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How can we show that the weak interaction violates the charge conjugation symmetry?

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  • $\begingroup$ Are you asking for a proof in the mathematical formalism, or are you asking for experimental evidence? And do you mean only the charge conjugation $C$, or the matter-antimatter conjugation $CP$? The first is kind of lame — we know that weak interactions strongly violate $P$, but approximately conserve $T$, so they must violate $C$ to conserve $CPT$. The fact that the universe has a matter-antimatter asymmetry implies $CP$ violation, which is an active area of research. $\endgroup$ – rob Aug 6 '14 at 17:14
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Weak interactions include only the left neutrinos (and right antineutrinos). It means that all neutrino-interaction terms in the Lagrangian also consist only the left particles (and right antiparticles), because $\bar{\Psi}\gamma^{\mu}\Phi_{R, L} = \bar{\Psi}_{R, L}\gamma^{\mu}\Phi_{R, L}$. It means that the charged current terms $L_{\int}^{CC} = g \bar{l}_{L} \gamma^{\mu}(\nu_{l})_{L}W_{\mu}$ breaks the C-invariance: neutrino interacts only with lepton, while the antineutrino interacts only with antilepton.

It can be easily understood from the definition of charge conjugation operation into the space of Dirac-type representations. For the arbitrary half-integer spin function $$ \Psi^{\mu_{1}...\mu_{n}} = \begin{pmatrix} \Psi_{a}^{\ \mu_{1}...\mu_{n}} \\ \Phi_{\dot {a}}^{\ \mu_{1}...\mu_{n}}\end{pmatrix}, $$ (here the number of the vector indice corresponds to the integer part of spin value; Dirac $\frac{1}{2}$ spinor has zero vector indices) $$ \hat{C} \Psi^{\mu_{1}...\mu_{n}} = \begin{pmatrix} \Phi_{a}^{\ \mu_{1}...\mu_{n}} \\ \Psi_{\dot {a}}^{\ \mu_{1}...\mu_{n}}\end{pmatrix}. $$ As it can be shown, $\hat {C} = \gamma_{2}K$ (I have neglected the phase), so $$ \frac{1 \pm \gamma_{5}}{2}\hat{C}\Psi = \left(\frac{1 \pm \gamma_{5}}{2} \right)\gamma_{2}K \Psi = \gamma_{2}K\left(\frac{1 \mp \gamma^{*}_{5}}{2} \right)\Psi^{*} = $$ $$ \hat{C}\left( \frac{1 \mp \gamma_{5}}{2} \right)\Psi. $$ Here I have used the Dirac representation of the gamma-matrices (or reps which are connected with it by the orthogonality matrix), in which $\gamma_{5}^{*} = \gamma_{5}$. Also I have used equality $[\gamma_{5}, \gamma_{\mu}]_{+} = 0$ which is hold in the each representation. The final equality means that if we acts on C-inverted spinor by the chirality-projector operator, the left-projector acts as right projector while right projector acts as the left one. It means that $\hat{C}$ changes the eigenstate of chirality projector to the "opposite" one.

This is the particle result of the statement given in this answer.

About C-conjugation operator.

The C-operator, in general, interchanges function from the left representation of the Dirac-type spinor to the right representation. We must to construct this specific structure, because as the irreducible representation of the Poincare group for half-integer spin has its own peculiarities (I don't want refine this statement since it will take a lot of space; in a few words, for half-integer spin representation we can't use only 4-vector indices). This operator combines complex conjugation (which is equal to charge conjugation for integer-spin representations) and the left and right representation "conjugation".

As it can be shown, this operator also changes the summary value of charge of the dirac-type field. I.e., if we construct the conserved $U(1)$ current and then act on it by charge-conjugation operator, we will change the sign of this operator (so particularly we will change the electric charge).

Let's have the simplest example - the Dirac spinor $\Psi = \begin{pmatrix} \psi_{a} & \kappa^{\dot {a}}\end{pmatrix}^{T}$. As the irrep of the Lorentz group it can be constructed as $\left( \frac{1}{2}, 0\right) \oplus \left( 0, \frac{1}{2}\right)$ (the first one refers to the $\psi_{a}$ while the second refers to the complex conjugated $\kappa^{\dot{a}}$). The charge conjugation operator (here I have restored the usual phase $i$) gives $$ \hat{C}\Psi = i\gamma_{2}\Psi^{*} = \begin{pmatrix} \kappa_{a} & \psi^{\dot{a}}\end{pmatrix} $$ The Dirac spinor current is equal to $j^{\mu} = \bar{\Psi}\gamma^{\mu}\Psi$, so if we act on $j^{0}$ by the C-operator, we will give $$ j^{0}_{c} = ((\hat{C}\Psi )^{\dagger} \hat{C}\Psi) = -\Psi^{T}\gamma_{2}\gamma_{2}\Psi^{*} = \Psi^{\dagger}\Psi . $$ Here I have used the fact that spinors are grassmanian, so $\Psi^{T}\Psi^{*} = -\Psi^{\dagger}\Psi$.

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  • $\begingroup$ Thank you very much, I did not get the last line, Could you expand it. $\endgroup$ – user55944 Aug 6 '14 at 19:51
  • $\begingroup$ @user55944 : excuse me, I have made the mistake in intermediate calculations. I'll fix in into a few minutes and I will add an explanation. $\endgroup$ – Andrew McAddams Aug 6 '14 at 20:03
  • $\begingroup$ What is K? and there is $i $ in the middle expression, I want to understand the C oper. more, for ex. $C(i) = -i $? and $CA_a = -A_a$? and C(e) =-e ? $\endgroup$ – user55944 Aug 6 '14 at 20:30
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    $\begingroup$ @user55944 : excuse me one more time. $K$ is the complex conjugation operator, $i$ is only the misprint. I will also add more info about specific structure of C-operator for Dirac-like representations into my answer. $\endgroup$ – Andrew McAddams Aug 6 '14 at 21:07
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Following the convention of Peskin & Schroeder- transformation rule of Dirac spinors $\psi(x,t)$ under $\cal{C}$ and $\cal{P}$ are given by, \begin{eqnarray} \cal{C}\psi(t,x)\cal{C}^{\dagger} & = & -i(\bar{\psi}\gamma^{0}\gamma^{2})^{\textbf{T}}\\ \cal{C}\bar{\psi}(t,x)\cal{C}^{\dagger} & = & -i(\gamma^{0}\gamma^{2}\psi)^{\textbf{T}}\\ \end{eqnarray} Lets study how $V^{\mu}\equiv\bar{\psi}\gamma^{\mu}\psi$ transform under charge conjugation $\cal{C}$.

\begin{eqnarray} {\cal{C}}V^{\mu}\cal{C}^{\dagger}& = & {\cal{C}}\bar{\psi}\gamma^{\mu}\psi\cal{C}^{\dagger}\\ & = & {\cal{C}}\bar{\psi}\cal{C}^{\dagger}{\cal{C}}\gamma^{\mu}\cal{C}^{\dagger}{\cal{C}}\psi\cal{C}^{\dagger}\qquad\boxed{\text{considering}~\cal{C}^{\dagger}=\cal{C}^{-1}}\\ & = & -i(\gamma^{0}\gamma^{2}\psi)^{\text{T}}\gamma^{\mu}(-i)(\bar{\psi}\gamma^{0}\gamma^{2})^{\text{T}} \\ & = & -\bar{\psi}\gamma^{\mu}\psi = -V^{\mu} \end{eqnarray} Likewise one can show that $A^{\mu}\equiv\bar{\psi}\gamma^{5}\gamma^{\mu}\psi$, transform under $\cal{C}$ operation as ${\cal{C}}A^{\mu}{\cal{C}}^{\dagger}= A^{\mu}$. Let us consider the weak interaction Lagrangian. \begin{equation} {\cal{L}_{\text{weak}}} \approx \frac{\text{G}_{F}}{\sqrt{2}}(V^{\mu}-A^{\mu})(V_{\mu}-A_{\mu}) = \frac{\text{G}_{F}}{\sqrt{2}}(V^2-2V^{\mu}A_{\mu}+A^{2}) \end{equation} It is suffice to study, how $V^2-2V^{\mu}A_{\mu}+A^{2}$ transform under $\cal{C}$ to check the invariance of the Lagrangian of weak interaction under charge conjugation. \begin{equation} V^2-2V^{\mu}A_{\mu}+A^{2} \xrightarrow{\cal{C}} V^2+2V^{\mu}A_{\mu}+A^{2} \neq {\cal{L}_{\text{weak}}} \end{equation} Therefore weak interaction Lagrangian is not invariant under charge conjugation.

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