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Observables are supposed to be independent of the scale $\mu$ introduced during regularization. Srednicki says so and yet the derivatives $\frac{d\alpha}{d\ln \mu}$ and $\frac{dm}{d\ln \mu}$ are taken to be nonzero (Eq.s 28.20 and 28.27 in Srednicki's book), where $\alpha$ is the coupling constant and $m$, the physical mass, both of which are observable. What am I missing?

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  • $\begingroup$ Coupling constants are not observable and renormalized masses are not physical masses. $\endgroup$ Commented Aug 22, 2022 at 23:03

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You are right, observables are supposed to be independent of the arbitrary renormalization scale $\mu$. But, why is everybody talking about $\frac{d\alpha}{d\ln \mu}$?

In reality, $\alpha(p)$ is momentum $p$ dependent. For the example of a scattering process, $\alpha(p)$ is related to the scattering amplitude, and $p$ is the momentum transfer of the incoming/outgoing particles. In other words, $\alpha(p)$ is "running" with momentum $p$, in stead of "running" with the arbitrary renormalization scale $\mu$. So where is this wicked and naughty $\frac{d\alpha}{d\ln \mu}$ coming from?

Let's look at a simplified example of $$ \alpha(p) = \ln(p/p_0) + \alpha_0. $$

$\alpha(p)$ is the solution of the differential equation ($\beta$-function) $$ \beta (\alpha) = \frac{d\alpha(p)}{d\ln p} = 1, $$ with the initial condition $$ \alpha(p)|_{p = p_0} = \alpha_0. $$

The "running with renormalization scale $\mu$" approach is tantamount to regarding $\alpha(p, p_0, \alpha_0)$ as the solution to an alternative differential equation (differentiating against the initial condition point $p_0$, which is the renormalization scale $\mu = p_0$) $$ \beta '(\alpha) = \frac{d\alpha(p, p_0, \alpha_0)}{d\ln p_0} = -1, $$ with the initial condition $$ \alpha(p_0)|_{p_0 = p} = \alpha_0. $$

For historical reasons, the above un-intuitive way of "running with renormalization scale $\mu$" was first introduced as the byproduct of dimensional regularization. Conceptually, it's wrong to invoke "running with renormalization scale $\mu$". The QFT text books can get away from this butchering of physics concept, because mathematically $p$ and $\mu$ usually appear in the form of $\ln(p/\mu)$ therefore differentiating by $\ln p$ or $\ln \mu$ is basically the same with only a sign change.

In my humble opinion, "running with momentum $p$ approach" should be taught in QFT text books, whereas "running with renormalization scale $\mu$" approach is baffling to new learners rather than clarifying.

For more details, please see another post here.


Added note:

Why do we say that the renormalization scale $\mu$ is arbitrary and non-physical, while the momentum $p$ of a particle is physical?

For the above simplified example: $$ \alpha(p) = \ln(p/p_0) + \alpha_0. $$ the initial condition is set at $p_0$ (corresponding to setting the renormalization scale to $\mu = p_0$) $$ \alpha(p)|_{p = p_0} = \alpha_0. $$

The selection of the initial point at $p_0$ is totally arbitrary. We can change the initial condition to be at $p_1$ (corresponding to setting the renormalization scale to $\mu = p_1$) and have $$ \alpha(p) = \ln(p/p_1) + \alpha_1. $$ where $$ \alpha(p)|_{p = p_1} = \alpha_1. $$ and $$\alpha_1 = \alpha_0 + \ln(p_1/p_0).$$

As we can verify in the above example, even though the arbitrary renormalization scale is changed from $\mu = p_0$ to $\mu = p_1$, the measurable quantity of $\alpha(p)$ remains the same.

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  • $\begingroup$ I agree that it is more meaningful to talk about the running of $\alpha$ with perhaps the momentum transfer scale. But the fake parameter $\mu$ introduced in dimensional regularization is used in an essential way in the derivation of the Callan-Symanzik equation (is there any experimental confirmation of CSE?). So we can't completely ignore $\frac{d}{d\ln \mu}$. Srednicki's statements appear to be inconsistent. In Ch. 28 he writes "physical observables must be independent of the fake parameter $\mu$"". After (28.12) he writes: " bare fields and parameters must be independent of $\mu$". $\endgroup$
    – Coriolis1
    Commented Aug 27, 2022 at 1:47
  • $\begingroup$ If both physical parameters and bare parameters are independent of $\mu$, where is the $\mu$ occurring in the counterterms going? $\endgroup$
    – Coriolis1
    Commented Aug 27, 2022 at 1:48
  • $\begingroup$ As I mentioned in the main text, the un-intuitive way of "running with renormalization scale $\mu$" was first introduced as the byproduct of dimensional regularization. Conceptually, it's wrong to invoke "running with renormalization scale $\mu$". One can get away from this butchering of physics concept, because mathematically $p$ and $\mu$ usually appear in the form of $\ln(p/\mu)$ therefore differentiating by $\ln p$ or $\ln \mu$ is basically the same with only a sign change. $\endgroup$
    – MadMax
    Commented Aug 29, 2022 at 14:41
  • $\begingroup$ I downvoted because it is simply not correct that "running with renormalization scale $\mu$" is "wrong." That is a perfectly standard and valid way to frame the renormalization group. $\endgroup$
    – Andrew
    Commented Aug 29, 2022 at 15:15
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    $\begingroup$ @Andrew, it's conceptually wrong, it only lucked out (from being totally wrong) by mere mathematical coincidence as I have explained in the main text. Maybe you don't agree with my take. but anything "standard" in physics text books does not automatically mean it is "correct". As long as I can remember, Ptolemy's Earth center model used to be 'standard' in text books. $\endgroup$
    – MadMax
    Commented Aug 29, 2022 at 15:27
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In perturbation theory, instead of an expansion purely in powers of the coupling constant $g$, you often find that the coupling constant is multiplied by "large logs" of the form $\log p/M$, where $M$ is some mass scale. If $p \ll M$ or $p \gg M$, then this log can be "large" (a very large or small number, not order 1). Since the "effective expansion parameter" is not really $g$ but $g \log p/M$, this can lead to a breakdown in perturbation theory if $\log p/M$ is large.

The renormalization group trick is that in some renormalization schemes, the parameter $M$ that appears in the log is a sliding renormalization scale $\mu$. Since this is an arbitrary parameter and nothing physical depends on it, we can choose $\mu$ so that the logs are under control. In other words, we choose $\mu$ so that perturbation theory converges as quickly as possible, for the given value of momentum transfer $p$.

The "cost" of using our freedom to choose $\mu$, is that the parameters in the Lagrangian become $\mu$ dependent, in such a way that physically observable quantities do not depend on $\mu$. The mass and coupling constant you refer to in your question are presumably these parameters in the Lagrangian, which are not directly measurable. The physical mass is defined by the pole in the propagator, which does not depend on $\mu$. Similarly the physical coupling is often defined in terms of a scattering amplitude; the electric charge can be defined in terms of the three point function $e\gamma \bar{e}$ in the limit that the photon momentum goes to zero.

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  • $\begingroup$ Invoking the specifics of perturbation theory and its convergence (and yes, it's the "standard" narrative you cherish) is not a particularly helpful angle of looking at the more general issue. Even for a "non-perturbative" theory, one still has to deal with "running", renormalization scale, and all that jazz. $\endgroup$
    – MadMax
    Commented Aug 29, 2022 at 16:00
  • $\begingroup$ @Andrew There is an intrinsic contradiction in the argument. You write "Since this is an arbitrary parameter and nothing physical depends on it, we can choose $\mu$ sot that the logs are under control. In other words, we choose $\mu$ so that perturbation theory converges as quickly as possible, for the given value of momentum transfer $p$." In other words, the arbitrary, physically meaningless parameter can convert a divergent series to a convergent series and conversely? The parameter $\mu$ can be hardly considered physically meaningless if it can have such a profound effect. $\endgroup$
    – Coriolis1
    Commented Aug 30, 2022 at 21:14
  • $\begingroup$ Secondly, the mass renormalization is done to make the pole in the propagator agree with the renormalized mass. So the suggestion that the physically measured mass is different from the renormalized mass is a bit baffling. $\endgroup$
    – Coriolis1
    Commented Aug 30, 2022 at 21:16
  • $\begingroup$ @Coriolis1 It's very subtle but there's no contradiction. The way I think about it is that there is really an infinite set of possible perturbative expansions, parameterized by $\mu$, that are all equivalent if you can sum each term in the series. However, since we can't sum each term in the series, the perturbative expansions are not equivalent in practice, because for some choices of $\mu$ the series converges much more quickly. Namely, if we choose $\mu$ to be on the same scale of the processes we are studying. Choosing $\mu$ "optimally" in this way links the fictitious scale (...) $\endgroup$
    – Andrew
    Commented Aug 30, 2022 at 21:28
  • $\begingroup$ (...) $\mu$ to the physical energy scale of the problem. As to your second point, this is a matter of definition I guess, but it is quite standard to define the renormalized mass as being different from the pole (or physical) mass. For example, in Eq 31 of arxiv.org/abs/hep-th/0701053, the renormalized mass is defined as the mass parameter after subtracting divergences, but this is not the same quantity that appears in the pole of the propagator. This boils down to the difference between the $\bar{MS}$ and $OS$ schemes (the source I'm quoting uses $\bar{MS}$). $\endgroup$
    – Andrew
    Commented Aug 30, 2022 at 21:33

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