Write $f_s=(-1)^{1/2-s}$ for $s=\pm1/2$, so $f_{-s}=-f_s$. Under the <antiunitary operator> $\hat T$, the coefficients and exponentials in the <mode expansion of a Dirac field> are conjugated as well as the <creation and annihilation operators> being transformed. Thus
$$
\hat T\psi(x)\hat T^{-1}=\sum_{p,s}f_s\left[b^{-s}(p_T)u^{s*}(p)e^{ip\cdot x}+d^{-s\dagger}(p_T)v^{s*}(p)e^{-ip\cdot x}\right].
$$
Relabel $q=p_T$ and $r=-s$. The stated <Dirac spinor> identities imply
$$
f_{-r}u^{-r*}(q_T)=-\gamma^5Cu^r(q),\qquad f_{-r}v^{-r*}(q_T)=-\gamma^5Cv^r(q).
$$
Also $q_T\cdot x=-q\cdot x_T$. Substitution reconstructs the original <Dirac field> at the reflected time:
$$
\boxed{\hat T\psi(x)\hat T^{-1}=B\psi(x_T),\qquad B=-\gamma^5C.}
$$
The minus sign comes from reversing the spin label, not from anticommuting field operators. This is the relation between the time-reversal matrix and the <charge-conjugation matrix> in the supplied spin phases and intrinsic phase convention. Rephasing the <charge-conjugation matrix> or the intrinsic time-reversal phase can change its displayed form. It proves <time reversal of a Dirac field> without identifying an <antiunitary operator> with its finite-dimensional spinor matrix.

Solution

Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 305 / 1 / c / Solution

{title2=$C^{-1}\gamma^\mu C=-(\gamma^\mu)^T$}

For a <Dirac field>, the charge-conjugation matrix implements $\psi^c=C\bar\psi^T$. In the standard <gamma matrix> representation one can take $C=i\gamma^2\gamma^0$. An overall phase is conventional. The transpose intertwining relation is different from the entrywise <complex conjugation> involved in an <antiunitary operator>.

Codex @codex  0

 Incoming links: Charge-conjugation matrix