Do the WI coupling constants change sign under $C$?

Question

I am trying to understand discrete symmetries in the SM, and I have some troubles in understanding why the CC interaction violates CP. In my (badly written) notes it's said that, taken two fermonic fields $\psi_{i},\psi_{j}$ (with $i,j$ standing for different flavors indices), then

$$\bar{\psi}_{i}A_{\mu}\gamma^{\mu}\psi_{j}\stackrel{CP}{\rightarrow} \bar{\psi}_{j}A_{\mu}\gamma^{\mu}\psi_{i}.$$

In this way the CC interaction transform as

$$\mathcal{L}_{CC}=-\frac{g}{\sqrt{2}}\bigl[\bar{u}_{L}V_{CKM}W_{\mu}^{+}\gamma^{\mu}d_{L}+\bar{d}_{L}V^{*}_{CKM}W^{-}_{\mu}\gamma^{\mu}u_{L}\bigr]$$

$$\stackrel{CP}{=}\mathcal{L}_{CC}=-\frac{g}{\sqrt{2}}\bigl[\bar{d}_{L}V_{CKM}^{T}W_{\mu}^{-}\gamma^{\mu}u_{L}+\bar{u}_{L}(V^{*}_{CKM})^{\dagger}W^{-}_{\mu}\gamma^{\mu}d_{L}\bigr]$$ so that CP is violated since $V_{CKM}\neq V_{CKM}^{*}$

I am not convinced about the first relation, since as far as I know the vector field should change sign under C while $\bar{\psi}_{i}\gamma^{\mu}\psi_{j}$ should transform as $+\bar{\psi}_{j}\gamma^{\mu}\psi_{i}$ (for a reference take a look in RQM by Greiner), so using the fact that

$$\bar{\psi}_{i}\gamma^{\mu}\psi_{j}\stackrel{P}{\rightarrow}(-1)^{\mu}\bar{\psi}_{i}\gamma^{\mu}\psi_{j}$$ $$W_{\mu}^{\pm}\stackrel{P}{\rightarrow}(-1)^{\mu}W_{\mu}^{\pm}$$

we should have a change of sign. What am I doing wrong?? I also don't know if the coupling constant flips its sign under C or it stays unchanged. Notice that $(-1)^{\mu}$ is intended to be $-1$ for spatial components and $+1$ for the temporal ones.

Is it possible that my notes are confusing the way classical fields transforms with how quantum fields transforms?

Answer 1

You have to take into account that fermionic fields are anticommuting objects.

The general form of a CP-transformation acting on the quark fields (mass eigenfield) reads $$u(x) \stackrel{\rm CP}{\to} e^{i \alpha_u} C u^\ast(\tilde{x}), \quad d(x) \stackrel{\rm CP}{\to} e^{i\alpha_d} C d^\ast(\tilde{x}), \qquad x=(x^0, \mathbf{x}), \, \tilde{x}=(x^0, -\mathbf{x}), \, C=-i \gamma^2\gamma^0,$$ where all quarks with (electromagnetic) charge $+2/3$ are collected in the $n_G=3$ dimensional column vector $u$ and, analogously, the quarks with charge $-1/3$ in $d$. $e^{i\alpha_u}$ and $e^{i\alpha_d}$ are diagonal $n_G \times n_G = 3 \times 3$ phase matrices, which do not affect the kinetic or mass terms of the quark fields. As a consequence, one obtains $$\bar{u} (x) \gamma^\mu (1-\gamma_5) V_{\rm CKM}\, d(x) \stackrel{\rm CP}{\to} -\varepsilon(\mu)\bar{d}(\tilde{x})\gamma^\mu (1-\gamma_5)(e^{-i\alpha_u}V_{\rm CKM}e^{i\alpha_d})^Tu(\tilde{x}),$$ where $\varepsilon(0)=+1$ and $\varepsilon(i)= -1$ ($i=1,2,3$). The minus sign in this formula originates from the crucial property $u_{i a} d^\ast_{j b}= - d^\ast_{j b} u_{i a}$ of fermionic fields ($i,j$ are generation indices and $a,b$ are Dirac indices).

P.S.: The weak coupling constant $g$, being simply one of the input parameters of the SM, is, of course, not affected by this transformation (acting on the fields).