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}

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

© Marcus R.A. Newman, a.k.a. "Prefetch". Available under CC BY-SA 4.0.