parameter μ dimensional regularization qed

paolorossi

in the dimensional regularization of the QED we introduct an arbitrary parameter μ with the dimension of a mass... so there are finite terms that are function of μ... so they are arbitrary??? how we can fix this parameter? what is the physical meaning of μ?

xepma

They are absorbed into the renormalized coupling constants and field renormalization.

paolorossi

a book to view this???

andrien

Do you mean for photon mass which is definitely taken zero at the end of calculation because the integrals are divergent without it and photon should not have any mass.

paolorossi

I speak about the parameter μ that we introduct in the extension of the 4 dimensional minkowsky space to a D dimensional space. This parameter is such that the Lagrangian density of interaction has the correct physical dimension M^D in D dimensions.

But when we return to 4 dimension some terms depend on μ. xepma say that , but how I can see that? can somebody suggest a book or paper?

(sorry for my english)

andrien

So what exactly is μ.

paolorossi

Is it a question?
μ is the arbitrary parameter, with mass dimension, which appears in the Lagrangian density of interaction in D dimesion

L_I = e μ ^2-D/2 [itex]\bar{\psi}[/itex][itex]\gamma[/itex]^{[itex]\mu[/itex]}[itex]\psi[/itex] A_{[itex]\mu[/itex]}

so that L_I has the correct dimension M^D (note that the fields have the dimension [ A_{[itex]\mu[/itex]} ] = M^D/2-1/2 , [ [itex]\psi[/itex] ] = M^D/2-1 ).
My question is: if the counter-terms are such that only the divergent terms are "eliminated" in the corrections to higher orders (propagators and vertex) , then there are finite terms that are function of μ. But μ is arbitrary. So how do you solve this? maybe there is a way to fix μ?

andrien

it is clear that when D=4(as is the case) ,μ does not play any roll after all.But only a tool to handle regularization in arbitrary dimension.(I thought it to be some photon mass type thing which is taken zero at end)

paolorossi

yes, what you say is correct, in fact, the bare interaction Lagrangian density will not have to depend on the parameter μ (in the limit D=4) ... so if we express the lagrangian in terms of physical quantity (renormalizated charge, mass , ... ) then these physical parameters must transform adequately under a change of the parameter μ, so that the theory (ie the amplitudes calculated) are independent from the choice of μ. that is, if we make the transformation

μ -> μ'

then, for example, the charge change as

e(μ) -> e(μ')=e'

so that, for example, the vertex (plus radiative corrections) remains unchanged. I saw that the set of this transformation forms the so-called " renormalization group" (RG). thanks anyway for answer!