Why can we freely disposal the renormalization conditions?by ndung200790 Tags: conditions, disposal, freely, renormalization 

#1
Jun711, 10:06 PM

P: 520

Please teach me this:
The parameters(mass,interaction constant) in classical Lagrangian can be freely changed in classical framwork,but how about in quantum framework?Then why we can freely arrange the renormalization conditions,because I think that we do not know whether the parameters can freely be changed in quantum framework. Thank you very much in advanced. 



#2
Jun711, 11:01 PM

P: 520

And why we can put the condition: 1PI=0 at square(p)= square(M)(spacelike momentum)(e.g in the renormalization of Phi4 theory)?




#3
Jun811, 12:29 AM

P: 520

In ordinary renormalization procedure,there are some relation with m(mass),that becomes singular at m=0.Then are there any wrong with this renormalization procedure or are there exist the redudancy configuration in case m=0,so in this case there is exist an infinity?




#4
Jun811, 03:18 AM

P: 520

Why can we freely disposal the renormalization conditions?
At the moment,I see that the ''classical'' parameters relate with the quantum factors,so we can change the ''classical'' parameters,then the quantum factors are dependently changed.So that we can freely disposal the renormalization conditions.




#5
Jun811, 03:39 AM

Sci Advisor
Thanks
P: 2,139

All the parameters in QFT are not given from theory but have to be fit to experiment. Sometimes relations between quantities are constraint by symmetry principles, particularly local gauge symmetry (WardTakahashi/SlavnovTaylor identities). You find a rather detailed explanation concerning basic renormalization theory in my QFT manuscript:
http://theorie.physik.unigiessen.de.../publ/lect.pdf 



#6
Jun811, 05:14 AM

Sci Advisor
PF Gold
P: 1,942

from which one is picked by the renormalization conditions to match a real system. 



#7
Jun811, 08:38 AM

P: 520

Thanks very much for all your helpfull answers!My English still weak,I am Vietnamese.
By the way,it seem that the WardTakahashi identity is accepted at the begining.They accept the corresponding symmetry for the diagrams as a proposition.But why we are permited to do that? 



#8
Jun811, 09:43 AM

Sci Advisor
PF Gold
P: 1,942

Thus the only question is why we are permitted to use a symmetric action. The answer to that is because it proved to work! 



#9
Jun811, 10:29 PM

P: 520

It seem that the symmetry of Lagrangian is not the same symmetry of correlation functions.Then the symmetry of Lagrangian is not the symmetry of corresponding diagrams.The symmetry of the Lagrangian is the same symmetry of correlation only happens when it also is the symmetry of the product of operators of fields at fix spacetime points.




#10
Jun911, 05:48 AM

P: 520

It seem that some UV cutoff violate WardTakahashi Identity,some make the dependence on m(mass) that become singularity at m=0.Then what is the best regulation(the cuttingoff the UV divergence)?




#11
Jun911, 06:31 AM

Sci Advisor
PF Gold
P: 1,942





#12
Jun911, 08:25 AM

P: 520

What is the error we meet when we use the ''not good'' regularity?How about the ''wrong'' when we use the ordinary renormalization, in this case, two types of counterterm ''entangle'' with each other(meaning the mass couterterm delta m and scale counterterm delta Z be subtracted at the same time).In the case there is a dependence on m(mass) and it becomes singularity at m=0(Smatrix=........log(....../square(m)))




#13
Jun911, 08:48 AM

P: 520

In Phi4 theory,at one loop pertubative,with dimension regularity,there is a singularity above(...log(.../square(m))).This problem is solved by renomalization group method.But how ''wrong'' is it with the ordinary renormalization procedure?




#14
Jun911, 08:53 AM

Sci Advisor
PF Gold
P: 1,942

Doing high accuracy computations in quantum field theory outside the welltrodden paths is always a mix of art and science! 


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