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17

You need nothing more than your understanding of $$ \int_{-\infty}^\infty f(x)\delta(x-a)dx=f(a) $$ Just treat one of the delta functions as $f(x)\equiv\delta(x-\lambda)$ in your problem. So it would be something like this: $$ \int\delta(x-\lambda)\delta(x-\lambda)dx=\int f(x)\delta(x-\lambda)dx=f(\lambda)=\delta(\lambda-\lambda) $$ So there you go.


16

Well, the Dirac delta function $\delta(x)$ is a distribution, also known as a generalized function. One can e.g. represent $\delta(x)$ as a limit of a rectangular peak with unit area, width $\epsilon$, and height $1/\epsilon$; i.e. $$\tag{1} \delta(x) ~=~ \lim_{\epsilon\to 0^+}\delta_{\epsilon}(x), $$ $$\tag{2} ...


16

You're totally right. The Wikipedia definition of the renormalization is obsolete i.e. it refers to the interpretation of these techniques that was believed prior to the discovery of the Renormalization Group. While the computational essence (and results) of the techniques hasn't changed much in some cases, their modern interpretation is very different ...


15

These are all good questions. Perhaps I can answer a few of them at once. The equation describing the violation of current conservation is $$\partial^\mu j_\mu=f(g)\epsilon^{\mu\nu\rho\sigma}F_{\mu\nu}F_{\rho\sigma}$$ where f(g) is some function of the coupling constant. It is not possible to write any other candidate answer by dimensional analysis and ...


15

This is a very interesting question which is usually overlooked. First of all, saying that "large scale physics is decoupled from the small-scale" is somewhat misleading, as indeed the renormalization group (RG) [in the Wilsonian sense, the only one I will use] tells us how to relate the small scale to the large scale ! But usually what people mean by that ...


14

Lack of convergence does not mean there is nothing mathematically rigorous one can extract from perturbation theory. One can use Borel summation. In fact, Borel summability of perturbation theory has been proved for some QFTs: by Eckmann-Magnen-Seneor for $P(\phi)$ theories in 2d, see this article. by Magnen-Seneor for $\phi^4$ in 3d, see this article. by ...


14

Nowadays there exists a more fundamental geometrical interpretation of anomalies which I think can resolve some of your questions. The basic source of anomalies is that classically and quantum-mechanically we are working with realizations and representations of the symmetry group, i.e., given a group of symmetries through a standard realization on some space ...


13

I have written a pedagogical article about renormalization and renormalization group and I would be happy to have your opinion about it. It is published in American Journal of Physics. You'll find it also on ArXiv: A hint of renormalization. B. Delamotte


11

Renormalization is absolutely not just a technical trick, it's a key part of understanding effective field theory and why we can compute anything without knowing the final microscopic theory of all physics. One good online source that explains a nice physical example is Joe Polchinski's "Effective field theory and the Fermi surface" (and you can also look up ...


10

I think you raise a very important question, but I think you make it sound more trivial than it is. The point is: a lot of physicists would like to have alternative expansions, but it is very difficult to come up with one. If you've got some suggestions, don't hesitate to put it forward. The standard expansion starts from the time evolution operator ...


10

By definition, a renormalizable quantum field theory (RQFT) has the following two properties (only the first one matters in regard to this question): i) Existence of a formal continuum limit: The ultraviolet cut-off may be taken to infinite, the physical quantities are independent of the regularization procedure (and of the renormalization subtraction ...


8

I'm not sure about it, but my understanding of this is that the $\int_\Lambda^\infty$ term is essentially constant between different processes, because whatever physics happens at high energies should not be affected by the low-energy processes we are able to control. That way, we can meaningfully calculate differences between two integrals, and the ...


8

Are you essentially asking about non-perturbative approaches to QFT? Lattice QCD (based on Monte-Carlo sampling) and various strong-couplig/weak-coupling dualities (like AdS/CFT) come to mind as most prominent examples. This is more of a hint than a real answer, of course.


8

Could this imply there is a formulation where that value comes naturally... This sentence implicitly assumes that analytic continuation is "unnatural". But the truth is the other way around: analytic continuation is one of the most natural mathematical procedures in physics. On the contrary, it's functions – especially functions of momenta or energy – ...


7

It took the insights of Wilson and Kadanoff to answer this question. Universality. It doesn't matter all that much what the precise details in the ultraviolet are. Under the renormalization group, only a small number of parameters are either relevant or marginal. All the rest are irrelevant. As long as you take care to match up the relevant and marginal ...


7

Suppose I want to show $$\int \delta(x-a)\delta(x-b) dx = \delta(a-b) $$ To do that , I need to show $$\int g(a)\int \delta(x-a)\delta(x-b) dx da = \int g(a)\delta(a-b) da$$ for any function $g(a)$. $$LHS = \int \int g(a) \delta(x-a)da \ \delta(x-b) dx $$ $$=\int g(x)\delta(x-b)dx $$ $$=g(b) $$ But RHS clearly = $g(b)$ too. The result follows putting ...


6

We put the permittivity $\varepsilon=1$ to one from now on. Let us first rephrase the question a bit. Instead of starting from the potential $$\Phi=\frac{1}{r} \qquad \mathrm{and} \qquad \Phi=\frac{1}{2r^2}, \qquad r\neq 0, $$ respectively, let us assume that the electric field has be given as $$\vec{E}=\frac{\vec{r}}{r^3} \qquad \mathrm{and} ...


6

Although, I do not know if a general proof exists, I think that the Casimir effect of a renormalizable quantum field theory should be completely understood by means of a theory of renormalization on manifolds with boundary. The key feature is that one cannot, in general, neglect the renormalization of the coupling constants in the boundary terms. Using this ...


6

In some sense, i understand this question of yours as regarding to more "mathematically precise" approaches to QFT: in the end of the day, your question implies "non-perturbative definitions of QFT" in a form or another — afterall, if you can use some other tool, why not turn the problem around and define your theory based on how you can use such tool? ...


6

Check this review on the cosmological constant problem for a nice discussion of what hierarchies look like in different regularizations. Here is the rough idea. In cutoff or Pauli-Villars regularization the counterterms are sensitive to the cutoff scale(s) $\Lambda$. But there is no such scale when using dim.reg. (only the renormalization point $\mu$, which ...


6

The role of holomorphic functions (and their generalizations in the form of holomorphic sections of vector bundles) in physics is invaluable. Please see for example the following review by B.C. Hall, discussing holomorphic methods in mathematical physics, especially in quantum mechanics. It should be emphasized that these theories cover important parts of ...


5

I'll just adress that point: ... $\int_{-\infty}^{\infty} = \int_{-\infty}^{\Lambda} + \int_{\Lambda}^{\infty}$, and then ignoring the second term. Is this really okay? The integrals we are dealing with look like this one: $$\int\frac{f(p_\mu)}{g(p_\mu)}d^4p$$ Or, more explicitly: $$ ...


5

Assuming that you can actually do the dimensionally regularized integrals, then yes, you can integrate iteratively. Normally, two-loop and higher-loop integrals are quite difficult and you need good tricks like turning them into differential equations or using Mellin-Barnes parametrization of the propagators (or even just Feynman or Schwinger ...


5

The answer is no, the generalized function (=distribution) $$ \lim_{\epsilon\rightarrow 0} \frac{\epsilon^2}{\epsilon^2+t^2}=0 \qquad\mathrm{a.e.} $$ is almost everywhere (a.e.) equal to the ordinary zero function $0:\mathbb{R}\to\mathbb{R}$ that sends $t\mapsto 0$. Proof. Consider a test function $f\in C^{\infty}_c(\mathbb{R})$, i.e., an infinitely ...


5

It looks like a delta-function. However, because $\epsilon / (\epsilon^2+t^2)$ - you should omit one $\epsilon$ in the numerator, to get the right integral equal to one, by the way - decreases too slowly as $|t|\to\infty$, as $1/t^2$, it will only work as the Dirac delta distribution for test functions that decrease at infinity or at least increase slower ...


5

Your question seems to come in two parts. First, it seems the use of functional determinants to represent (formally) the result of taking a path integral may be new to you. Is that so? If it is, then I would suggest reading about this idea first. It's broader than zeta regularization. Once you're comfortable with that, then I would suggest thinking ...


5

In the case of equal masses, there is an analytical solution (of this diagram known by the name "the two loop sunrise diagram" for the obvious reason) in terms of hypergeometric functions given by O.V. Tarasov (equation 4.32). There is also a numerical method given by: Pozzorini and Remiddi. In the case of unequal masses Müller-Stach, Weinzierl and Zayadeh ...



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