Research Paper Notes on Perturbative QCD

Research Paper Notes on Perturbative QCD

本文档除了包括推导, 疑惑 外,做 读书重点 的记录 $$$$

文献列表

 * Factorization of Hard Processes in QCD, arXiv:hep-ph/0409313v1, by John C. Collins, Davison E. Soper, George Sterman
 * New macroscopic forces, Phys. Rev. D 30 (1984) 130, J. E. Moody and Frank Wilczek

Factorization of Hard Processes in QCD, arXiv:hep-ph/0409313v1, by John C. Collins, Davison E. Soper, George Sterman
这是介绍标准微扰QCD因子化的综述文章.

本文档除了包括推导, 疑惑 外,做 读书重点 的记录 $$$$

New macroscopic forces, Phys. Rev. D 30 (1984) 130, J. E. Moody and Frank Wilczek
(1-2)

这两个是标量和赝标量相互作用恒等式,等式右边便于在非相对论极限下获得相应的表达式.

注意到这里仅仅考虑正粒子间的相互作用,而不考虑正负粒子湮灭的情况.同时初末态粒子四动量$$p_i, p_f$$都在质壳上.这样利用定义$$\sigma^{\mu\nu}=\frac{i}{2}[\gamma^\mu,\gamma^\nu]$$,$$\vec{\Sigma}=\begin{pmatrix}\vec{\sigma}&0\\0&\vec{\sigma}\end{pmatrix}$$,并利用$$\gamma$$函数的关系,如Mandl一书附录(A.19a)即$$\not{A}\not{B}=AB-i\sigma^{\alpha\beta}A_\alpha B_\beta=2AB-\not{B}\not{A}$$,(A.19b)即$$\not{A}\not{A}=A^2$$,和(A.25-26)中对质壳上的正粒子的关系$$\not{p}_iu_r(p_i)=M u_r(p_i), \bar{u}_r(p_f)\not{p}_f=\bar{u}_r(p_f) M$$,和定义$$p^\mu=\frac12(p^\mu_i+p^\mu_f), q^\mu=p^\mu_f-p^\mu_i$$,我们有:


 * $$\frac{p^\mu p_\mu}{M^2}-i\frac{p_\mu q_\nu}{2M^2}\sigma^{\mu\nu}=\frac{p_{\mu i}+p_{\mu f}}{2}\frac{p_{\nu i}+p_{\nu f}}{2}\frac{g^{\mu\nu}}{M^2}-i\frac{p_{\mu i}+p_{\mu f}}{2}(p_{\mu f}-p_{\mu f})\frac{\sigma^{\mu\nu}}{2M^2}=\frac{M^2+M^2+2p_{\mu i}p_{\nu f}g^{\mu\nu}}{4M^2}+\frac{-i\sigma^{\mu\nu}(p_{\mu i}p_{\nu f}-p_{\mu f}p_{\nu i}-p_{\mu i}p_{\nu i}+p_{\mu f}p_{\nu f})}{4M^2}=\frac{M^2+M^2+2p_{\mu i}p_{\nu f}g^{\mu\nu}}{4M^2}+\frac{-i\sigma^{\mu\nu}(p_{\mu i}p_{\nu f}-p_{\mu f}p_{\nu i})}{4M^2}$$
 * $$=\frac{M^2+M^2+2p_{\mu i}p_{\nu f}g^{\mu\nu}}{4M^2}+\frac{-i\sigma^{\mu\nu}p_{\mu i}p_{\nu f}+i\sigma^{\mu\nu}p_{\mu f}p_{\nu i}}{4M^2}

=\frac{M^2+M^2+2p_{\mu i}p_{\nu f}g^{\mu\nu}}{4M^2}+\frac{-i\sigma^{\mu\nu}p_{\mu i}p_{\nu f}+i\sigma^{\nu\mu}p_{\nu f}p_{\mu i}}{4M^2} =\frac{M^2+M^2+2p_{\mu i}p_{\nu f}g^{\mu\nu}}{4M^2}+\frac{-i\sigma^{\mu\nu}p_{\mu i}p_{\nu f}+i\sigma^{\nu\mu}p_{\mu i}p_{\nu f}}{4M^2} =\frac{M^2+M^2+2p_{\mu i}p_{\nu f}g^{\mu\nu}}{4M^2}+\frac{-i\sigma^{\mu\nu}p_{\mu i}p_{\nu f}-i\sigma^{\mu\nu}p_{\mu i}p_{\nu f}}{4M^2}$$
 * $$=\frac{M^2+M^2+2(p_{\mu i}p_{\nu f}g^{\mu\nu}-i\sigma^{\mu\nu}p_{\mu i}p_{\nu f})}{4M^2}

=\frac{2M^2+2\not{p}_{\mu i}\not{p}_{\nu f}}{4M^2}=\frac{M^2+\not{p}_{\mu i}\not{p}_{\nu f}}{2M^2}\to 1$$ 其中最后一步利用了动量在质壳上和正粒子的运动方程.这就是(1)的结果.

类似的 $$\frac{i\gamma_5(\not{p}_f-\not{p}_i)}{2M}=\frac{-i\gamma_5\not{p}_f-i\gamma_5\not{p}_i}{2M}$$ 其中利用了$$\gamma$$矩阵对易关系(A.8)即$$[\gamma^\mu,\gamma^5]_+=0$$,而交换位置是为了把$$\not{p}_{f,i}$$作用到适当动量的波函数上去.同时注意到$$q^\mu$$只有空间分量时
 * $$\gamma^0\gamma^5\gamma_\mu=\begin{pmatrix}0&1\\-1&0\end{pmatrix} \begin{pmatrix}0&\sigma_\mu\\\bar{\sigma}_\mu&0\end{pmatrix}

=\begin{pmatrix}\bar{\sigma}_\mu&0\\0&-{\sigma}_\mu\end{pmatrix}\to \begin{pmatrix}\vec{\sigma}&0\\0&\vec{\sigma}\end{pmatrix}=\vec{\Sigma}$$. 除了一个负号外,这就是(2)的结果.

本文档除了包括推导, 疑惑 外,做 读书重点 的记录 $$$$