Physics Asked by user178466 on May 19, 2021
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?
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 $det square$.
Answered by J.G. on May 19, 2021
Faddeev-Poppov ghosts are brought into the picture when adding the gauge-fixing term $$1=int mathcal{D}alpha (ncdot A^a)left|text{det}frac{delta(ncdot A^a)}{delta alpha}right|$$ , 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 $$delta(ncdot A^a)=n^{mu}partial_{mu}alpha^a$$ 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.
Answered by Kenny H on May 19, 2021
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
$$ mathcal L_{GF} = -frac{1}{2xi} (S)^2 $$
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.
Answered by Luca Naterop on May 19, 2021
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