以下で定義される\(\sigma_x, \sigma_y, \sigma_z\)を Pauli 行列という。
\[
\sigma_x =
\left(
\begin{array}{cc}
0 & 1 \\
1 & 0 \\
\end{array}
\right),\ \ \
\sigma_y =
\left(
\begin{array}{cc}
0 & – {\rm i} \\
{\rm i} & 0 \\
\end{array}
\right),\ \ \
\sigma_z =
\left(
\begin{array}{cc}
1 & 0 \\
0 & -1 \\
\end{array}
\right)
\]
Pauli 行列は以下の関係式を満たすことを示せ。
(1)
\[\begin{align}
\sigma_x^2 = \sigma_y^2 &= \sigma_z^2 = \sigma_0 \\
[\sigma_x, \sigma_y] &= 2 {\rm i} \sigma_z \\
[\sigma_y, \sigma_z] &= 2 {\rm i} \sigma_x \\
[\sigma_z, \sigma_x] &= 2 {\rm i} \sigma_y \\
\left\{\sigma_x, \sigma_y\right\} &= \left\{\sigma_y, \sigma_z\right\} = \left\{\sigma_z, \sigma_x\right\} = 0
\end{align}\]
ここに、\(\sigma_0\)は2行2列の単位行列であり、\([\cdot,\cdot], \left\{\cdot,\cdot\right\}\)は各々\([A, B] = AB – BA, \left\{A, B\right\} = AB + BA\)で定義される、交換関係と反交換関係である。
(2)
\[\begin{align}
\exp\left[- {\rm i}\frac{\varphi}{2}\sigma_x\right] &= \cos\frac{\varphi}{2} \sigma_0 – {\rm i} \sin\frac{\varphi}{2} \sigma_x \\
\exp\left[- {\rm i}\frac{\varphi}{2}\sigma_y\right] &= \cos\frac{\varphi}{2} \sigma_0 – {\rm i} \sin\frac{\varphi}{2} \sigma_y \\
\exp\left[- {\rm i}\frac{\varphi}{2}\sigma_z\right] &= \cos\frac{\varphi}{2} \sigma_0 – {\rm i} \sin\frac{\varphi}{2} \sigma_z
\end{align}\]
ここに行列の指数関数\(\exp[A]\)は以下で定義される。
\[
\exp[A] = \sum_{n = 0}^{\infty} \frac{1}{n!}A^n
\]
\[\begin{align}
\sigma_x^2 &=
\left(
\begin{array}{cc}
0 & 1 \\
1 & 0 \\
\end{array}
\right)
\left(
\begin{array}{cc}
0 & 1 \\
1 & 0 \\
\end{array}
\right)
=
\left(
\begin{array}{cc}
1 & 0 \\
0 & 1 \\
\end{array}
\right)
= \sigma_0 \\
\sigma_y^2 &=
\left(
\begin{array}{cc}
0 & – {\rm i} \\
{\rm i} & 0 \\
\end{array}
\right)
\left(
\begin{array}{cc}
0 & – {\rm i} \\
{\rm i} & 0 \\
\end{array}
\right)
=
\left(
\begin{array}{cc}
1 & 0 \\
0 & 1 \\
\end{array}
\right)
= \sigma_0 \\
\sigma_z^2 &=
\left(
\begin{array}{cc}
1 & 0 \\
0 & -1 \\
\end{array}
\right)
\left(
\begin{array}{cc}
1 & 0 \\
0 & -1 \\
\end{array}
\right)
=
\left(
\begin{array}{cc}
1 & 0 \\
0 & 1 \\
\end{array}
\right)
= \sigma_0
\end{align}\]
\[\begin{align}
\sigma_x \sigma_y &=
\left(
\begin{array}{cc}
0 & 1 \\
1 & 0 \\
\end{array}
\right)
\left(
\begin{array}{cc}
0 & – {\rm i} \\
{\rm i} & 0 \\
\end{array}
\right)
=
{\rm i} \left(
\begin{array}{cc}
1 & 0 \\
0 & -1 \\
\end{array}
\right)
= {\rm i} \sigma_z \\
\sigma_y \sigma_x &=
\left(
\begin{array}{cc}
0 & – {\rm i} \\
{\rm i} & 0 \\
\end{array}
\right)
\left(
\begin{array}{cc}
0 & 1 \\
1 & 0 \\
\end{array}
\right)
=
– {\rm i} \left(
\begin{array}{cc}
1 & 0 \\
0 & -1 \\
\end{array}
\right)
= – {\rm i} \sigma_z \\
\sigma_y \sigma_z &=
\left(
\begin{array}{cc}
0 & – {\rm i} \\
{\rm i} & 0 \\
\end{array}
\right)
\left(
\begin{array}{cc}
1 & 0 \\
0 & -1 \\
\end{array}
\right)
=
{\rm i} \left(
\begin{array}{cc}
0 & 1 \\
1 & 0 \\
\end{array}
\right)
= {\rm i} \sigma_x \\
\sigma_z \sigma_y &=
\left(
\begin{array}{cc}
1 & 0 \\
0 & -1 \\
\end{array}
\right)
\left(
\begin{array}{cc}
0 & – {\rm i} \\
{\rm i} & 0 \\
\end{array}
\right)
=
– {\rm i} \left(
\begin{array}{cc}
0 & 1 \\
1 & 0 \\
\end{array}
\right)
= – {\rm i} \sigma_x \\
\sigma_z \sigma_x &=
\left(
\begin{array}{cc}
1 & 0 \\
0 & -1 \\
\end{array}
\right)
\left(
\begin{array}{cc}
0 & 1 \\
1 & 0 \\
\end{array}
\right)
=
\left(
\begin{array}{cc}
0 & 1 \\
-1 & 0 \\
\end{array}
\right)
= {\rm i} \sigma_y \\
\sigma_x \sigma_z &=
\left(
\begin{array}{cc}
0 & 1 \\
1 & 0 \\
\end{array}
\right)
\left(
\begin{array}{cc}
1 & 0 \\
0 & -1 \\
\end{array}
\right)
=
\left(
\begin{array}{cc}
0 & -1 \\
1 & 0 \\
\end{array}
\right)
= -{\rm i} \sigma_y
\end{align}\]
以上の関係式から、示すべき式は明らかに成立することが分かる。
(2)
\[\begin{align}
\exp\left[- {\rm i}\frac{\varphi}{2} \sigma_x\right] &=
\sum_{n = 1}^{\infty} \frac{1}{n!}\left(- {\rm i} \frac{\varphi}{2} \sigma_x\right)^n \\
&=
\sum_{k = 0}^{\infty} \frac{1}{(2 k)!} \left(- {\rm i} \frac{\varphi}{2} \sigma_x\right)^{2 k}
+
\sum_{k = 0}^{\infty} \frac{1}{(2 k + 1)!} \left(- {\rm i} \frac{\varphi}{2} \sigma_x\right)^{2 k + 1} \\
&=
\sum_{k = 0}^{\infty} \frac{(-1)^k}{(2 k)!} \left(\frac{\varphi}{2}\right)^{2 k} \sigma_0\
– {\rm i} \sum_{k = 0}^{\infty} \frac{(-1)^k}{(2 k + 1)!}\left(\frac{\varphi}{2}\right)^{2k + 1} \sigma_x \\
&= \cos\frac{\varphi}{2} \sigma_0 – {\rm i} \sin\frac{\varphi}{2} \sigma_x
\end{align}\]
ここで(1)の結果である\(\sigma_x^2 = \sigma_0\)を使った。
\(\sigma_y, \sigma_z\)についても\(\sigma_y^2 = \sigma_0, \sigma_z^2 = \sigma_0\)が成立するので、全く同様の計算が成り立つ。
