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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 ...

2

Massive scalar fields don't have singularities in their counterterms, however they have singularities when you take $m^2 \to 0$. For example, consider the counterterm $\delta_{\lambda}$ for the massive scalar field action $$L = \frac{(\partial_{\mu} \phi_r)^2}{2} - \frac{(m \phi_r)^2}{2} - \frac{\lambda \phi_r^4}{4!} + \frac{\delta_Z (\partial_{\mu} \phi_r)^... 2 The beta function of a generic theory is typically expected to be asymptotic too. (Note the "generic": some theories with extra symmetry, e.g., enough supersymmetry, have convergent or even vanishing beta functions). For example, by instanton considerations, the beta function of \phi^4 theory is expected to be \beta_n(g)\sim(-1)^nn!\,n^{7/2}\times\text{... 2 The additional term to the QED Lagrangian is simply J_\mu A^\mu with J_\mu an external current. There is no counter-term associated with this term. The external current couples with the renormalized field A^\mu. Intuitively, the external current produces a field that can interact with electrons. The interaction contains the combined effects of the ... 2 Indeed, the eigenstates of the unperturbed Hamiltonian (\hat{H}_0) are not the eigenstates of the perturbation (\hat{H}_1) - otherwise one wouldn't need the perturbation theory, since the perturbing Hamiltonian is diagonal: it does not mix the states and merely adds a correction to the energy levels of \hat{H}_0. Let me further note that frequently ... 2 From what you suggest, the book does not claim that every quadratic Hamiltonian has an exact solution. Moreover, it’s not clear what is mean by “exact”. If you consider the Hamiltonian with arbitrary quadratics$$ \hat H=\sum_{ij} c_{ij} \hat L_i\hat L_j $$you could find an arbitrarily accurate numerical solution by diagonalizing within each (2\ell+1)\... 1 Considering the perturbation is quadratic we can not say that the full operator is linear, howeveryou can take as an Ansatz the same decomposition of h as usual:$$h = h^{(0)}+\lambda h^{(1)} + \lambda^2h^{(2)}+\cdots and each order is smaller because by assumption $\lambda\ll 1$, while checking what are the conditions needed. If one then solves for the $... 1 Generally, the non-diagonal elements are complex numbers, i.e. they may have a phase: $$\hat{H} = \begin{bmatrix} E_1 & \Delta e^{i\phi}\\ \Delta e^{-i\phi} & E_2\\ \end{bmatrix}.$$ In your case the phase is$\phi = \pi$. From the point of view of the energy levels the phase is irrelevant, as they are determined in terms ... 1 Actually, it is very simple if I am not mistaken. The idea is that any odd permutation of the$\tau$'s on the Wilson line brings a minus sign because of interchanging two generators in the trace, while the limits of integration stay the same. Adding all the diagrams together is therefore the same as integrating over the whole domain, with the$\epsilon (\...

1

This is a "how-to-make-my-QM-textbook-mathematically-rigorous" question. I usually refer these to Kato's Perturbation Theory for Linear Operators. For "reasonable" $H$ and $V$ it can be shown that $H+\lambda V$ has an eigenfunction $\Psi(\lambda)$ that varies smoothly with $\lambda$ for "small" $\lambda$. Whether $\langle \Psi(0) \Psi(\lambda) \rangle \ne 0$...

1

It’s built in the physics of the perturbation approach. Perturbation theory makes sense when the dominant term to the exact ground state is the unperturbed ground state, else it is not a perturbation. More to the point, one expects that the largest overlap $\vert \langle \Psi_0\vert \phi_k\rangle\vert$ occurs for $k=0$. if this is NOT the case, it’s no ...

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Look up the term "precession of the perehelion of Mercury's orbit" in any general relativity text. The term "gravitational force" is a thorny one in general relativity, and it does not map well to the formalism of the theory, but as mentioned in the comments, the Post-Newtonian expansion is the simplest way to apply that concept.

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