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Let $A^{\alpha}_\mu$ be the gauge field of a Yang-Mills theory where $\alpha$ is the gauge index of generators for some Lie algebra with structure constant $C_{\alpha \beta}^\gamma$ and $\mu$ is the space-time index.

According to Wiki article, the field strength tensor is given by \begin{equation} F^\alpha_{\mu \nu}:=\partial_\mu A^\alpha_\nu - \partial_\nu A^\alpha_\mu + C_{\alpha \beta}^\gamma A^\alpha_\mu A^\beta_\nu \end{equation} and the Lagrangian is \begin{equation} L_{YM} := F^{\alpha \mu \nu} F^\alpha_{\mu \nu} \text{ up to an overall normalization factor } \end{equation}

Now, expanding every term in the above $L_{YM}$, we roughly have \begin{equation} L_{YM} \simeq (\partial_\mu A_\nu^\alpha)^2 - \Bigl[C_{\alpha \beta}^\gamma A^\alpha_\mu A^\beta_\nu\partial_\mu A^\gamma_\nu \Big] + \Bigl[ C_{\alpha \beta}^\gamma C_{\alpha' \beta'}^\gamma A^\alpha_\mu A^\beta_\nu A^{\alpha' \mu} A^{\beta' \nu} \Bigr] \end{equation} where I have omitted numerical factors in each term.

Obviously the first term $(\partial_\mu A_\nu^\alpha)^2$ can be interpreted as the "free theory part". On the other hands, the two terms in $[ \cdots]$ are "self-interacting parts" Moreover, the third term in $L_{YM}$ above looks like a "quartic" interaction.

Now my question is that

What sort of interaction does the second term $-C_{\alpha \beta}^\gamma A^\alpha_\mu A^\beta_\nu\partial_\mu A^\gamma_\nu$ stand for? Is it some sort of "cubic" interaction because there are only three factors of $A$? However, presence of the derivative $\partial_\mu$ seems to complicate things...

Could anyone please clarify for me?

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    $\begingroup$ I don't know what sort of answer you are looking for. Yes, it's a cubic interaction with a derivative. What specifically do you want to know about this? Also, why does the presence of a derivative complicate things for you? That is a very standard type of interaction term. $\endgroup$
    – Prahar
    Commented Jun 2 at 13:00
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    $\begingroup$ Let's try a simpler example where you might worry about a cubic. Consider the Lagrangian $L=\frac12(\dot{A}+CA^2)^2$ with momentum $\dot{A}+CA^2$ and Hamiltonian $H=\frac12(p-CA^2)^2+\frac12C^2A^4$. There's nothing unstable about that. $\endgroup$
    – J.G.
    Commented Jun 2 at 13:32
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    $\begingroup$ @Keith - There is no such statement that theories with cubic interactions are unstable. This is something you have to analyze case-by-case. For example, the theory with interaction $L_{int} = ( \phi + \phi^2)^2$ is completely stable even though it has a cubic interaction term. $\endgroup$
    – Prahar
    Commented Jun 2 at 14:21
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    $\begingroup$ @Prahar Did you mean $L_\text{int}=(\dot{\phi}+\phi^2)^2$? $\endgroup$
    – J.G.
    Commented Jun 2 at 15:03
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    $\begingroup$ @Keith since $L=\frac12p^2$ with $p=\dot{A}+CA^2$, $H+L=p\dot{A}=p^2-CA^2p$ so $H=\frac12p^2-CA^2p$. Now complete the square. $\endgroup$
    – J.G.
    Commented Jun 2 at 15:05

1 Answer 1

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For what it's worth, if one calculates the symmetric SEM tensor $T^{\mu\nu}$ of the Yang-Mills (YM) theory with a compact gauge group $G$, one may check that the 00-component $T^{00}$ (=the energy-density) is non-negative, i.e. the energy spectrum is bounded from below, i.e. YM theory has a stable vacuum, even if this is not immediately apparent from the form of the cubic and quartic interaction terms in the action.

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  • $\begingroup$ What you explained applies for both quantum and classical theories? $\endgroup$
    – Keith
    Commented Jun 2 at 19:05
  • $\begingroup$ Well I guess I asked quite a trivial question with the above comment. Thank you! $\endgroup$
    – Keith
    Commented Jun 4 at 11:44

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