3
$\begingroup$

When quantizing YM we start from the gauge fixed path integral (to remove redundancy of integrating over Gauge symmetric configurations) $$\int \mathcal{D}A \delta(G(A)) \text{det} \Delta_{FP}e^{i\int d^4x \mathcal{L}_{YM}} \tag{1}$$ Now, one would rewrite the delta and introduce ghosts to the theory by raising the resulting term into the exponential. Which would then be BRST symmetric. I assume BEFORE we did this, the BRST symmetry was hidden in the Gauge symmetry of the Lagrangian.

But where is the BRST symmetry in the above gauge fixed integral? We haven't introduced any ghosts particles yet. Adding a source term would change the above integral into a generating functional defining our whole theory WITHOUT any ghosts appearing.

$\endgroup$
3
$\begingroup$

We do not start from the gauge fixed path integral in the BRST construction. What you describe (once one adds the missing Faddeev-Popov determinant) is the original Faddeev-Popov trick to get the ghosts, not the systematic BRST construction. The (Hamiltonian) BRST construction crucially first introduces the ghosts as parts of the extended phase space, and then chooses a gauge-fixing fermion and a BRST-closed correction to the Hamiltonian (the latter is not there for closed constraint algebras) to achieve a BRST-invariant action.

Even in the Lagrangian/path integral form, the BRST transformation involves mixing the ghosts and the gauge fields. The version of the BRST symmetry without ghosts is the gauge symmetry itself. There is nothing hidden about the ghost-free action - the "unghostly" gauge symmetry is equivalent to the "ghostly" BRST symmetry. The whole point of the BRST procedure is that it turns out to be more useful (e.g. due to the powerful nilpotency constraint on the BRST operator) to describe the gauge symmetry by adjoining the ghosts to the phase space and demanding BRST invariance on the extended space instead of gauge invariance on the original space.

$\endgroup$
2
$\begingroup$

The BRST symmetry cannot be seen without introducing auxiliary variables. The fastest way to realize the BRST symmetry is to "exponentiate" the delta function $$\delta(G)~=~\int \!{\cal D} B ~\exp\left[iB_{\alpha}G^{\alpha}\right]$$ and the Faddeev-Popov (FP) determinant $$\det\Delta ~=~\int \!{\cal D} c ~{\cal D} \bar{c} ~\exp\left[\bar{c}_{\alpha} \Delta^{\alpha}{}_{\beta} ~c^{\beta}\right]$$ by introducing Lagrange multipliers $B_{\alpha}$ and FP Grassmann-odd ghosts $c^{\alpha}$ and antighosts $\bar{c}_{\alpha}$ in the action, see. e.g. this Phys.SE post (for abelian gauge group).

And voila: the extended action (with the auxiliary variables $B$, $c$, $\bar{c}$) has BRST symmetry!

Of course the BRST symmetry encodes nothing but the original gauge symmetry, cf. this Phys.SE post.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.