Tag Info

0

Each coupling and each field normalization factor potentially gets quantum corrections due to the integration over high-momentum modes You can always consider strongly connected diagrams with appropriate external legs (e.g. two electrons for the electron normalization constant or two electrons and a photon for charge normalization). In other words, on tree ...

1

I recently stumbled upon a good comment about this in Jared Kaplan's AdS/CFT notes Any quantum field theory which has hope of having an UV-completion can be viewed as as effective theory at point in the RG flow from an UV complete theory. Field theories at the UV fixed point are conformal. Hence all 'well-defined' field theories are either CFTs or points ...

1

For your first question: the physical mass is the measured mass. If $m$ is not the measured mass, don't call it the physical mass, because it isn't. If someone does, correct them. :-) For your second question, I think we should look at the physical interpretation of all those mathematical manipulations. The hierarchy problem (as I understand it) comes from ...

2

Yes tensor network algorithms have been developed to describe braiding of anyons, Abelian and non-Abelian. The networks are constructed from tensors that explicitly conserve topological charge and the braiding, fusion, and recoupling data are taken as input to the algorithms. This reference: http://arxiv.org/pdf/1311.0967.pdf describes how to use the Time ...

0

Putting the fermions on-shell or off-shell doesn't change the divergent part of counterterms. In the renormalization schemes, the counter terms are determined to cancel the divergencies. You can put $p^2=m^2$ or $p^2=-\mu^2$ or etc in the diagrams to determine the counter terms, but notice that the derived renormalization constants, $Z_1$, $Z_2$, etc, should ...

3

Consider for example, simple $\lambda \phi^3$-theory with Lagrangian $$\mathcal{L}=-1\frac{1}{2}\left|\partial_\mu\phi\right|^2-\frac{m^2}{2}\phi^2+\frac{\lambda}{3!}\phi^3.$$ One can say that the $\lambda \phi^3$ term renormalizes the mass term, because the regularization and renormalization of the divergence of the one-loop diagram will lead to a ...

1

I think the confusion is due to a lack of mathematically precise definitions of what is quantum field theory? what is one trying to construct and how? etc. There are a lot of vague notions used in the physics literature: the partition function (which does not make much sense in infinite volume), the effective action,...but the bottom line is the collection ...

0

If a theory contains divergences, there is no need to assume it is plain wrong. It probably won't be the full theory of everything, but that's the only criticism that can be leveled at it. It can still be in perfect agreement with experiments below very-high-energy. The issue with QED is that QED is not a complete theory. In the real world, QED is just a ...

0

For example, when we meassure Higgs boson mass to be 125 GeV, do we think about renormalized or pole mass? Pole mass is the physical mass and independent of any renormalization scheme we use to subtract any infinite parts of the loop corrections. It is what we observe. Should the mass of the Higgs change if it is produced at higher energies? So ...

3

Schwartz is simply noting that the $\beta$-function has a generic expansion in QED of the form (29) where $\beta_{0,1,2}$ are some numbers that can be computed by explicitly calculating the various Feynman diagrams. For instance the leading $\epsilon/2$ is the tree-level result in $d=4-\epsilon$ dimensions. This can be easily seen as follows. In ...

1

He is writing the $\beta$ function as a perturbation series in the small parameter $\alpha$, $$\beta(\alpha) = \sum_{n=1}^\infty c_n \alpha^n.$$ By convention, it's common to write this expansion in the form he has given. The first coefficient is always $c_1 = - 4\pi\epsilon$ (where $\epsilon$ is the dimensional regularization parameter). The remaining terms ...

0

$Λ_{QCD}$ is measured in processes where the strong coupling constant and other measurables vary with momentum scale $Q$. For instance, evolution of nucleon structure functions measured in lepton-nucleon deep-inelastic scattering, heavy quarkonia decays, collider jet physics, electroweak physics at the Z, ... Most results are in the 200 to 300 MeV range. ...

1

No. Gauge invariance is not a real physical symmetry but a mathematical property of the formalism while renormalization is more deep property related to the scaling of the coupling constant. One can think interaction that breaks gauge but is still renormalizable. It is fact that QED and QCD are gauge theories but gravity is a gauge theory and not ...

5

Yes, that is the Higgs potential of the Standard Model. Note that a $\phi^3$ term is forbidden by symmetry (it would not be an $\mathrm{SU}(2)$ scalar), and $\mathcal{O}(\phi^5)$ terms would be non-renormalizable, so this is really the only potential we can write down that does not need other new physics.

3

In the lecture notes, the Feynman diagrams for the two examples of $\phi^3$ in $D=6$ and $\phi^6$ in $D=3$ are mixed up. You can easily identify this by noticing that in a scalar $\phi^n$ theory $n$ lines meet in all vertices. The example graph for $\phi^6$ theory in the lecture notes (which is wrongly listed as an example of $\phi^3$ in $D=6$) has six ...

1

It turns out to be a mistake in this version of the book (2nd edition), I checked the latest version (3rd edition) and it has been corrected there. Naturally, the corrected one would be $$R(\vec{K}^*+\vec{k})=R(\vec{K}^*)+R'(\vec{K}^*)\vec{k}+...$$ which after linearization simply becomes $\vec{K}^*+R'(\vec{K}^*)\vec{k}$, and then we have a linearized ...

3

There are at least two answers possible to give, but both, in the end, amount to the same thing: There is no "right" way to fix the energy scale of a process, but that doesn't matter, except that your perturbation theory will probably break if you choose the scale badly. The old answer: The renormalization scale is arbitrarily defined to fix some parameters ...

1

You go to the center of mass frame to find that $\sum_i \vec{p}=\vec{0}$, and the total momentum four vector is thus $$P_{\text{tot}}^{\mu}=\left(\frac{1}{c}\sum_{i}E_i^{\text{COM}}, \vec{0}\right)$$ then we define the energy scale covariantly as $\mu=\sqrt{-s}$ where $s$ is the mandelstam variable $s\equiv P_{\text{tot}}^{\mu}P_{\text{tot}\mu}$ in units of ...

Top 50 recent answers are included