核磁共振的例子

原子的核心具有磁矩 $\vec{\mu}$ ，它与核心的固有角动量 (称为自旋 $\vec{J}$ ，其值用 $J$ 表示) 相关: $\vec{\mu}=g \vec{J}$ ，其中 $g$ 是一个常数。假设核心被放置在一个沿 $\mathrm{Oz}$ 方向的均匀磁场中。

哈密顿算子

$\widehat H = \frac{\widehat p}{2m}+\widehat V = 0-\vec{B} \cdot \overrightarrow{\widehat{\mu}} =-g B \widehat{J}_z$

哈密顿算子的本征态

设$\widehat{J}^2$的本征值为$\hbar^2 j(j+1)$，$\widehat J_z$的本征值为$m_j$。如此有：

$\widehat{H}\left|j, m_j\right\rangle=-g B \hbar m_j\left|j, m_j\right\rangle$

因此，哈密顿算子有$2j+1$个本征态，其本征能量为$E_{m_j}=-g B \hbar m_j$。

确定$\overrightarrow{\widehat{\mu}} \wedge \overrightarrow{\widehat{\mu}}$

$\overrightarrow{\widehat{\mu}} \wedge \overrightarrow{\widehat{\mu}} = g^2 \widehat J\wedge\widehat J$

$\begin{aligned}\overrightarrow{\widehat{J}} \wedge \overrightarrow{\widehat{J}} &= \overrightarrow{\widehat{J}}_x \overrightarrow{\widehat{J}}_y - \overrightarrow{\widehat{J}}_y \overrightarrow{\widehat{J}}_x + \overrightarrow{\widehat{J}}_z \overrightarrow{\widehat{J}}_x - \overrightarrow{\widehat{J}}_x \overrightarrow{\widehat{J}}_z + \overrightarrow{\widehat{J}}_y \overrightarrow{\widehat{J}}_z - \overrightarrow{\widehat{J}}_z \overrightarrow{\widehat{J}}_y\& = i\hbar\overrightarrow{\widehat{J}}_z+i\hbar\overrightarrow{\widehat{J}}_y+i\hbar\overrightarrow{\widehat{J}}_x = i\hbar\overrightarrow{\widehat{J}}\end{aligned}$

$\overrightarrow{\vec{\mu}} \wedge \overrightarrow{\hat{\mu}}=g \overrightarrow{\widehat{J}} \wedge g \overrightarrow{\widehat{J}}=i \hbar g^2 \overrightarrow{\widehat{J}}=i \hbar g \vec{\mu}$

计算磁矩的三个分量均值对时间的导数

根据埃伦费斯特原理：

$\frac{d}{d t}\langle A\rangle=\frac{1}{i \hbar}\langle[A, H]\rangle+\left\langle\frac{\partial A}{\partial t}\right\rangle$

由于角动量算子不含时，所以磁矩算子也不含时，三个维度均值的导数：

$\begin{aligned} & \frac{\mathrm{d}\left\langle\widehat{\mu_x}\right\rangle}{\mathrm{d} t}=\frac{1}{i \hbar}\left\langle\left[\widehat{\mu_x}, \widehat{H}\right]\right\rangle=\frac{1}{i \hbar}\left\langle\left[g \widehat{J}_x,-g B \widehat{J_z}\right]\right\rangle=\frac{g^2 B}{i \hbar}\left\langle\left[\widehat{J}_z, \widehat{J}_x\right]\right\rangle=g^2 B\left\langle\widehat{J}_y\right\rangle=g B\left\langle\widehat{\mu_y}\right\rangle \\ & \frac{\mathrm{d}\left\langle\widehat{\mu_y}\right\rangle}{\mathrm{d} t}=\frac{1}{i \hbar}\left\langle\left[\widehat{\mu_y}, \widehat{H}\right]\right\rangle=\frac{1}{i \hbar}\left\langle\left[g \widehat{J}_y,-g B \widehat{J}_z\right]\right\rangle=-\frac{g^2 B}{i \hbar}\left\langle\left[\widehat{J}_y, \widehat{J}_z\right]\right\rangle=-g^2 B\left\langle\widehat{J}_x\right\rangle=-g B\left\langle\widehat{\mu_x}\right\rangle \\ & \frac{\mathrm{d}\left\langle\widehat{\mu_z}\right\rangle}{\mathrm{d} t}=\frac{1}{i \hbar}\left\langle\left[\widehat{\mu_z}, \widehat{H}\right]\right\rangle=\frac{1}{i \hbar}\left\langle\left[g \widehat{J}_z,-g B \widehat{J}_z\right]\right\rangle=-\frac{g^2 B}{i \hbar}\left\langle\left[\widehat{J}_z, \widehat{J_z}\right]\right\rangle=0 \end{aligned}$

讨论Ω$\Omega$$\frac{\mathrm{d}\langle\overrightarrow{\hat{\mu}}\rangle}{\mathrm{d} t}=\vec{\Omega} \wedge\langle\overrightarrow{\widehat{\mu}}\rangle$

$\frac{\mathrm{d}\langle\overrightarrow{\widehat{\mu}}\rangle}{\mathrm{d} t}=\left(\begin{array}{c}\frac{\mathrm{d}\left\langle\widehat{\mu_x}\right\rangle}{\mathrm{d} t} \\\frac{\mathrm{d}\left\langle\mu_y\right\rangle}{\mathrm{d} t} \\\frac{\mathrm{d}\left\langle\mu_z\right\rangle}{\mathrm{d} t}\end{array}\right)=g B\left(\begin{array}{c}\left\langle\widehat{\mu_y}\right\rangle \\-\left\langle\widehat{\mu_x}\right\rangle \\0\end{array}\right)=\left(\begin{array}{c}0 \\0 \\-g B\end{array}\right) \wedge\left(\begin{array}{c}\left\langle\widehat{\mu_x}\right\rangle \\\left\langle\widehat{\mu_y}\right\rangle \\\left\langle\widehat{\mu_z}\right\rangle\end{array}\right)=\vec{\Omega} \wedge\langle\overrightarrow{\hat{\mu}}\rangle$

$E{m_j}-E{m_j+1}=-g B \hbar m_j+g B \hbar\left(m_j+1\right)=g B \hbar=|\hbar \vec{\Omega}|$

$Ω$ 的范数对应于两个连续能量子能级之间的差异

讨论

根据第4问的结果，我们有:

$\frac{\mathrm{d}^2\left\langle\widehat{\mu_x}\right\rangle}{\mathrm{d} t^2}=g B \frac{\mathrm{d}\left\langle\widehat{\mu_y}\right\rangle}{\mathrm{d} t}=-g^2 B^2\left\langle\widehat{\mu_x}\right\rangle \Leftrightarrow \frac{\mathrm{d}^2\left\langle\widehat{\mu_x}\right\rangle}{\mathrm{d} t^2}+g^2 B^2\left\langle\widehat{\mu_x}\right\rangle=0$

同样地:

$\frac{\mathrm{d}^2\left\langle\widehat{\mu_y}\right\rangle}{\mathrm{d} t^2}=-g B \frac{\mathrm{d}\left\langle\widehat{\mu_x}\right\rangle}{\mathrm{d} t}=-g^2 B^2\left\langle\widehat{\mu_y}\right\rangle \Leftrightarrow \frac{\mathrm{d}^2\left\langle\widehat{\mu_y}\right\rangle}{\mathrm{d} t^2}+g^2 B^2\left\langle\widehat{\mu_y}\right\rangle=0$

因此，可以写出:

$\left\{\begin{array}{l} \left\langle\widehat{\mu_x}\right\rangle=\left\langle\widehat{\mu_x}\right\rangle_0 \cos \left(g B t+\varphi_x\right) \\ \left\langle\widehat{\mu_y}\right\rangle=\left\langle\widetilde{\mu_y}\right\rangle_0 \sin \left(g B t+\varphi_y\right) \\ \left.\widehat{\mu_z}\right\rangle=\left\langle\widetilde{\mu_z}\right\rangle_0 \end{array}\right.$

因此，向量 $\langle\vec{\mu}\rangle$ 围绕 $\vec{B}$ 方向进行进动，这种进动被称为拉莫尔进动，其角频率为 $\Omega=g B$ 。通过测量 $\Omega$可以实验性地确定核的特征常数 $g$ 。因此，可以给每种类型的核关联一个特定的共振频率，对于给定的 $\vec{B}$值。例如，对于氢 $H$ ，共振频率调至 $42.57 \mathrm{MHz}$ ，对于磷 $P$ 是 $17.24 \mathrm{MHz}$ ，对于碳 $C$ 是 $10.70 \mathrm{MHz}$ 。因此，这种技术被用于检测极低浓度的化学元素。在成像（$MRI$）中，组织的对比度获得依赖于测量局部由非均匀 $\vec{B}$ 场激发的磁矩的松弛时间。