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The four quantum field theories (QCD, QED, QFD, and EWT) unite quantum mechanics and special relativity. They are all fully understood, complete, and proven. In your quote for the standard model, there are not four distinct field theories, the electroweak has united the electromagnetic and the weak in one field theory, the electroweak theory. The ...


3

If you have another model, then use it to generate a set of predictions, and let's compare those to the predictions we get in the world we see. If you think there's some ether filling the universe, then write down some properties it has, and make some predictions with those properties. Talking in vague generalities will only lead you down a rabbit hole of ...


2

I think this would be tricky, since any force mediator (at least from conventional thinking) must have a three-valent vertex, two of which are the charged object and one of them is the force carrier. If the force carrier is a fermion, I don't think this combination can be Lorentz invariant (spin zero combination).


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Some terms in $H_0$ can be absorbed into the definition of $\epsilon$. To understand the validity of this substitution, you must go back to chapter 7 and calculate according to eqn. (7.2)-(7.6). You will find "we can absorb the positive coefficient into $\epsilon$ to get..."


2

No, the gauge current need not be gauge invariant, since it carries a group index in non-abelian theories. You should recall that both sides of the Yang-Mills equation (and therefore the current itself) are Lie-algebra valued and therefore transform in the adjoint representation. Not even the field strength $F^a_{\mu\nu}$ is gauge invariant, but transforms ...


1

No. Feynman diagrams are made by summing over the perturbative contributions of quantum amplitudes. They cannot hold non-perturbative information.


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As others pointed out, the statement probably means that if you want to have a nonzero contribution from creation and annihilation operators acting on vacuum, you need to apply the creation operator first, since $\hat{a}|0\rangle = 0$, in other words, you cannot annihilate if you have nothing. In the question you link to, the situation is different, as that ...


1

'Square root of geometry' A Dirac spinor field $\psi^\alpha(x)$ under Lorentz transformations behaves as, $$\psi^\alpha(x)\to S[\Lambda]^\alpha_\beta \psi^\beta(\Lambda^{-1}x)$$ where $\Lambda = \exp(\frac{1}{2} \Omega_{\rho\sigma}\mathcal{M}^{\rho\sigma})$ and $S[\Lambda] = \exp(\frac{1}{2}\Omega_{\rho\sigma}\mathcal{M}^{\rho\sigma})$, where ...


1

The kinetic term $$\mathcal{L}_\text{fermion} = i\overline{\psi}\gamma^\mu \partial_\mu \psi$$ changes too, by precisely the right quantity to cancel the change in the interaction term. Thus, the total Lagrangian is invariant, and this is what matters.


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As @RobinEkman mentioned, the kinetic term changes as well. This can be easily computed \begin{equation} \begin{split} {\cal L}_D = i {\bar \psi} \gamma^\mu \partial_\mu \psi &\to i {\bar \psi} e^{- i \alpha} \gamma^\mu \partial_\mu \left( e^{i \alpha} \psi \right) = i {\bar \psi} \gamma^\mu \partial_\mu \psi - \partial_\mu \alpha ( {\bar \psi} ...



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