# Tag Info

## Hot answers tagged non-perturbative

5

QED is non-perturbative because it is not defined by the sum over Feynman diagrams. Standard QFT does not rely on the perturbative expansion to define scattering amplitudes, it only uses it to compute them. String theory through CFT, on the other hand, defines the string scattering amplitude through the sum over worldsheets. This is not a perturbative ...

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1) As you know, $$\tag 1 \theta\epsilon^{\mu \nu \alpha \beta}F_{\mu \nu}F_{\alpha \beta} = \theta\partial_{\mu}K^{\mu},$$ where $K_{\mu}$ is the so-called Chern-Simons class. The Feynman diagrams method tells us that the term $(1)$ defines the diagram which corresponds to the two-photon (or two-three-four non-abelian bosons) vertex $V_{A}$ (where the ...

3

There are many phenomenon in quantum field theory that falls outside the understanding of perturbative analysis. To understand the nonperturbative physics is to understand the full dynamics of the theory whereas perturbation theory is not reliable. The basic examples are 1) dynamical symmetry breaking (supersymmetry or chiral symmetry) which is usually ...

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Perturbative theory is understood as an application of the Taylor expansion around a linearity $g=0$ (unperturbed theory) in an underlying theory: $$f(g)=\sum_{n}A_n g^n$$ The unperturbed theory could be, for example, an Hydrogen Atom for the underlying QED. Then, the QED perturbative theory is a prescription of how inserting $g\neq0$ corrections of the ...

3

In general, for computing the scattering cross-section it is sufficient to use the pole approximation of the corresponding Green function in Heisenberg representation (details can be found in Weinberg's QFT Vol. 1, paragraph 10). This is nonperturbative approach, since poles are invisible in perturbation theory: they define bound states masses. To find them, ...

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In general the theory with nontrivial chiral structure isn't invariant under chiral transformations of the fermion field; corresponding phenomena is called anomaly, and it leads to nonconservation of corresponding current. Formally it is related to the fact that there is no way to define bot gauge invarian and chiral invariant regularization for such class ...

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Rather tautologically, a non-perturbative effect is one that is invisible to perturbation theory. An effect is invisible to perturbation theory exactly if it is in a non-analytic part of the partition function with respect to the coupling constant $g$. Observe that perturbation theory is essentially the Taylor expansion of the partition function $Z$ (or ...

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1) $\exp(-1/g)$ is not necessarily related to bound states. In the standard QM double well problem it is the splitting, not the binding energy, that is $O(\exp(-1/g))$. In conformal field theories instantons can give $\exp(-1/g)$ effects even though there are no bound states at all. 2) Instantons are one source of $\exp(-1/g)$ effects, but there are others. ...

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The bare coupling vanishes in the continuum limit, because the renormalized quantities are associated with measurements over very large scales (compared to the lattice spacing). One way to think about this is that since QCD is confining, the coupling at the scale of the lattice spacing must vanish if the coupling at the continuum scale is to remain finite. ...

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The baryon non-conservation in the quantum theory is caused by the chiral anomaly, which leads to the baryon current $j_B^\mu = \sum_i \bar q_i \gamma^\mu q_i$ being not conserved as $$\partial_\mu j_B^\mu \propto F\wedge F$$ where $F$ is the field strength tensor of the electroweak field. Now, $\int \partial_\mu j_B^\mu \propto \int F\wedge F$ is a ...

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Here is one possible way to evaluate this expression using FeynCalc: << FeynCalc` SUNF[a1, a2, a3] SUNF[a4, a2, a7] SUNF[a7, a8, a1] SUNF[a5, a6, a3] * SUNF[a9, a4, a5] SUNF[a8, a9, a6] // SUNSimplify[#, Explicit -> True, SUNNToCACF -> False] & // Simplify and the answer is: 1/4 SUNN^3 (-1 + SUNN^2) i.e. what the OP got from doing the ...

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