Categories: Physics, Quantum mechanics.

# Larmor precession

Consider a stationary spin-1/2 particle, placed in a magnetic field with magnitude $B$ pointing in the $z$-direction. In that case, its Hamiltonian $\hat{H}$ is given by:

\begin{aligned} \hat{H} = - \gamma B \hat{S}_z = - \frac{\hbar}{2} \gamma B \hat{\sigma_z} \end{aligned}

Where $\gamma = - q / m$ is the gyromagnetic ratio, and $\hat{\sigma}_z$ is the Pauli spin matrix for the $z$-direction. Since $\hat{H}$ is proportional to $\hat{\sigma}_z$, they share eigenstates $\Ket{\downarrow}$ and $\Ket{\uparrow}$. The respective eigenenergies $E_{\downarrow}$ and $E_{\uparrow}$ are as follows:

\begin{aligned} E_{\downarrow} = \frac{\hbar}{2} \gamma B \qquad E_{\uparrow} = - \frac{\hbar}{2} \gamma B \end{aligned}

Because $\hat{H}$ is time-independent, the general time-dependent solution $\Ket{\chi(t)}$ is of the following form, where $a$ and $b$ are constants, and the exponentials are “twiddle factors”:

\begin{aligned} \Ket{\chi(t)} = a \exp(- i E_{\downarrow} t / \hbar) \: \Ket{\downarrow} \:+\: b \exp(- i E_{\uparrow} t / \hbar) \: \Ket{\uparrow} \end{aligned}

For our purposes, we can safely assume that $a$ and $b$ are real, and then say that there exists an angle $\theta$ satisfying $a = \sin(\theta / 2)$ and $b = \cos(\theta / 2)$, such that:

\begin{aligned} \Ket{\chi(t)} = \sin(\theta / 2) \exp(- i E_{\downarrow} t / \hbar) \: \Ket{\downarrow} \:+\: \cos(\theta / 2) \exp(- i E_{\uparrow} t / \hbar) \: \Ket{\uparrow} \end{aligned}

Now, we find the expectation values of the spin operators $\expval{\hat{S}_x}$, $\expval{\hat{S}_y}$, and $\expval{\hat{S}_z}$. The first is:

\begin{aligned} \matrixel{\chi}{\hat{S}_x}{\chi} &= \frac{\hbar}{2} \begin{bmatrix} a \exp(i E_{\downarrow} t / \hbar) \\ b \exp(i E_{\uparrow} t / \hbar) \end{bmatrix}^{\mathrm{T}} \cdot \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \cdot \begin{bmatrix} a \exp(- i E_{\downarrow} t / \hbar) \\ b \exp(- i E_{\uparrow} t / \hbar) \end{bmatrix} \\ &= \frac{\hbar}{2} \begin{bmatrix} a \exp(i E_{\downarrow} t / \hbar) \\ b \exp(i E_{\uparrow} t / \hbar) \end{bmatrix}^{\mathrm{T}} \cdot \begin{bmatrix} b \exp(- i E_{\uparrow} t / \hbar) \\ a \exp(- i E_{\downarrow} t / \hbar) \end{bmatrix} \\ &= \frac{\hbar}{2} \Big( a b \exp(i (E_{\downarrow} \!-\! E_{\uparrow}) t / \hbar) + b a \exp(i (E_{\uparrow} \!-\! E_{\downarrow}) t / \hbar) \Big) \\ &= \frac{\hbar}{2} \cos(\theta/2) \sin(\theta/2) \Big( \exp(i \gamma B t) + \exp(- i \gamma B t) \Big) \\ &= \frac{\hbar}{2} \cos(\gamma B t) \Big( \cos(\theta/2) \sin(\theta/2) + \cos(\theta/2) \sin(\theta/2) \Big) \\ &= \frac{\hbar}{2} \sin(\theta) \cos(\gamma B t) \end{aligned}

The other two are calculated in the same way, with the following results:

\begin{aligned} \matrixel{\chi}{\hat{S}_y}{\chi} = - \frac{\hbar}{2} \sin(\theta) \sin(\gamma B t) \qquad \matrixel{\chi}{\hat{S}_z}{\chi} = \frac{\hbar}{2} \cos(\theta) \end{aligned}

The result is that the spin axis is off by $\theta$ from the $z$-direction, and is rotating (or precessing) around the $z$-axis at the Larmor frequency $\omega$:

\begin{aligned} \boxed{ \omega = \gamma B } \end{aligned}

## References

1. D.J. Griffiths, D.F. Schroeter, Introduction to quantum mechanics, 3rd edition, Cambridge.