I have been reading 'Quantum Field Theory and the Standard Model' by Schwartz and have gotten stuck on a line of reasoning in Section 10.2.2.

I understand that we can construct a (right-handed) four-vector $V_R^\mu$ given by, $$ V_R^\mu = (\psi_R^\dagger\psi_R, \psi_R^\dagger\vec{\sigma}\psi_R) $$ Schwartz then says that (eq. 10.54) $$ \psi_{R}^{\dagger} \partial_{t} \psi_{R}+\psi_{R}^{\dagger} \partial_{j} \sigma_{j} \psi_{R} $$ is therefore a Lorentz invariant. This can be proven using the same methodology as here, but how does this result follow from the construction of $V_R^\mu$?

I expect that $\partial_\mu V_R^\mu$ is Lorentz invariant but this isn't equivalent to the expression given by Schwartz.

Any help is appreciated!

I have been reading 'Quantum Field Theory and the Standard Model' by Schwartz and have gotten stuck on a line of reasoning in Section 10.2.2.

I understand that we can construct a (right-handed) four-vector $V_R^\mu$ given by, $$ V_R^\mu = (\psi_R^\dagger\psi_R, \psi_R^\dagger\vec{\sigma}\psi_R). $$ Schwartz then says that (eq. 10.54) $$ \psi_{R}^{\dagger} \partial_{t} \psi_{R}+\psi_{R}^{\dagger} \partial_{j} \sigma_{j} \psi_{R} $$ is therefore a Lorentz invariant. This can be proven using the same methodology as here, but how does this result follow from the construction of $V_R^\mu$?

I expect that $\partial_\mu V_R^\mu$ is Lorentz invariant but this isn't equivalent to the expression given by Schwartz.

Any help is appreciated!