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ShayanJ

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2-What is the situation with QCD? Does it have conformal symmetry at a classical level? If yes, does it survive renormalization?

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ShayanJ

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2-What is the situation with QCD? Does it have conformal symmetry at a classical level? If yes, does it survive renormalization?

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The intuitive answer is simple: If you have conformal symmetry there shouldn't be any energy (or equivalently length) scale in the model, and this is the case, e.g., for QED or QCD with massless matter particles (since the gauge bosons of un-Higgsed gauge theories are massless by local gauge symmetry). Now, if you calculate radiative corrections, i.e., Feynman diagrams with loops, they diverge (for self-energy and vertex diagrams). You have to subtract the infinities in the usual way in the renormalization procedure. You can do this independently from any regularization scheme by using BPHZ renormalization, i.e., subtracting the divergent parts directly from the integrands of the loop integrals. Now since the theory is massless, you cannot choose the usual BPHZ subtraction point, with all external momenta of the diverging diagrams set to 0, because then you'd get additional infrared singularities, but you have to subtract at a point where the external momenta are chosen spacelike, and this implies that you are forced to introduce a scale, the renormalization scale, and this breakes scale invariance and thus conformal symmetry, which implies that the corresponding Ward-Takahashi identity, it.e., the vanishing of the trace of the energy-momentum tensor, is violated. That's why this anomaly is also known as the "trace anomaly".

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ShayanJ

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ShayanJ

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So does that mean that we still do not know whether QCD has conformal symmetry or not?

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samalkhaiat

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Any field theory with conserved symmetric and traceless energy-momentum tensor is conformally invariant. This happens in theories with no dimension-full parameters (coupling constants). However, in QFT’s, quantization introduce a scale (the UV-cut off) and coupling “constants” run with energy. This introduces a scale which breaks conformal symmetry. But, as it is always the case, classical symmetry casts a shadow on the quantum theory and, therefore, remains a powerful predictive tool. This happens even in ordinary QM: In atomic physics, we continue to label the states [itex]Y_{lm}(\theta , \phi)[/itex] by the eigen value [itex]l[/itex] of the [itex]SO(3)[/itex]-Casimir even though the spin-orbit coupling breaks rotational symmetry.

2-What is the situation with QCD? Does it have conformal symmetry at a classical level? If yes, does it survive renormalization?

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To fully appreciate the predictive power of the conformal group in QFT’s you need to be familiar with RG and [itex]\beta[/itex]-function: the topology of RG flow is controlled by fixed points. Fixed points are those points in the (coupling parameter)-space that have vanishing [itex]\beta[/itex]-function. If [itex]\beta[/itex] is zero, clearly the coupling is a constant, i.e., it is scale invariant and does not change with energy scale. A fixed point [itex]g_{\ast}[/itex] of the RG, therefore, corresponds to a scale-invariant (and as far as we are currently understand, conformally-invariant) QFT.

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