,\mu_x \\ \dot{\mu}_z & = 0 \end{cases} \Rightarrow \begin{cases} \dot{\mu}_x & = \omega_L\,\mu_y \\ \dot{\mu}_y & = -\omega_L\,\mu_x \\ \dot{\mu}_z & = 0 \end{cases}$$

where $\omega_L=\gamma_0 H$ is called the Larmor frequency.

The solution of the equations is

\begin{aligned} \mu_x & = \mu\sin\theta\,\cos(\omega_L t) \\ \mu_y & = -\mu\sin\theta\,\sin(\omega_L t) \\ \mu_z & = \mu \cos\theta\; \text{(=const.)} \end{aligned}

where $\mu=|\bm{\mu}|$ is constant and $\theta$ is the constant angle between $\bm{\mu}$ and $\bm{H}$.

• The component of $\bm{\mu}$ parallel to $\bm{H}$ remains constant.
• The projection of $\bm{\mu}$ on the plane perpendicular to $\bm{H}$ is $(\mu_x,\mu_y)$ and it rotates.
• The moment $\bm{\mu}$ performs precession around $\bm{H}$.