Following my class on RQM, we wanted to evaluate the unpolarised cross section for the following process $$ e^+ e^- \rightarrow \mu^+ \mu^- $$ In doing so, whenever summing over spinor indices, a trace appears and I cannot understand why. For example, summing over spins we got $$ \sum_{\sigma,\sigma'}\bar{v}_a(p',\sigma')\gamma^\mu_{ab}u_b(p, \sigma)\bar{u}_c(p,\sigma)\gamma^\nu_{cd}v_d(p',\sigma') = (\not p\,'-m)_{da}\,\gamma^\mu_{ab}(\not p+m)_{bc}\,\gamma^\nu_{cd}\\=\text{tr}\left((\not p\,'-m)\gamma^\mu(\not p+m)\gamma^\nu\right)$$ I understand where the $(\not p \pm m)$ factors come from but not how, in the last step, the matrix product becomes a trace!

Being this process heavily used in QED calculations I wanted to understand it to the fullest. So my question is

How do traces pop-up when calculating products with spinors and gamma matrices?

  • 1
    $\begingroup$ The expression $(\not p\,'-m)_{da}\,\gamma^\mu_{ab}(\not p+m)_{ab}\,\gamma^\nu_{cd}$ is wrong (the index $a$ appears thrice). In any case, $A_{ij}B_{ji}=(AB)_{ii}\equiv \operatorname{tr}(AB)$. $\endgroup$ Nov 24, 2019 at 15:00
  • $\begingroup$ @AccidentalFourierTransform that was easy.. thank you! $\endgroup$ Nov 24, 2019 at 15:13

2 Answers 2


Answer: Casimir trick

In a typical collider experiment the momentum vectors are $$ p_1=\begin{pmatrix}E\\0\\0\\p\end{pmatrix}\qquad p_2=\begin{pmatrix}E\\0\\0\\-p\end{pmatrix}\qquad p_3=\begin{pmatrix} E\\ \rho\sin\theta\cos\phi\\ \rho\sin\theta\sin\phi\\ \rho\cos\theta \end{pmatrix} \qquad p_4=\begin{pmatrix} E\\ -\rho\sin\theta\cos\phi\\ -\rho\sin\theta\sin\phi\\ -\rho\cos\theta \end{pmatrix} $$

where $E$ is beam energy, $p=\sqrt{E^2-m^2}$, $\rho=\sqrt{E^2-M^2}$, $m$ is electron mass $0.51\,\text{MeV}$, and $M$ is muon mass $106\,\text{MeV}$. The spinors are \begin{gather*} u_{11}=\begin{pmatrix}E+m\\0\\p\\0\end{pmatrix}\quad v_{21}=\begin{pmatrix}-p\\0\\E+m\\0\end{pmatrix}\quad u_{31}=\begin{pmatrix}E+M\\0\\p_3^z\\p_3^x+ip_3^y\end{pmatrix}\quad v_{41}=\begin{pmatrix}p_4^z\\p_4^x+ip_4^y\\E+M\\0\end{pmatrix} \\ u_{12}=\begin{pmatrix}0\\E+m\\0\\-p\end{pmatrix}\quad v_{22}=\begin{pmatrix}0\\p\\0\\E+m\end{pmatrix}\quad u_{32}=\begin{pmatrix}0\\E+M\\p_3^x-ip_3^y\\-p_3^z\end{pmatrix}\quad v_{42}=\begin{pmatrix}p_4^x-ip_4^y\\-p_4^z\\0\\E+M\end{pmatrix} \end{gather*}

The last digit in a spinor subscript is 1 for spin up and 2 for spin down. Note that the spinors are not individually normalized. Instead, a combined spinor normalization constant $N=(E+m)^2(E+M)^2$ will be used where needed.

This is the probability density for muon production. Symbol $s=(p_1+p_2)^2=4E^2$, symbol $s_j$ selects the spin of spinor $j$, and $e$ is electron charge. \begin{equation*} |\mathcal{M}(s_1,s_2,s_3,s_4)|^2 =\frac{e^4}{s^2}\frac{1}{N}\left|(\bar{u}_3\gamma_\mu v_4)(\bar{v}_2\gamma^\mu u_1)\right|^2 \end{equation*}

The expected probability density $\langle|\mathcal{M}|^2\rangle$ is computed by summing $|\mathcal{M}|^2$ over all spin states and dividing by the number of inbound states. There are four inbound states. \begin{align*} \langle|\mathcal{M}|^2\rangle &=\frac{1}{4}\sum_{s_1=1}^2\sum_{s_2=1}^2\sum_{s_3=1}^2\sum_{s_4=1}^2|\mathcal{M}(s_1,s_2,s_3,s_4)|^2 \\ &=\frac{e^4}{4s^2}\sum_{s_1=1}^2\sum_{s_2=1}^2\sum_{s_3=1}^2\sum_{s_4=1}^2 \frac{1}{N}\left|(\bar{u}_3\gamma_\mu v_4)(\bar{v}_2\gamma^\mu u_1)\right|^2 \end{align*}

Another way to compute $\langle|\mathcal{M}|^2\rangle$ is to use the Casimir trick. \begin{equation*} \langle|\mathcal{M}|^2\rangle =\frac{e^4}{4s^2} \mathop{\rm Tr}\left((\not p_3+M)\gamma^\mu(\not p_4-M)\gamma^\nu\right) \mathop{\rm Tr}\left((\not p_2-m)\gamma_\mu(\not p_1+m)\gamma_\nu\right) \end{equation*}

Here is a third way to compute $\langle|\mathcal{M}|^2\rangle$. \begin{equation*} \langle|\mathcal{M}|^2\rangle =\frac{e^4}{4s^2} \left( 32 (p_1\cdot p_3) (p_2\cdot p_4) + 32 (p_1\cdot p_4) (p_2\cdot p_3) + 32 m^2 (p_3\cdot p_4) + 32 M^2 (p_1\cdot p_2) + 64 m^2 M^2 \right) \end{equation*}

For the momentum vectors given above the result is \begin{equation*} \langle|\mathcal{M}|^2\rangle =e^4\left(1+\cos^2\theta+\frac{m^2+M^2}{E^2}\sin^2\theta+\frac{m^2M^2}{E^4}\cos^2\theta\right) \end{equation*}

The Stanford Linear Collider had a collision energy of $2E=91$ GeV. For beam energies such as SLC where $E\gg M$ the above equation can be approximated as $$ \langle|\mathcal{M}|^2\rangle=e^4(1+\cos^2\theta) $$

The differential cross section is $$ \frac{d\sigma}{d\Omega} =\frac{\langle|\mathcal{M}|^2\rangle}{64\pi^2s} =\frac{e^4}{256\pi^2E^2}(1+\cos^2\theta) $$

Recall that $e^2=4\pi\alpha$ hence \begin{equation*} \frac{d\sigma}{d\Omega}=\frac{\alpha^2}{16E^2}(1+\cos^2\theta) \end{equation*}

The total cross section calculation requires the following definite integral. $$ \int_\Omega(1+\cos^2\theta)\,d\Omega =\int_0^{2\pi}\int_0^\pi(1+\cos^2\theta)\sin\theta\,d\theta\,d\phi =\frac{8}{3}\int_0^{2\pi}d\phi =\frac{16\pi}{3} $$

Hence the total cross section is $$ \sigma =\int_\Omega d\sigma =\int_\Omega\frac{\alpha^2}{16E^2}(1+\cos^2\theta)\,d\Omega =\frac{\alpha^2}{16E^2}\frac{16\pi}{3} =\frac{\pi\alpha^2}{3E^2} $$


It's essentially just matrix algebra. I would refer you to appendix A.3 of Quantum Field Theory by Mandl and Shaw. It might be a good exercise to fully work through the algebra and compute the trace.

  • $\begingroup$ I'll check the reference as soon as I can. Thank you! $\endgroup$ Nov 24, 2019 at 15:13

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