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Beta function in Euclidean and Minkowskian QFT 
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#1
Feb1610, 03:38 PM

P: 63

Hi!
I have a question regarding the renormalization group Beta function, i.e., [tex]\beta = \mu \frac{dg_R}{d \mu}[/tex] where [tex]g_R[/tex] is the renormalized coupling constant and [tex] \mu [/tex] the renormalization scale. My question in a nutshell: are the Beta functions calculated for QFT and, respectively, Euclidean QFT exactly the same (maybe only in the limit where [tex] \epsilon \rightarrow 0[/tex] , refering to Dimensional Regularization)? I.e., if I perform the Wick rotation to the Euclidean theory, does the beta function change? If so, can one conclude the Minkowskian Beta function directly from the Euclidean one? I would expect that one has to rotate back to Minkowskian spacetime somehow. But how to do this for the betafunction?? To be more explicit: For simplicity I restrict myself to phi^4 theory. I will refer to Ryder's book "Quantum field theory", Kleinert's book "Quantum Field theory and Particle physics" and ZinnJustin's book "QFT and Crit. Phen." One can calculate [tex] \beta [/tex] for Minkowskian QFT and for Euclidean QFT. Ryder (using Minkowskian QFT) obtains in Dimensional Regularization [tex]\epsilon = 4d [/tex] (page 328) [tex] \beta = \epsilon g_R \mu^{\epsilon} + \frac{3}{16 \pi^2} g_R^2 + O(g_R^3) [/tex] Kleinert (using Euclidean QFT) obtains in Dimensional Regularization (formula (21.54)) [tex] \beta =  \epsilon g + 3 g^2 +O(g^3) [/tex] I guess in his notation [tex] g \equiv g_R /(4\pi) [/tex] , but I can't find this statement in his book. As a third reference there's ZinnJustin (using Euclidean QFT and Dimensional Regularization, too): (chapter 9.3) [tex] \beta = \epsilon g_R + \frac{3}{16 \pi^2} g_R^2 + O(g_R^3) [/tex] What seems to be different in Euclidean and Minkowskian case is the sign of the leading term [tex] \pm \epsilon g_R [/tex] , which however vanishes if one carries out renormalization, i.e., [tex] \epsilon \rightarrow 0[/tex]. Is this true, or is the + a typo in Ryder's book? If the + is true: is this the only difference between the Euclidean and the Minkowskian Beta function? Best regards, Martin 


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