2
$\begingroup$

In the book it says that in Yang-Mills theory with axial gauge: $n_{\mu}A^{\mu}=0$ using Faddeev-Popov ghosts are needless. Does anyone know how to prove this?

$\endgroup$
2
  • 2
    $\begingroup$ What book? I.e., provide a full reference. $\endgroup$ Commented Dec 21, 2017 at 18:18
  • $\begingroup$ Perform the usual FP trick and plug the axial gauge condition, observe the ghosts decouple.. $\endgroup$ Commented Dec 22, 2017 at 10:20

3 Answers 3

2
$\begingroup$

Faddeev-Poppov ghosts are brought into the picture when adding the gauge-fixing term $$1=\int \mathcal{D}\alpha (n\cdot A^a)\left|\text{det}\frac{\delta(n\cdot A^a)}{\delta \alpha}\right|$$ , where I've used the axial gauge fixing condition you are interested in. For abelian gauge theories, the determinant term will contribute a partial derivative, but for general non-abelian gauge theories it will contribute a covariant derivative. This covariant derivative depends on the gauge field and thus cannot be moved outside the integral and absorbed into the normalization. The nice thing about the axial gauge is that you will get $$\delta(n\cdot A^a)=n^{\mu}\partial_{\mu}\alpha^a$$ for the determinant term. Thus there is no dependence on the gauge-field and you can absorb the term into the normalization of the path integral.

$\endgroup$
0
$\begingroup$

It's for the same reason they're unnecessary in Abelian theories. The FP-ghost term multiplies $e^{iS}$ by $\exp-\int d^4 x \bar{c}\partial_\mu D^\mu c=\det \partial_\mu D^\mu$, a determinant that can be cancelled from the numerator and denominator of operator means in the path-integral formalism as long as it's spacetime-constant. And if the interaction is Abelian or in the axial gauge, this reduces to $\det \square$.

$\endgroup$
0
$\begingroup$

The gist of the Faddev-Popov procedure is that a gauge condition of the form

$$ G(A_\mu) = S-w(x) $$

(where in the modified Lorentz gauge $ S= \partial_\mu A^a_\mu $ or in the axial gauge $S= n_\mu A_\mu $) will produce a gauge fixing Lagrangian of the form

$$ \mathcal L_{GF} = -\frac{1}{2\xi} (S)^2 $$

and - with $(A')_\mu^a$ the gauge-transformed field - a ghost Lagrangian of the form

$$ \mathcal L_{ghost} = \bar c^a (\frac{\delta}{\delta \alpha^c} G((A')_\mu^a))c^c $$

up to some constants from the functional derivative that are being absorbed into the ghost fields. In the last equation, the $A_\mu^a$ term is the gauge-transformed field. We know that the gauge field transforms with

$$ A_\mu^a \to (A')_\mu ^a = A_\mu^a + \frac{1}{g}D_\mu ^{ac}\alpha^c$$

where $D_\mu ^{ac}= \partial _\mu \delta ^{ac} + g f^{abc} A_\mu ^b $ is the covariant derivative acting on a field in the adjoint representation . Taking the axial gauge and performing the functional derivative we end up with a ghost Lagrangian

$$ \mathcal L_{ghost}= \bar{c}^a n_\mu (\partial _\mu \delta ^{ac} + g f^{abc} A_\mu^b )c^a.$$

Here we can see that if $n_\mu A_\mu ^a = 0$ there is no more an interaction between ghosts and gauge field.

$\endgroup$
1
  • $\begingroup$ But why would $n^\mu A_\mu^a = 0$? By the time we reach your last equation in Fadeev-Popov procedure, either you have already got rid of the Dirac delta imposing the gauge condition (hence setting $n^\mu A_\mu^a = 0$) and in this case that condition should not apply; either you have not got rid of the Dirac delta yet, in which case this amounts to imposing the gauge since the beginning, which defeats the purpose of the whole Fadeev-Popov procedure. Am I missing something? Also I believe you commited a typo: the second ghost is contracted with the $b$, which otherwise remains uncontracted. $\endgroup$
    – Albert
    Commented Aug 14, 2023 at 17:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.