= Solution
Use the metric $g^{\mu\nu}=\operatorname{diag}(1,-1,-1,-1)$ and the usual <Dirac basis>, with $(\gamma^0)^\dagger=\gamma^0$ and $(\gamma^i)^\dagger=-\gamma^i$. In this solution the sign of the <chirality matrix> is the one specified in the paper, $\gamma^5=-i\gamma^0\gamma^1\gamma^2\gamma^3$. Define $t_0=1$ and $t_i=-1$, with no summation in componentwise transformation formulas.
Complex conjugation changes the explicit $-i$ to $+i$. Three spatial <gamma matrices> each contribute another minus sign, so
$$
B\gamma^{5*}B^{-1}=i\gamma^0(-\gamma^1)(-\gamma^2)(-\gamma^3)=-i\gamma^0\gamma^1\gamma^2\gamma^3=\boxed{\gamma^5}.
$$
Also $(\gamma^5)^2=1$ and $\{\gamma^5,\gamma^\mu\}=0$ by the <Clifford algebra>.
The <quantum time-reversal operator> is <antiunitary>: it conjugates numerical coefficients, including the <Dirac spinors> and plane-wave exponentials in the <mode expansion of a Dirac field>. To fix the spin phase explicitly, put $\epsilon_s=(-1)^{1/2-s}$ and use
$$
\hat T b^s(p)\hat T^{-1}=\epsilon_s b^{-s}(p_T),\qquad
\hat T d^{s\dagger}(p)\hat T^{-1}=\epsilon_s d^{-s\dagger}(p_T).
$$
This convention gives $\hat T^2=-1$ on a one-fermion state, since $\epsilon_s\epsilon_{-s}=-1$. In the transformed expansion, change variables from $(p,s)$ to $(p_T,-s)$. The <Lorentz-invariant phase-space measure> is unchanged, and $p_T\cdot x=-p\cdot x_T$. The coefficient of $b^s(p)e^{-ip\cdot x_T}$ is therefore
$$
\epsilon_{-s}u^{-s*}(p_T)=-\epsilon_su^{-s*}(p_T)=\gamma^5Cu^s(p).
$$
The same calculation applies to the antiparticle coefficient. Thus, with this explicitly fixed spin convention,
$$
\boxed{B=\gamma^5C,\qquad \hat T\psi(x)\hat T^{-1}=B\psi(x_T).}
$$
The <spin phase in the Dirac time-reversal matrix> matters here: a common change of the one-particle time-reversal phase replaces $B$ by $-B$; the phase-independent relation is $B\propto\gamma^5C$. It changes neither the defining conjugation property nor any bilinear result below. Normalize $B$ to be a <unitary matrix>. In the <Dirac basis>, $\gamma^0$ is real and commutes with $B$, so transforming the <Dirac adjoint> gives
$$
\hat T\bar\psi(x)\hat T^{-1}=(B\psi(x_T))^\dagger\gamma^{0*}
=\psi^\dagger(x_T)B^{-1}\gamma^0
=\boxed{\bar\psi(x_T)B^{-1}}.
$$
Here complex conjugation of the matrix defining the <Dirac adjoint> is essential. An explicit realization consistent with the sign of $\gamma^5$ used here is $C=i\gamma^2\gamma^0$ and $B=\gamma^1\gamma^3$; it has $B^{-1}=-B$, so conjugation by $B$ and by $B^{-1}$ coincides. The usual changes of <Dirac spinor> basis by <unitary matrices> carry the adjoint and time-reversal matrix with them.
For the <charge-conjugation matrix>, the <gamma matrix adjoint and transpose identities> give $\gamma^{\mu T}=t_\mu\gamma^{\mu*}$. Consequently $B\gamma^{\mu T}B^{-1}=\gamma^\mu$. Since $C=\gamma^5B$ in the chosen phase convention,
$$
C\gamma^{\mu T}C^{-1}=\gamma^5B\gamma^{\mu T}B^{-1}\gamma^5
=\gamma^5\gamma^\mu\gamma^5=\boxed{-\gamma^\mu}.
$$
This proof uses a temporal <Hermitian matrix> and spatial <skew-Hermitian matrices> explicitly; the transpose identity is preserved when $C$ is transformed appropriately with the <gamma matrices>.
For the <Fermi interaction>, let $V_{pn}^\mu=\bar p\gamma^\mu n$ and $A_{pn}^\mu=\bar p\gamma^\mu\gamma^5 n$. In the displayed <Dirac basis>, the inverse version of the conjugation relation also holds. <Antiunitarity>, $B^{-1}\gamma^{\mu*}B=t_\mu\gamma^\mu$ and $B^{-1}\gamma^{5*}B=\gamma^5$ give
$$
\hat T V_{pn}^\mu(x)\hat T^{-1}=t_\mu V_{pn}^\mu(x_T),\qquad
\hat T A_{pn}^\mu(x)\hat T^{-1}=t_\mu A_{pn}^\mu(x_T).
$$
The leptonic <weak charged current> has the same component signs. In the contraction of leptonic and hadronic currents, the two $t_\mu$ factors cancel. Its two independent operators thus keep their form while their coefficients become $g_V^*$ and $g_A^*$. The Hermitian-conjugate term transforms separately; its presence does not remove a relative complex phase between the vector and axial couplings.
With all intrinsic phases fixed to one and a real positive <Fermi constant>, invariance requires real coefficients in that convention. A common phase of $g_V$ and $g_A$ can instead be absorbed into a rephasing of the nucleon fields, and hence into their intrinsic time-reversal phases. The convention-independent condition, for $g_V\ne0$, is
$$
\boxed{\operatorname{Im}(g_Ag_V^*)=0,\qquad g_A/g_V\in\mathbb R.}
$$
This is the <relative weak phase condition for time reversal>. If one coefficient vanishes, there is no relative phase to constrain; the remaining common phase can be removed. A nonreal ratio violates <time-reversal symmetry> even though the <Lagrangian density> includes its Hermitian conjugate.
Back to article page
