# Tag Info

## New answers tagged renormalization

0

There are two different ways of dealing with infinities: There is renormalization which has nothing to do with substituting some value for the cutoff. Rather, parameters of the Lagrangian are expressed in terms of measurable quantities which effectively hides the divergences. Another way of trying to get information out of these infinities is to treat a ...

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The source of confusion that prompted this question was that people seem to mean lots of things when they talk about things "flowing." Amplitudes for various processes depending on the momenta of the involved particles Coupling constants for a theory changing when we change the scale we probe the system at Coupling constants varying with the cutoff of a ...

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I'll try to explain why there could be a critical line and not just a critical point, and hopefully that will answer your question. If you think about the Ising model, we have the standard Hamiltonian: $$-\beta H = J_1\sum_{<i,j>}s_i s_j + h\sum_{i}s_i$$ where $\sum_{<i,j>}$ is a sum over nearest neighbors. This model ...

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The canonical example for MPS (in fact, the first MPS ever) is the AKLT model (http://journals.aps.org/prl/abstract/10.1103/PhysRevLett.59.799, https://projecteuclid.org/euclid.cmp/1104161001). The 2nd reference also discusses the 2D (=PEPS) version of the state. Another example of an exact MPS/PEPS model are (nearest-neighbor) RVB states (https://arxiv.org/...

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In fact, analytical continuation in some sense is subtracting off the divergence using local counter terms. A very clear treatment has been made in Terry Tao's blog: https://terrytao.wordpress.com/2010/04/10/the-euler-maclaurin-formula-bernoulli-numbers-the-zeta-function-and-real-variable-analytic-continuation/ (See section 2) Note the result of the ...

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Your example has no counterterms to cancel the 1/r singularity! This explains the discrepancy. In a correctly regularized expression, the counterterms are not introduced in an ad hoc way but by renormalizing some constants in the original problem that gives rise to the series. These modify all terms in the sum and make the divergent part vanish for ...

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I don't know much about QFT at all, but I have read some discussions of Casimir's effect in Matthew Schwartz's QFT textbook and I hope I can offer some valuable ideas about why the different regulators agree in some cases but not others. Essentially, in Schwartz's treatment, there is a class of regulators that yield the same physical result, namely the $-\... 3 If you look at the first part in your expression after integrating out$k^0$, you will find it divergent, i.e. $$\lim_{L\to\infty}\int_{L^3} \frac{d^3k}{(2\pi)^3}\left[\gamma^\mu\frac{i(p\!\!\!\big /-k\!\!\!\big /+m_0)}{2(p^0+|p|)|k|}\gamma_\mu\frac{-i}{-2|k|}|_{k^0=-|k|}\right]=\infty$$ as one can direct count the power of$|k|\$ in the integrand. This ...

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