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6

That's a completely valid diagram, but it's higher order in perturbation theory. The 3-point vertex comes with a factor of $e$, and the 4-point vertex comes with a factor of $e^2$, so your diagram is order $e^4$, which is higher than the order $e^2$ contributions calculated in standard textbooks. Note that for the purposes of computing corrections to the ...

3

You do not need QED for this, Newtonian dynamics of two point charges will suffice, since the theory still has time reversal symmetry. The solution to your conundrum is that there is a difference between a symmetry of a physical law, and a symmetry of a particular solution admitted by that law. For example, the hydrogen atom Hamiltonian has spherical ...

2

The idea is that the quadratic part of the action should be invertible and the inverse gives the propagator of the field. Eq. 9.52 is trying to solve for the inverse $\tilde{D}^{\nu\rho}_F$, but you cannot because $-k^2g_{\mu\nu} + k_\mu k_\nu$ is singular. That's the exact same as saying that the quadratic action vanishes for too many field configurations ...

2

The additional term to the QED Lagrangian is simply $J_\mu A^\mu$ with $J_\mu$ an external current. There is no counter-term associated with this term. The external current couples with the renormalized field $A^\mu$. Intuitively, the external current produces a field that can interact with electrons. The interaction contains the combined effects of the ...

2

The electrical power does not only come from a flow of electrons, but also because something pulls them : an electric field. We model this by defining the electric power received by a component as being the product of the current flowing through it by the voltage applied to it (in the opposite direction, by convention). As for whence the energy came from ...

2

It is because the Ward identity applies to S-matrix elements, for which particles must be on mass shell. A simple calculation shows that with only one vertex, energy-momentum conservation is not consistent with the mass shell condition for all particles.

2

The elephant in the room is the following question: How can we treat $\frac{e^2}{4\pi^2\epsilon}$ as subleading as compared to $1$? In dimensional regularization the parameter $\epsilon$ is supposed to be small. The brief answer is that renormalization is first-and-foremost a perturbative formal power series in the coupling constant $e^2$. Secondly, each ...

2

This are what are called bilinear covariants. Any object of the form $$\bar\psi\Gamma\psi$$ where $\Gamma = \{1, \gamma^\mu, \sigma^{\mu\nu}, \gamma_5, \gamma^\mu\gamma_5\}$, transform in a different way. Note that these are $16$ linearly independent $4\times 4$ matrices, this means that the furnish a basis for all the Clifford algebra. Now you can see ...

2

If by "nuclear transitions" you mean changing the internal state of the nucleus (besides realigning its net spin), then the answer is unlikely, but potentially not impossible. The fundamental issue here is that typical energy scale of the nucleus is so high compared electronic energies. The electron-nuclear spin coupling (hyperfine splitting) is already a ...

1

This is indeed a Lorentz vector. We are used to seeing a 4-vector which is a representation of group $SO(3,1)$(i.e. the Lorentz group). But we can write the same group locally as $SO(3,1)\sim SU(2)\times SU(2)$. Now the spinors $\psi$ are spin-$1/2$ representations of $SU(2)$ group. Thus we can say the left and right spinor basicallt are the $(1/2,0)$ and $(... 1 In all aspects of nature, movement (and more general stuff too) arises due to what physicists call potential differences. That's why one calls them potential: Because they have de potential of creating kinetic energy. In the case of gravitational potential energy, for instance, it can be converted to kinetic energy by getting rid of normal reaction forces (i.... 1 As per @Qmechanic's suggestion, you should have gotten $$0= {\epsilon \over 2} e \Bigg(1+{e^2 \over {12\pi ^2 \epsilon}}\Bigg)+\mu {\partial e \over \partial \mu } \Bigg(1+{3e^2 \over {12\pi ^2 \epsilon}}\Bigg),$$ i.e.$$\mu {\partial e \over \partial \mu }=-{\epsilon \over 2}e \frac{1+{e^2 \over {12\pi ^2 \epsilon}}}{1+{3e^2 \over {12\pi ^2 \epsilon}}}... 1 I'll talk about nonrelativistic quantum. The fundamental object at play here is the rotation group$SO(3)$. (If you want to talk about spin 1/2 objects, you need to talk about$SU(2)\$, but let's start slowly.) You correctly described how a scalar field transforms, and you want to know about how vector fields transform. For now, forget about fields of any ...

1

This is not a full answer but hopefully it is helpful to you. The S-matrix is defined as a change of basis between the 'in' and 'out' states, both span the same Hilbert space. So, by definition, if an S-matrix exists, it must be unitary. It remains to show that the 'in' and 'out' states can be constructed for a given theory. Certain restrictions on the ...

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