Larmor Precession

Place a magnetic dipole in a uniform magnetic field and it experiences a torque that makes its angular-momentum vector sweep out a cone — this is Larmor precession. For an electron spin in a field $B$ pointing along $\hat{z}$ , the Hamiltonian is

$H = -\boldsymbol{\mu} \cdot \mathbf{B} = -\mu_z B,$

where the magnetic moment operator is $\boldsymbol{\mu} = -g_s \mu_B \mathbf{S}/\hbar$ with $g_s \approx 2$ and $\mu_B = e\hbar/(2m_e)$ the Bohr magneton.

Deriving the Larmor frequency

Schrodinger's equation for the spin state $|\chi(t)\rangle$ gives the time evolution

$i\hbar \frac{d}{dt}|\chi\rangle = H|\chi\rangle.$

Because $H$ is diagonal in the $\hat{z}$ eigenbasis — $H|+\rangle = +\mu_B B|+\rangle$ (with $g_s = 2$ ) and $H|-\rangle = -\mu_B B|-\rangle$ — the two energy eigenstates just pick up phases:

$|\chi(t)\rangle = a\,e^{-i E_+ t/\hbar}|+\rangle + b\,e^{-i E_- t/\hbar}|-\rangle,$

where $E_{\pm} = \pm \mu_B B$ . The relative phase between the two components advances at the rate

$\omega_0 = \frac{E_+ - E_-}{\hbar} = \frac{2\mu_B B}{\hbar}.$

Substituting $\mu_B = e\hbar/(2m_e)$ :

$\omega_0 = \frac{2 \cdot \frac{e\hbar}{2m_e} \cdot B}{\hbar} = \frac{eB}{m_e}.$

In the literature you will also see the classical Larmor frequency

$\omega_L = \frac{eB}{2m_e},$

which is exactly half of $\omega_0$ . The factor of two arises because the electron's $g$ -factor $g_s \approx 2$ doubles the coupling compared to a classical spinning charge. In SI units with $e = 1.602 \times 10^{-19}$ C and $m_e = 9.109 \times 10^{-31}$ kg, a field of $B = 1\,\text{T}$ gives

\approx 8.794 \times 10^{10}\,\text{rad/s}.$$ That corresponds to a precession frequency $f_L = \omega_L/(2\pi) \approx 14\,\text{GHz}$ — deep in the microwave range, which is why electron spin resonance (ESR) experiments use microwave sources. ## Expectation values precess, wavefunctions rotate It is instructive to track the expectation value $\langle S_x\rangle(t)$. Starting from a state pointing along $+\hat{x}$ — i.e., $a = b = 1/\sqrt{2}$ — one finds $$\langle S_x\rangle(t) = \frac{\hbar}{2}\cos(\omega_0 t), \quad \langle S_y\rangle(t) = \frac{\hbar}{2}\sin(\omega_0 t), \quad \langle S_z\rangle(t) = 0.$$ The spin vector rotates uniformly in the $xy$-plane at angular frequency $\omega_0 = eB/m_e$. This is quantum Larmor precession: the same steady conical sweep you would expect classically, but the magnitude of the angular momentum is fixed at $\hbar/2$ and any actual measurement still yields only $\pm\hbar/2$. <Callout type="tip"> The precession axis is always parallel to **B**. Tipping the initial spin away from $\hat{z}$ changes the opening angle of the cone but not the precession frequency — a property that makes NMR and ESR spectroscopy possible. </Callout> ## Try it This is a numerical exercise — your code should `return` a number. Compute the classical Larmor angular frequency $\omega_L = eB/(2m_e)$ in rad/s for $B = 1\,\text{T}$. The constants are already defined for you. <RunnableExample />