I'm considering the following Lagrangian:
$$\mathcal{L}=\partial_\mu \phi^*\partial^\mu\phi + \chi^\dagger i\overline{\sigma}^\mu\partial_\mu\chi +F^*F+m\phi F+\frac{i}{2}m\chi^T\sigma^2\chi +h.c.$$
where $\epsilon$ is an infinitesimal spinor, $\chi$ is a spinor, and $F$ and $\phi$ are scalars. All spinors anticommute, $\sigma^2$ is the second Pauli spin matrix, $\sigma = (1,\vec{\sigma})$, and $\overline{\sigma} = (1, -\vec{\sigma})$.
I want to show that this is invariant under
$$\delta\phi = -i\epsilon^T\sigma^2\chi$$
$$\delta\chi = \epsilon F+\sigma^\mu\partial_\mu\phi\sigma^2\epsilon^*$$
$$\delta F = -i\epsilon^\dagger \overline{\sigma}^\mu\partial_\mu\chi$$
Doing the variation to first order, I need to show sums such as
$$ -i \epsilon^T \sigma^2\partial_\mu \phi^* \partial^\mu \chi + i\epsilon^T \sigma^2 \sigma^\mu \partial_\mu \phi^* \overline{\sigma}^\nu \partial_\nu\chi =0$$
or that it is a total derivative. How can I show this?
I saw from a reference online saying that by the anticommutation relations of the Pauli matrices
$$\sigma^\mu\overline{\sigma}^\nu\partial_\mu\phi^*\partial_\nu\chi=\partial_\mu\phi^*\partial^\mu\chi$$
If true, this would take care of things. But I don’t see how this is true.
Best Answer
To prove the last equation consider the term, $$ \sigma^\mu\overline{\sigma}^\nu\partial_{\mu}\phi^{\ast}\partial_\nu\chi, \tag{1}\label{1} $$ which can be written as $$ \frac{1}{2}\sigma^\mu\overline{\sigma}^\nu\partial_{\mu}\phi^{\ast}\partial_\nu\chi+\frac{1}{2}\sigma^\mu\overline{\sigma}^\nu\partial_{\mu}\phi^{\ast}\partial_\nu\chi. \tag{2}\label{2} $$ Now, using the inverse product rule twice on the second term the above takes the following form $$ \frac{1}{2}\sigma^\mu\overline{\sigma}^\nu\partial_{\mu}\phi^{\ast}\partial_\nu\chi+\frac{1}{2}\sigma^\mu\overline{\sigma}^\nu\partial_\nu\phi^{\ast}\partial_\mu\chi+\underbrace{\frac{1}{2}\partial_\mu\left(\sigma^\mu\overline{\sigma}^\nu\phi^{\ast}\partial_\nu\chi\right)-\frac{1}{2}\partial_\nu\left(\sigma^\mu\overline{\sigma}^\nu\phi^{\ast}\partial_\mu\chi\right)}_{\text{boundary terms}}, \tag{3}\label{3} $$ or $$ \frac{1}{2}\sigma^\mu\overline{\sigma}^\nu\partial_{\mu}\phi^{\ast}\partial_\nu\chi+\frac{1}{2}\sigma^\mu\overline{\sigma}^\nu\partial_\nu\phi^{\ast}\partial_\mu\chi+\text{boundary terms}. \tag{4}\label{4} $$ The next step is to rename $\mu\leftrightarrow\nu$ only in the second term to get $$ \frac{1}{2}\left(\sigma^\mu\overline{\sigma}^\nu+\sigma^\nu\overline{\sigma}^\mu\right)\partial_{\mu}\phi^{\ast}\partial_\nu\chi+\text{boundary terms}. \tag{5}\label{5} $$ From there expand the sums explicitly as $$ \frac{1}{2}\left(\sigma^0\overline{\sigma}^0+\sigma^0\overline{\sigma}^0\right)\partial_{0}\phi^{\ast}\partial_0\chi+\frac{1}{2}\left(\sigma^0\overline{\sigma}^i+\sigma^i\overline{\sigma}^0\right)\partial_{0}\phi^{\ast}\partial_i\chi+\frac{1}{2}\left(\sigma^i\overline{\sigma}^0+\sigma^0\overline{\sigma}^i\right)\partial_{i}\phi^{\ast}\partial_0\chi+\frac{1}{2}\left(\sigma^i\overline{\sigma}^j+\sigma^j\overline{\sigma}^i\right)\partial_{i}\phi^{\ast}\partial_j\chi+\text{boundary terms}, \tag{6}\label{6} $$ or $$ \frac{1}{2}2I_2\partial_{0}\phi^{\ast}\partial_0\chi+\frac{1}{2}\left(-\sigma^i+\sigma^i\right)\partial_{0}\phi^{\ast}\partial_i\chi++\frac{1}{2}\left(\sigma^i-\sigma^i\right)\partial_{i}\phi^{\ast}\partial_0\chi-\frac{1}{2}\underbrace{\{\sigma^i,\sigma^j\}}_{2\delta^{ij}I_{2}}\partial_{i}\phi^{\ast}\partial_j\chi+\text{boundary terms}, \tag{7}\label{7} $$ or $$ \frac{1}{2}2I_2\partial_{0}\phi^{\ast}\partial_0\chi-\frac{1}{2}2I_2\partial_{i}\phi^{\ast}\partial_i\chi+\text{boundary terms}, \tag{8}\label{8} $$ where $I_{2}=\text{diag}(1,1)$. From there it is easy to observe that the above can be written as $$ \frac{1}{2}2I_{2}\eta^{\mu\nu}\partial_{\mu}\phi^{\ast}\partial_\nu\chi+\text{boundary terms}=\partial_{\mu}\phi^{\ast}\partial^\mu\chi+\text{boundary terms}. \tag{9}\label{9} $$ Combining \eqref{1} and \eqref{9} leads to $$ \sigma^\mu\overline{\sigma}^\nu\partial_{\mu}\phi^{\ast}\partial_\nu\chi=\partial_{\mu}\phi^{\ast}\partial^\mu\chi+\text{boundary terms}, \tag{10}\label{10} $$ as you wanted, and from \eqref{5} and \eqref{9} $$ \sigma^\mu\overline{\sigma}^\nu+\sigma^\nu\overline{\sigma}^\mu=2I_{2}\eta^{\mu\nu}, \tag{11}\label{11} $$ which is another important result.
I hope this helped!