
#1
May1610, 09:16 AM

P: 411

Hi,
I'm reading through CH27 of Srednicki at the moment, and struggling to understand a couple of concepts. 1) He states that in the MS (bar) scheme the location of the pole in exact propagator is no longer when [tex] k^2=m^2 [/tex], where m is Lagrangian parameter usually thought of as mass. I think I understand this because we no longer are imposing the conditions [tex] \Pi(m^2)=0 [/tex] etc, as in the OS scheme, so of course there is no pole in the exact prop when [tex] k^2=m^2 [/tex]. However he then goes on to say the physical mass, [tex]m_{ph}[/tex] is defined by the location of the pole: [tex] k^2=m^{2}_{ph}[/tex]. Why is the physical mass defined this way? what's so special about the place where there is a pole in exact propagator? How do these things tie in with the LehmanKallen form of the exact propagator that clearly shows there must be a pole when [tex] k^2=m^2[/tex], is this 'm' in the LehmanKallen formula [tex] m_{ph}[/tex]? 2) He then states the LSZ formula must be corrected by mutliplying it's RHS by a factor of [tex] \tfrac{1}{\sqrt{R}}[/tex] where R is the residue of the pole, and the reason he gives is that it is the field [tex] \tfrac{\phi(x)}{\sqrt{R}} [/tex] that has unit amplitude to create a one particle state. I have no idea why this is, and it would be really great if anyone could explain some more. 3) My final question is how he gets to 27.12 just by taking the log of 27.11: He starts with, [tex] m^{2}_{ph}=m^2[1+\tfrac{5}{12}\alpha(ln(\mu^2/m^2)+c')+O(\alpha^2)] [/tex] Now taking logs: [tex]2ln(m_{ph})=2ln(m)+ln[1+\tfrac{5}{12}\alpha(ln(\mu^2/m^2)+c')+O(\alpha^2)] [/tex] My only thought is that in the second term perhaps you could write [tex] 1+\tfrac{5}{12}\alpha(ln(\mu^2/m^2)+c')+O(\alpha^2) =exp(\tfrac{5}{12}\alpha(ln(\mu^2/m^2)+c')) [/tex] since the second term is second order anyway, it kind of doesnt matter if its the real [tex] O(\alpha^2) [/tex] term form or not. Then you would have [tex] ln[1+\tfrac{5}{12}\alpha(ln(\mu^2/m^2)+c')+O(\alpha^2)]=ln( exp(\tfrac{5}{12}\alpha(ln(\mu^2/m^2)+c')))=\tfrac{5}{12}\alpha(ln(\mu^2/m^2)+c'))[/tex] and you recover 27.12? 


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