For $x'=\Lambda x$, a <Dirac spinor> transforms as
$$
\psi'(x')=S[\Lambda]\psi(x).
$$
Hence
$$
\partial'_\mu\psi'(x')
=(\Lambda^{-1})^\nu{}_\mu S[\Lambda]\partial_\nu\psi(x).
$$
Using the result of part i, the <Dirac adjoint> transforms as
$$
\boxed{\overline\psi'(x')=\overline\psi(x)S[\Lambda]^{-1}}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 301 / 3 / ii / Solution

{c}
{title2=$\bar\psi=\psi^\dagger\gamma^0$}

The Dirac adjoint of a spinor is $\bar\psi=\psi^\dagger\gamma^0$. It transforms contragrediently, making $\bar\psi\psi$ a Lorentz scalar and $\bar\psi\gamma^\mu\psi$ a Lorentz vector.


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