# Square of invariant matrix element involving the Levi-Civita symbol

Consider the invariant matrix element

$$\mathcal M=2ige\frac{\epsilon^{\mu \nu \rho \sigma}k_{1 \mu}\epsilon_\nu(k_1,s_1)\bar{u}(p_2,r_2)\gamma_\sigma u(p_1,r_1)q_\rho}{q^2}, \quad q=k_2-k_1$$ corresponding to the following diagram for the process $$e^-+\gamma\to e^-+\pi^0$$:

My goal is to compute the quantity $$|\mathcal M|^2$$. However, I'm not sure what is the best way to go about this calculation, as I haven't done a lot of algebra involving these objects.

A trick that I've seen before is to rewrite $$\mathcal M$$ after noticing that since $$\nu$$ is the index of a photon polarization, it must be spatial, and therefore

\begin{align*} \mathcal M=\epsilon^{0ijk}k_{10}\epsilon_i(k_1)\bar u(p_2)\gamma_ju(p_1)q_k+\epsilon^{ij0k}k_{1i}\epsilon_j(k_1)\bar u(p_2)\gamma_0 u(p_1)q_k+\epsilon^{ijk0}k_{1i}\epsilon_j(k_1)\bar u(p_2)\gamma_k u(p_1)q_0 \\ =\epsilon^{ijk}\left(-|\mathbf k_1|\epsilon_i(k_1)\bar u(p_2)\gamma_ju(p_1)q_k+-k_{1i}\epsilon_j(k_1)\bar u(p_2)\gamma_0 u(p_1)q_k+k_{1i}\epsilon_j(k_1)\bar u(p_2)\gamma_k u(p_1)q_0\right) \end{align*}

but this doesn't seem to lead to something much simpler. Any suggestions?

• I know it's usually the most complicated but have you tried just multiplying the complex conjugate and summing over spins/polarizations? The epsilons will collapse to kroneker deltas. Commented Dec 9, 2022 at 17:13

As a general rule, you should avoid to select a specific reference frame doing such type of calculation. You may still do so only at the END of your calculation. Just use $$\sum_{r_1} \epsilon^\nu (k_1,r_1)\epsilon^{\bar{\nu}}(k_1,r_1) = -g^{\nu \bar{\nu}}$$. As already remarked in the comment of Triatticus, this leads to $$\epsilon^{\nu \mu \rho \sigma} \epsilon_\nu^{\, \, \, \bar{\mu} \bar{\rho} \bar{\sigma}}$$, which can be written as a sum of products of Kronecker deltas. Analogogously use $$\sum_s u(p,s) \bar{u}(p,s) = p {\! \! \!/}+m$$ to obtain $${\rm Tr}(p_2{\! \! \! \! \!/}+m) \gamma^\sigma (p_1 {\! \! \! \!/}+m) \gamma^{\bar{\sigma}}$$. Combining these pieces will give you the result after a few lines.
• Let me try a simpler case. Suppose i want to calculate $$|\epsilon^{\mu\nu\sigma\lambda}k_{1\mu}k_{2\lambda}\epsilon_\nu^*(\mathbf k_1,\alpha)\epsilon_\sigma^*(\mathbf k_2,\beta)|^2.$$ Then summing over $\alpha, \beta$ and using your first relation I'd get $$\epsilon^{\mu\nu\sigma\lambda}\epsilon^{\bar\mu\bar\nu\bar\sigma\bar\lambda}k_{1\mu}k_{2\lambda}k_{1\bar\mu}k_{2\bar\lambda}\eta_{\nu\bar\nu}\eta_{\nu\bar\nu},$$ but this should be just $k_1^2\cdot k_2^2$. Is this correct, or I am missing some factor? Commented Dec 9, 2022 at 22:30
• Your second $\eta_{\nu \bar{\nu}}$ should read $\eta_{\sigma \bar{\sigma}}$. The (corrected) expression contains a product of two $\epsilon$ tensors, where two pairs of indices are contracted, leaving only $\mu, \lambda, \bar{\mu}, \bar{\lambda}$ as free indices. What does this give as a sum of products two Kronecker deltas? Commented Dec 9, 2022 at 22:49
• Right, I think I'm just missing a factor of $2$. Commented Dec 9, 2022 at 23:00
• My last hint: $\epsilon^{\mu \lambda \nu \sigma} \epsilon^{\bar{\mu} \bar{\lambda}}_{\, \, \, \, \nu \sigma} = C (\eta^{\mu \bar{\mu} } \eta^{\lambda \bar{\lambda}}-\eta^{\mu \bar{\lambda}} \eta^{\lambda \bar{\mu}})$, where the constant $C$ can easily be determined. Commented Dec 9, 2022 at 23:35
• OK, but what is really missing in your final result (apart from the factor 2)? By the way, $k_1^2=k_2^2=0$. Commented Dec 10, 2022 at 6:57