3 ディラック方程式(SKI)

\def\slashchar#1{\not\!#1}

[演習3.1]

$[p_1,x_1] = -i, [p_1,x_2] = 0, [p_1,x_3] = 0$
$[p_2,x_1] = 0, [p_2,x_2] = -i, [p_2,x_3] = 0$
$[p_3,x_1] = 0, [p_3,x_2] = 0, [p_3,x_3] = -i$
であるから、
$\frac{dx_1}{dt} = [H,x_1] = i\alpha_1[p_1,x_1] = \alpha_1$
$\frac{dx_2}{dt} = [H,x_2] = i\alpha_2[p_2,x_2] = \alpha_2$
$\frac{dx_3}{dt} = [H,x_3] = i\alpha_3[p_3,x_3] = \alpha_3$
よって、
$\frac{d{\bf x}}{dt} = {\bf \alpha}$

[演習3.2]

$[p_1,L_1] = [p_1,x_2p_3-x_3p_2] = [p_1,x_2p_3] - [p_1,x_3p_2]$
$=x_2p_1p_3 - x_2p_3p_1 - x_3p_1p_2 + x_3p_2p_1 = 0$
$[p_1,L_2] = [p_1,x_3p_1-x_1p_3] = [p_1,x_3p_1] - [p_1,x_1p_3]$
$=x_3p_1^2 - x_3p_1^2 - \frac{1}{i}p_3 - x_1p_1p_3 + x_1p_3p_1 =ip_3$
$[p_1,L_3] = [p_1,x_1p_2-x_2p_1] = [p_1,x_1p_2] - [p_1,x_2p_1]$
$=\frac{1}{i}p_2 + x_1p_1p_2 - x_1p_2p_1 - x_2p_1^2 + x_2p_1^2 = -ip_2$
$[p_2,L_1] = -ip_3$
$[p_2,L_2] = 0$
$[p_2,L_3] = ip_1$
$[p_3,L_1] = ip_2$
$[p_3,L_2] = -ip_1$
$[p_3,L_3] = 0$
よって、
$[H,L_1] = \alpha_1[p_1,L_1]+\alpha_2[p_2,L_1]+\alpha_3[p_3,L_1]$
$= -i\alpha_2p_3 + i\alpha_3p_2$
$[H,L_2] = i\alpha_1p_3 - i\alpha_3p_1$
$[H,L_3] = -i\alpha_1p_2 + i\alpha_2p_1$
より、
$[H,{\bf L}] = -i({\bf \alpha} \times {\bf p})$

また、
$[\alpha_1, \Sigma_1] = \left[ \begin{array} {cc} 0 & \sigma_1 \\ \sigma_1 & 0 \end{array} \right] \left[ \begin{array} {cc} \sigma_1 & 0 \\ 0 & \sigma_1 \end{array} \right] - \left[ \begin{array} {cc} \sigma_1 & 0 \\ 0 & \sigma_1 \end{array} \right]\left[ \begin{array} {cc} 0 & \sigma_1 \\ \sigma_1 & 0 \end{array} \right] = 0$
$[\alpha_1, \Sigma_2] = 2i\alpha_3$
$[\alpha_1, \Sigma_3] = -2i\alpha_2$
$[\alpha_2, \Sigma_1] = -2i\alpha_3$
$[\alpha_2, \Sigma_2] = 0$
$[\alpha_2, \Sigma_3] = 2i\alpha_1$
$[\alpha_3, \Sigma_1] = 2i\alpha_2$
$[\alpha_3, \Sigma_2] = -2i\alpha_1$
$[\alpha_3, \Sigma_3] = 0$
$[\beta , \Sigma_1] = [\beta , \Sigma_2] = [\beta , \Sigma_3] = 0$
より、
$[H, \Sigma_1] = [\alpha_1, \Sigma_1]p_1 + [\alpha_2, \Sigma_1]p_2 + [\alpha_3, \Sigma_1]p_3 + [\beta ,\Sigma_1]m$
$= -2i\alpha_3p_2 + 2i\alpha_2p_3$
$[H, \Sigma_2] = 2i\alpha_3p_1 - 2i\alpha_1p_3$
$[H, \Sigma_3] = 2i\alpha_1p_2 - 2i\alpha_2p_1$
だから、
$[H, {\bf \Sigma}/2] = i({\bf \alpha} \times {\bf p})$

よって、
$[H, {\bf L} + {\bf \Sigma}/2] = 0$

[演習3.3]
(a) $j \neq k$ のとき、
$\alpha_j\alpha_k = \left[ \begin{array} {cc} 0 & \sigma_j \\ \sigma_j & 0 \end{array} \right]\left[ \begin{array} {cc} 0 & \sigma_k \\ \sigma_k & 0 \end{array} \right]$
$= \left[ \begin{array} {cc} \sigma_j\sigma_k & 0 \\ 0 & \sigma_j\sigma_k \end{array} \right] = \left[ \begin{array} {cc} i\sigma_i & 0 \\ 0 & i\sigma_i \end{array} \right] = i\Sigma_i$

(b)
$\rho_1\rho_1 = (-i\alpha_1\alpha_2\alpha_3)(-i\alpha_1\alpha_2\alpha_3)$
$=-\alpha_1\alpha_2\alpha_3\alpha_1\alpha_2\alpha_3$
$=\alpha_1\alpha_2\alpha_1\alpha_3\alpha_2\alpha_3$
$=\alpha_1\alpha_1\alpha_2\alpha_2\alpha_3\alpha_3=1$
同様に
$\rho_1\rho_2 = i\rho_3$
$\rho_1\rho_3 = -i\rho_2$
$\rho_2\rho_1 = -i\rho_3$
$\rho_2\rho_2 = 1$
$\rho_2\rho_3 = i\rho_1$
$\rho_3\rho_1 = i\rho_2$
$\rho_3\rho_2 = -i\rho_1$
$\rho_3\rho_3 = 1$
から、
$\rho_i\rho_j+\rho_j\rho_i = 2\delta_{ij}$
$[\rho_i,\rho_j] = 2i\rho_k$
は示せる。
$[\Sigma_1,\rho_1] = \Sigma_1\rho_1 - \rho_1\Sigma_1$
$=(-i\alpha_2\alpha_3)(-i\alpha_1\alpha_2\alpha_3) -(-i\alpha_1\alpha_2\alpha_3)(-i\alpha_2\alpha_3)$
$=-\alpha_2\alpha_3\alpha_1\alpha_2\alpha_3+\alpha_1\alpha_2\alpha_3\alpha_2\alpha_3 = 0$
同様にして、
$[\Sigma_i,\rho_j] = 0$
が示せる。

[演習3.4]
$[H,{\bf \Sigma}] = 2i({\bf \alpha}\times {\bf p})$
であるから、これと $\frac{{\bf p}}{|{\bf p}|}$ の内積をとると、
$[H,\frac{{\bf \Sigma} \cdot {\bf p}}{|{\bf p}|} = 0$
すなわち、
$[H,h]=0$
である。

[演習3.5]

わからん。さっぱりわからん。

[演習3.6]
$\frac{{\bf p}}{|{\bf p}|}=(\sin\theta\cos\phi ,\sin\theta\sin\phi ,\cos\theta )$
であるから、
$h=\left[ \begin{array} {cc} {\bf \sigma}_1 & 0 \\ 0 & {\bf \sigma}_1 \end{array} \right]\sin\theta\cos\phi +\left[ \begin{array} {cc} {\bf \sigma}_2 & 0 \\ 0 & {\bf \sigma}_2 \end{array} \right]\sin\theta\sin\phi +\left[ \begin{array} {cc} {\bf \sigma}_3 & 0 \\ 0 & {\bf \sigma}_3 \end{array} \right]\cos\theta$
２次元行列部分のみを考えると、
$h=\left[ \begin{array} {cc} \cos\theta & \sin\theta\cos\phi - i\sin\theta\sin\phi \\ \sin\theta\cos\phi + i\sin\theta\sin\phi & - \cos\theta \end{array} \right]$
$= \left[ \begin{array} {cc} \cos\theta & \sin\theta e^{-i\phi} \\ \sin\theta e^{i\phi} & - \cos\theta \end{array} \right]$
$h$ の固有値を $k$ とすると、
$|h - kI| =\left| \begin{array} {cc} \cos\theta - k & \sin\theta e^{-i\phi} \\ \sin\theta e^{i\phi} & - \cos\theta - k \end{array} \right|$
$= k^2 - 1 = 0$
よって、固有値は $\pm 1$ であり、それぞれの固有状態は、
$\chi_+ = \left[ \begin{array} {c} \cos (\theta /2)\exp (-i\phi /2) \\ \sin (\theta /2)\exp (i\phi /2) \end{array} \right]$
$\chi_- = \left[ \begin{array} {c} - \sin (\theta /2)\exp (-i\phi /2) \\ \cos (\theta /2)\exp (i\phi /2) \end{array} \right]$
である。

[演習3.7]
$({\bf \sigma}\cdot{\bf p})^2\chi_r = (E + m)({\bf \sigma}\cdot{\bf p})\chi_r$
$=(E+m)(E-m)\chi_r = (|E|^2-m^2)\chi_r$
であるから、 $E&gt;0$ のとき、
$({\bf \alpha}\cdot{\bf p}+\beta m)\psi (x) = \left[ \begin{array} {cc} m & {\bf \sigma}\cdot {\bf p} \\ {\bf \sigma}\cdot {\bf p} & -m \end{array} \right]\left[ \begin{array} {c} \chi_r \\ \frac{({\bf \sigma}\cdot{\bf p})}{|E|+m}\chi_r \end{array} \right]$
$=\left[ \begin{array} {c} m\chi_r + \frac{({\bf \sigma}\cdot{\bf p})^2}{|E|+m}\chi_r \\ ({\bf \sigma}\cdot{\bf p})\cdot\frac{|E|}{|E|+m}\chi_r \end{array} \right]$
$=|E|\left[ \begin{array} {c} \chi_r \\ \frac{({\bf \sigma}\cdot{\bf p})}{|E|+m}\chi_r \end{array} \right]$
となる。 $E&lt;0$ についても、同様に計算できる。

[演習3.8]
$r=1,2,s=1,2$ の場合だけを考えるらしい。つまり、 $E&gt;0$ の粒子。
$u_r^+(p)u_s(p) = N^2 \left[ \begin{array} {cc} \chi_r^+ & \chi_r^+ \frac{({\bf \sigma}\cdot{\bf p})}{|E|+m} \end{array} \right]\left[ \begin{array} {c} \chi_s \\ \frac{({\bf \sigma}\cdot{\bf p})}{|E|+m}\chi_s \end{array} \right]$
$= (|E|+ m)\chi_r^+ \left[1+\frac{({\bf \sigma}\cdot{\bf p})^2}{(|E|+m)^2}\right]\chi_s$
$= (|E|+ m)\chi_r^+ \left[1+\frac{(|E|+ m)(|E|- m)}{(|E|+m)^2}\right]\chi_s$
$= (|E|+ m + |E|- m)\chi_r^+\chi_s^ = 2E\delta_{rs}$
また（これ、誤植だと思うんだけど $v_2(p)=u_3(-p)$ だよね？）
$v_1(p)=u_4(-p)=N\left[ \begin{array} {c} \frac{({\bf \sigma}\cdot{\bf p})}{|E|+m}\chi_2 \\ \chi_2 \end{array} \right]$
$v_2(p)=u_3(-p)=N\left[ \begin{array} {c} \frac{({\bf \sigma}\cdot{\bf p})}{|E|+m}\chi_1 \\ \chi_1 \end{array} \right]$
をまとめると、
$v_r(p) = N\left[ \begin{array} {c} \frac{({\bf \sigma}\cdot{\bf p})}{|E|+m}\chi_{3-r} \\ \chi_{3-r} \end{array} \right]$
となる。すると、
$v_r^+(p)v_s(p) = N^2\left[ \begin{array} {cc} \chi_{3-r}^+ \frac{({\bf \sigma}\cdot{\bf p})}{|E|+m} & \chi_{3-r}^+ \end{array} \right]\left[ \begin{array} {c} \frac{({\bf \sigma}\cdot{\bf p})}{|E|+m}\chi_{3-s} \\ \chi_{3-s} \end{array} \right]$
$2E\chi_{3-r}\chi_{3-s} = 2E\delta_{rs}$
さらに、
$\bar{u}_r(p)u_s(p) = N^2 \left[ \begin{array} {cc} \chi_r^+ & \chi_r^+ \frac{({\bf \sigma}\cdot{\bf p})}{|E|+m} \end{array} \right]\left[ \begin{array} {cc} 1 & 0 \\ 0 & -1 \right]\left[ \begin{array} {c} \chi_s \\ \frac{({\bf \sigma}\cdot{\bf p})}{|E|+m}\chi_s \end{array} \right]$
$= (|E|+ m) \left[ \begin{array} {cc} \chi_r^+ & \chi_r^+ \frac{({\bf \sigma}\cdot{\bf p})}{|E|+m} \end{array} \right]\left[ \begin{array} {c} \chi_s \\ - \frac{({\bf \sigma}\cdot{\bf p})}{|E|+m}\chi_s \end{array} \right]$
$= (|E|+ m) \chi_r^+ \left[1-\frac{({\bf \sigma}\cdot{\bf p})^2}{(|E|+m)^2}\right]\chi_s$
$= (|E|+ m) \chi_r^+ \left[1-\frac{(|E|+ m)(|E|- m)}{(|E|+m)^2}\right]\chi_s$
$= (|E|+ m - |E|+ m)\chi_r^+\chi_s = 2m\delta_{rs}$
および、
$\bar{v}_r(p)v_s(p) = N^2\left[ \begin{array} {cc} \chi_{3-r}^+ \frac{({\bf \sigma}\cdot{\bf p})}{|E|+m} & \chi_{3-r}^+ \end{array} \right]\left[ \begin{array} {cc} 1 & 0 \\ 0 & -1 \right]\left[ \begin{array} {c} \frac{({\bf \sigma}\cdot{\bf p})}{|E|+m}\chi_{3-s} \\ \chi_{3-s} \end{array} \right]$
$= - 2m\chi_{3-r}\chi_{3-s} = - 2m\delta_{rs}$
同様の計算で、
$\bar{u}_r(p)v_s(p) = \bar{v}_r(p)u_s(p) = 0$
も示せる。

[演習3.9]

$(\slashchar{p} - m)u(p) = (\gamma^{\mu}p_{\mu}-m)u(p)$
$=(\gamma^{\mu}i\partial_{\mu}-m)u(p)$
$=(\beta i\partial_0 + \beta\alpha^k i\partial_k - \beta^2 m)u(p)$
$=\beta (i\partial_0 + \alpha^k i\partial_k - \beta m)u(p)$
$=\beta (E - {\bf \alpha}\cdot{\bf p} - m\beta )u(p)$
$=\beta (E - E)u(p) = 0$
また、
$(\slashchar{p} + m)v(p) = (\gamma^{\mu}p_{\mu} + m)u(-p)$
$=(\gamma^{\mu}i\partial_{\mu} + m)u(-p)$
$=(\beta i\partial_0 + \beta\alpha^k i\partial_k + \beta^2 m)u(-p)$
$=\beta (E + {\bf \alpha}\cdot{\bf -p} + m\beta)u(-p)$
$=\beta (E + (-E)) = 0$
さらに、
$\bar{u}(p)(\slashchar{p}-m) = u^+\gamma_0(\slashchar{p}-m)$
$=u^+\gamma_0(\gamma^{\mu}p_{\mu} - m)$
$=u^+\gamma_0(\gamma^0 i\partial_0 + \gamma^k i\partial_k -m)$
$=\left[(- \gamma^{0+}i\partial_0 - \gamma^{k+} i\partial_k -m)\gamma_0^+u \right]^+$
$=\left[(- \beta i\partial_0 + \beta\alpha^k i\partial_k - m)\beta u \right]^+$
$=\left[(- i\partial_0 + \beta\alpha^k\beta i\partial_k - m\beta )u \right]^+$
$=\left[(- i\partial_0 - \beta\beta\alpha^k i\partial_k - m\beta )u \right]^+$
$=\left[(- E + {\bf \alpha}\cdot {\bf p} - m\beta )u \right]^+$
あれ？なんか符号がちがうのかうまくいかない。
$\bar{v}(p)(\slashchar{p}+m) = v^+(p)\gamma_0(\gamma^{\mu}p_{\mu} + m)$
$=u^+(-p)\gamma_0(\gamma^0 i\partial_0 + \gamma^k i\partial_k + m)$
$=\left[(- \gamma^{0+} i\partial_0 - \gamma^{k+} i\partial_k + m)\gamma_{0+}u(-p) \right]^+$
$=\left[(- \beta i\partial_0 + \beta\alpha^k i\partial_k + m)\beta u(-p) \right]^+$
$=\left[(- i\partial_0 - \alpha^k i\partial_k + m)u(-p) \right]^+$
$=\left[(- E - {\bf \alpha}\cdot{\bf - p} + m)u(-p) \right]^+$
あれ？これもどこかで符号が違うのかなあ？

[演習3.10]

$\Lambda_+^2u(p) = \frac{1}{4m^2}(m+\slashchar{p})(m+\slashchar{p})u(p)$
$= \frac{1}{4m^2}(m+\slashchar{p})\cdot 2mu(p)$
$= \frac{1}{2m}(m+\slashchar{p})u(p) = \Lambda_+u(p)$

$\Lambda_-^2v(p) = \frac{1}{4m^2}(m-\slashchar{p})(m-\slashchar{p})v(p)$
$= \frac{1}{4m^2}(m-\slashchar{p})\cdot 2mv(p)$
$= \frac{1}{2m}(m-\slashchar{p})v(p) = \Lambda_-v(p)$

$(\Lambda_++\Lambda_-)u(p) = \frac{1}{2m}[(m+\slashchar{p})+(m-\slashchar{p})]u(p)$
$=\frac{1}{2m}[2mu(p) + 0] = u(p)$

$(\Lambda_++\Lambda_-)v(p) = \frac{1}{2m}[(m+\slashchar{p})+(m-\slashchar{p})]v(p)$
$=\frac{1}{2m}[0 + 2mv(p)] = v(p)$

[演習3.11]

${\bf \sigma}\cdot{\bf p} = \left[ \begin{array} {cc} 0 & 1 \\ 1 & 0 \end{array} \right]p_1 + \left[ \begin{array} {cc} 0 & -i \\ i & 0 \end{array} \right]p_2 + \left[ \begin{array} {cc} 1 & 0 \\ 0 & -1 \end{array} \right]p_3$
$= \left[ \begin{array} {cc} p_3 & p_1-ip_2 \\ p_1+ip_2 & -p_3 \end{array} \right]$
だから、E>0のとき、
$u_1(p) = N \left[ \begin{array} {c} 1 \\ 0 \\ \frac{p_3}{E+m} \\ \frac{p_1 + ip_2}{E+m} \end{array} \right]$
$u_2(p) = N \left[ \begin{array} {c} 0 \\ 1 \\ \frac{p_1 - ip_2}{E+m} \\ -\frac{p_3}{E+m} \end{array} \right]$

$\sum_{r=1,2}u_r(p)\bar{u}_r(p) = u_1(p)\bar{u}_1(p) + u_2(p)\bar{u}_2(p)$
$= N \left[ \begin{array} {c} 1 \\ 0 \\ \frac{p_3}{E+m} \\ \frac{p_1 + ip_2}{E+m} \end{array} \right] N\left[ \begin{array} {cccc} 1 & 0 & \frac{p_3}{E+m} & \frac{p_1 - ip_2}{E+m} \end{array} \right]\left[ \begin{array} {cc} 1 & 0 \\ 0 & -1 \end{array} \right]$
$+ N \left[ \begin{array} {c} 0 \\ 1 \\ \frac{p_1 - ip_2}{E+m} \\ -\frac{p_3}{E+m} \end{array} \right] N \left[ \begin{array} {cccc} 0 & 1 & \frac{p_1 + ip_2}{E+m} & -\frac{p_3}{E+m} \end{array} \right]\left[ \begin{array} {cc} 1 & 0 \\ 0 & -1 \end{array} \right]$
$=N^2 \left[\begin{array}{cccc} 1 & 0 & \frac{-p_3}{E+m} & \frac{-p_1+ip_2}{E+m} \\ 0 & 1 & \frac{-p_1-ip_2}{E+m} & \frac{p_3}{E+m} \\ \frac{p_3}{E+m} & \frac{p_1-ip_2}{E+m} & \frac{-p_1^2-p_2^2-p_3^2}{(E+m)^2} & 0 \\ \frac{p_1+ip_2}{E+m} & \frac{p_3}{E+m} & 0 & \frac{-p_1^2-p_2^2-p_3^2}{(E+m)^2} \end{array} \right]$
ここで、
$E^2 - (p_1^2+p_2^2+p_3^2) = m^2$ , $N^2 = E + m$
なので、整理すると、
$= \left[\begin{array}{cc} E+m & -{\bf \sigma}\cdot{\bf p} \\ {\bf \sigma}\cdot{\bf p} & m-E \end{array}\right]$
$= m + E\beta - {\bf \gamma}\cdot{\bf p} = m+\gamma^{\mu}p_{\mu} = m + \slashchar{p}$

また、
$v_1(p) = u_4(-p) = N \left[ \begin{array} {c} \frac{-p_1 + ip_2}{-E+m} \\ \frac{p_3}{-E+m} \\ 0 \\ 1 \end{array} \right]$
$v_2(p) = N \left[ \begin{array} {c} \frac{-p_3}{-E+m} \\ \frac{-p_1 - ip_2}{-E+m} \\ 1 \\ 0 \right]$
$N^2 = -E+m$
を使って、E<0も同様に計算できる。

[演習3.12]
i,j=1,2,3のとき、
$\sigma^{ij} = (i/2)[\gamma^i,\gamma^j] = (i/2)(\beta\alpha_i\beta\alpha_j - \beta\alpha_j\beta\alpha_i)$
$=(i/2)(- \beta\beta\alpha_i\alpha_j + \beta\beta\alpha_j\alpha_i) = (i/2)(-2i\alpha_k)=\alpha_k$
また
$\sigma^{0i} = (i/2)[\gamma^0,\gamma^i] = (i/2)(\beta\beta\alpha_i - \beta\alpha_i\beta)$
$=(i/2)(2\alpha_i)=i\alpha_i$

$L_k = \sigma_{ij} = g_{i\rho}g_{j\sigma}\sigma^{\rho\sigma} = (-\delta_{i\rho})(-\delta_{j\sigma})\sigma^{\rho\sigma} = \sigma^{ij} = \alpha_k$
$N_k = \sigma_{0k} = g_{0\rho}g_{k\sigma}\sigma^{\rho\sigma} = (\delta_{0\rho})(-\delta_{j\sigma})\sigma^{\rho\sigma} = -\sigma^{0k} = -i\alpha_k$
とすると、(2.63)の関係式を満たす。

[演習3.13]

(3.55)の導出

$\left(1+\frac{i}{4}\sigma^{\mu\nu}\epsilon_{\mu\nu}\right)\gamma^{\mu}\left(1-\frac{i}{4}\sigma^{\mu\nu}\epsilon_{\mu\nu}\right) = (\delta^{\mu}_{\;\nu}+\epsilon^{\mu}_{\!\nu})\gamma^{\nu}$
展開して、 $\epsilon$ の二次の項を無視すると、
$\gamma^{\mu} + \left(\frac{i}{4}\sigma^{\rho\sigma}\epsilon_{\rho\sigma}\right)\gamma^{\mu} - \frac{i}{4} \gamma^{\mu}(\sigma^{\rho\sigma}\epsilon_{\rho\sigma}) = \gamma^{\mu} + \epsilon^{\mu}_{\;\nu}\gamma^{\nu}$
整理すると、
$\frac{i}{4}\left(\sigma^{\rho\sigma}\epsilon_{\rho\sigma}\right)\gamma^{\mu} - \frac{i}{4}\gamma^{\mu}\left(\sigma^{\rho\sigma}\epsilon_{\rho\sigma}\right) = g^{\mu\rho}\epsilon_{\rho\sigma}\gamma^{\sigma}$
$\epsilon_{\rho\sigma}$ の係数を比較して、
$\frac{i}{4}[\gamma^{\mu},\sigma^{\rho\sigma}] = -g^{\mu\rho}\gamma^{\sigma}$
$\rho$ と $\sigma$ を入れ替えると、
$\frac{i}{4}[\gamma^{\mu},\sigma^{\sigma\rho}] = -g^{\mu\sigma}\gamma^{\rho}$
$-\frac{i}{4}[\gamma^{\mu},\sigma^{\rho\sigma}] = -g^{\mu\sigma}\gamma^{\rho}$
引き算して、
$g^{\mu\rho}\gamma^{\sigma} -g^{\mu\sigma}\gamma^{\rho} = - \frac{i}{2}[\gamma^{\mu},\sigma^{\rho\sigma}]$
$= \frac{1}{2i}[\gamma^{\mu},\sigma^{\rho\sigma}]$

(3.56)が(3.55)を満たすことの証明

$[A,[B,C]] = [B,[A,C]] - [C,[A,B]]$
を使う。
$\frac{1}{2i}[\gamma^{\mu},\sigma^{\rho\sigma}] = \frac{1}{2i}[\gamma^{\mu},\frac{i}{2}[\gamma^{\rho},\gamma^{\sigma}]]$
$=\frac{1}{4}[\gamma^{\mu},[\gamma^{\rho},\gamma^{\sigma}]]$
$=\frac{1}{4}[\gamma^{\rho},[\gamma^{\mu},\gamma^{\sigma}]] - \frac{1}{4}[\gamma^{\sigma},[\gamma^{\mu},\gamma^{\rho}]]$
$=\frac{1}{2i}[\gamma^{\rho},\frac{i}{2}[\gamma^{\mu},\gamma^{\sigma}]] - \frac{1}{2i}[\gamma^{\sigma},\frac{i}{2}[\gamma^{\mu},\gamma^{\rho}]]$
$=\frac{1}{2i}[\gamma^{\rho},\sigma^{\mu\sigma}]-\frac{1}{2i}[\gamma^{\sigma},\sigma^{\mu\rho}]$
$=(g^{\rho\mu}\gamma^{\sigma} - g^{\rho\sigma}\gamma^{\mu}) - (g^{\sigma\mu}\gamma^{\rho}-g^{\sigma\rho}\gamma^{\mu})$
$=g^{\rho\mu}\gamma^{\sigma} - g^{\sigma\mu}\gamma^{\rho}$
$=g^{\mu\rho}\gamma^{\sigma} - g^{\mu\sigma}\gamma^{\rho}$

[演習3.14]

$\epsilon_{\mu\nu}^+ = - \epsilon_{\nu\mu} = \epsilon_{\mu\nu}$
にはなりそう。
あとは、
$\sigma^{\mu\nu +} = \gamma_0\sigma^{\mu\nu}\gamma_0$
らしいけど、これはなんでかなあ？

$\gamma_0S^+\gamma_0 = \gamma_0\gamma_0 + \frac{i}{4}\gamma_0\sigma^{\mu\nu +}\gamma_0\epsilon_{\mu\nu +}$
$= 1 + \frac{i}{4} \gamma_0\gamma_0\sigma^{\mu\nu}\gamma_0\gamma_0\epsilon_{\mu\nu}$
$= 1 + \frac{i}{4} \sigma^{\mu\nu} \epsilon_{\mu\nu} = S^{-1}$

また、
$\gamma^5S =i\gamma^0\gamma^1\gamma^2\gamma^3\left(1-\frac{i}{4}\sigma^{\mu\nu}\epsilon_{\mu\nu}\right)$
$=i\gamma^0\gamma^1\gamma^2\gamma^3 - \frac{i}{4}i\gamma^0\gamma^1\gamma^2\gamma^3\sigma^{\mu\nu}\epsilon_{\mu\nu}$
$=i\gamma^0\gamma^1\gamma^2\gamma^3 - \frac{i}{4}i\gamma^0\gamma^1\gamma^2\gamma^3 \frac{i}{2}(\gamma^{\mu}\gamma^{\nu}-\gamma^{\nu}\gamma^{\mu})\epsilon_{\mu\nu}$
$=i\gamma^0\gamma^1\gamma^2\gamma^3 - \frac{i}{4}\frac{i}{2}(\gamma^{\mu}\gamma^{\nu}-\gamma^{\nu}\gamma^{\mu})i\gamma^0\gamma^1\gamma^2\gamma^3\epsilon_{\mu\nu}$
$=\left(1-\frac{i}{4}\sigma^{\mu\nu}\epsilon_{\mu\nu}\right)i\gamma^0\gamma^1\gamma^2\gamma^3$
$=S\gamma^5$

[演習3.15]

$N_1 = M_{01} = \sigma_{01}/2 = - \frac{i}{2} \alpha_1$
$\alpha_1^{2n} = 1, \alpha_1^{2n+1} = \alpha_1$
を利用して、
$S = \exp (-iN_1\eta ) = \sum^{\infty}_{n=0} \frac{1}{n!}(-iN_1\eta )^n$
$= \sum^{\infty}_{n=0}\frac{1}{n!}\left(-\frac{1}{2}\alpha_1\eta\right)^n$
$= \sum_{n=0}^{\infty}\frac{1}{2n!}\left(-\frac{1}{2}\alpha_1\eta\right)^{2n}+\sum_{n=0}^{\infty}\frac{1}{(2n+1)!}\left(-\frac{1}{2}\alpha_1\eta\right)^{2n+1}$
$= \sum_{n=0}^{\infty}\frac{1}{2n!}\left(\frac{1}{2}\eta\right)^{2n}-\sum_{n=0}^{\infty}\frac{1}{(2n+1)!}\left(\frac{1}{2}\eta\right)^{2n+1}\alpha_1$
$= \cosh \left(\frac{\eta}{2}\right)-\sinh \left(\frac{\eta}{2}\right)\alpha_1$
$= \left[\begin{array}{cc} \cosh\left(\frac{\eta}{2}\right) & -\sinh \left(\frac{\eta}{2}\right)\sigma_1 \\ -\sinh \left(\frac{\eta}{2}\right)\sigma_1 & \cosh\left(\frac{\eta}{2}\right) \end{array}\right]$

[演習3.16]

$[{\bf \alpha}\cdot{\bf p}]^{2n} = [(\alpha_1p_1+\alpha_2p_2+\alpha_3p_3)^2]^n$
$=[p_1^2+p_2^2+p_3^2]^n$
および
$[{\bf \alpha}\cdot{\bf p}]^{2n+1} = [(\alpha_1p_1+\alpha_2p_2+\alpha_3p_3)^2]^n(\alpha_1p_1+\alpha_2p_2+\alpha_3p_3)$
$=[p_1^2+p_2^2+p_3^2]^n{\bf \alpha}\cdot{\bf p}$
であるから、
$S = \exp(\frac{\eta}{2}{\bf \alpha}\cdot{\bf n}$
$=\sum_{n=0}^{\infty}\frac{1}{(2n)!}\left(\frac{\eta}{2|{\bf p}|}{\bf \alpha}\cdot{\bf p}\right)^{2n} + \sum_{n=0}^{\infty}\frac{1}{(2n+1)!}\left(\frac{\eta}{2|{\bf p}|}{\bf \alpha}\cdot{\bf n}\right)^{2n+1}$
$=\sum_{n=0}^{\infty}\frac{1}{(2n)!}\left(\frac{\eta}{2|{\bf p}|}\right)^{2n}(p_1^2+p_2^2+p_3^2)^n + \sum_{n=0}^{\infty}\frac{1}{(2n+1)!}\left(\frac{\eta}{2|{\bf p}|}\right)^{2n+1}(p_1^2+p_2^2+p_3^2)^n{\bf \alpha}\cdot{\bf p}$
$=\sum_{n=0}^{\infty}\frac{1}{(2n)!}\left(\frac{\eta}{2}\right)^{2n} + \sum_{n=0}^{\infty}\frac{1}{(2n+1)!}\left(\frac{\eta}{2}\right)^{2n+1}{\bf \alpha}\cdot{\bf n}$
$=\left[ \begin{array}{cc} \cosh \left(\frac{\eta}{2}\right) & \sinh \left(\frac{\eta}{2}\right){\bf \alpha}\cdot{\bf n} \\ \sinh \left(\frac{\eta}{2}\right){\bf \alpha}\cdot{\bf n} & \cosh \left(\frac{\eta}{2}\right) \end{array}\right]$
となる。

$w_{1,2} = \left[\begin{array}{c} \chi_{1,2} & 0 \end{array}\right]$
$w_{3,4} = \left[\begin{array}{c} 0 & \chi_{1,2} \end{array}\right]$
である。

$Sw_1 = \left[\begin{array}{c} \cosh\left(\frac{\eta}{2}\right)\chi_{1} & \sinh\left(\frac{\eta}{2}\right)({\bf \sigma}\cdot{\bf n})\chi_{1} \end{array}\right]$

ここで、
$\cosh\left(\frac{\eta}{2}\right) = \sqrt{\frac{E+m}{2m}},\sinh\left(\frac{\eta}{2}\right) = \sqrt{\frac{E-m}{2m}}$
なんだそうだ。（なんでだろ？）
それを使うと、

$Sw_1 = \left[\begin{array}{c} \sqrt{\frac{E+m}{2m}}\chi_{1} & \sqrt{\frac{E-m}{2m}}({\bf \sigma}\cdot{\bf n})\chi_{1} \end{array}\right]$
$= \sqrt{\frac{E+m}{2m}}\left[\begin{array}{c} \chi_{1} & \frac{{\bf \sigma}\cdot{\bf p}}{E+m}\chi_{1} \end{array}\right]$

＝＝式(3.59)の証明＝＝
(3.48)と(3.57)を使う

$\psi^{\prime}(x^{\prime}) = S\psi (x)$
両辺の共役をとると
$\psi^{\prime +}(x^{\prime}) = \psi^+(x)S^+$
右から $\gamma_0$ を掛けて変形
$\psi^{\prime +}(x^{\prime})\gamma_0 = \psi^+(x)\gamma_0\gamma_0S^+\gamma_0$
$\bar{\psi}^{\prime}(x^{\prime}) = \bar{\psi}(x)S^{-1}$
＝＝＝＝＝＝＝＝＝＝＝

[演習3.17]
(3.38)と(3.44)を使う

$\partial_{\mu}j^{\mu} = \partial_{\mu}(\bar{\psi}\gamma^{\mu}\psi )$
$= (\partial_{\mu}\bar{\psi})\gamma^{\mu}\psi + \bar{\psi}\gamma^{\mu}(\partial_{\mu}\psi )$
$= -\frac{m}{i}\bar{\psi}\psi + \frac{m}{i}\bar{\psi}\psi = 0$

[演習3.18]
パリティ変換の場合、 $S=\gamma^0$ であるから、
$\bar{\psi}^{\prime} = \bar[\psi}\gamma^0, \psi^{\prime} = \gamma^0\psi$
よって、
$\bar{\psi}^{\prime}\psi^{\prime}=\bar{\psi}\gamma^0\gamma^0\psi = \bar{\psi}\psi$

$i\bar{\psi}^{\prime}\gamma^5\psi^{\prime} = i\bar{\psi}\gamma^0\gamma^5\gamma^0\psi$
$=-i\bar{\psi}\gamma^5\gamma^0\gamma^0\psi =-i\bar{\psi}\gamma^5\psi$

$\bar{\psi}^{\prime}\gamma^{\mu}\psi^{\prime} =\bar{\psi}\gamma^0\gamma^{\mu}\gamma^0\psi$
$\mu = 0$ のとき、
$=\bar{\psi}\gamma^0\gamma^0\gamma^0\psi = \bar{\psi}\gamma^0\psi$
$\mu = 1,2,3$ のとき、
$=\bar{\psi}\gamma^0\gamma^k\gamma^0\psi = -\bar{\psi}\gamma^k\gamma^0\gamma^0\psi$
$=-\bar{\psi}\gamma^k\psi$
以上より、
$=\bar{\psi}\gamma_{\mu}\psi$

$\bar{\psi}^{\prime}\gamma^5\gamma^{\mu}\psi^{\prime} =\bar{\psi}\gamma^0\gamma^5\gamma^{\mu}\gamma^0\psi$
$\mu = 0$ のとき、
$=\bar{\psi}\gamma^0\gamma^5\gamma^0\gamma^0\psi = \bar{\psi}\gamma^0\gamma^5\psi$
$=-\bar{\psi}\gamma^5\gamma^0\psi$
$\mu = 1,2,3$ のとき、
$=\bar{\psi}\gamma^0\gamma^5\gamma^k\gamma^0\psi = \bar{\psi}\gamma^5\gamma^k\gamma^0\gamma^0\psi$
$=\bar{\psi}\gamma^5\gamma^k\psi$
以上より、
$=-\bar{\psi}\gamma^5\gamma_{\mu}\psi$

$\bar{\psi}^{\prime}\sigma^{\mu\nu}\psi^{\prime} = \bar{\psi}\gamma^0\sigma^{\mu\nu}\gamma^0\psi$
$\mu < \nu$ で考えれば十分である。
$\mu = 0$ のとき、
$=\bar{\psi}\gamma^0\sigma^{0k}\gamma^0\psi$
$=\bar{\psi}\gamma^0\frac{i}{2}(\gamma^0\gamma^k-\gamma^k\gamma^0)\gamma^0\psi$
$=\frac{i}{2}\bar{\psi}\gamma^0\gamma^0\gamma^k\gamma^0\psi - \frac{i}{2}\bar{\psi}\gamma^0\gamma^k\gamma^0\gamma^0\psi$
$=- \frac{i}{2}\bar{\psi}\gamma^0\gamma^0\gamma^0\gamma^k\psi + \frac{i}{2}\bar{\psi}\gamma^0\gamma^0\gamma^k\gamma^0\psi$
$=- \frac{i}{2}\bar{\psi}\gamma^0\gamma^k\psi + \frac{i}{2}\bar{\psi}\gamma^k\gamma^0\psi$
$=- \bar{\psi}\frac{i}{2}[\gamma^0,\gamma^k]\psi = - \bar{\psi}\sigma^{0k}\psi$
$\mu = 1,2,3$ のとき、
$=\bar{\psi}\gamma^0\sigma^{jk}\gamma^0\psi$
$=\bar{\psi}\gamma^0\frac{i}{2}(\gamma^j\gamma^k-\gamma^k\gamma^j)\gamma^0\psi$
$=\frac{i}{2}\bar{\psi}\gamma^0\gamma^j\gamma^k\gamma^0\psi - \frac{i}{2}\bar{\psi}\gamma^0\gamma^k\gamma^j\gamma^0\psi$
$=\frac{i}{2}\bar{\psi}\gamma^0\gamma^0\gamma^j\gamma^k\psi - \frac{i}{2}\bar{\psi}\gamma^0\gamma^0\gamma^k\gamma^j\psi$
$=\frac{i}{2}\bar{\psi}\gamma^j\gamma^k\psi - \frac{i}{2}\bar{\psi}\gamma^k\gamma^j\psi$
$=\bar{\psi}\sigma^{jk}\psi$
以上より、
$=\bar{\psi}\sigma_{\mu\nu}\psi$

[演習3.19]

$({\bf \sigma}\cdot{\bf p}^{\prime})({\bf \sigma}\cdot{\bf p})\chi_r =(|E|+m)(|E|-m)\chi_r$ を忘れないこと。
$\bar{u}(p^{\prime}) = N^+ \left[\begin{array}{cc} \chi_r^+ & - \chi_r^+ \frac{{\bf \sigma}\cdot{\bf p}}{E+m} \end{array}\right]$
$u(p) = N \left[ \begin{array}{c} \chi_s \\ \frac{{\bf \sigma}\cdot{\bf p}}{E+m} \chi_s \end{array} \right]$

$<p^{\prime}|1|p> = N^+ \left[\begin{array}{cc} \chi_r^+ & \chi_r^+ \frac{{\bf \sigma}\cdot{\bf p}^{\prime}}{E+m} \end{array}\right]N \left[ \begin{array}{c} \chi_s \\ \frac{{\bf \sigma}\cdot{\bf p}}{E+m} \chi_s \end{array} \right]$
$= |N|^2 \chi_r^+ \left[\begin{array}{cc} 1 & -\frac{{\bf \sigma}\cdot{\bf p}^{\prime}}{E+m} \end{array}\right]\left[ \begin{array}{c} 1 \\ \frac{{\bf \sigma}\cdot{\bf p}}{E+m} \end{array} \right]\chi_s$
$= (E+m) \chi_r^+ \left( 1 - \frac{({\bf \sigma}\cdot{\bf p}^{\prime})({\bf \sigma}\cdot{\bf p})}{(E+m)^2} \right) \chi_s$
$= \chi_r^+ \left( (E+m) - (E-m) \right)\chi_s$
$= \chi_r^+ 2m \chi_s$

$<p^{\prime}|\gamma^0|p> = N^+ \left[\begin{array}{cc} \chi_r^+ & \chi_r^+ \frac{{\bf \sigma}\cdot{\bf p}^{\prime}}{E+m} \end{array}\right] \left[\begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array}\right] N \left[ \begin{array}{c} \chi_s \\ \frac{{\bf \sigma}\cdot{\bf p}}{E+m} \chi_s \end{array} \right]$
$= |N|^2 \chi_r^+ \left[\begin{array}{cc} 1 & \frac{{\bf \sigma}\cdot{\bf p}^{\prime}}{E+m} \end{array}\right]\left[ \begin{array}{c} 1 \\ \frac{{\bf \sigma}\cdot{\bf p}}{E+m} \end{array} \right]\chi_s$
$= (E+m) \chi_r^+ \left( 1 + \frac{({\bf \sigma}\cdot{\bf p}^{\prime})({\bf \sigma}\cdot{\bf p})}{(E+m)^2} \right) \chi_s$
$= \chi_r^+ \left( (E+m) + (E-m) \right)\chi_s$
$= \chi_r^+ 2E \chi_s = \chi_r^+ 2(m+p^2) \chi_s \sim \chi_r^+ 2m \chi_s$

$<p^{\prime}|\gamma^k|p> = N^+ \left[\begin{array}{cc} \chi_r^+ & \chi_r^+ \frac{{\bf \sigma}\cdot{\bf p}^{\prime}}{E+m} \end{array}\right] \left[\begin{array}{cc} 0 & \sigma_k \\ -\sigma_k & 0 \end{array}\right] N \left[ \begin{array}{c} \chi_s \\ \frac{{\bf \sigma}\cdot{\bf p}}{E+m} \chi_s \end{array} \right]$
$= |N|^2 \chi_r^+ \left[\begin{array}{cc} 1 & -\frac{{\bf \sigma}\cdot{\bf p}^{\prime}}{E+m} \end{array}\right]\left[\begin{array}{cc} 0 & \sigma_k \\ -\sigma_k & 0 \end{array}\right]\left[ \begin{array}{c} 1 \\ \frac{{\bf \sigma}\cdot{\bf p}}{E+m} \end{array} \right]\chi_s$
$= (E+m) \chi_r^+ \left[ \frac{{\bf \sigma}\cdot{\bf p}^{\prime}}{E+m}\sigma_k + \sigma_k\frac{{\bf \sigma}\cdot{\bf p}}{E+m} \right]\chi_s$
$= \chi_r^+ \left[{\bf \sigma}\cdot{\bf p}^{\prime}\sigma_k + \sigma_k{\bf \sigma}\cdot{\bf p}\right]\chi_s$
$= \chi_r^+ \left[(\sigma_ip_i^{\prime}+\sigma_jp_j^{\prime}+\sigma_kp_k^{\prime})\sigma_k + \sigma_k (\sigma_ip_i+\sigma_jp_j+\sigma_kp_k)\right]\chi_s$
$= \chi_r^+ \left[\sigma_i\sigma_kp_i^{\prime}+\sigma_j\sigma_kp_j^{\prime}+\sigma_k\sigma_kp_k^{\prime} + \sigma_k\sigma_ip_i+\sigma_k\sigma_jp_j+\sigma_k\sigma_kp_k\right]\chi_s$
$= \chi_r^+ \left[-i\sigma_jp_i^{\prime}+i\sigma_ip_j^{\prime}+p_k^{\prime} +i\sigma_jp_i - i\sigma_ip_j + p_k\right]\chi_s$
$= \chi_r^+ \left[p_k + p_k^{\prime} + i\sigma_i(p_j^{\prime}-p_j) - i\sigma_j(p_j^{\prime}-p_j)\right]\chi_s$
$= \chi_r^+ \left[{\bf p} + {\bf p}^{\prime} + i{\bf \sigma}\times({\bf p}^{\prime} - {\bf p})\right]_k\chi_s$

$<p^{\prime}|\gamma^0\gamma^5|p> = \left[\begin{array}{cc} \chi_r^+ & -\chi_r^+ \frac{{\bf \sigma}\cdot{\bf p}^{\prime}}{E+m} \end{array}\right] \left[\begin{array}{cc} 0 & 1 \\ -1 & 0 \end{array}\right] N \left[ \begin{array}{c} \chi_s \\ \frac{{\bf \sigma}\cdot{\bf p}}{E+m} \chi_s \end{array} \right]$
$= |N|^2 \chi_r^+ \left[\begin{array}{cc} \frac{{\bf \sigma}\cdot{\bf p}^{\prime}}{E+m} & 1 \end{array}\right]\left[ \begin{array}{c} 1 \\ \frac{{\bf \sigma}\cdot{\bf p}}{E+m} \end{array} \right]\chi_s$
$= (E+m) \chi_r^+ \left[\frac{{\bf \sigma}\cdot{\bf p}^{\prime}}{E+m} + \frac{{\bf \sigma}\cdot{\bf p}}{E+m}\right]\chi_s$
$= \chi_r^+ \left[{\bf \sigma}\cdot{\bf p}^{\prime}+{\bf \sigma}\cdot{\bf p}\right]\chi_s$
$= \chi_r^+ \left[{\bf \sigma}\cdot({\bf p}+{\bf p}^{\prime})\right]\chi_s$

$<p^{\prime}|\gamma^k\gamma^5|p> = N^+ \left[\begin{array}{cc} \chi_r^+ & \chi_r^+ \frac{{\bf \sigma}\cdot{\bf p}^{\prime}}{E+m} \end{array}\right] \left[\begin{array}{cc} \sigma_k & 0 \\ 0 & -\sigma_k \end{array}\right] N \left[ \begin{array}{c} \chi_s \\ \frac{{\bf \sigma}\cdot{\bf p}}{E+m} \chi_s \end{array} \right]$
$= |N|^2 \chi_r^+ \left[\begin{array}{cc} \sigma_k & \sigma_k\frac{{\bf \sigma}\cdot{\bf p}^{\prime}}{E+m} \end{array}\right]\left[ \begin{array}{c} 1 \\ \frac{{\bf \sigma}\cdot{\bf p}}{E+m} \end{array} \right]\chi_s$
$= (E+m) \chi_r^+ \left[ \sigma_k + \sigma_k\frac{({\bf \sigma}\cdot{\bf p}^{\prime})^2}{(E+m)^2} \right]\chi_s$
$= \chi_r^+ \left[ \sigma_k(E+m) + \sigma_k (E-m) \right]\chi_s$
$= \chi_r^+ 2E\sigma_k \chi_s = \chi_r^+ 2(m+p^2)\sigma_k \chi_s \sim \chi_r^+ 2m\sigma_k \chi_s$

$<p^{\prime}|\sigma_{ij}|p>$
これは、よくわからないよう。
$g_{ik}g_{jl}\sigma^{kl}$
にすればいいと思うのだがこれだと４成分計算しないといけなくなるようなー？
するのかな？

$<p^{\prime}|\sigma^{k0}|p> =\left[\begin{array}{cc} \chi_r^+ & -\chi_r^+ \frac{{\bf \sigma}\cdot{\bf p}^{\prime}}{E+m} \end{array}\right](i/2)\left[\begin{array}{cc} 0 & -2\sigma_k \\ -2\sigma_k & 0 \end{array}\right] N \left[ \begin{array}{c} \chi_s \\ \frac{{\bf \sigma}\cdot{\bf p}}{E+m} \chi_s \end{array} \right]$
$= |N|^2(i/2) \chi_r^+ \left[\begin{array}{cc} -2\frac{{\bf \sigma}\cdot{\bf p}^{\prime}}{E+m}\sigma_k & -2\sigma_k \end{array}\right]\left[ \begin{array}{c} 1 \\ \frac{{\bf \sigma}\cdot{\bf p}}{E+m} \end{array} \right]\chi_s$
$= (E+m) (i/2)\chi_r^+ \left[ -2\frac{{\bf \sigma}\cdot{\bf p}^{\prime}}{E+m}\sigma_k -2\sigma_k\frac{{\bf \sigma}\cdot{\bf p}}{E+m} \right]\chi_s$
$= \chi_r^+(i/2) \left[-2{\bf \sigma}\cdot{\bf p}^{\prime}\sigma_k-2\sigma_k{\bf \sigma}\cdot{\bf p}\right]\chi_s$
$= (-i) \chi_r^+ \left[{\bf \sigma}\cdot{\bf p}^{\prime}\sigma_k - \sigma_k{\bf \sigma}\cdot{\bf p}\right]\chi_s$
$= (-i) \chi_r^+ \left[(\sigma_ip_i^{\prime}+\sigma_jp_j^{\prime}+\sigma_kp_k^{\prime})\sigma_k - \sigma_k(\sigma_ip_i+\sigma_jp_j+\sigma_kp_k)\right]\chi_s$
$= (-i) \chi_r^+ \left[\sigma_i\sigma_kp_i^{\prime}+\sigma_j\sigma_kp_j^{\prime}+\sigma_k\sigma_kp_k^{\prime} - \sigma_k\sigma_ip_i-\sigma_k\sigma_jp_j-\sigma_k\sigma_kp_k\right]\chi_s$
$= (-i) \chi_r^+ \left[-\sigma_jp_i^{\prime}+\sigma_ip_j^{\prime} + p_k^{\prime} - \sigma_jp_i +\sigma_ip_j - p_k\right]\chi_s$
$= (-i) \chi_r^+ \left[ p_k^{\prime} - p_k +\sigma_i(p_j^{\prime}+p_j) -\sigma_j(p_i^{\prime}+p_i)\right]\chi_s$
$= (-i) \chi_r^+ \left[ {\bf p}_k^{\prime} - {\bf p}_k + {\bf \sigma}\times ({\bf p}^{\prime} +{\bf p})\right]_k \chi_s$
あれ、ちょっと違うな。計算間違いしたかな？

$<p^{\prime}|\gamma^5|p>$
これも同じように計算。

[演習3.20]

$\bar{u}_1^T = [u_1^+\gamma_0]^T = \gamma_0^Tu_1^*$
$= \left[\begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array}\right] N \left[\begin{array}{c} \chi_1^* \\ \frac{{\bf \sigma}^*\cdot{\bf p}}{|E|+m}\chi_1^* \end{array}\right]$
$= N \left[\begin{array}{c} \chi_1^* \\ - \frac{{\bf \sigma}^*\cdot{\bf p}}{|E|+m}\chi_1^* \end{array}\right]$
ここで、
$({\bf \sigma}^*\cdot{\bf p})\chi_1^* = \left[ \left[\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}\right]p_1 + \left[\begin{array}{cc} 0 & i \\ -i & 0 \end{array}\right]p_2 + \left[\begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array}\right]p_3 \right] \left[ \begin{array}{c} 1 \\ 0 \end{array} \right]$
$= \left[ \begin{array}{c} p_3 \\ p_1 - ip_2 \end{array} \right]$
であるから、
$\bar{u}_1^T = N \left[ \begin{array}{c} 1 \\ 0 \\ -\frac{p_3}{|E|+m} \\ -\frac{p_1-ip_2}{|E|+m} \end{array} \right]$
となる。
また、
$-({\bf \sigma}\cdot{\bf p})\chi_2 = \left[ \left[\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}\right]p_1 + \left[\begin{array}{cc} 0 & -i \\ i & 0 \end{array}\right]p_2 + \left[\begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array}\right]p_3 \right] \left[ \begin{array}{c} 0 \\ 1 \end{array} \right]$
$= \left[ \begin{array}{c} p_1-ip_2 \\ -p_3 \end{array} \right]$
であるから、
$C\bar{u}_1^T = \left[\begin{array}{cccc} 0 & 0 & 0 & -1 \\ 0 & 0 & 1 & 0 \\ 0 & -1 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{array}\right] N \left[ \begin{array}{c} 1 \\ 0 \\ -\frac{p_3}{|E|+m} \\ -\frac{p_1-ip_2}{|E|+m} \end{array} \right]$
$= N \left[ \begin{array}{c} \frac{p_1-ip_2}{|E|+m} \\ -\frac{p_3}{|E|+m} \\ 0 \\ 1 \end{array} \right]$
$= u_4(-p) = v_1$

他も同様に計算できる。

＝＝式(3.101)(3.102)の説明＝＝
$N=\psi + C\bar{\psi}^T = \psi + C\gamma_0^T\psi^*$
$=\left[\begin{array}{c} \chi_r \\ \frac{{\bf \sigma}\cdot{\bf p}}{|E|+m}\chi_r \end{array}\right] + \left[\begin{array}{cc} 0 & -i\sigma_2 \\ -i\sigma_2 & 0 \end{array}\right]\left[\begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array}\right]\left[\begin{array}{c} \chi_r^* \\ \frac{{\bf \sigma}^*\cdot{\bf p}}{|E|+m}\chi_r^* \end{array}\right]$
$=\left[\begin{array}{c} \chi_r + i\sigma_2\frac{{\bf \sigma}^*\cdot{\bf p}}{|E|+m}\chi_r^* \\ -i\sigma_2\chi_r^* + \frac{{\bf \sigma}\cdot{\bf p}}{|E|+m}\chi_r \end{array}\right]$

これを変形すると、
$N_1 = \left[\begin{array}{c} \chi_r + i\sigma_2\frac{{\bf \sigma}^*\cdot{\bf p}}{|E|+m}\chi_r^* \\ -i\sigma_2\left( \chi_r + i\sigma_2\frac{{\bf \sigma}\cdot{\bf p}}{|E|+m}\chi_r \right)^* \end{array}\right]$
$N_2 = \left[\begin{array}{c} i\sigma_2\left( -i\sigma_2\chi_r^* + \frac{{\bf \sigma}\cdot{\bf p}}{|E|+m}\chi_r \right)^* \\ -i\sigma_2\chi_r^* + \frac{{\bf \sigma}\cdot{\bf p}}{|E|+m}\chi_r \end{array}\right]$

よって、
$\eta = \chi_r + i\sigma_2\frac{{\bf \sigma}^*\cdot{\bf p}}{|E|+m}\chi_r^*$
$\xi = -i\sigma_2\chi_r^* + \frac{{\bf \sigma}\cdot{\bf p}}{|E|+m}\chi_r$
とすればよい。

ワイル表示で、
$(\gamma^{\mu}i\partial_{\mu} - m) = \left[\begin{array}{cc} 0 & 1 \\ 1 & 0\end{array}\right]i\partial_0 + \left[\begin{array}{cc} 0 & i{\bf \sigma}\cdot{\bf \nabla} \\ -i{\bf \sigma}\cdot{\bf \nabla} & 0 \end{array}\right]- m\left[\begin{array}{cc} 1 & 0 \\ 0 & 1\end{array}\right]$
$=\left[\begin{array}{cc} -m & i\partial_0+i{\bf \sigma}\cdot{\bf \nabla} \\ i\partial_0-i{\bf \sigma}\cdot{\bf \nabla} \end{array} & -m\right]$
だから、
$(\gamma^{\mu}i\partial_{\mu} - m)N_1 = \left[\begin{array}{cc} -m & i\partial_0+i{\bf \sigma}\cdot{\bf \nabla} \\ i\partial_0-i{\bf \sigma}\cdot{\bf \nabla} \end{array} & -m\right]\left[\begin{array}{c} \eta \\ -i\sigma_2\eta^* \end{array}\right]$
$= \left[\begin{array}{c} i\partial_0(-i\sigma_2\eta^*)+i{\bf \sigma}\cdot{\bf \nabla}(-i\sigma_2\eta^*)-m\eta \\ i\partial_0\eta -i{\bf \sigma}\cdot{\bf \nabla}\eta + im\sigma_2\eta^* \end{array}\right] = 0$
$(\gamma^{\mu}i\partial_{\mu} - m)N_2 = \left[\begin{array}{cc} -m & i\partial_0+i{\bf \sigma}\cdot{\bf \nabla} \\ i\partial_0-i{\bf \sigma}\cdot{\bf \nabla} \end{array} & -m\right]\left[\begin{array}{c} i\sigma\xi^* \\ \xi \end{array}\right]$
$= \left[\begin{array}{c} i\partial_0\xi+i{\bf \sigma}\cdot{\bf \nabla}\xi-m(i\sigma_2\xi^*) \\ i\partial_0(i\sigma_2\xi^*) -i{\bf \sigma}\cdot{\bf \nabla}(i\sigma_2\xi^*) - m\xi \end{array}\right] = 0$
＝＝＝＝＝＝＝＝＝＝＝

[演習3.21]
(a) 上の説明の最後の式をちょちょいと変形するとわかる。
(b) よくわかりません。

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最終更新：2014年03月29日 16:52

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3 ディラック方程式(SKI)