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Symmetries of the action must be considered without use of the equations of motion. An on-shell symmetry is a vacuous notion - if you use the equations of motion, as you do when using eq. (5) to conclude that $\square f = 0$ should be a symmetry, then any infinitesimal transformation of the action will leave the action invariant, precisely because the ...

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As long as the ghosts are not external particles they can give non-zero contributions to amplitudes, including for boson self-energies. Recall that the Feynman diagrams are just a graphical representation of a mathematical perturbation series.

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First, some clarifications. It does not make sense to talk about "primary first class constraints." You refer to "primary" and "secondary" constraints while going through the Dirac procedure. After the procedure terminates, there is no distinction between primary and secondary constraints; indeed the difference depends on how ...

4

The action of $\Box$ on $x^\mu$ is defined by treating $x^\mu$ as a scalar (even though its not). Then, \begin{align} \Box x^\mu &= g^{\alpha\beta} \nabla_\alpha \nabla_\beta x^\mu \\ &= g^{\alpha\beta} [ \partial_\alpha \partial_\beta x^\mu - \Gamma^\lambda_{\alpha\beta} \partial_\lambda x^\mu ] \\ &= - g^{\alpha\beta} \Gamma^\lambda_{\alpha\...

3

As my2cts said the equation is not invertible. Here's a quick proof. First we go to momentum space by Fourier transforming both sides of the equation $$(-k^2 \eta^{\mu\nu}+ k^\mu k^\nu )\tilde A_\nu(k) = \frac{4\pi}{c} \tilde j(k)^\mu$$ Now the solution for $\tilde A(k)$ would be simply given by solving the equation algebrically and then doing an inverse ...

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Electric and magnetic fields are related to potentials this way $$\vec{B}=\vec{\nabla}\times\vec{A} \tag{1}$$ $$\vec{E}=-\vec{\nabla}\phi-\frac{\partial \vec{A}}{\partial t}. \tag{2}$$ A gauge transformation for those potentials is $$\phi´=\phi-\frac{\partial f(\vec{x},t)}{\partial t} \tag{3}$$ $$\vec{A}´=\vec{A}+\vec{\nabla}f(\vec{x},t), \tag{4}$$ which ...

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Independence of gauge-fixing is easiest to see in the BRST formulation. It turns out that all of the gauge-fixing (including the $\xi$-parameter) can be tucked away inside a BRST-exact part of the action, where it cannot affect BRST-invariant (=physical) observables, cf. e.g. my Phys.SE answer here.

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You might have also come across this problem when trying to invert $k^2P^{\mu\nu} \rightarrow (1/k^2)P_{\mu\nu}$. The nullspace equation for the projector, $P^{\mu\nu}(k)k_\nu = 0$, is a linear algebra equation rather than something that stems from the equations of motion. Generally, whenever an equation involves a Fourier transform from position space to ...

2

Gauge condition is any human imposed restriction on the functions $\varphi(\mathbf x,t), \mathbf A(\mathbf x,t)$ that does not change electric and magnetic field implied by those potential functions. For any single physical situation, one can use either potential functions obeying the Coulomb gauge condition or those obeying the Lorenz gauge condition. Both ...

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The reason for this is that Gauß' law is not an equation of motion, but a constraint. The Lagrangian $$L = -\frac{1}{4}\int F^{\mu\nu}F_{\mu\nu}\mathrm{d}^4 x$$ gives us the canonical momenta $\pi^\mu = F^{\mu0}$ and hence a primary constraint $\pi^0\approx 0$ ($\approx$ means an equality that hold on the constraint surface/upon use of the equations of ...

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There is no relation at all between the particle fields $\phi$ that are sections of associated bundles and the gauge field $A$. $A$ is the field of the gauge bosons, $\phi$ is a field of a particle charged under the interaction with that boson. The Standard Model does not contain particles charged in the adjoint of some gauge group, so there is no analogue ...

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This argument is still valid, for this expression of $j^\mu$ as for any other current distribution. The reason is that the LHS is invariant under $A^\mu \rightarrow A^\mu +\partial^\mu f$ for any function $f$, so there is no hope of find a general solution without fixing the gauge.

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By assuming the Lorenz gauge you also have restricted $\psi$ to obey $$\square \psi = 0 \,.$$ $\square = \partial_\mu\partial^\mu=\partial_t^2- \Delta$ up to a minus sign depending on your Minkowski metric convention. This solves your problem. By the way, I recommend to use covariant notation.

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Since you edited your question after my initial comment, I think that I now know what you were trying to ask. Any metric perturbation $h_{\mu\nu}$ (in any gauge) can be decomposed uniquely into several pieces, which correspond to the irreducible representations of $SO(3)$ as outlined here: h_{00} = -2\phi \\ h_{0i} = \beta_i + \partial_i \lambda \\ h_{ij} =...

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If your question is, as the title suggests, whether the second equation can be inverted, the answer is no. Inversion requires 'gauge fixing'. An operator can be inverted only if it establishes a one to one relation between solution and source. The operator in your second equation assigns many potentials to a single source and hence cannot be inverted.

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A gauge transformation is any transformation on $A = (\phi, \vec{A})$ that does not change any physical observable, c.q. $F$. So any transformation $A \to A + d \lambda$ with $\lambda$ any scalar field. A gauge fix is, within this gauge freedom, a particular choice for $A$. This choice can be either complete ($A$ completely fixed, no further gauge ...

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It seems you are essentially asking two questions which aren't really related. One of the two can be phrased as: given a Lie algebra-valued 1-form $A$ (whether it's a connection or not is irrelevant to this question), does there exist a spacetime vector field $v$ such that on some patch $U$ of spacetime we may obtain $A(v)=\phi$ for an arbitrary Lie algebra-...

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Indeed, non-observable quantities are not necessarily gauge invariant. Consider first the effective action, $S_{\textrm{eff}} = \int d^{4}x\, \mathcal{L}_{\textrm{eff}}(x)$. It's straightforward to see that this will be gauge invariant, but this in turn implies that the effective Lagrangian, $\mathcal{L}_{\textrm{eff}}(x)$, is only gauge invariant up to ...

1

The Coulomb gauge condition $\nabla\cdot A$ does not uniquely specify a gauge. For any given field configuration, there are still infinitely many choices of $A$ which have that property. One says that there is still *residual gauge freedom * after imposing the Coulomb gauge condition. Normally this freedom goes away if you demand that the fields vanish at ...

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Yes, it is a new constraint, but it does not follow from Maxwell's equations. Like any gauge condition, it's an optional constraint, which you can choose to obey if it makes your life easier, and disobey if it doesn't. In a way, it's a tradeoff. You can choose to take the Lorenz gauge; if you do that, the equations for the fields will become simpler, but you ...

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Just to add to Cosmas' answer. I think people are confusing gauge transformation and gauge fixing. What we are doing here is gauge fixing, which means that we constrain the fields from an arbitrary choice $(\phi\in\mathbb{C}, A_\mu)$ to a gauge fixed choice $(\phi_0\in\mathbb{R}, A_\mu)$. The point of doing this is that we remove redundancy due to gauge ...

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