Spin in a Magnetic Field

A spinning charge carries a magnetic moment, so a spin-½ particle in an external magnetic field has its energy split according to its spin orientation. This is the Zeeman effect, and its dynamics — Larmor precession — are the foundation of nuclear magnetic resonance, MRI, and qubit control.

The magnetic moment and the Hamiltonian

The magnetic moment of a spin-½ is proportional to its spin,

\boldsymbol{\mu} = \gamma\,\mathbf{S},

where $\gamma$ is the gyromagnetic ratio. The energy of a moment in a field $\mathbf{B}$ is $-\boldsymbol{\mu}\cdot\mathbf{B}$ , so the Hamiltonian is

H = -\boldsymbol{\mu}\cdot\mathbf{B} = -\gamma\,\mathbf{S}\cdot\mathbf{B}.

Choosing the field along $z$ , $\mathbf{B} = B\hat{z}$ ,

H = -\gamma B\,S_z = -\frac{\gamma B \hbar}{2}\,\sigma_z.

Energy eigenstates and the splitting

Since $H$ is proportional to $\sigma_z$ , its eigenstates are $|0\rangle$ and $|1\rangle$ with energies

E_\uparrow = -\frac{\gamma B \hbar}{2}, \qquad E_\downarrow = +\frac{\gamma B \hbar}{2}.

(For $\gamma > 0$ the spin-up state is the lower-energy ground state — the moment prefers to align with the field.) The gap between them is

\Delta E = E_\downarrow - E_\uparrow = \gamma B \hbar = \hbar\omega_L, \qquad \omega_L \equiv \gamma B,

defining the Larmor angular frequency $\omega_L$ .

Larmor precession

A spin not aligned with the field is a superposition of the two energy eigenstates. Under time evolution $e^{-iHt/\hbar}$ the relative phase between them winds at rate $\omega_L$ , which makes the Bloch vector precess about the field axis at the Larmor frequency:

\langle S_x\rangle(t) = \langle S_x\rangle(0)\cos\omega_L t + \langle S_y\rangle(0)\sin\omega_L t,

with $\langle S_z\rangle$ constant. The transverse spin rotates exactly like a classical gyroscope in a field — one of the cleanest correspondences between quantum and classical mechanics.

Try it

For $\gamma = 5$ and $B = 0.4$ (with $\hbar = 1$ ), compute the Larmor angular frequency $\omega_L = \gamma B$ . It should equal $2$ , which is also the energy splitting $\Delta E = \hbar\omega_L$ .

Run your code to see the quantum state.