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From what I now understand of renormalization it is really a reparametrization of the theory in terms of measurable quantities instead of the 'inobservable bare quantities' that follow the Lagrangian; at least that is one interpretation of what is going on. The originally divergent physical quantities in the theory thus are rendered finite by absorbing the divergences into the relations between the bare and the measurable quantities.
For example in ##\phi^4##-theory one is reparametrizing the four-point vertex function ##\Gamma^{(4)}(p_i)## in terms of a renormalized coupling constant ##g_R## which is defined by 'measuring' ##\Gamma^{(4)}## at a certain energy/momentum scale (or renormalization point) ##\mu## and this together with a renormalization of the mass renders it finite (at least to one-loop corrections).
As I understand it so far the renormalization group expresses the fact that it should not matter at which renormalization point, and thus which value of ##g_R## you choose to reparametrize your theory; physical quantities should not be affected by that choice. (This is called the renormalization group law.) Or at least that should be the case if we were working with the exact theory and not a perturbation expansion of it.
According to this paper it does have an effect on your theory to a finite order of your perturbation expansion what parametrization you choose. I therefore wonder what the interpretation and use of equations like
$$ \mu \frac{\partial g_R}{\partial \mu} = \beta(g_R) $$
or the so called Callan-Symanzik equation is. The thing I so far think they express is that if you have already reparametrized your theory at say, ##\mu_0##, obtaining a renormalized coupling ##g_0##, the equation above expresses how ##g_R## must change in order for the theory to remain invariant when the renormalization point ## \mu## is changing. Is this correct?
If it is then it seems like when you choose your first renormalization point ##\mu_0## you generally get 'another theory' than if you had choosen ##\mu_0'## and that with these two initial renormalization points one might end up with different ##\beta##-functions (for the differential equation above) expressing how the renormalized coupling must change to leave the theory corresponding to ##\mu_0## or ##\mu_0'## invariant. It seems very unsatisfactory to get different ##\phi^4## theories depending on which renormalization point you choose.
However I suspect that I have gotten something wrong here?(In the Perimeter QFT II video lectures (http://pirsa.org/displayFlash.php?id=11110011) it seems like the equation above is derived for phi-four theory by demanding that ##\Gamma(g_R(\mu_2); \mu_2) = \Gamma(g_R(\mu_1); \mu_1)##. The lecturer then goes on to say that depending on what you measure for the first renormalization point you are 'describing a different theory'; it is only the same theory if the renormalized coupling constants lies on the line described by the equation ##\Gamma(g_R(\mu_2); \mu_2) = \Gamma(g_R(\mu_1); \mu_1)## as a function of the renormalization point ##\mu##.=
For example in ##\phi^4##-theory one is reparametrizing the four-point vertex function ##\Gamma^{(4)}(p_i)## in terms of a renormalized coupling constant ##g_R## which is defined by 'measuring' ##\Gamma^{(4)}## at a certain energy/momentum scale (or renormalization point) ##\mu## and this together with a renormalization of the mass renders it finite (at least to one-loop corrections).
As I understand it so far the renormalization group expresses the fact that it should not matter at which renormalization point, and thus which value of ##g_R## you choose to reparametrize your theory; physical quantities should not be affected by that choice. (This is called the renormalization group law.) Or at least that should be the case if we were working with the exact theory and not a perturbation expansion of it.
According to this paper it does have an effect on your theory to a finite order of your perturbation expansion what parametrization you choose. I therefore wonder what the interpretation and use of equations like
$$ \mu \frac{\partial g_R}{\partial \mu} = \beta(g_R) $$
or the so called Callan-Symanzik equation is. The thing I so far think they express is that if you have already reparametrized your theory at say, ##\mu_0##, obtaining a renormalized coupling ##g_0##, the equation above expresses how ##g_R## must change in order for the theory to remain invariant when the renormalization point ## \mu## is changing. Is this correct?
If it is then it seems like when you choose your first renormalization point ##\mu_0## you generally get 'another theory' than if you had choosen ##\mu_0'## and that with these two initial renormalization points one might end up with different ##\beta##-functions (for the differential equation above) expressing how the renormalized coupling must change to leave the theory corresponding to ##\mu_0## or ##\mu_0'## invariant. It seems very unsatisfactory to get different ##\phi^4## theories depending on which renormalization point you choose.
However I suspect that I have gotten something wrong here?(In the Perimeter QFT II video lectures (http://pirsa.org/displayFlash.php?id=11110011) it seems like the equation above is derived for phi-four theory by demanding that ##\Gamma(g_R(\mu_2); \mu_2) = \Gamma(g_R(\mu_1); \mu_1)##. The lecturer then goes on to say that depending on what you measure for the first renormalization point you are 'describing a different theory'; it is only the same theory if the renormalized coupling constants lies on the line described by the equation ##\Gamma(g_R(\mu_2); \mu_2) = \Gamma(g_R(\mu_1); \mu_1)## as a function of the renormalization point ##\mu##.=
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