2
votes
Reasoning behind the Lorenz gauge
Your are confusing two things
The condition on $\chi$ such that the gauge transformed field
$$A'_\mu := A_\mu +\partial_\mu \chi$$
is in Lorenz gauge. This is indeed $\Box \chi = -\partial^\mu A_\mu$...
- 1,088
2
votes
Functional derivative of gauge fixing condition - Peskin QFT page 295
Hints:
Functional/variational differentiation
$$\frac{\delta\alpha(x)}{\delta\alpha(y)}=\delta^4(x\!-\!y)$$
here acts on different/independent spacetime points $x$ and $y$; not the same spacetime ...
- 213k
2
votes
BRST charge action on fields
Eqs. (1) & (2) are the standard way a Noether charge $Q$ generates a symmetry transformation, cf. e.g. this & this related Phys.SE posts.
More generally, an operator $\hat{F}$ is transformed ...
- 213k
1
vote
Reasoning behind the Lorenz gauge
If you start out with a potential that does not obey the Lorenz condition you can fix that with a gauge transformation as you correctly state. Then you still have the remaining freedom of a gauge ...
- 26.6k
1
vote
Accepted
Proving gauge transformation of non-abelian field strength
Specifically regarding the derivative. By definition:
$$
U(x) = \exp(i \alpha_a(x) T^{a}) = \sum_{n=0}^{\infty} \frac{(i \alpha_a(x) T^a)^n}{n!}
$$
Taking derivative yields
$$
\partial U = \sum_{n=0}^...
- 438
1
vote
Proving gauge transformation of non-abelian field strength
You are making heavy weather of something easy. I'll use slightly different conventions but the result is the same.
Start by defining the gauge-covariant derivative
$$
\nabla_\mu=\partial_\mu+A_\mu.
$...
- 56.5k
1
vote
Accepted
What's the form of EL equation for KG field with gauge covariant derivative?
Remember EL equations are derived from the least-action principle:
$$
\frac{\delta S}{\delta\phi} = \frac{\delta \int{d^4x \mathcal{L}}}{\delta \phi}=0
$$
Expanding this out using your Lagrangian ...
- 36
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