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 Sunset Feb16-10 03:38 PM

Beta function in Euclidean and Minkowskian QFT

Hi!

I have a question regarding the renormalization group Beta function, i.e.,
$$\beta = \mu \frac{dg_R}{d \mu}$$

where $$g_R$$ is the renormalized coupling constant and $$\mu$$ 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 $$\epsilon \rightarrow 0$$ , 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 space-time somehow. But how to do this for the beta-function??

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 Zinn-Justin's book "QFT and Crit. Phen."

One can calculate $$\beta$$ for Minkowskian QFT and for Euclidean QFT.
Ryder (using Minkowskian QFT) obtains in Dimensional Regularization $$\epsilon = 4-d$$ (page 328)
$$\beta = \epsilon g_R \mu^{\epsilon} + \frac{3}{16 \pi^2} g_R^2 + O(g_R^3)$$

Kleinert (using Euclidean QFT) obtains in Dimensional Regularization (formula (21.54))
$$\beta = - \epsilon g + 3 g^2 +O(g^3)$$
I guess in his notation $$g \equiv g_R /(4\pi)$$ , but I can't find this statement in his book.

As a third reference there's Zinn-Justin (using Euclidean QFT and Dimensional Regularization, too): (chapter 9.3)
$$\beta = -\epsilon g_R + \frac{3}{16 \pi^2} g_R^2 + O(g_R^3)$$

What seems to be different in Euclidean and Minkowskian case is the sign of the leading term $$\pm \epsilon g_R$$ , which however vanishes if one carries out renormalization, i.e., $$\epsilon \rightarrow 0$$. 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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