# Slavnov operator on the Lagrangian, does it always cancel? With example [duplicate]

I have a long Lagrangian when I apply the Slavnov operator all terms cancel except for the Gauge fixing term and the ghost term. I am using an unusual gauge fixing condition, $$F=(\partial_\mu + \frac{\lambda}{2}A_\mu) A^\mu.$$ My question is should the Slavnov operator always cancel out no matter what gauge condition is used?

$$\mathcal{L}_{eff} = -\frac{1}{4} F_{\mu\nu}F^{\mu\nu} -\frac{1}{2\zeta} \left( \partial_\mu A^\mu + \frac{\lambda}{2} A_\mu A^\mu \right)^{2} + \overline{\omega} \left(\partial_\mu \partial^\mu + \lambda A_\mu \partial^\mu \right) \omega.$$

Where I have used $$sA_\mu = \partial_\mu\omega$$, $$s\overline{\omega}=\dfrac{\partial_\mu A^\mu + \frac{\lambda}{2} A_\mu A^\mu}{\zeta}$$ and $$s\omega = 0$$.

When I apply $$s\mathcal{L}_{eff}$$ I should expect to obtain $$0$$, right?

But I obtain

$$s\mathcal{L}_{eff}= -\frac{1}{\zeta} F\left( \partial_\mu \partial^\mu + \lambda A_\mu \partial^\mu \right) + \frac{F}{\zeta}\left(\partial_\mu \partial^\mu + \lambda A_\mu \partial^\mu \right)\omega +\overline{\omega}\lambda \partial_\mu \omega \partial^\mu \omega = \overline{\omega}\lambda \partial_\mu \omega \partial^\mu \omega$$

Any ideas?

• I should add that the gauge fixing condition is $F=\partial_\mu + \frac{\lambda}{2}A_\mu A^\mu$ – pablo bilbao Feb 8 at 18:00
• Hi Pablo, in the future, whenever you need to add more information to your question, you should edit the post itself. I've added the gauge fixing condition for you. – Urb Feb 8 at 18:25
• Hi pablo bilbao. Welcome to Phys.SE. Please don't repost a question in a new entry. Instead, you are supposed to edit the original question within the original entry. – Qmechanic Feb 8 at 19:26
• Possible duplicate by OP: physics.stackexchange.com/q/613197/2451 – Qmechanic Feb 8 at 19:29
• @Qmechanic I am very very sorry I will leep that in mind for next time. – pablo bilbao Feb 8 at 22:22