(1)

パウリ行列は明らかにエルミート行列である。

(2)

パウリ行列はエルミート行列であり、
\begin{align*} I & =\sigma_{k}^{2}\\ & =\sigma_{k}\sigma_{k}\\ & =\sigma_{k}^{*}\sigma_{k} \end{align*} となるので、パウリ行列はユニタリ行列となる。

(3)

\begin{align*} \sigma_{1}^{2} & =\left(\begin{array}{cc} 0 & 1\\ 1 & 0 \end{array}\right)^{2}\\ & =I \end{align*} \begin{align*} \sigma_{2}^{2} & =\left(\begin{array}{cc} 0 & -i\\ i & 0 \end{array}\right)^{2}\\ & =I \end{align*} \begin{align*} \sigma_{3}^{2} & =\left(\begin{array}{cc} 1 & 0\\ 0 & -1 \end{array}\right)^{2}\\ & =I \end{align*} \begin{align*} \sigma_{1}\sigma_{2} & =\left(\begin{array}{cc} 0 & 1\\ 1 & 0 \end{array}\right)\left(\begin{array}{cc} 0 & -i\\ i & 0 \end{array}\right)\\ & =i\left(\begin{array}{cc} 1 & 0\\ 0 & -1 \end{array}\right)\\ & =i\sigma_{3} \end{align*} \begin{align*} \sigma_{2}\sigma_{3} & =\left(\begin{array}{cc} 0 & -i\\ i & 0 \end{array}\right)\left(\begin{array}{cc} 1 & 0\\ 0 & -1 \end{array}\right)\\ & =i\left(\begin{array}{cc} 0 & 1\\ 1 & 0 \end{array}\right)\\ & =i\sigma_{1} \end{align*} \begin{align*} \sigma_{3}\sigma_{1} & =\left(\begin{array}{cc} 1 & 0\\ 0 & -1 \end{array}\right)\left(\begin{array}{cc} 0 & 1\\ 1 & 0 \end{array}\right)\\ & =i\left(\begin{array}{cc} 0 & -i\\ i & 0 \end{array}\right)\\ & =i\sigma_{2} \end{align*} \begin{align*} \sigma_{2}\sigma_{1} & =\left(\begin{array}{cc} 0 & -i\\ i & 0 \end{array}\right)\left(\begin{array}{cc} 0 & 1\\ 1 & 0 \end{array}\right)\\ & =-i\left(\begin{array}{cc} 1 & 0\\ 0 & -1 \end{array}\right)\\ & =-i\sigma_{3} \end{align*} \begin{align*} \sigma_{3}\sigma_{2} & =\left(\begin{array}{cc} 1 & 0\\ 0 & -1 \end{array}\right)\left(\begin{array}{cc} 0 & -i\\ i & 0 \end{array}\right)\\ & =-i\left(\begin{array}{cc} 0 & 1\\ 1 & 0 \end{array}\right)\\ & =-i\sigma_{3} \end{align*} \begin{align*} \sigma_{1}\sigma_{3} & =\left(\begin{array}{cc} 0 & 1\\ 1 & 0 \end{array}\right)\left(\begin{array}{cc} 1 & 0\\ 0 & -1 \end{array}\right)\\ & =-i\left(\begin{array}{cc} 0 & -i\\ i & 0 \end{array}\right)\\ & =-i\sigma_{2} \end{align*} これより、
\begin{align*} \sigma_{j}\sigma_{k} & =\begin{cases} I\delta_{jk} & j=k\\ i\sum_{l=1}^{3}\epsilon_{jkl}\sigma_{l} & j\ne k \end{cases}\\ & =I\delta_{jk}+i\sum_{l=1}^{3}\epsilon_{jkl}\sigma_{l} \end{align*} また、
\begin{align*} \sigma_{1}\sigma_{2}\sigma_{3} & =i\sigma_{3}\sigma_{3}\\ & =iI \end{align*} となるので与式は成り立つ。

(4)

(4-1)

\begin{align*} \left[\sigma_{j},\sigma_{k}\right] & =\sigma_{j}\sigma_{k}-\sigma_{k}\sigma_{j}\\ & =I\delta_{jk}+i\sum_{l=1}^{3}\epsilon_{jkl}\sigma_{l}-\left(I\delta_{kj}+i\sum_{l=1}^{3}\epsilon_{kjl}\sigma_{l}\right)\\ & =i\sum_{l=1}^{3}\epsilon_{jkl}\sigma_{l}-i\sum_{l=1}^{3}\epsilon_{kjl}\sigma_{l}\\ & =i\sum_{l=1}^{3}\epsilon_{jkl}\sigma_{l}+i\sum_{l=1}^{3}\epsilon_{jkl}\sigma_{l}\\ & =2i\sum_{l=1}^{3}\epsilon_{jkl}\sigma_{l} \end{align*}

(4-2)

\begin{align*} \left\{ \sigma_{j},\sigma_{k}\right\} & =\sigma_{j}\sigma_{k}+\sigma_{k}\sigma_{j}\\ & =I\delta_{jk}+i\sum_{l=1}^{3}\epsilon_{jkl}\sigma_{l}+\left(I\delta_{kj}+i\sum_{l=1}^{3}\epsilon_{kjl}\sigma_{l}\right)\\ & =I\delta_{jk}+i\sum_{l=1}^{3}\epsilon_{jkl}\sigma_{l}+\left(I\delta_{kj}-i\sum_{l=1}^{3}\epsilon_{jkl}\sigma_{l}\right)\\ & =2I\delta_{jk} \end{align*} となるので与式は成り立つ。

(5)

\[ \sigma_{j}\sigma_{k}=I\delta_{jk}+i\sum_{l=1}^{3}\epsilon_{jkl}\sigma_{l} \] より、両辺に\(\epsilon_{jkm}\)を掛けて、\(j,k\)について和をとると、
\begin{align*} \sum_{j,k}\epsilon_{jkm}\sigma_{j}\sigma_{k} & =\sum_{j,k}\epsilon_{jkm}\left(I\delta_{jk}+i\sum_{l=1}^{3}\epsilon_{jkl}\sigma_{l}\right)\\ & =I\sum_{j,k}\epsilon_{jkm}\delta_{jk}+i\sum_{l=1}^{3}\sum_{j,k}\epsilon_{jkm}\epsilon_{jkl}\sigma_{l}\\ & =I\sum_{k}\epsilon_{kkm}\delta_{kk}+i\sum_{l=1}^{3}2!\delta_{ml}\sigma_{l}\\ & =0+i2!\sigma_{m}\\ & =2i\sigma_{m} \end{align*} となるので、両辺\(2i\)で割ると、
\begin{align*} \sigma_{m} & =\frac{1}{2i}\sum_{j,k}\epsilon_{jkm}\sigma_{j}\sigma_{k}\\ & =-\frac{i}{2}\sum_{j,k}\epsilon_{jkm}\sigma_{j}\sigma_{k} \end{align*} となる。
従って題意は成り立つ。

(6)

\begin{align*} \tr\sigma_{1} & =\tr\left(\begin{array}{cc} 0 & 1\\ 1 & 0 \end{array}\right)\\ & =0 \end{align*} \begin{align*} \tr\sigma_{2} & =\tr\left(\begin{array}{cc} 0 & -i\\ i & 0 \end{array}\right)\\ & =0 \end{align*} \begin{align*} \tr\sigma_{3} & =\tr\left(\begin{array}{cc} 1 & 0\\ 0 & -1 \end{array}\right)\\ & =0 \end{align*} \begin{align*} \det\sigma_{1} & =\det\left(\begin{array}{cc} 0 & 1\\ 1 & 0 \end{array}\right)\\ & =-1 \end{align*} \begin{align*} \det\sigma_{2} & =\det\left(\begin{array}{cc} 0 & -i\\ i & 0 \end{array}\right)\\ & =-1 \end{align*} \begin{align*} \det\sigma_{3} & =\det\left(\begin{array}{cc} 1 & 0\\ 0 & -1 \end{array}\right)\\ & =-1 \end{align*} 単位行列については、
\begin{align*} \tr\sigma_{0} & =\tr\left(\begin{array}{cc} 1 & 0\\ 0 & 1 \end{array}\right)\\ & =2 \end{align*} \begin{align*} \det\sigma_{0} & =\det\left(\begin{array}{cc} 1 & 0\\ 0 & 1 \end{array}\right)\\ & =1 \end{align*} となるので与式は成り立つ。

(7)

\begin{align*} \sigma_{j}\sigma_{k}\sigma_{j} & =\left(\delta_{jk}I_{2}+i\sum_{l=1}^{3}\epsilon_{jkl}\sigma_{l}\right)\sigma_{j}\\ & =\delta_{jk}\sigma_{j}+i\sum_{l=1}^{3}\epsilon_{jkl}\sigma_{l}\sigma_{j}\\ & =\delta_{jk}\sigma_{k}+i\sum_{l=1}^{3}\epsilon_{jkl}\left(\delta_{lj}I_{2}+i\sum_{m=1}^{3}\epsilon_{ljm}\sigma_{m}\right)\\ & =\delta_{jk}\sigma_{k}+i\sum_{l=1}^{3}\epsilon_{jkl}\delta_{lj}I_{2}-\sum_{l=1}^{3}\sum_{m=1}^{3}\epsilon_{jkl}\epsilon_{ljm}\sigma_{m}\\ & =\delta_{jk}\sigma_{k}+i\epsilon_{jkj}I_{2}-\sum_{m=1}^{3}\sum_{l=1}^{3}\epsilon_{jkl}\epsilon_{jml}\sigma_{m}\\ & =\delta_{jk}\sigma_{k}-\sum_{m=1}^{3}\left(\delta_{jj}\delta_{km}-\delta_{jm}\delta_{kj}\right)\sigma_{m}\\ & =\delta_{jk}\sigma_{k}-\sigma_{k}+\delta_{kj}\sigma_{j}\\ & =\left(2\delta_{jk}-1\right)\sigma_{k} \end{align*}

(8)

\begin{align*} 0 & =\det\left(\lambda I-\sigma_{1}\right)\\ & =\det\left(\left(\begin{array}{cc} \lambda & 0\\ 0 & \lambda \end{array}\right)-\left(\begin{array}{cc} 0 & 1\\ 1 & 0 \end{array}\right)\right)\\ & =\det\left(\begin{array}{cc} \lambda & -1\\ -1 & \lambda \end{array}\right)\\ & =\left(\lambda-1\right)\left(\lambda+1\right) \end{align*} となるので固有値は\(\pm1\)となる。
このとき、固有ベクトルは、
\begin{align*} \ker\left(\pm I-\sigma_{1}\right) & =\ker\left(\begin{array}{cc} \pm1 & -1\\ -1 & \pm1 \end{array}\right)\\ & =\ker\left(\begin{array}{cc} 1 & \mp1\\ -1 & \pm1 \end{array}\right)\\ & =\ker\left(\begin{array}{cc} 1 & \mp1\\ 0 & 0 \end{array}\right)\\ & =c_{1}\left(\begin{array}{c} 1\\ \pm1 \end{array}\right) \end{align*} となるので規格化すると、
\begin{align*} \boldsymbol{a}_{k,\pm} & =\frac{1}{\left|c_{1}\right|\sqrt{1^{2}+\left(\pm1\right)^{2}}}c_{1}\left(\begin{array}{c} 1\\ \pm1 \end{array}\right)\\ & =\frac{e^{i\Arg\left(c_{1}\right)}}{\sqrt{2}}\left(\begin{array}{c} 1\\ \pm1 \end{array}\right)\\ & =\frac{1}{\sqrt{2}}\left(\begin{array}{c} 1\\ \pm1 \end{array}\right) \end{align*} となる。
最後で\(\Arg\left(c_{1}\right)=0\)を選んだ。
\(\sigma_{2},\sigma_{3}\)についても同様である。

(9)

(9-1)

\begin{align*} \exp\left(ia\sigma_{k}\right) & =\sum_{j=0}^{\infty}\frac{1}{j!}\left(ia\sigma_{k}\right)^{j}\\ & =\sum_{j=0}^{\infty}\frac{1}{\left(2j\right)!}\left(ia\sigma_{k}\right)^{2j}+\sum_{j=0}^{\infty}\frac{1}{\left(2j+1\right)!}\left(ia\sigma_{k}\right)^{2j+1}\\ & =I\sum_{j=0}^{\infty}\frac{1}{\left(2j\right)!}\left(-a^{2}\right)^{j}+\sigma_{k}\sum_{j=0}^{\infty}\frac{1}{\left(2j+1\right)!}\left(ia\right)^{2j+1}\\ & =I\sum_{j=0}^{\infty}\left(-1\right)^{j}\frac{a^{2j}}{\left(2j\right)!}+i\sigma_{k}\sum_{j=0}^{\infty}\left(-1\right)^{j}\frac{a^{2j+1}}{\left(2j+1\right)!}\\ & =I\cos a+i\sigma_{k}\sin a \end{align*}

(9-2)

\begin{align*} \exp\left(i\boldsymbol{a}\cdot\boldsymbol{\sigma}\right) & =\sum_{j=0}^{\infty}\frac{1}{j!}\left(i\boldsymbol{a}\cdot\boldsymbol{\sigma}\right)^{j}\\ & =\sum_{j=0}^{\infty}\frac{1}{\left(2j\right)!}\left(i\boldsymbol{a}\cdot\boldsymbol{\sigma}\right)^{2j}+\sum_{j=0}^{\infty}\frac{1}{\left(2j+1\right)!}\left(i\boldsymbol{a}\cdot\boldsymbol{\sigma}\right)^{2j+1}\\ & =\sum_{j=0}^{\infty}\frac{\left(-1\right)^{j}}{\left(2j\right)!}\left(\boldsymbol{a}\cdot\boldsymbol{\sigma}\right)^{2j}+i\sum_{j=0}^{\infty}\frac{\left(-1\right)^{j}}{\left(2j+1\right)!}\left(\boldsymbol{a}\cdot\boldsymbol{\sigma}\right)^{2j+1}\\ & =\sum_{j=0}^{\infty}\frac{\left(-1\right)^{j}}{\left(2j\right)!}\left(\sum_{l,m}a_{l}\sigma_{l}a_{m}\sigma_{m}\right)^{j}+i\sum_{j=0}^{\infty}\frac{\left(-1\right)^{j}}{\left(2j+1\right)!}\boldsymbol{a}\cdot\boldsymbol{\sigma}\left(\sum_{l,m}a_{l}\sigma_{l}a_{m}\sigma_{m}\right)^{j}\\ & =\sum_{j=0}^{\infty}\frac{\left(-1\right)^{j}}{\left(2j\right)!}\left(\sum_{l,m}a_{l}a_{m}\left(I\delta_{lm}+i\sum_{n=1}^{3}\epsilon_{lmn}\sigma_{n}\right)\right)^{j}+i\sum_{j=0}^{\infty}\frac{\left(-1\right)^{j}}{\left(2j+1\right)!}\boldsymbol{a}\cdot\boldsymbol{\sigma}\left(\sum_{l,m}a_{l}a_{m}\left(I\delta_{lm}+i\sum_{n=1}^{3}\epsilon_{lmn}\sigma_{n}\right)\right)^{j}\\ & =\sum_{j=0}^{\infty}\frac{\left(-1\right)^{j}}{\left(2j\right)!}\left(\sum_{m}a_{m}^{2}I+i\sum_{l,m,n}\left(\frac{1}{2}\left(\epsilon_{lmn}-\epsilon_{mln}\right)\sigma_{n}\right)\right)^{j}+i\sum_{j=0}^{\infty}\frac{\left(-1\right)^{j}}{\left(2j+1\right)!}\boldsymbol{a}\cdot\boldsymbol{\sigma}\left(\sum_{m}a_{m}^{2}I+i\sum_{l,m,n}\left(\frac{1}{2}\left(\epsilon_{lmn}-\epsilon_{mln}\right)\sigma_{n}\right)\right)^{j}\\ & =I\sum_{j=0}^{\infty}\frac{\left(-1\right)^{j}}{\left(2j\right)!}\left(\sum_{m}a_{m}^{2}\right)^{j}+i\sum_{j=0}^{\infty}\frac{\left(-1\right)^{j}}{\left(2j+1\right)!}\frac{\boldsymbol{a}}{\left|\boldsymbol{a}\right|}\cdot\boldsymbol{\sigma}\left|\boldsymbol{a}\right|\left(\sum_{m}a_{m}^{2}\right)^{j}\\ & =I\sum_{j=0}^{\infty}\frac{\left(-1\right)^{j}}{\left(2j\right)!}\left(\sqrt{\sum_{m}a_{m}^{2}}\right)^{2j}+i\frac{\boldsymbol{a}}{\left|\boldsymbol{a}\right|}\cdot\boldsymbol{\sigma}\sum_{j=0}^{\infty}\frac{\left(-1\right)^{j}}{\left(2j+1\right)!}\left|\boldsymbol{a}\right|\left(\sqrt{\sum_{m}a_{m}^{2}}\right)^{2j}\\ & =I\sum_{j=0}^{\infty}\frac{\left(-1\right)^{j}}{\left(2j\right)!}\left(\left|\boldsymbol{a}\right|\right)^{2j}+i\frac{\boldsymbol{a}}{\left|\boldsymbol{a}\right|}\cdot\boldsymbol{\sigma}\sum_{j=0}^{\infty}\frac{\left(-1\right)^{j}}{\left(2j+1\right)!}\left(\left|\boldsymbol{a}\right|\right)^{2j+1}\\ & =I\cos\left|\boldsymbol{a}\right|+i\frac{\boldsymbol{a}}{\left|\boldsymbol{a}\right|}\sin\left|\boldsymbol{a}\right| \end{align*}

(10)

(10-1)

\begin{align*} \left(\boldsymbol{\sigma}\cdot\boldsymbol{a}\right)\left(\boldsymbol{\sigma}\cdot\boldsymbol{b}\right) & =\sum_{j,k}\sigma_{j}a_{j}\sigma_{k}b_{k}\\ & =\sum_{j,k}a_{j}b_{k}\sigma_{j}\sigma_{k}\\ & =\sum_{j,k}a_{j}b_{k}\left(\frac{1}{2}\left\{ \sigma_{j},\sigma_{k}\right\} +\frac{1}{2}\left[\sigma_{j},\sigma_{k}\right]\right)\\ & =\sum_{j,k}a_{j}b_{k}\left(\frac{1}{2}\left\{ \sigma_{j},\sigma_{k}\right\} +\frac{1}{2}\left[\sigma_{j},\sigma_{k}\right]\right)\\ & =\sum_{j,k}a_{j}b_{k}\left(\frac{1}{2}\cdot2I\delta_{jk}+\frac{1}{2}\cdot2i\sum_{l=1}^{3}\epsilon_{jkl}\sigma_{l}\right)\cmt{\because\left\{ \sigma_{j},\sigma_{k}\right\} =2I\delta_{jk},\left[\sigma_{j},\sigma_{k}\right]=2i\sum_{l=1}^{3}\epsilon_{jkl}\sigma_{l}}\\ & =I\sum_{j,k}a_{j}b_{k}\delta_{jk}+i\sum_{j,k,l}a_{j}b_{k}\epsilon_{jkl}\sigma_{l}\\ & =I\sum_{j}a_{j}b_{j}+i\sum_{j,k,l}\epsilon_{ljk}\sigma_{l}a_{j}b_{k}\\ & =I\boldsymbol{a}\cdot\boldsymbol{b}+i\boldsymbol{\sigma}\cdot\left(\boldsymbol{a}\times\boldsymbol{b}\right) \end{align*}

(10-2)

\begin{align*} \left(\boldsymbol{\sigma}\times\boldsymbol{\sigma}\right) & =\sum_{j,k}\epsilon_{jkl}\sigma_{j}\sigma_{k}\boldsymbol{e}_{l}\\ & =\sum_{j,k}\epsilon_{jkl}\left(I\delta_{jk}+i\sum_{l=1}^{3}\epsilon_{jkl}\sigma_{l}\right)\boldsymbol{e}_{l}\cmt{\because\sigma_{j}\sigma_{k}=I\delta_{jk}+i\sum_{l=1}^{3}\epsilon_{jkl}\sigma_{l}}\\ & =i\sum_{j,k,l}\epsilon_{jkl}\epsilon_{jkl}\sigma_{l}\boldsymbol{e}_{l}\\ & =i\sum_{,l}2\delta_{l.l}\sigma_{l}\boldsymbol{e}_{l}\\ & =2i\boldsymbol{\sigma} \end{align*}

(11)

アルファベットは\(1,2,3\)、ギリシャ文字は\(0,1,2,3\)とする。
\begin{align*} \left\langle \sigma_{j},\sigma_{k}\right\rangle & =\tr\left(\sigma_{j}^{*}\sigma_{k}\right)\\ & =\tr\left(\sigma_{j}\sigma_{k}\right)\\ & =\tr\left(I\delta_{jk}+i\sum_{l=1}^{3}\epsilon_{jkl}\sigma_{l}\right)\\ & =\delta_{jk}\tr\left(I\right)+i\sum_{l=1}^{3}\epsilon_{jkl}\tr\left(\sigma_{l}\right)\\ & =2\delta_{jk} \end{align*} \begin{align*} \left\langle \sigma_{0},\sigma_{k}\right\rangle & =\left\langle I,\sigma_{k}\right\rangle \\ & =\tr\left(I^{*}\sigma_{k}\right)\\ & =\tr\left(\sigma_{k}\right)\\ & =0 \end{align*} \begin{align*} \left\langle \sigma_{k},\sigma_{0}\right\rangle & =\overline{\left\langle \sigma_{0},\sigma_{k}\right\rangle }\\ & =0 \end{align*} \begin{align*} \left\langle \sigma_{0},\sigma_{0}\right\rangle & =\left\langle I,I\right\rangle \\ & =\tr\left(I^{*}I\right)\\ & =\tr\left(I\right)\\ & =2 \end{align*} これより、
\[ \left\langle \sigma_{\mu},\sigma_{\nu}\right\rangle =2\delta_{\mu\nu} \] となるので与式は成り立つ。

(12)

1次独立性

\[ O=\sum_{\mu=0}^{3}a_{\mu}\sigma_{\mu} \] に右から\(\sigma_{\nu}\)を掛けてトレースをとると、
\begin{align*} \tr\left(O\sigma_{\nu}\right) & =\tr\left(\sum_{\mu=0}^{3}a_{\mu}\sigma_{\mu}\sigma_{\nu}\right)\\ & =\sum_{\mu=0}^{3}a_{\mu}\tr\left(\sigma_{\mu}\sigma_{\nu}\right)\\ & =\sum_{\mu=0}^{3}a_{\mu}\tr\left(\sigma_{\mu}^{*}\sigma_{\nu}\right)\\ & =\sum_{\mu=0}^{3}a_{\mu}\left\langle \sigma_{\mu},\sigma_{\nu}\right\rangle \\ & =\sum_{\mu=0}^{3}a_{\mu}2\delta_{\mu\nu}\\ & =a_{\nu} \end{align*} となるので、\(a_{\mu}=0\)となる。
従って、\(\sigma_{\mu},\mu\in\left\{ 0,1,2,3\right\} \)は独立である。

全域性

任意の
\[ \left(\begin{array}{cc} a & b\\ c & d \end{array}\right)\in M_{2}\left(\mathbb{C}\right) \] に対し、
\begin{align*} \left(\begin{array}{cc} a & b\\ c & d \end{array}\right) & =\sum_{\mu=0}^{3}a_{\mu}\sigma_{\mu}\\ & =a_{0}\left(\begin{array}{cc} 1 & 0\\ 0 & 1 \end{array}\right)+a_{1}\left(\begin{array}{cc} 0 & 1\\ 1 & 0 \end{array}\right)+a_{2}\left(\begin{array}{cc} 0 & -i\\ i & 0 \end{array}\right)+a_{3}\left(\begin{array}{cc} 1 & 0\\ 0 & -1 \end{array}\right)\\ & =\left(\begin{array}{cc} a_{0}+a_{3} & a_{1}-a_{2}i\\ a_{1}+a_{2}i & a_{0}-a_{3} \end{array}\right) \end{align*} となるので、
\[ \left(\begin{array}{c} a\\ b\\ c\\ d \end{array}\right)=\left(\begin{array}{cccc} 1 & 0 & 0 & 1\\ 0 & 1 & -i & 0\\ 0 & 1 & i & 0\\ 1 & 0 & 0 & -1 \end{array}\right)\left(\begin{array}{c} a_{0}\\ a_{1}\\ a_{2}\\ a_{3} \end{array}\right) \] である。
これは
\begin{align*} \det\left(\begin{array}{cccc} 1 & 0 & 0 & 1\\ 0 & 1 & -i & 0\\ 0 & 1 & i & 0\\ 1 & 0 & 0 & -1 \end{array}\right) & =\det\left(\begin{array}{cccc} 1 & 0 & 0 & 1\\ 0 & 1 & -i & 0\\ 0 & 0 & 2i & 0\\ 0 & 0 & 0 & -2 \end{array}\right)\\ & =-2\cdot2\det\left(\begin{array}{cccc} 1 & 0 & 0 & 1\\ 0 & 1 & -i & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{array}\right)\\ & =-4\det\left(\begin{array}{cccc} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{array}\right)\\ & =-4\\ & \ne0 \end{align*} なので、
\[ \left(\begin{array}{c} a_{0}\\ a_{1}\\ a_{2}\\ a_{3} \end{array}\right)=\left(\begin{array}{cccc} 1 & 0 & 0 & 1\\ 0 & 1 & -i & 0\\ 0 & 1 & i & 0\\ 1 & 0 & 0 & -1 \end{array}\right)^{-1}\left(\begin{array}{c} a\\ b\\ c\\ d \end{array}\right) \] とできるので\(a_{0},a_{1},a_{2},a_{3}\)が求まり全域性が成り立つ。

直交性

\(\sigma_{\mu};\mu\in\left\{ 0,1,2,3\right\} \)はフロベニウス内積に対し、
\[ \left\langle \sigma_{\mu},\sigma_{\nu}\right\rangle =2\delta_{\mu\nu} \] となるので直交する。

-

これらより、\(\sigma_{\mu},\mu\in\left\{ 0,1,2,3\right\} \)は1次独立性・全域性が成り立ちフロベニウス内積に対し直交しているので、直交基底となる。

(13)

\(\sigma_{\mu},\mu\in\left\{ 0,1,2,3\right\} \)は複素2次正方行列空間\(M_{2}\left(\mathbb{C}\right)\)の基底であるので、任意の\(A\in M_{2}\left(\mathbb{C}\right)\)は、
\[ A=\sum_{\mu=0}^{3}a_{\mu}\sigma_{\mu} \] の形で表される。
この両辺に右から\(\sigma_{\nu}\)を掛けてトレースを作用させると、
\begin{align*} \tr\left(A\sigma_{\nu}\right) & =\tr\left(\left(\sum_{\mu=0}^{3}a_{\mu}\sigma_{\mu}\right)\sigma_{\nu}\right)\\ & =\sum_{\mu=0}^{3}a_{\mu}\tr\left(\sigma_{\mu}\sigma_{\nu}\right)\\ & =\sum_{\mu=0}^{3}a_{\mu}\tr\left(\sigma_{\mu}^{*}\sigma_{\nu}\right)\\ & =\sum_{\mu=0}^{3}a_{\mu}\left\langle \sigma_{\mu},\sigma_{\nu}\right\rangle \\ & =\sum_{\mu=0}^{3}a_{\mu}2\delta_{\mu\nu}\\ & =2a_{\nu} \end{align*} となるので、
\[ a_{\mu}=\frac{1}{2}\tr\left(A\sigma_{\mu}\right) \] となる。
故に題意は成り立つ。