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- Thread starter eljose79
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selfAdjoint

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The Callan-Symanzik equation is not a part of renormalization per se. It is a key part of **Renormalization Group** theory, which is someting different erected on the same base. In Kaku's Quantum Field Theory textbook he discusses the derivation of the Callan-Symanzik relation on p. 486, in the chapter QCD and the renormalization group.

"We begin with the obvious identity that the derivative of a propagator with respect to the unroneomalized mass=squared, simply squares the propagator.,,,

"Now assume that [the propagator] occurs in some vertex function .. of arbitrary order. From a field theory point of view, the squaring of the propagator (with the same momentum) is equivalent to the insertion of an operator φ^{2}(x) in the diagram with zero momentum.

"This means that the derivative of an arbitrary vertex function with respect to [unrenomalized mass squared] yields another vertex function where φ^{2}(x) has been inserted.

"We now make the transition from the unrenormalized vertices to the renormalized ones. This means the introduction of yet another renormalized constant Z_{φ2} to renormalize the insertion of the composite field operator."

He then does the math, simple algebra and calculus. The result is a partial differential equation stating that the linear combination of the derivatives of a vertex function with respect to normalized mass and to the coupling stength is equal to the renormalized mass squared times the same (renormalized) vertex function with φ squared inserted.

I won't reproduce the equations here, but you can see how they were derived by working back and forth between the unrenomalized Feynmann diagrams and the renormalized ones.

"We begin with the obvious identity that the derivative of a propagator with respect to the unroneomalized mass=squared, simply squares the propagator.,,,

"Now assume that [the propagator] occurs in some vertex function .. of arbitrary order. From a field theory point of view, the squaring of the propagator (with the same momentum) is equivalent to the insertion of an operator φ

"This means that the derivative of an arbitrary vertex function with respect to [unrenomalized mass squared] yields another vertex function where φ

"We now make the transition from the unrenormalized vertices to the renormalized ones. This means the introduction of yet another renormalized constant Z

He then does the math, simple algebra and calculus. The result is a partial differential equation stating that the linear combination of the derivatives of a vertex function with respect to normalized mass and to the coupling stength is equal to the renormalized mass squared times the same (renormalized) vertex function with φ squared inserted.

I won't reproduce the equations here, but you can see how they were derived by working back and forth between the unrenomalized Feynmann diagrams and the renormalized ones.

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