三浦と窮理とブログ

主に物理学・数学について自分が勉強してきたことについて書いていきます.誰かの役に立ってくれれば嬉しいです.

ニュートリノ・電子散乱の荷電カレントのフィルツ変換の導出

次の教科書の11.2節より,SU(2)×U(1) 電弱理論における,電子ニュートリノ $\nu _ e$・電子 $e$ の弾性散乱を考える.

Gauge Theory of Elementary Particle Physics (Oxford Science Publications)

Gauge Theory of Elementary Particle Physics (Oxford Science Publications)

散乱振幅 $T(\nu _ e e \to \nu _ e e) $において,荷電カレント相互作用の寄与は以下で表される.

\begin{equation} \frac{g^2}{8M_W^2} [\bar{\nu}_e\gamma_\mu (1-\gamma_5)e][\bar{e}\gamma_\mu (1-\gamma_5)\nu_e] \end{equation}

これがフィルツ変換により,

\begin{equation} \frac{g^2}{8M_W^2} [\bar{\nu}_e\gamma_\mu (1-\gamma_5)\nu_e][\bar{e}\gamma_\mu (1-\gamma_5)e] \end{equation}

に式変形できることを示す.

フィルツ変換による式変形

これは,スピノル添え字の組み換え

\begin{equation} \label{eq:354a} [\gamma _ \mu (1-\gamma _ 5)] _ {\alpha\beta} [\gamma ^ \mu (1-\gamma _ 5)] _ {\delta \epsilon} =- [\gamma _ \mu (1-\gamma _ 5)] _ {\alpha \epsilon} [\gamma ^ \mu (1-\gamma _ 5)] _ {\delta\beta } \end{equation}

と,フェルミオンの反可換性 $ e ^ \beta \bar{e} ^ \delta \nu _ e ^ \epsilon=- \nu _ e ^ \epsilon \bar{e} ^ \delta e ^ \beta $ より成り立つことが分かる.

\eqref{eq:354a}の導出

フィルツ変換そのものの導出手順の把握も兼ねて,2通りの方法で導こう.

直接フィルツ変換をする場合

Fierz変換の導出手順を直接適用する..

$ [\gamma _ \mu (1-\gamma _ 5)] _ {\alpha\beta} [\gamma ^ \mu (1-\gamma _ 5)] _ {\delta \epsilon} =: (\Gamma _ {\alpha\epsilon}) _ {\delta\beta} $を,添え字$ \delta,\beta $の$ 4\times 4 $行列16個が$ \alpha\epsilon $でラベル付けられているとして見る. $ \{\Gamma _ i\} _ i =\{I,\gamma ^ \mu,\sigma ^ {\mu\nu},\gamma ^ \mu\gamma _ 5,\gamma _ 5\} ,(i=S,V,T,A,P)$は$ 4\times 4 $行列の完全系を成すので,

\begin{align} (\Gamma _ {\alpha\epsilon}) _ {\delta\beta} &= \hat{g} _ S (I) _ {\delta\beta} + (\hat{g} _ V) _ \mu (\gamma ^ \mu) _ {\delta\beta} + (\hat{g} _ T) _ {\mu\nu} (\sigma ^ {\mu\nu}) _ {\delta\beta} + (\hat{g} _ A) _ \mu(\gamma ^ \mu\gamma _ 5) _ {\delta\beta} + \hat{g} _ P (\gamma _ 5) _ {\delta\beta} \label{eq:345firez} \\ \hat{g} _ i &= \frac{1}{\mathrm{tr} (\Gamma _ i \Gamma _ i)} \mathrm{tr} (\Gamma _ {\alpha\epsilon} \Gamma _ i) = \frac{1}{\mathrm{tr} (\Gamma _ i \Gamma _ i)} (\Gamma _ {\alpha\epsilon}) _ {\delta\beta} (\Gamma _ i) _ {\beta\delta} \end{align}

と展開できる.分母の$ \mathrm{tr} (\Gamma _ i \Gamma _ i) $は$ i $について和を取ってないことに注意.

\begin{align} \mathrm{tr} (\Gamma _ 0 \Gamma _ 0) &= 4 \\ \mathrm{tr} (\Gamma _ i \Gamma _ i) &=-4 \quad (i=1,2,3) \\ \frac{\gamma ^ \mu}{\mathrm{tr}(\gamma ^ \mu \gamma ^ \mu) } &= (\frac{\gamma ^ 0}{\mathrm{tr}(\gamma ^ 0 \gamma ^ 0) }, \frac{\gamma ^ i}{\mathrm{tr}(\gamma ^ i \gamma ^ i) }) = \frac{1}{4}(\gamma ^ 0,-\gamma ^ i) =\frac{\gamma _ \mu}{4} \\ \sigma ^ {\mu\nu} &:= \frac{i}{2}[\gamma ^ \mu,\gamma ^ \nu]= \frac{i}{2} (2\gamma ^ \mu\gamma ^ \nu - \{\gamma ^ \mu,\gamma ^ \nu\}) =i(\gamma ^ \mu\gamma ^ \nu - g ^ {\mu\nu}) \\ \gamma _ \nu\gamma ^ \mu\gamma ^ \nu &= \gamma _ \nu(2g ^ {\mu\nu}-\gamma ^ \nu\gamma ^ \mu) =-2 \gamma ^ \mu \end{align}

を後に使う.

\begin{align} \hat{g} _ S &= \frac{1}{\mathrm{tr} (II)} (\Gamma _ {\alpha\epsilon}) _ {\delta\beta} (I) _ {\beta\delta} = \frac{1}{4} [\gamma _ \mu (1-\gamma _ 5)] _ {\alpha\beta} [\gamma ^ \mu (1-\gamma _ 5)] _ {\delta \epsilon} (I) _ {\beta\delta} = \frac{1}{4} [\gamma _ \mu (1-\gamma _ 5)\gamma ^ \mu (1-\gamma _ 5)] _ {\alpha \epsilon} \\ &= \frac{1}{4} [\gamma _ \mu\gamma ^ \mu - \gamma _ \mu\gamma ^ \mu\gamma _ 5 -\gamma _ \mu\gamma _ 5\gamma ^ \mu + \gamma _ \mu\gamma _ 5\gamma ^ \mu \gamma _ 5] _ {\alpha \epsilon} =0 \\ (\hat{g} _ V) _ \mu &= \frac{1}{\mathrm{tr} (\gamma ^ \mu\gamma ^ \mu)} (\Gamma _ {\alpha\epsilon}) _ {\delta\beta} (\gamma ^ \mu) _ {\beta\delta} = \frac{1}{\mathrm{tr} (\gamma ^ \mu\gamma ^ \mu)} [\gamma _ \nu (1-\gamma _ 5)\gamma ^ \mu\gamma ^ \nu (1-\gamma _ 5)] _ {\alpha \epsilon} \\ &= \frac{1}{\mathrm{tr} (\gamma ^ \mu\gamma ^ \mu)} [\gamma _ \nu\gamma ^ \mu\gamma ^ \nu - \gamma _ \nu\gamma ^ \mu\gamma ^ \nu\gamma _ 5 -\underbrace{\gamma _ \nu\gamma _ 5\gamma ^ \mu\gamma ^ \nu} _ {=\gamma _ \nu\gamma ^ \mu\gamma ^ \nu\gamma _ 5} + \underbrace{\gamma _ \nu\gamma _ 5\gamma ^ \mu\gamma ^ \nu \gamma _ 5} _ {=\gamma _ \nu\gamma ^ \mu\gamma ^ \nu}] _ {\alpha \epsilon} \\ &= \frac{2}{\mathrm{tr} (\gamma ^ \mu\gamma ^ \mu)} [\gamma _ \nu\gamma ^ \mu\gamma ^ \nu - \gamma _ \nu\gamma ^ \mu\gamma ^ \nu\gamma _ 5 ] _ {\alpha \epsilon} = \frac{-4}{\mathrm{tr} (\gamma ^ \mu\gamma ^ \mu)} [\gamma ^ \mu (1- \gamma _ 5) ] _ {\alpha \epsilon} \\ &= - [\gamma _ \mu (1- \gamma _ 5) ] _ {\alpha \epsilon} \\ (\hat{g} _ T) _ {\mu\nu} &= \frac{1}{\mathrm{tr} (\sigma ^ {\mu\nu}\sigma ^ {\mu\nu})}(\Gamma _ {\alpha\epsilon}) _ {\delta\beta} (\sigma ^ {\mu\nu}) _ {\beta\delta} = \frac{i}{\mathrm{tr} (\sigma ^ {\mu\nu}\sigma ^ {\mu\nu})} [\gamma _ \lambda (1-\gamma _ 5)(\gamma ^ \mu\gamma ^ \nu - g ^ {\mu\nu})\gamma ^ \lambda (1-\gamma _ 5)] _ {\alpha \epsilon} \\ &= \frac{i}{\mathrm{tr} (\sigma ^ {\mu\nu}\sigma ^ {\mu\nu})} [\gamma _ \lambda (1-\gamma _ 5)\gamma ^ \mu\gamma ^ \nu\gamma ^ \lambda (1-\gamma _ 5) - g ^ {\mu\nu}\underbrace{\gamma _ \lambda (1-\gamma _ 5)\gamma ^ \lambda (1-\gamma _ 5)} _ {=0}] _ {\alpha \epsilon} \\ &= \frac{i}{\mathrm{tr} (\sigma ^ {\mu\nu}\sigma ^ {\mu\nu})} [\gamma _ \lambda\gamma ^ \mu\gamma ^ \nu\gamma ^ \lambda - \gamma _ \lambda\gamma ^ \mu\gamma ^ \nu\gamma ^ \lambda\gamma _ 5 -\gamma _ \lambda\gamma _ 5\gamma ^ \mu\gamma ^ \nu\gamma ^ \lambda + \gamma _ \lambda\gamma _ 5\gamma ^ \mu\gamma ^ \nu\gamma ^ \lambda \gamma _ 5] _ {\alpha \epsilon} \\ &=0 \\ (\hat{g} _ A) _ \mu &= \frac{1}{\mathrm{tr} (\gamma ^ \mu\gamma _ 5\gamma ^ \mu\gamma _ 5)} (\Gamma _ {\alpha\epsilon}) _ {\delta\beta} (\gamma ^ \mu\gamma _ 5) _ {\beta\delta} = -\frac{1}{\mathrm{tr} (\gamma ^ \mu\gamma ^ \mu)} [\gamma _ \nu (1-\gamma _ 5)\gamma ^ \mu\underbrace{\gamma _ 5\gamma ^ \nu (1-\gamma _ 5)} _ {=\gamma ^ \nu (1-\gamma _ 5)}] _ {\alpha \epsilon} \\ &= [\gamma _ \mu (1- \gamma _ 5) ] _ {\alpha \epsilon} \\ \hat{g} _ P &= \frac{1}{\mathrm{tr} (\gamma _ 5\gamma _ 5)} (\Gamma _ {\alpha\epsilon}) _ {\delta\beta} (\gamma _ 5) _ {\beta\delta} = \frac{1}{\mathrm{tr} (\gamma _ 5\gamma _ 5)} [\gamma _ \mu (1-\gamma _ 5)\underbrace{\gamma _ 5\gamma ^ \mu (1-\gamma _ 5)} _ {=\gamma ^ \mu (1-\gamma _ 5)}] _ {\alpha \epsilon} =0 \end{align}

よって,式\eqref{eq:345firez}は,

\begin{align} [\gamma _ \mu (1-\gamma _ 5)] _ {\alpha\beta} [\gamma ^ \mu (1-\gamma _ 5)] _ {\delta \epsilon} &= (\hat{g} _ V) _ \mu (\gamma ^ \mu) _ {\delta\beta} + (\hat{g} _ A) _ \mu(\gamma ^ \mu\gamma _ 5) _ {\delta\beta} \\ &= -[\gamma ^ \mu (1- \gamma _ 5) ] _ {\alpha \epsilon} [ \gamma ^ \mu (1- \gamma _ 5)] _ {\delta\beta} \end{align}

となる. //

付録Aのフィルツ変換公式を適用する場合

付録(A.19)に特別な場合についてのフィルツ変換公式があるので,それを適用する.

\begin{equation} (\eqref{eq:354a}の左辺)= ( \gamma _ \mu ) _ {\alpha\beta}( \gamma ^ \mu ) _ {\delta \epsilon}-( \gamma _ \mu ) _ {\alpha\beta}(\gamma ^ \mu \gamma _ 5 ) _ {\delta \epsilon}- ( \gamma _ \mu\gamma _ 5 ) _ {\alpha\beta}( \gamma ^ \mu ) _ {\delta \epsilon} +( \gamma _ \mu\gamma _ 5 ) _ {\alpha\beta}( \gamma ^ \mu \gamma _ 5) _ {\delta \epsilon} \end{equation}

上の式において,

\begin{align} (\gamma _ \mu) _ {\alpha\beta} (\gamma ^ \mu) _ {\delta \epsilon} &+ (\gamma _ \mu\gamma _ 5) _ {\alpha\beta} (\gamma ^ \mu \gamma _ 5) _ {\delta \epsilon} = - (\gamma _ \mu) _ {\alpha\epsilon} (\gamma ^ \mu) _ {\delta \beta} - (\gamma _ \mu\gamma _ 5) _ {\alpha\epsilon} (\gamma ^ \mu \gamma _ 5) _ {\delta \beta} \label{eq:cheng354gg} \\ ( \gamma _ \mu ) _ {\alpha\beta}(\gamma ^ \mu \gamma _ 5 ) _ {\delta \epsilon} &= ( \gamma _ \mu ) _ {\alpha\beta}(\gamma ^ \mu) _ {\delta\sigma}( \gamma _ 5 ) _ {\sigma \epsilon} \\ &=[(I) _ {\alpha\sigma}(I) _ {\delta\beta}-\frac{1}{2}(\gamma _ \mu) _ {\alpha\sigma}(\gamma ^ \mu) _ {\delta\beta} - \frac{1}{2}(\gamma _ \mu\gamma _ 5) _ {\alpha\sigma}(\gamma ^ \mu\gamma _ 5) _ {\delta\beta} - (\gamma _ 5) _ {\alpha\sigma}(\gamma _ 5) _ {\delta\beta}]( \gamma _ 5 ) _ {\sigma \epsilon} \label{eq:cheng354g5} \\ & = (\gamma _ 5 ) _ {\alpha\epsilon}(I) _ {\delta\beta}-\frac{1}{2}(\gamma _ \mu\gamma _ 5 ) _ {\alpha\epsilon}(\gamma ^ \mu) _ {\delta\beta} - \frac{1}{2}(\gamma _ \mu) _ {\alpha\epsilon}(\gamma ^ \mu\gamma _ 5) _ {\delta\beta} - (I) _ {\alpha\epsilon}(\gamma _ 5) _ {\delta\beta} \\ ( \gamma _ \mu\gamma _ 5 ) _ {\alpha\beta}( \gamma ^ \mu ) _ {\delta \epsilon} &= ( \gamma _ \mu ) _ {\alpha\sigma}( \gamma _ 5 ) _ {\sigma \beta} (\gamma ^ \mu) _ {\delta\epsilon} = ( \gamma _ \mu ) _ {\alpha\sigma} (\gamma ^ \mu) _ {\delta\epsilon}( \gamma _ 5 ) _ {\sigma \beta} \\ &=[(I) _ {\alpha\epsilon}(I) _ {\delta\sigma}-\frac{1}{2}(\gamma _ \mu) _ {\alpha\epsilon}(\gamma ^ \mu) _ {\delta\sigma} - \frac{1}{2}(\gamma _ \mu\gamma _ 5) _ {\alpha\epsilon}(\gamma ^ \mu\gamma _ 5) _ {\delta\sigma} - (\gamma _ 5) _ {\alpha\epsilon}(\gamma _ 5) _ {\delta\sigma}]( \gamma _ 5 ) _ {\sigma \beta} \\ & = (I ) _ {\alpha\epsilon}(\gamma _ 5) _ {\delta\beta}-\frac{1}{2}(\gamma _ \mu ) _ {\alpha\epsilon}(\gamma ^ \mu\gamma _ 5) _ {\delta\beta} - \frac{1}{2}(\gamma _ \mu\gamma _ 5) _ {\alpha\epsilon}(\gamma ^ \mu) _ {\delta\beta} - (\gamma _ 5) _ {\alpha\epsilon}(I) _ {\delta\beta} \\ ( \gamma _ \mu ) _ {\alpha\beta}(\gamma ^ \mu \gamma _ 5 ) _ {\delta \epsilon} &+ ( \gamma _ \mu\gamma _ 5 ) _ {\alpha\beta}( \gamma ^ \mu ) _ {\delta \epsilon} = - (\gamma _ \mu) _ {\alpha\epsilon}(\gamma ^ \mu\gamma _ 5) _ {\delta\beta} -(\gamma _ \mu\gamma _ 5 ) _ {\alpha\epsilon}(\gamma ^ \mu) _ {\delta\beta} \end{align}

である.式\eqref{eq:cheng354gg}は,フィルツ変換(A.19)より,

\begin{equation} \begin{pmatrix} \hat{g} _ S\\ \hat{g} _ V\\ \hat{g} _ T\\ \hat{g} _ A\\ \hat{g} _ P \end{pmatrix} =\frac{1}{4} \begin{pmatrix}1 & 4 & 12 & -4 & 1\\ 1 & -2 & 0 & -2 & -1\\ \frac{1}{2} & 0 & -2 & 0 & \frac{1}{2}\\ -1 & -2 & 0 & -2 & 1\\ 1 & -4 & 12 & 4 & 1\end{pmatrix} \begin{pmatrix} 0\\ 1\\ 0\\ 1\\ 0 \end{pmatrix} = \begin{pmatrix} 0\\ -1\\ 0\\ -1\\ 0 \end{pmatrix} \end{equation}

を用い,式\eqref{eq:cheng354g5}も

\begin{equation} \begin{pmatrix} \hat{g} _ S\\ \hat{g} _ V\\ \hat{g} _ T\\ \hat{g} _ A\\ \hat{g} _ P \end{pmatrix} =\frac{1}{4} \begin{pmatrix}1 & 4 & 12 & -4 & 1\\ 1 & -2 & 0 & -2 & -1\\ \frac{1}{2} & 0 & -2 & 0 & \frac{1}{2}\\ -1 & -2 & 0 & -2 & 1\\ 1 & -4 & 12 & 4 & 1\end{pmatrix} \begin{pmatrix} 0\\ 1\\ 0\\ 0\\ 0 \end{pmatrix} = \begin{pmatrix} 1\\ -\frac{1}{2}\\ 0\\ -\frac{1}{2}\\ -1 \end{pmatrix} \end{equation}

を用いた.