问题及解答

$O(3)$ 是由所有保持内积不变或等价的保持 $x_1^2+x_2^2+x_3^2$ 不变的线性变换构成的群. 也称三维旋转群. $O(3)$ 中的元素一一对应到正交矩阵. 因此 $O(3)$ 也可定义为

\[O(3)=\{A\in\mathcal{M}_{3\times 3}(\mathbb{R})\cong\mathbb{R}^{3\times 3}\mid A^T A=I\}.\]

其中 $\mathcal{M}_{3\times 3}(\mathbb{R})$ 是指所有 $3\times 3$ 矩阵的集合, $I$ 指单位矩阵.

易见 $O(3)$ 中的方阵, 其行列式要么是 $+1$, 要么是 $-1$. 前者这样的矩阵所对应的线性变换称为纯旋转(变换)(pure rotation), 其构成的集合记为

\[SO(3)=\{A\in O(3)\mid\det{A}=1\},\]

容易验证它是 $O(3)$ 的一个子群.

后者这样的矩阵所对应的线性变换描述了一个复合变换: 旋转+反射.

酉群 $U(n)$ 定义为:

\[U(n)=\{A\in\mathcal{M}_{3\times 3}(\mathbb{C})\mid A^*A=I_n\},\]

其中 $A^*$ 指矩阵 $A$ 的共轭转置(也叫 Hermitian conjugate), 即 $A^*=\bar{A}^T$, $I_n$ 是 $n$ 阶单位矩阵.

\[SU(n)=\{A\in U(n)\mid \det{A}=1\}.\]

任取矩阵 $g\in SO(3)$, $g=(g_{kl})_{3\times 3}$. 设 $\mathbf{x},\mathbf{x'}$ 是两个向量, $\mathbf{x}$ 在 $g$ 对应的正交变换下变到 $\mathbf{x'}$, 即 $\mathbf{x'}=g\mathbf{x}$. 若记 $\mathbf{x}=(x_1,x_2,x_3)^T$, $\mathbf{x'}=(x'_1,x'_2,x'_3)^T$, 则

\[x'_k=g_{kl}x_l,\]

这里采用了 Einstein 求和记号约定.

对每个向量 $\mathbf{x}=(x_1,x_2,x_3)^T$, 对应一个 $2\times 2$ 的 Hermitian 矩阵 $P$, 如下定义

\[P=x_k\sigma^k=x_1\sigma^1+x_2\sigma^2+x_3\sigma^3=
\begin{pmatrix}
x_3 & x_1+i x_2\\
x_1-i x_2 & -x_3
\end{pmatrix},
\]

其中

\[
\sigma_1=
\begin{pmatrix}
0 & 1\\
1 & 0
\end{pmatrix},\quad
\sigma_2=
\begin{pmatrix}
0 & i\\
-i & 0
\end{pmatrix},\quad
\sigma_3=
\begin{pmatrix}
1 & 0\\
0 & -1
\end{pmatrix},
\]

是 Pauli spin 矩阵.

于是对于 $\mathbf{x'}$,

\[P'=x'_l\sigma^l.\]

而由 $P'$ 与 $P$ 的定义, 存在 $u\in SU(2)$, 使得

\[P'=uPu^*,\]

这里 $u^*$ 是指 $u$ 的 Hermitian 共轭, 即 $u^*=\bar{u}^T$.

根据计算, 可以得出 $u\in SU(2)$ 与 $g\in SO(3)$ 之间的关系:

\[
g_{rs}=\frac{1}{2}\text{Tr}(\sigma^r u\sigma^s u^*),\tag{1}
\]

\[
u=\pm(1+\sigma^r\sigma^s g_{rs})/2(1+\text{Tr}g)^{1/2}.
\]

Claim 1. $SU(2)$ 中的矩阵形如

\[
u=\begin{pmatrix}z_1 & z_2\\ -\overline{z_2} & \overline{z_1}\end{pmatrix},\quad |z_1|^2+|z_2|^2=1.
\]

Pf. 假设 $u=\begin{pmatrix}z_1 & z_2\\ w_1 & w_2\end{pmatrix}\in SU(2)$, 则 $\det u=1$, 即 $z_1w_2-z_2w_1=1$, 并且

\[I_2=
\begin{pmatrix}
\overline{z_1} & \overline{w_1}\\
\overline{z_2} & \overline{w_2}
\end{pmatrix}
\begin{pmatrix}
z_1 & z_2\\
w_1 & w_2
\end{pmatrix}=
\begin{pmatrix}
|z_1|^2+|w_1|^2 & \overline{z_1}z_2+\overline{w_1}w_2\\
z_1\overline{z_2}+w_1\overline{w_2}& |z_2|^2+|w_2|^2
\end{pmatrix},
\]

因此,

\[|z_1|^2+|w_1|^2=1=|z_2|^2+|w_2|^2,\quad\overline{z_1}z_2+\overline{w_1}w_2=0,\quad z_1w_2-z_2w_1=1.\]

在 $\overline{z_1}z_2+\overline{w_1}w_2=0$ 两边乘以 $z_1$, 并且注意到 $z_1w_2=1+z_2w_1$, 有

\[
\begin{split}
& z_1\overline{z_1}z_2+z_1\overline{w_1}w_2=0\\
\Rightarrow & |z_1|^2 z_2+\overline{w_1}(1+z_2w_1)=0\\
\Rightarrow & |z_1|^2 z_2+\overline{w_1}+z_2|w_1|^2=0\\
\Rightarrow & z_2(|z_1|^2+|w_1|^2)+\overline{w_1}=0\\
\Rightarrow & w_1=-\overline{z_2}.
\end{split}
\]

类似的, 在 $z_1\overline{z_2}+w_1\overline{w_2}=0$ 两边乘以 $z_2$, 并且注意到 $z_2w_1=z_1w_2-1$, 有

\[
\begin{split}
& z_2z_1\overline{z_2}+z_2w_1\overline{w_2}=0\\
\Rightarrow & z_1|z_2|^2+(z_1w_2-1)\overline{w_2}=0\\
\Rightarrow & z_1|z_2|^2+z_1|w_2|^2-\overline{w_2}=0\\
\Rightarrow & z_1(|z_2|^2+|w_2|^2)-\overline{w_2}=0\\
\Rightarrow & w_2=\overline{z_1}.
\end{split}
\]

因此 $SU(2)$ 中的矩阵形如

\[
u=\begin{pmatrix}z_1 & z_2\\ -\overline{z_2} & \overline{z_1}\end{pmatrix},\quad |z_1|^2+|z_2|^2=1.
\]

Q.E.D. of Claim 1.

Reference:

Moshe Carmeli, Shimon Malin, Theory of Spinors, An Introduction.

下面我们计算 $\sigma^r u\sigma^s u^*$. 这里假设

\[
u=\begin{pmatrix}z_1 & z_2\\ -\overline{z_2} & \overline{z_1}\end{pmatrix},\quad |z_1|^2+|z_2|^2=1.
\]

其中 $z_1=a+ib$, $z_2=c+id$, $a^2+b^2+c^2+d^2=1$.

为了方便, 首先单独计算 $\sigma^r u$ 与 $\sigma^s u^*$.

\[
\begin{aligned}
\sigma^1 u&=
\begin{pmatrix}0 & 1\\ 1 & 0\end{pmatrix}
\begin{pmatrix}z_1 & z_2\\ -\overline{z_2} & \overline{z_1}\end{pmatrix}
=\begin{pmatrix}-\overline{z_2} & \overline{z_1}\\ z_1 & z_2\end{pmatrix},\\
\sigma^2 u&=
\begin{pmatrix}0 & i\\ -i & 0\end{pmatrix}
\begin{pmatrix}z_1 & z_2\\ -\overline{z_2} & \overline{z_1}\end{pmatrix}
=i\begin{pmatrix}-\overline{z_2} & \overline{z_1}\\ -z_1 & -z_2\end{pmatrix},\\
\sigma^3 u&=
\begin{pmatrix}1 & 0\\ 0 & -1\end{pmatrix}
\begin{pmatrix}z_1 & z_2\\ -\overline{z_2} & \overline{z_1}\end{pmatrix}
=\begin{pmatrix}z_1 & z_2\\ \overline{z_2} & -\overline{z_1}\end{pmatrix};
\end{aligned}
\]

\[
\begin{aligned}
\sigma^1 u^*&=
\begin{pmatrix}0 & 1\\ 1 & 0\end{pmatrix}
\begin{pmatrix}\overline{z_1} & -z_2\\ \overline{z_2} & z_1\end{pmatrix}
=\begin{pmatrix}\overline{z_2} & z_1\\ \overline{z_1} & -z_2\end{pmatrix},\\
\sigma^2 u^*&=
\begin{pmatrix}0 & i\\ -i & 0\end{pmatrix}
\begin{pmatrix}\overline{z_1} & -z_2\\ \overline{z_2} & z_1\end{pmatrix}
=i\begin{pmatrix}\overline{z_2} & z_1\\ -\overline{z_1} & z_2\end{pmatrix},\\
\sigma^3 u^*&=
\begin{pmatrix}1 & 0\\ 0 & -1\end{pmatrix}
\begin{pmatrix}\overline{z_1} & -z_2\\ \overline{z_2} & z_1\end{pmatrix}
=\begin{pmatrix}\overline{z_1} & -z_2\\ -\overline{z_2} & -z_1\end{pmatrix}.
\end{aligned}
\]

于是

\[
\begin{split}
\sigma^1 u\sigma^1 u^*&=\begin{pmatrix}-\overline{z_2} & \overline{z_1}\\ z_1 & z_2\end{pmatrix}
\begin{pmatrix}\overline{z_2} & z_1\\ \overline{z_1} & -z_2\end{pmatrix}\\
&=\begin{pmatrix}\overline{z_1}^2-\overline{z_2}^2 & -z_1\overline{z_2}-\overline{z_1}z_2\\ z_1\overline{z_2}+\overline{z_1}z_2 & z_1^2-z_2^2\end{pmatrix},
\end{split}
\]

\[
\begin{split}
\sigma^1 u\sigma^2 u^*&=\begin{pmatrix}-\overline{z_2} & \overline{z_1}\\ z_1 & z_2\end{pmatrix}
i\begin{pmatrix}\overline{z_2} & z_1\\ -\overline{z_1} & z_2\end{pmatrix}\\
&=i\begin{pmatrix}-\overline{z_1}^2-\overline{z_2}^2 & -z_1\overline{z_2}+\overline{z_1}z_2\\ z_1\overline{z_2}-\overline{z_1}z_2 & z_1^2+z_2^2\end{pmatrix},
\end{split}
\]

\[
\begin{split}
\sigma^1 u\sigma^3 u^*&=\begin{pmatrix}-\overline{z_2} & \overline{z_1}\\ z_1 & z_2\end{pmatrix}
\begin{pmatrix}\overline{z_1} & -z_2\\ -\overline{z_2} & -z_1\end{pmatrix}\\
&=\begin{pmatrix}-2\overline{z_1}\overline{z_2} & |z_2|^2-|z_1|^2\\ |z_1|^2-|z_2|^2  & -2z_1 z_2\end{pmatrix},
\end{split}
\]

\[
\begin{split}
\sigma^2 u\sigma^1 u^*&=i\begin{pmatrix}-\overline{z_2} & \overline{z_1}\\ -z_1 & -z_2\end{pmatrix}
\begin{pmatrix}\overline{z_2} & z_1\\ \overline{z_1} & -z_2\end{pmatrix}\\
&=i\begin{pmatrix}\overline{z_1}^2-\overline{z_2}^2 & -z_1\overline{z_2}-\overline{z_1}z_2\\ -z_1\overline{z_2}-\overline{z_1}z_2  & -z_1^2+z_2^2\end{pmatrix},
\end{split}
\]

\[
\begin{split}
\sigma^2 u\sigma^2 u^*&=i\begin{pmatrix}-\overline{z_2} & \overline{z_1}\\ -z_1 & -z_2\end{pmatrix}
i\begin{pmatrix}\overline{z_2} & z_1\\ -\overline{z_1} & z_2\end{pmatrix}\\
&=(-1)\begin{pmatrix}-\overline{z_1}^2-\overline{z_2}^2 & -z_1\overline{z_2}+\overline{z_1}z_2\\ -z_1\overline{z_2}+\overline{z_1}z_2  & -z_1^2-z_2^2\end{pmatrix}\\
&=\begin{pmatrix}\overline{z_1}^2+\overline{z_2}^2 & z_1\overline{z_2}-\overline{z_1}z_2\\ z_1\overline{z_2}-\overline{z_1}z_2  & z_1^2+z_2^2\end{pmatrix},
\end{split}
\]

\[
\begin{split}
\sigma^2 u\sigma^3 u^*&=i\begin{pmatrix}-\overline{z_2} & \overline{z_1}\\ -z_1 & -z_2\end{pmatrix}
\begin{pmatrix}\overline{z_1} & -z_2\\ -\overline{z_2} & -z_1\end{pmatrix}\\
&=i\begin{pmatrix}-2\overline{z_1}\overline{z_2} & |z_2|^2-|z_1|^2\\ -|z_1|^2+|z_2|^2 & 2 z_1 z_2\end{pmatrix},
\end{split}
\]

\[
\begin{split}
\sigma^3 u\sigma^1 u^*&=\begin{pmatrix}z_1 & z_2\\ \overline{z_2} & -\overline{z_1}\end{pmatrix}
\begin{pmatrix}\overline{z_2} & z_1\\ \overline{z_1} & -z_2\end{pmatrix}\\
&=\begin{pmatrix}z_1\overline{z_2}+\overline{z_1}z_2 & z_1^2-z_2^2\\ \overline{z_2}^2-\overline{z_1}^2 & z_1\overline{z_2}+\overline{z_1}z_2\end{pmatrix},
\end{split}
\]

\[
\begin{split}
\sigma^3 u\sigma^2 u^*&=\begin{pmatrix}z_1 & z_2\\ \overline{z_2} & -\overline{z_1}\end{pmatrix}
i\begin{pmatrix}\overline{z_2} & z_1\\ -\overline{z_1} & z_2\end{pmatrix}\\
&=i\begin{pmatrix}z_1\overline{z_2}-\overline{z_1}z_2 & z_1^2+z_2^2\\ \overline{z_2}^2+\overline{z_1}^2 & z_1\overline{z_2}-\overline{z_1}z_2\end{pmatrix},
\end{split}
\]

\[
\begin{split}
\sigma^3 u\sigma^3 u^*&=\begin{pmatrix}z_1 & z_2\\ \overline{z_2} & -\overline{z_1}\end{pmatrix}
\begin{pmatrix}\overline{z_1} & -z_2\\ -\overline{z_2} & -z_1\end{pmatrix}\\
&=\begin{pmatrix} |z_1|^2-|z_2|^2 & -2z_1 z_2\\ 2\overline{z_1 z_2}  & |z_1|^2-|z_2|^2\end{pmatrix},
\end{split}
\]

令

\[
\begin{split}
g_{11}&=\frac{1}{2}\text{Tr}(\sigma^1 u\sigma^1 u^*)\\
&=\frac{1}{2}(\overline{z_1}^2-\overline{z_2}^2+z_1^2-z_2^2)\\
&=\frac{1}{2}\bigl(z_1^2+\overline{z_1}^2-(z^2+\overline{z_2}^2)\bigr)\\
&=\frac{1}{2}(2(a^2-b^2)-2(c^2-d^2))\\
&=a^2-b^2-c^2+d^2;
\end{split}
\]

\[
\begin{split}
g_{12}&=\frac{1}{2}\text{Tr}(\sigma^1 u\sigma^2 u^*)\\
&=\frac{1}{2}i(-\overline{z_1}^2-\overline{z_2}^2+z_1^2+z_2^2)\\
&=\frac{1}{2}i(z_1^2-\overline{z_1}^2+z^2-\overline{z_2}^2)\\
&=\frac{1}{2}i(4iab+4icd)\\
&=-2(ab+cd);
\end{split}
\]

\[
\begin{split}
g_{13}&=\frac{1}{2}\text{Tr}(\sigma^1 u\sigma^3 u^*)\\
&=\frac{1}{2}(-2\overline{z_1 z_2}-2z_1 z_2)\\
&=-(z_1 z_2+\overline{z_1 z_2})\\
&=-2(ac-bd);
\end{split}
\]

\[
\begin{split}
g_{21}&=\frac{1}{2}\text{Tr}(\sigma^2 u\sigma^1 u^*)\\
&=\frac{1}{2}(i\overline{z_1}^2-i\overline{z_2}^2-i z_1^2+i z_2^2)\\
&=\frac{i}{2}\bigl(-(z_1^2-\overline{z_1}^2)+z_2^2-\overline{z_2}^2\bigr)\\
&=\frac{i}{2}(-4iab+4icd)\\
&=2(ab-cd);
\end{split}
\]

\[
\begin{split}
g_{22}&=\frac{1}{2}\text{Tr}(\sigma^2 u\sigma^2 u^*)\\
&=\frac{1}{2}(\overline{z_1}^2+\overline{z_2}^2+z_1^2+z_2^2)\\
&=\frac{1}{2}(z_1^2+\overline{z_1}^2+z_2^2+\overline{z_2}^2)\\
&=a^2-b^2+c^2-d^2;
\end{split}
\]

\[
\begin{split}
g_{23}&=\frac{1}{2}\text{Tr}(\sigma^2 u\sigma^3 u^*)\\
&=\frac{1}{2}i(-2\overline{z_1 z_2}+2z_1 z_2)\\
&=i(z_1 z_2-\overline{z_1 z_2})\\
&=-2(bc+ad);
\end{split}
\]

\[
\begin{split}
g_{31}&=\frac{1}{2}\text{Tr}(\sigma^3 u\sigma^1 u^*)\\
&=\frac{1}{2}2(z_1\overline{z_2}+\overline{z_1}z_2)\\
&=z_1\overline{z_2}+\overline{z_1}z_2\\
&=2(ac+bd);
\end{split}
\]

\[
\begin{split}
g_{32}&=\frac{1}{2}\text{Tr}(\sigma^3 u\sigma^2 u^*)\\
&=\frac{1}{2}i2(z_1\overline{z_2}-\overline{z_1}z_2)\\
&=i(z_1\overline{z_2}-\overline{z_1}z_2)\\
&=2(ad-bc);
\end{split}
\]

\[
\begin{split}
g_{33}&=\frac{1}{2}\text{Tr}(\sigma^3 u\sigma^3 u^*)\\
&=\frac{1}{2}2(|z_1|^2-|z_2|^2)\\
&=|z_1|^2-|z_2|^2\\
&=a^2+b^2-c^2-d^2.
\end{split}
\]

由上面的 $g_{rs}$ 组成的矩阵 $g$ 为

\[
g=\begin{pmatrix}
a^2-b^2-c^2+d^2 & -2(ab+cd) & -2(ac-bd)\\
2(ab-cd) & a^2-b^2+c^2-d^2 & -2(ad+bc)\\
2(ac+bd) & 2(ad-bc) & a^2+b^2-c^2-d^2
\end{pmatrix}.
\]

容易验证 $g\in SO(3)$. (验证这三个列向量互相正交且模长均为 1, 这并不是太繁. 行列式计算建议按第一行展开, 通过详细运算, 结果是 $(a^2+b^2+c^2+d^2)^2=1$. 参见问题804)

根据 Pauli spin 矩阵的性质(见问题805), 对于任意 $g=(g_{rs})\in SO(3)$, $\sigma^r\sigma^s g_{rs}$ 为

\[
\begin{split}
\sigma^r\sigma^s g_{rs}&=\sigma^1\sigma^1 g_{11}+\sigma^1\sigma^2 g_{12}+\sigma^1\sigma^3 g_{13}\\
&\quad+\sigma^2\sigma^1 g_{21}+\sigma^2\sigma^2 g_{22}+\sigma^2\sigma^3 g_{23}\\
&\quad+\sigma^3\sigma^1 g_{31}+\sigma^3\sigma^2 g_{32}+\sigma^3\sigma^3 g_{33}\\
&=Ig_{11}-i\sigma^3 g_{12}+i\sigma^2 g_{13}\\
&\quad+i\sigma^3 g_{21}+I g_{22}-i\sigma^1 g_{23}\\
&\quad-i\sigma^2 g_{31}+i\sigma^1 g_{32}+I g_{33}\\
&=\begin{pmatrix}g_{11}& 0\\ 0&g_{11}\end{pmatrix}-i\begin{pmatrix}g_{12}& 0\\ 0&-g_{12}\end{pmatrix}+\begin{pmatrix}0& -g_{13}\\ g_{13}& 0\end{pmatrix}\\
&\quad i\begin{pmatrix}g_{21}& 0\\ 0&-g_{21}\end{pmatrix}+\begin{pmatrix}g_{22}& 0\\ 0&g_{22}\end{pmatrix}-i\begin{pmatrix}0& g_{23}\\ g_{23}& 0\end{pmatrix}\\
&\quad -\begin{pmatrix}0 &-g_{31}\\ g_{31}& 0\end{pmatrix}+i\begin{pmatrix}0 &g_{32}\\ g_{32}& 0\end{pmatrix}+\begin{pmatrix}g_{33}& 0\\ 0&g_{33}\end{pmatrix}\\
&=\begin{pmatrix}g_{11}+g_{22}+g_{33}&g_{31}-g_{13}\\ g_{13}-g_{31}&g_{11}+g_{22}+g_{33}\end{pmatrix}+i\begin{pmatrix}g_{21}-g_{12}&g_{32}-g_{23}\\ g_{32}-g_{23}&g_{12}-g_{21}\end{pmatrix}
\end{split}
\]

考虑矩阵 $I+\sigma^r\sigma^s g_{rs}$, 即

\[
\begin{pmatrix}1+g_{11}+g_{22}+g_{33}+i(g_{21}-g_{12})&g_{31}-g_{13}+i(g_{32}-g_{23})\\ g_{13}-g_{31}+i(g_{32}-g_{23})&1+g_{11}+g_{22}+g_{33}+i(g_{12}-g_{21})\end{pmatrix}
\]

记之为

\[
A=\begin{pmatrix}a+ib & c+id\\ -c+id &a-ib\end{pmatrix}.
\]

\[
\begin{split}
A^*A&=\begin{pmatrix}a-ib & -c-id\\ c-id &a+ib\end{pmatrix}
\begin{pmatrix}a+ib & c+id\\ -c+id &a-ib\end{pmatrix}\\
&=\begin{pmatrix}a^2+b^2+c^2+d^2 & 0\\ 0 &a^2+b^2+c^2+d^2\end{pmatrix}\\
&=(a^2+b^2+c^2+d^2)I_2
\end{split}
\]