The Yang-Mills Lagrangian gauge invariant under an $SU(N)$ tranformation can be written as

$${\cal L} = -\frac{1}{4}F_{\mu\nu}^i F^{i\ \mu\nu} \tag1$$

(Sum over $i$ implicit)

This Lagrangian contains a term of the form

$${\cal L'} = -g·f_{ijk}A_\mu^i A_\nu^j\partial^\mu A^{k\ \nu} \tag2$$

$f_{ijk}$ are the structure constants.

Under charge conjugation, the self-adjoint gauge field transforms as

$$A_\mu \rightarrow -A_\mu \tag3$$

And therefore Eq. (2) isn't charge conjugation (${\cal C}$) invariant while the other terms in the complete Lagrangian (Eq. (1)) are invariant. This implies that the QCD Lagrangian isn't ${\cal C}$-invariant.

But is this correct or what am I misunderstanding?

Also, you can't just pick $A_\mu \rightarrow A_\mu$ because in that case the couplings to matter, i.e., $J_\mu A^\mu = \bar{\psi}\gamma_\mu \psi A^\mu$ wouldn't be ${\cal C}$-invariant since $J_\mu \rightarrow -J_\mu$; as you can check in Invariance of the QED Lagrangian under charge conjugation


2 Answers 2


Just like the asker's transformation in the question you link, your transformations are wrong. Charge conjugation is literal conjugation - you are replacing all fields that transform in a non-trivial representation of the gauge group by fields that transform in its conjugate representation. It just so happens that this transformation is the same as "inverting the sign" for a $\mathrm{U}(1)$ gauge field, but this is not true for a more general non-Abelian field. Conjugation means taking the representation $\rho(g)$ of the group $T^a$ and replacing it by $\overline{\rho(g)}$, where $\bar{}$ is the transpose in the representation space. Given that a group element is written as an exponential $\exp(\mathrm{i}A^a T^a)$ for $T^a$ the representations of generators of the algebra, you can see that this sends $\rho(T^a)$ to $-\overline{T^a}$.

For a $\mathrm{U}(1)$-theory, the $T^a$ is just equal to the identity on the one-dimensional representation space, meaning you can express the charge conjugation map as $A^a \mapsto -A^a$ map on the coefficients $A^a$, explaining the origin of the minus sign in the Abelian case.

  • $\begingroup$ Then, what's the transformation for these fields? $\endgroup$
    – Vicky
    Feb 2, 2019 at 11:46
  • $\begingroup$ @Vicky I gave the recipe for figuring it out, but I don't think it is useful to write down the explicit transformation of the field components. It is much easier to convince oneself that he Lagrangian is invariant under the more abstract notion of conjugation I outlined here than to muck around with formulae. $\endgroup$
    – ACuriousMind
    Feb 2, 2019 at 11:51

Let me try to bring some physical intuition to what happens under charge conjugation. In the case of Yang-Mills with SU(N) the charge conjugation on the gauge fields acts explicitly in the following manner $$ \mathcal{C} A_\mu^i \mathcal{C} \,T^i = - A_\mu^j (T^j)^T , $$ where $T^j$ are the generators of $SU(N)$, more specifically you can compute the transformation by expressing $T^i$ in the fundamental representation $$ \mathcal{C} A_\mu^i \mathcal{C} = - 2 \text{tr}(T^i (T^j)^T) A_\mu^j = -M^{ij}A_\mu^j, $$ since $M$ is a symmetric matrix it can always be diagonalized. Specifically in the case of $U(1)$ the only generator is the identity, and that is why the photon gets transformed into the negative of itself. In the case of SU(2) then $M^{ij}$ will look like $$ M = \begin{bmatrix}1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{bmatrix}. $$ You can imagine that there is a basis where two of the gauge bosons are exchanged under charge conjugation, where the matrix will be $$ M = \begin{bmatrix}0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{bmatrix}, $$ which is what the intuition would tell us about how the $Z$ boson only flips sign like the photon under a charge conjugation, but the $W^\pm$ bosons will get exchanged. Finally, to address more specifically your question, this transformation induces a flip sing in the structure constants $f_{ijk}$, so that the cross term of the Lagrangian that you were worried about is charge conjugation invariant.

The flip in sign of the structure constants arises since $$ \mathcal{L}^\prime=\text{tr}(\partial_\mu A^i_\nu T^i[A^{j\mu}T^j,A^{k\mu}T^k])\to \text{tr}(-\partial_\mu A^i_\nu (T^i)^T[-A^{j\mu}(T^j)^T,-A^{k\mu}(T^k)^T])\\ =-\partial_\mu A^i_\nu A^{j\mu}A^{k\mu}\text{tr}\{(T^i)^T[(T^j)^T,(T^k)^T]\}=-\partial_\mu A^i_\nu A^{j\mu}A^{k\mu}(-f_{jki})/2\,. $$


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