8

Here's a slick way to derive the Ostragradsky instability in a higher derivative theory, which we can use to illustrate the main point. Let's consider the following higher derivative theory for a scalar field$^\star$: \begin{equation} S = \int {\rm d}^4 x \left(-\frac{1}{2} (\partial \phi)^2 + \frac{1}{M^8} (\square \phi)^4 \right) \end{equation} where $M$ ...


6

There are two perspectives (or rather two ways of saying the same thing), according to whom you ask, they might prefer one or the other: Non-renormalizable theories need an infinite number of counterterms. While that is perturbatively ok, it leads to a theory that is ultimately non predictive as all these counterterms need to be fixed by infinitely many ...


5

Quantum field theory is a very delicate thing. The history of its development is also intricately intertwined with perturbation theory because for most of its development, there were very few techniques for answering questions non-perturbatively. As a result, large swaths of the language are still tied to perturbation theory. So while the speaker you ...


4

Being IR or UV is not a property of the fixed point by itself but rather in reference to a trajectory of the RG. Excluding the possibility of limit cycles, a complete RG trajectory must start at a UV fixed point and end at an IR fixed point. Often the UV fixed point is the trivial or Gaussian one whereas nontrivial fixed points like WF in 3d arise as IR ...


2

At the end of the day, everything about RG flows, in any number of coupling-dimensions, can be understood by considering general flow equations on a manifold. A flow is defined only by a vector field which is defined everywhere on the manifold. A fixed point of such a flow is always a location where the vector field vanishes. In the special case of the RG ...


2

Below I give a general sketch of how the arguments go. For the details, see the short book written by Zamolodchikov and Zamolodchikov "Conformal Field Theory and Critical Phenomena in Two-Dimensional Systems." They are only interested in the two-dimensional case because that's where the C-Theorem can be proved, but the part of the argument I ...


2

There's no reason to expect that a true, valid-everywhere ToE would require regularization. One of the main arguments in favor of string theory is that it does not.


2

There are various views on this topic as far as I understand but I would put forth here my understanding. The basic reason as to why we would expect a quantum theory of gravity or any theory of everything to be renormalizable nonetheless is not because we expect some other theory behind the curtain from which we want to shield ourselves but because we expect ...


2

What may be wrong in Wilsonian concept of renormalisation? It is known that some theories cannot have a standard Wilsonian (ie, weakly coupled) UV completion. Here are some references: [1] https://arxiv.org/abs/hep-th/0602178 [2] https://arxiv.org/abs/1601.04068 [3] https://arxiv.org/abs/1710.09611 Which approaches may replace Wilsonian RG procedure? You ...


2

Your confusion arises from the following fact: In the situation that you describe, the beta function vanishes at some finite value of the coupling. But it does not vanish at a finite value of the energy! When you go down in energy from the UV to the IR, the coupling increases from zero to its fixed point value. But it never exactly reaches the fixed-point ...


1

All degrees of freedom contribute to correlations. Coarse-graining does not change the correlation functions and other statistical moments of a systems---it changes the Hamiltonian/action, introducing new interaction terms such that the statistical moments are preserved. Moreover, if the correlation length is infinite after renormalizing a model, then the ...


1

The second part of the question is about the Feynman propagators $G(p)$, and whether they are also annihilated by $(\gamma^\mu p_\mu - m)$. This is not the case; however applying this operator to $G$ does cancel the pole associated with that propagator, yielding a simple constant. So in your example, this eliminates $q_1$ and $q_2$ from the expression of the ...


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