---
title: "Larmor precession"
firstLetter: "L"
publishDate: 2021-07-02
categories:
- Physics
- Quantum mechanics

date: 2021-07-02T15:48:41+02:00
draft: false
markup: pandoc
---

# Larmor precession

Consider a stationary spin-1/2 particle,
placed in a [magnetic field](/know/concept/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.