Tuesday, 14 March 2017

lagrangian formalism - Beta function in lambda0phi4 theory


For a real scalar field ϕ after performing all the 1-loop renormalization for dimensional regulator d=4ϵ, ϵ0+, I have found that the renormalized coupling λ can be related to the bare one by


λ(1+3(4π)2ϵλp)=λ0


I'm stuck trying to get the beta function from that equation. We call beta function to



β(λ)=μλμ


Taking into account that


λ=λpμϵ,[λp]=0,[μ]=1


My problem is that any way I use to get β(λ) from Eq. (1) gives a dependence on ϵ, but in Peskin it is used a different way to solve this and the solution is


β(λ)=3λ2(4π)2


How can I get the beta function via Eq. (1)?



Answer



The idea is that the bare quantities explicitly do not depend on μ, thus one has the equation 0=μdλ0dμ=(μ+β(λp)λp)(μϵλpZλ)=ϵμϵλpZλ+μϵβ(λp)d(λpZλ)dλp.

Where Zλ=1+3λp(4π)2ϵ.
This determines the β function as β(λp)=ϵλpZλZλ+λpdZλdλp.
Obviously you need to do this perturbatively in λ, which means that even if λ multiplies a pole in ϵ (which is divergent), you still series expand it as if it was small. If Zλ1+λ2pz/ϵ one has β(λp)=ϵλp+zλ2p1+2zλp/ϵ=zλ2p+O(ϵ)=3λ2p(4π)2.
A consistency check is that the result should never be divergent as ϵ0, this can be proved with the Callan-Symanzik equations.


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