# Tag Info

4

It's not a sufficient explanation. There are asymptotically free theories which are not strongly coupled in the IR. The rate at which the coupling gets strong is important. In QCD, it seems to get strong very quickly near the confinement scale, so that beyond a certain scale, you only see hadrons. It is not really understood how this works. The ...

1

Alex Nelson's answer is much better that mine, but it doesn't address your question at all. (a) Does it form a group? No, it doesn't. See bellow, to find out what does 'look like' (but isn't) a group. (b) What are the elements of the group? The group-like structure is the following. Being sloppy, effective action satisfies the folowing semigroup ...

7

There are really several questions here: (a) What is the renormalization group? Specifically the law of composition, etc. (b) How does the equation the OP gave relate to this? Short Answer It's a semigroup (see references below). The equation you wrote, $$\tag{1} \left[\mu \frac{\partial}{\partial \mu} + \beta \frac{\partial}{\partial g} + m \gamma_{m^2} ... 1 Let us suppose that that the Standard Model is an effective field theory, valid below a scale \Lambda, and that its bare parameters are set at the scale \Lambda by a fundamental, UV-complete theory, maybe string theory. The logarithmic corrections to bare fermion masses if \Lambda\sim M_P is a few percent of their masses. The quadratic correction to ... 1 Starting from $$\tag1 \mathcal{P}\sim\frac{ \mathrm{i} }{ p ^2 - m _0 ^2 + M ^2 ( p ^2 )}$$ and as Jeff points out, by the Optical theorem*, M^2 (for a particle that decays) can have a nonzero imaginary part. Hence one defines the physical mass m of the particle, not through m ^2 - m _0 ^2 - M ^2 ( m ^2 ) ... 2 You are certainly correct that there are other terms in the sum. However, the derivative term is zero by the renormalization conditions for the scalar field and the other terms are assumed to be small (when p^2  is far from m^2 the diagrams as small anyways). For completeness here the derivation: By performing an infinite sum over  1PI  diagrams we ... 2 Actually, in renormalization of QED, there is no demand to put the fermions of vertex diagram on mass shell. Renormalization procedure is usually performed on the level of Green functions with general four-momenta of outer legs. Note that the off-shell propagator you mention is connected to vertex function via Ward-Takahashi identity$$ (p'-p)_{\mu} ...

4

To answer this we first need to be clear about why this wavefunction renormalization arises. For simplicity we focus on $\phi^4$ theory. For a free field we have, \phi ( x ) \left| 0 \right\rangle = \int \frac{ d ^3 p }{ (2\pi)^3 } \frac{1}{ 2 E _{ {\mathbf{p}} } } e ^{ - i {\mathbf{p}} \cdot {\mathbf{x}} } \left| {\mathbf{p}} ...

1

Although the answer given by soliton is sufficient, I've found a way to explicitly evaluate this integral (in case anybody might be interested). Let us start from equation $(2)$ in the original message: I = \int\limits^1_0 \mathrm{d} x \int \frac{d^d p'_\text{E}}{(2 \pi)^d} \frac{1}{\left[p_\text{E}'^2 + q_\text{E}^2 x(1-x) + m^2 \right]^2} ...

3

It does not really say that. All the fields (not just scalars) are finite at any given point in space. The correct statement (in the context of the mentioned paper) is that for a scalar field, the correction to the squared mass term involves the average over all space of another field $\langle|A(x)|^2\rangle$ where <> indicates averaging over all space. ...

0

This can be solved by adding the non-electromagnetic energy $E_{p}$ of the Poincaré stresses to $E_{em}$, the electron's total energy $E_{tot}$ now becomes: $$\frac{E_{tot}}{c^{2}}=\frac{E_{em}+E_{p}}{c^{2}}=\frac{E_{em}+\frac{E_{em}}{3}}{c^{2}}=\frac{4}{3}\times\frac{E_{em}}{c^{2}}=\frac{4}{3}m_{es}=m_{em}$$ Thus the missing 4/3 factor is restored when ...

2

It can be seen to follow from a more general statement, namely: "if a parameter in the theory is such that the symmetry gets enhanced when it vanishes, then at every order in perturbation theory the corrections to this parameter will be proportional to its bare value". This is because perturbation theory respects the symmetry of the classical theory. If ...

4

A good analogy for the difference between the two can be given in terms of two other examples of anomalies, that are possibly more familiar. Consider a field theory with a global symmetry, take $U(1)$ for simplicity. At the classical level, the equations of motion lead to the existence of a conserved current (Noether's theorem). At the quantum level, the ...

4

The UV divergences have (most the time) nothing to do with perturbation theory (or, stated otherwise, free particles). They are also present in non-perturbative approach (see for example Non-Perturbative Renormalization Group, or Exact Renormalization Group). Divergences, or better, cut-off dependence of observables means that the quantity you are looking ...

1

This is basically second order perturbation theory. We are looking at diagrams where the external momenta are much smaller than the mass $M$. In Feynmann diagrams then the propogator for the heavy Higgs is basically $1/M^2$. This is a small parameter, so the leading contribution to the, say, $\ell \ell \rightarrow \phi \phi$ process is just a tree level ...

4

You can prove a general formula $$\int\frac{d^dp_{\mathrm{E}}}{(2\pi)^d} \frac{(p_{\mathrm{E}}^2)^m}{(p_{\mathrm{E}}^2+\Delta)^n}= \frac{1}{(4\pi)^{d/2}}\frac{\Gamma(m+d/2)\Gamma(n-m-d/2)}{\Gamma(d/2)\Gamma(n)} \left(\frac{1}{\Delta}\right)^{n-m-d/2},\quad n>m+d/2$$ by using Gaussian integral and Euler integral of the first kind.

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