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Now I will calculate the amplitude for the decay $\pi^{+}\rightarrow l^{+}\nu$ at leading order. The effective Lagrangian for semileptonic weak interaction is \begin{eqnarray*} \Delta\mathcal{L} & = & -\frac{4G_{F}}{\sqrt{2}}\bigg((\bar{l}_{L}\gamma^{\mu}\nu_{L})(\bar{u}_{L}\gamma_{\mu}d_{L})+(\bar{\nu}_{L}\gamma^{\mu}l_{L})(\bar{d}_{L}\gamma_{\mu}u_{L})\bigg) \end{eqnarray*}

With \begin{eqnarray*} & & j^{\mu a}=\bar{Q}\gamma^{\mu}\tau^{a}Q,\ \ j^{\mu5a}=\bar{Q}\gamma^{\mu}\gamma^{5}\tau^{a}Q,\ \ Q=\left(\begin{array}{cc} u & d\end{array}\right)^{T}\\ & & j^{\mu-}=j^{\mu1}-ij^{\mu2}=\bar{Q}\gamma^{\mu}\tau^{-}Q,\ \ j^{\mu5-}=j^{\mu51}-ij^{\mu52}=\bar{Q}\gamma^{\mu}\gamma^{5}\tau^{-}Q \end{eqnarray*}

we can see \begin{eqnarray*} j^{\mu-}-j^{\mu5-} & = & \bar{Q}\gamma^{\mu}(1-\gamma^{5})\tau^{-}Q\\ & = & 2\bar{Q}_{L}\gamma^{\mu}\tau^{-}Q_{L}\\ & = & 2\left(\begin{array}{cc} \bar{u}_{L} & \bar{d}_{L}\end{array}\right)\gamma^{\mu}\left(\begin{array}{cc} 0 & 0\\ 1 & 0 \end{array}\right)\left(\begin{array}{c} u_{L}\\ d_{L} \end{array}\right)\\ & = & 2\bar{d}_{L}\gamma^{\mu}u_{L} \end{eqnarray*}

i.e. \begin{eqnarray*} & & \bar{d}_{L}\gamma^{\mu}u_{L}=\frac{1}{2}(j^{\mu-}-j^{\mu5-}) \end{eqnarray*}

The decay amplitude is

\begin{eqnarray*} & & i\mathcal{M}[\pi^{+}(p)\rightarrow l^{+}(k)+\nu(q)]\cdot(2\pi)^{4}\delta(p-k-q)\\ & = & \langle l^{+}(k)\nu(q)|i\int d^{4}x\Delta\mathcal{L}|\pi^{+}(p)\rangle\\ & = & -\frac{4iG_{F}}{\sqrt{2}}\int d^{4}x\langle l^{+}(k)\nu(q)|(\bar{\nu}_{L}\gamma^{\mu}l_{L})(\bar{d}_{L}\gamma_{\mu}u_{L})|\pi^{+}(p)\rangle\\ & = & -\frac{4iG_{F}}{\sqrt{2}}\int d^{4}x\langle l^{+}(k)\nu(q)|[\bar{\nu}\gamma^{\mu}\frac{1}{2}(1-\gamma^{5})l]\frac{1}{2}(j^{\mu-}-j^{\mu5-})|\pi^{+}(p)\rangle\\ & = & \frac{iG_{F}}{\sqrt{2}}\int d^{4}x\langle l^{+}(k)\nu(q)|\bar{\nu}\gamma^{\mu}(1-\gamma^{5})l|0\rangle\cdot\langle0|j^{\mu5-}|\pi^{+}(p)\rangle\\ & = & \frac{iG_{F}}{\sqrt{2}}\int d^{4}x\bigg(\bar{u}(q)e^{iq\cdot x}\gamma^{\mu}(1-\gamma^{5})v(k)e^{ik\cdot x}\bigg)\cdot\langle0|j^{\mu5-}|\pi^{+}(p)\rangle \end{eqnarray*}

In order to obtain $i\mathcal{M}$, we must calculate$\langle0|j^{\mu5-}|\pi^{+}(p)\rangle$. The formula that could be used is list below,

\begin{eqnarray*} & & \langle0|j^{\mu5a}(x)|\pi^{b}(p)\rangle=-ip^{\mu}f_{\pi}\delta^{ab}e^{-ip\cdot x} \end{eqnarray*}

My question is: How to calculate $\langle0|j^{\mu5-}|\pi^{+}(p)\rangle$? The following is my calculation, which I do not know whether it is right or not.

\begin{eqnarray*} \langle0|j^{\mu5-}(x)|\pi^{+}(p)\rangle & = & \frac{1}{\sqrt{2}}\langle0|[j^{\mu51}(x)-ij^{\mu52}(x)]|[\pi^{1}(p)+i\pi^{2}(p)]\rangle\\ & = & \frac{1}{\sqrt{2}}\langle0|j^{\mu51}(x)|\pi^{1}(p)\rangle+\frac{1}{\sqrt{2}}\langle0|j^{\mu52}(x)|\pi^{2}(p)\rangle\\ & = & -i\sqrt{2}p^{\mu}f_{\pi}e^{-ip\cdot x} \end{eqnarray*}

Is the decomposition of $|\pi^{+}(p)\rangle$ into $|\pi^{1}(p)\rangle+i|\pi^{2}(p)\rangle$ right?

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  • $\begingroup$ Good question, but it's kind of long. If you do wind up editing the question for some other reason, while you're at it, perhaps you could pare down the math a bit to focus on the core of what you want to ask? I think that would help readers understand the question better. (But again, it seems okay in its current form too.) $\endgroup$
    – David Z
    Nov 1, 2016 at 21:48
  • $\begingroup$ How did $|0><0|$ appear in the amplitude derivation? Seems as if you used a decomposition over a state basis, but then where is the sum? $\endgroup$ Nov 27, 2016 at 1:00
  • $\begingroup$ Looks OK, assuming you have been mindful of normalizations. $\endgroup$ Dec 4, 2016 at 20:36

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