Hi!(adsbygoogle = window.adsbygoogle || []).push({});

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 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 [tex] \beta [/tex] for Minkowskian QFT and for Euclidean QFT.

Ryder (using Minkowskian QFT) obtains in Dimensional Regularization [tex]\epsilon = 4-d [/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 Zinn-Justin (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

**Physics Forums - The Fusion of Science and Community**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Beta function in Euclidean and Minkowskian QFT

Can you offer guidance or do you also need help?

Draft saved
Draft deleted

Loading...

Similar Threads - Beta function Euclidean | Date |
---|---|

Current conservation beta function | May 18, 2015 |

Strong force and beta function | Jan 5, 2015 |

Invariance of beta function in dimensional regularization? | Dec 15, 2013 |

How to determine SU(3) multiplicity in beta function | Jun 20, 2013 |

How to determine quantum numbers for beta functions? | Jun 18, 2013 |

**Physics Forums - The Fusion of Science and Community**