1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
|
---
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.
|
