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Assume that we have two 1/2-spin particles with four-momenta $p$ and $p'$. Particle Dirac spinors satisfy the completeness relation $$ \sum_{s=1}^2u_s(p)\overline{u}_s(p)=\not p+m $$ My goal now is to find an equivalent expression for $$ \sum_{s=1}^2u_s(p)\overline{u}_s(p') $$ My attempt:

We have $$ u_s(p)=\sqrt{E+m} \begin{pmatrix} \phi_s \\ \frac{\vec{\sigma}\cdot\vec{p}}{E+m}\phi_{s} \\ \end{pmatrix} \quad\text{and}\quad \overline{u}_s(p')=\sqrt{E'+m'} \begin{pmatrix} \phi^T_s & -\phi^T_{s}\frac{\vec{\sigma}\cdot\vec{p'}}{E'+m'} \\ \end{pmatrix} $$ with $$ \phi_1= \begin{pmatrix} 1 \\ 0 \\ \end{pmatrix} \quad\text{and}\quad \phi_2= \begin{pmatrix} 0 \\ 1 \\ \end{pmatrix} $$ Then (and working in the Dirac-Pauli representation) \begin{align} \sum_{s=1}^2u_s(p)\overline{u}_s(p')&=\sqrt{(E+m)(E'+m')}\sum_{s=1}^2 \begin{pmatrix} \phi_s\phi^T_s & -\frac{\vec{\sigma}\cdot\vec{p'}}{E'+m'}\phi_s\phi^T_s \\ \frac{\vec{\sigma}\cdot\vec{p}}{E+m}\phi_s\phi^T_s & -\frac{(\vec{\sigma}\cdot\vec{p})(\vec{\sigma}\cdot\vec{p'})}{(E+m)(E'+m')}\phi_s\phi^T_s \\ \end{pmatrix} \\ &=\sqrt{(E+m)(E'+m')} \begin{pmatrix} 1 & -\frac{\vec{\sigma}\cdot\vec{p'}}{E'+m'} \\ \frac{\vec{\sigma}\cdot\vec{p}}{E+m} & -\frac{(\vec{\sigma}\cdot\vec{p})(\vec{\sigma}\cdot\vec{p'})}{(E+m)(E'+m')} \\ \end{pmatrix} \\ \end{align} where i have used $$ \sum_{s=1}^2\phi_s\phi^T_s=1\quad\text{: a 2x2 identity matrix} $$ Here is where i ask for help, because i am not sure how to proceed with the above expression. Of course, any other way to find a relationship is welcome.

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    $\begingroup$ Since $\sigma_i \sigma_j = \delta_{ij} I_2 + i\varepsilon_{ijk}\sigma_k$ with $I_2$ the $2\times 2$ identity matrix and implicit summation over $k$ in the second term, we have (with all summation implicit) $\left(\vec{\sigma}\cdot\vec{p}\right)\left(\vec{\sigma}\cdot\vec{p}'\right)=\sigma_i\sigma_jp_ip'_j=\vec{p}\cdot\vec{p}'I_2+\vec{\sigma}\cdot\vec{p}\times\vec{p}'$. $\endgroup$ – J.G. Sep 30 '16 at 6:23
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    $\begingroup$ My answetr to this question physics.stackexchange.com/q/237908 may be of some help. $\endgroup$ – Lewis Miller Oct 1 '16 at 14:25
  • $\begingroup$ @LewisMiller I'm breaking the above matrix as a linear combination of $\{1,\gamma^\mu,\gamma^5,\gamma^\mu\gamma^5,\sigma^{\mu\nu}\}$. I just take this approach for the case $p=p'$ and it worked fine. Thank you. $\endgroup$ – SNC92 Oct 1 '16 at 22:24
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Ok, so here I found an answer. I'm gonna start rewriting the last expression of the question: $$ \begin{align} \sum_{s=1}^2u_s(p)\overline{u}_s(p') &= \sqrt{(E+m)(E'+m')} \begin{pmatrix} 1 & -\frac{\vec{\sigma}\cdot\vec{p'}}{E'+m'} \\ \frac{\vec{\sigma}\cdot\vec{p}}{E+m} & -\frac{(\vec{\sigma}\cdot\vec{p})(\vec{\sigma}\cdot\vec{p'})}{(E+m)(E'+m')} \\ \end{pmatrix} \\ \frac{1}{\sqrt{ss'}}\sum_{s=1}^2u_s(p)\overline{u}_s(p') &= \Omega \end{align} $$ where I have defined for convenience $$ s:=E+m \quad\quad s':=E'+m' \quad\quad \Omega:= \begin{pmatrix} 1 & -\frac{\vec{\sigma}\cdot\vec{p'}}{s'} \\ \frac{\vec{\sigma}\cdot\vec{p}}{s} & -\frac{(\vec{\sigma}\cdot\vec{p})(\vec{\sigma}\cdot\vec{p'})}{ss'} \\ \end{pmatrix} $$ Now, I know that $\Omega\in M(4,\mathbb{C})$ and a basis for this linear space is given by the set $$ \{1,\gamma^\mu,\gamma^5,\gamma^\mu\gamma^5\}_{\mu=0}^3\cup\{\sigma^{\mu\nu}\}_{\mu,\nu=0\; (\mu<\nu)}^3 $$ (where $\sigma^{\mu\nu}:=i/2[\gamma^\mu,\gamma^\nu]$) so that $\Omega$ can be written as $$ \Omega=a+b_\mu\gamma^\mu+c\gamma^5+d_\mu\gamma^\mu\gamma^5+e_{\mu\nu}\sigma^{\mu\nu} $$ and where the coefficients are given by

  • $a=\frac{1}{4}tr(\Omega)$
  • $b_\mu=\frac{1}{4}tr(\Omega\gamma_\mu)$
  • $c=\frac{1}{4}tr(\Omega\gamma_5)$
  • $d_\mu=\frac{1}{4}tr(\Omega\gamma_5\gamma_\mu)$
  • $e_{\mu\nu}=\frac{1}{8}tr(\Omega\sigma_{\mu\nu})$

With this, it follows that: $$ \begin{align} a&=\frac{1}{2}\left(1-\frac{\vec{p}\cdot\vec{p}'}{ss'}\right) \\ b_0&=\frac{1}{2}\left(1+\frac{\vec{p}\cdot\vec{p}'}{ss'}\right) \\ b_j&=-\frac{1}{2}\left(\frac{p_j}{s}+\frac{p'_j}{s'}\right)\quad j=1,2,3 \\ c&=0 \\ d_0&=0 \\ d_j&=\frac{i}{2ss'}(\vec{p}\times\vec{p}')_j\quad j=1,2,3 \\ e_{0j}&=-\frac{i}{4}\left(\frac{p_j}{s}-\frac{p'_j}{s'}\right)\quad j=1,2,3 \\ e_{jk}&=-\frac{i\varepsilon_{jkl}}{4ss'}(\vec{p}\times\vec{p}')_l\quad j=1,2,3\;\;\text{and}\;\; j<k \end{align} $$ Therefore, $$ \frac{2}{\sqrt{ss'}}\sum_{s=1}^2u_s(p)\overline{u}_s(p')=1-\frac{\vec{p}\cdot\vec{p}'}{ss'}+\left(1+\frac{\vec{p}\cdot\vec{p}'}{ss'}\right)\gamma^0-\left(\frac{p_j}{s}+\frac{p'_j}{s'}\right)\gamma^j+\frac{i}{ss'}(\vec{p}\times\vec{p}')_j\gamma^j\gamma^5-\frac{i}{2}\left(\frac{p_j}{s}-\frac{p'_j}{s'}\right)\sigma^{0j}-\frac{i\varepsilon_{jkl}}{2ss'}(\vec{p}\times\vec{p}')_l\sigma^{jl} $$ or, using the following identities: $$ \sigma^{0j}=i\gamma^0\gamma^j\quad\quad \varepsilon_{jkl}\varepsilon^{jkn}=2\delta_l^n \quad\quad\sigma^{jk}=\begin{pmatrix} \varepsilon^{jkn}\sigma_n & 0 \\ 0 & \varepsilon^{jkn}\sigma_n \\ \end{pmatrix} =\varepsilon^{jkn}\sigma_nI_4 $$ then $$ \frac{2}{\sqrt{ss'}}\sum_{s=1}^2u_s(p)\overline{u}_s(p')=1-\frac{\vec{p}\cdot\vec{p}'}{ss'}+\left(1+\frac{\vec{p}\cdot\vec{p}'}{ss'}\right)\gamma^0-\left(\frac{p_j}{s}+\frac{p'_j}{s'}\right)\gamma^j+\frac{i}{ss'}(\vec{p}\times\vec{p}')_j\gamma^j\gamma^5+\frac{1}{2}\left(\frac{p_j}{s}-\frac{p'_j}{s'}\right)\gamma^0\gamma^j-\frac{i}{ss'}(\vec{p}\times\vec{p}')\cdot\vec{\sigma}I_4 $$ As a particular case, if we assume now that both spinors correspond to the same 1/2-spin particle, then $p=p'$, $ss'=(E+m)^2$, $\vec{p}\cdot\vec{p}'=E^2-m^2$ and $\vec{p}\times\vec{p}'=0$ and the above expression reduces to $$ \begin{align} \frac{2}{E+m}\sum_{s=1}^2u_s(p)\overline{u}_s(p')&=1-\frac{E^2-m^2}{(E+m)^2}+\left(1+\frac{E^2-m^2}{(E+m)^2}\right)\gamma^0-\frac{2}{E+m}p_j\gamma^j \\ 2\sum_{s=1}^2u_s(p)\overline{u}_s(p')&=(E+m)-(E-m)+(E+m+E-m)\gamma^0-2p_j\gamma^j \\ \sum_{s=1}^2u_s(p)\overline{u}_s(p')&=m+E\gamma^0-p_j\gamma^j\quad\quad\text{but}\quad p_\mu=(p_0,-\vec{p})=(E,-p_x,-p_y,-p_z) \\ \sum_{s=1}^2u_s(p)\overline{u}_s(p')&=m+p_\mu\gamma^\mu \\ \sum_{s=1}^2u_s(p)\overline{u}_s(p')&=\not p+m \end{align} $$

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