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
|
---
title: "Larmor precession"
sort_title: "Larmor precession"
date: 2021-07-02
categories:
- Physics
- Quantum mechanics
layout: "concept"
---
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.
|
