- #1
Thor Shen
- 17
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Note from mfb: I fixed the LaTeX formulas
http://image.keyan.cc/data/bcs/2015/0428/w84h1446064_1430226899_628.jpg
I met some trouble by using latex on this version, so I try to add the image from the pdf,
the Latex code is as follow if any help.
The lowest order lagrangian for ChPT is
[itex]\mathcal{L}^{(2)}=\frac{f^2}{12}\langle(\partial\Phi \Phi-\Phi\partial\Phi)^2+M\Phi^4\rangle[/itex]
where the isospin limit is assumed and the $\Phi$ is made of the pseudo Goldstone boson fields,
$$\begin{pmatrix}
\frac{\pi^0}{\sqrt{2}}+\frac{\eta}{\sqrt{6}} & \pi^+ & K^+\\
\pi^-& \frac{-\pi^0}{\sqrt{2}}+\frac{\eta}{\sqrt{6}} & K^0\\
K^-& \overline{K}^0&\frac{-2\eta}{\sqrt{6}}\\
\end{pmatrix}$$Let us consider the physical reaction $$I=0~\pi^+(\vec{k})\pi^-(\vec{p})\leftrightarrow\pi^+(\vec{k'})\pi^-(\vec{p'})$$,where the k,p,k' and p' are the momentums of the pion, respectively. The relevant lagrangian can be derived from the above formula,
$$\begin{eqnarray}\nonumber
\mathcal{L}_{\pi\pi\pi\pi}&=&2\partial\pi^+\partial\pi^+\pi^-\pi^- -2\partial\pi^+\partial\pi^-\pi^+\pi^--2\partial\pi^-\partial\pi^+\pi^+\pi^- \\ \nonumber
&&+2\partial\pi^-\partial\pi^-\pi^+\pi^+ +2m^2_{\pi}\pi^+\pi^+\pi^-\pi^-
\end{eqnarray}$$
and the amplitudes of this reaction,
$$\begin{eqnarray}\nonumber
V(s,t,u)&=&\langle\pi^+\pi^-|H|\pi^+\pi^-\rangle \\ \nonumber
&=&-\langle\pi^+\pi^-|\mathcal{L}|\pi^+\pi^-\rangle \\
=-\frac{1}{6f^2}&\{&\langle\pi^+(k')\pi^-(p')|\partial\pi^+\partial\pi^+\pi^-\pi^-|\pi^+(k)\pi^-(p)\rangle\\
&-&\langle\pi^+(k')\pi^-(p')|\partial\pi^+\partial\pi^-\pi^+\pi^-|\pi^+(k)\pi^-(p)\rangle\\
&-&\langle\pi^+(k')\pi^-(p')|\partial\pi^-\partial\pi^+\pi^+\pi^-|\pi^+(k)\pi^-(p)\rangle\\
&+&\langle\pi^+(k')\pi^-(p')|\partial\pi^-\partial\pi^-\pi^+\pi^+|\pi^+(k)\pi^-(p)\rangle\\
&+&\langle\pi^+(k')\pi^-(p')|m^2_{\pi}\pi^+\pi^+\pi^-\pi^-|\pi^+(k)\pi^-(p)\rangle\}.
\end{eqnarray}$$
Finally, the result can be expressed as $$-\frac{1}{3f^2}(s+t-2u+2m^2_{\pi})$$ with the Mandelstam variables $$s=(k+p)^2$,$t=(k-k')^2$$ and $$u=(k-p')^2$$, that is $$-\frac{1}{6f^2}(2(k-k')^2-2(k-p')^2-2(k-p')^2+2(k+p)^2+4m^2_{\pi})$$. Then how to get the answers by the five amplitudes, respectively?
http://image.keyan.cc/data/bcs/2015/0428/w84h1446064_1430226899_628.jpg
I met some trouble by using latex on this version, so I try to add the image from the pdf,
the Latex code is as follow if any help.
The lowest order lagrangian for ChPT is
[itex]\mathcal{L}^{(2)}=\frac{f^2}{12}\langle(\partial\Phi \Phi-\Phi\partial\Phi)^2+M\Phi^4\rangle[/itex]
where the isospin limit is assumed and the $\Phi$ is made of the pseudo Goldstone boson fields,
$$\begin{pmatrix}
\frac{\pi^0}{\sqrt{2}}+\frac{\eta}{\sqrt{6}} & \pi^+ & K^+\\
\pi^-& \frac{-\pi^0}{\sqrt{2}}+\frac{\eta}{\sqrt{6}} & K^0\\
K^-& \overline{K}^0&\frac{-2\eta}{\sqrt{6}}\\
\end{pmatrix}$$Let us consider the physical reaction $$I=0~\pi^+(\vec{k})\pi^-(\vec{p})\leftrightarrow\pi^+(\vec{k'})\pi^-(\vec{p'})$$,where the k,p,k' and p' are the momentums of the pion, respectively. The relevant lagrangian can be derived from the above formula,
$$\begin{eqnarray}\nonumber
\mathcal{L}_{\pi\pi\pi\pi}&=&2\partial\pi^+\partial\pi^+\pi^-\pi^- -2\partial\pi^+\partial\pi^-\pi^+\pi^--2\partial\pi^-\partial\pi^+\pi^+\pi^- \\ \nonumber
&&+2\partial\pi^-\partial\pi^-\pi^+\pi^+ +2m^2_{\pi}\pi^+\pi^+\pi^-\pi^-
\end{eqnarray}$$
and the amplitudes of this reaction,
$$\begin{eqnarray}\nonumber
V(s,t,u)&=&\langle\pi^+\pi^-|H|\pi^+\pi^-\rangle \\ \nonumber
&=&-\langle\pi^+\pi^-|\mathcal{L}|\pi^+\pi^-\rangle \\
=-\frac{1}{6f^2}&\{&\langle\pi^+(k')\pi^-(p')|\partial\pi^+\partial\pi^+\pi^-\pi^-|\pi^+(k)\pi^-(p)\rangle\\
&-&\langle\pi^+(k')\pi^-(p')|\partial\pi^+\partial\pi^-\pi^+\pi^-|\pi^+(k)\pi^-(p)\rangle\\
&-&\langle\pi^+(k')\pi^-(p')|\partial\pi^-\partial\pi^+\pi^+\pi^-|\pi^+(k)\pi^-(p)\rangle\\
&+&\langle\pi^+(k')\pi^-(p')|\partial\pi^-\partial\pi^-\pi^+\pi^+|\pi^+(k)\pi^-(p)\rangle\\
&+&\langle\pi^+(k')\pi^-(p')|m^2_{\pi}\pi^+\pi^+\pi^-\pi^-|\pi^+(k)\pi^-(p)\rangle\}.
\end{eqnarray}$$
Finally, the result can be expressed as $$-\frac{1}{3f^2}(s+t-2u+2m^2_{\pi})$$ with the Mandelstam variables $$s=(k+p)^2$,$t=(k-k')^2$$ and $$u=(k-p')^2$$, that is $$-\frac{1}{6f^2}(2(k-k')^2-2(k-p')^2-2(k-p')^2+2(k+p)^2+4m^2_{\pi})$$. Then how to get the answers by the five amplitudes, respectively?
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