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The change of the overall phase of the wave function, $\left|\Phi \right> \to e^{i\phi}\left|\Phi \right>$ is unphysical since they give the same expectation value for any observable operator. So, they represent the same physical state: indeed the space of physical states is not the Hilbert space, but rather it is the space of equivalence classes of ...


4

When you introduce a new field to make the Lagrangian gauge invariant, then you are at liberty to choose the transformation behaviour of the new field such that the Lagrangian becomes gauge invariant. If $\sigma\mapsto\sigma+\alpha$ leads to an invariant Lagrangian, then you are free to choose $\sigma$ as a field transforming such. In the situation you ...


1

Suppose I have a Lagrangian density $\mathcal{L}(\phi^\mu,\sigma)$ depending on vector fields $\phi^\mu$ and their derivatives and a scalar field $\sigma$ and its derivatives. If I make a gauge transformation $\phi^\mu\rightarrow \phi^\mu+\partial^\mu\alpha$ does the field $\sigma$ transform? I've seen notes claiming $\sigma\rightarrow \sigma + ...


5

Regardless of renormalizability, the term that you wrote down $(gA_\mu A^\mu \phi)$ does not describe photons because it is not gauge invariant. This would be a theory of a massless vector boson with three dynamical propagating degrees of freedom (two transverse and one longitudinal), which is inconsistent with Lorentz invariance and irrelevant to ...


0

This gauge invariant vector potential increases without bound as long as there exists a static electric field. Indeed, even when the electric field is removed, there appears to be no mechanism by which the gauge invariant vector potential disappears. Static electric field has zero transversal component; entire field is longitudinal. The unbounded ...



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