Beta function in Euclidean and Minkowskian QFT

[Sunset]

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 $$\epsilon \rightarrow 0$$ , referring 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?


Martin

 

[eljose79]

Hi Martin,

Thank you for your question regarding the renormalization group Beta function. The answer to your question is yes, the Beta functions calculated for QFT and Euclidean QFT are exactly the same, as long as the same renormalization scheme is used. This is because the Wick rotation from Minkowskian to Euclidean space does not change the underlying physics, it only changes the mathematical representation.

In terms of the specific example you mentioned, the sign difference in the leading term of the Beta function is not a typo and is indeed the only difference between the Minkowskian and Euclidean Beta functions. This is because the sign difference arises from the definition of the renormalized coupling constant, which is different in Minkowskian and Euclidean QFT. However, as you correctly pointed out, this sign difference is irrelevant in the limit of \epsilon \rightarrow 0 and does not affect the physical predictions of the theory.

In order to obtain the Minkowskian Beta function from the Euclidean one, one can simply perform the Wick rotation back to Minkowskian space. This is equivalent to analytically continuing the Euclidean Beta function to Minkowskian space. This can be done by replacing the Euclidean momentum p_E with its Minkowskian counterpart p_M = i p_E, which then leads to the same Beta function as calculated in Minkowskian QFT.

I hope this answers your question. Let me know if you have any further queries.