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I am studying the textbook An introduction to Quantum Field Theory by Peskin and Schroeder, and am not able to understand a seemly important and straightforward result on running coupling constant. It is probably a stupid one but I just cannot figure it out.

After some calculations, one obtains the expression for the running coupling in $\phi^4$ theory, namely, Eq.(12.82) on P.422 of Chapter 12 which reads $$\bar{\lambda}(p)=\frac{\lambda}{1-(3\lambda/16\pi^2)\log(p/M)}.$$ Owing to Eq.(12.69) which gives the following relation between the Coleman's hydrodynamic-bacteriological equation and the Callan-Symanzik equation for $\phi^4$ theory $$\log(p/M) \leftrightarrow t,$$ $$\lambda \leftrightarrow x,$$ $$-\beta(\lambda) \leftrightarrow v(x),$$ we have $$\bar{\lambda}(\log(p/M);\lambda) \leftrightarrow \bar{x}(t;x).$$

Now, according to the interpretation above Eq.(12.70), $\bar{x}$ is the initial position (at $t=0$) of the element which is at $x$ at time $t$. To be more specific, one emphasizes that $\bar{x}(t;x)$ is not the position at time $t$. Similarly, I understand that $\bar{\lambda}(p)\equiv\bar{\lambda}(\log(p/M);\lambda)$ is not the coupling constant at $p$. $\bar{\lambda}$ is the value of the coupling constant at renormalization energy scale $p=M$, which evolves to $\lambda$ at scale $p$.

If the above understanding is correct, Eq.(12.82) cannot be interpreted as written in the textbook

The running coupling constant goes to zero at a logarithmic rate as $p\rightarrow 0$.

This is because, again, $\bar{\lambda}$ itself corresponds to the coupling fixed at $M$, it does not run.

On the contrary, the inverse function $\lambda(\log(p/M);\bar{\lambda})$ shall describe the running coupling, as analogically $x(t;\bar{x})$ will do. In fact, similarly to Eq.(12.70) (to be more explicit, I changed $d$ to $\partial$) $$\frac{\partial}{\partial t'}\bar{x}(t';x)=-v(\bar{x}),$$ which states that if an element arrives at a given $x$ but $\Delta t$ later, its departure position at initial time $t=0$ should be shifted to the left (indicated by the negative sign) by an amount $v(\bar{x})\Delta t$, where $v(\bar{x})$ is the fluid velocity at the initial position. On the other hand, one has $$\frac{\partial}{\partial t'}x(t';\bar{x})=v(x).$$ The above equation states that for an element starts at $\bar{x}$ at $t=0$, its position is shifted to the right (corresponding to the $+$ sign before the velocity) by an amount $v(x)\Delta t$ , if we let it evolve for $\Delta t$ longer. Therefore, it seems the solution of $\lambda(\log(p/M);\bar{\lambda})$ is simply (12.82) by exchanging $\lambda$ with $\bar{\lambda}$ and adding an extra $(-1)$ before $\log(p/M)$. But then the conclusion will be completely the opposite?! What am I missing? Many thanks for the explanation.

Edit

Regarding NowIGetToLearnWhatAHeadIs's answer, I try to refine my doubts as follows:

  1. Why $p\rightarrow 0$ is interpreted as going to small energy scale? Imagine that one would like to study the running coupling of $\phi^4$ theory at a (small) scale $p$? Intuitively, I understand that this can be achieved experimentally by carrying out a collision between few (two to two, to be exact) $\phi$s with energy roughly at $E\sim p$, and then measuring the cross section of the process. In this context, the coupling we talk about ($\bar{\lambda}$) should be evaluated at the corresponding momentum $M$ (not $p$).

  2. If we want to study the coupling as a function of energy scale for the same theory, the renormalization condition should be something given in the first place, and it should not be modified afterwards. For instance, one can study $\lambda$ as a function of $p$ for given $\bar{\lambda}$ and $M$. I understand, as pointed out, that $\bar{\lambda}$ indeed can be viewed as a function of $p$. However, if one evaluates $\bar{\lambda}$ while varies $p$ for given $\lambda$ and $M$, it seems to me that we are no longer talking about the same theory. (c.f. in fact, fix $\lambda$ and $p$ and study $\bar\lambda$ as a function of $M$ is also ok, but not mix variables from different pairs.)

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3 Answers 3

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My interpretation is that you are right when you say

I understand that $\bar{\lambda}(p)\equiv\bar{\lambda}(\log(p/M);\lambda)$ is not the coupling constant at $p$. $\bar{\lambda}$ is the value of the coupling constant at renormalization energy scale $p=M$, which evolves to $\lambda$ at scale $p$.

However, I don't think you are on the same wavelength (no pun intended) as Peskin and Schroeder (P&S) when you say

This is because, again, $\bar{\lambda}$ itself corresponds to the coupling fixed at M, it does not run.

I think P&S call $\bar{\lambda}(p)\equiv\bar{\lambda}(\log(p/M);\lambda)$ (actually, they write $\bar{\lambda}(p)\equiv\bar{\lambda}(p;\lambda)$) the running coupling constant, not because of its dependence on $M$, but because of its dependence on $p$ when the coupling constant $\lambda$ at the momentum scale $p$ is held fixed. For example, when they introduce the term "running coupling constant", they say

We will refer to $\bar{\lambda}(p)$ as the running coupling constant.

In the above quote the dependence of $\bar{\lambda}$ on the momentum $p$ is emphasized.

I think if you reread the section with that understanding, everything makes sense. In particular, it makes sense when they say

The running coupling constant goes to zero at a logarithmic rate as $p \to 0$.

This makes sense because if you put the scale $p$ way into the infrared by putting $p \to 0$ and you hold the coupling constant $\lambda$ at this scale $p$ fixed, then the coupling constant $\bar{\lambda}$ at the scale $M$ must get very small, because as you integrate out the wavenumbers between $M$ and $p$, the coupling constant $\bar{\lambda}$ gets multiplied by a large factor to give the renormalized coupling $\lambda$.

If P&S had really meant instead for $\bar{\lambda}$ to be the coupling at $p$ and $\lambda$ to be the coupling at $M$, then you what have the opposite (and therefore incorrect) situation. You would have that as $p\to 0$ with the coupling constant at $M$ fixed, that the coupling constant at $p$ goes to zero, contrary to the fact that $\lambda$ is a relevant parameter. Therefore P&S must really have meant that $\bar{\lambda}$ is the coupling at $M$.

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  • $\begingroup$ Thanks for the answer! If I understood you correctly, you meant that we can look at $\bar{\lambda}$ as how it depends on $p$, that is, find the coupling at energy scale $M$ by using different measured coupling $\lambda$ at given $p$. I (do willingly) understand this, in the same way that $\bar{x}$ is a function of $t$, as I explained my understanding in the question. $\endgroup$
    – gamebm
    Oct 2, 2017 at 0:22
  • $\begingroup$ Now, in this context, my confusion is the following. Why letting $p\rightarrow 0$ is interpreted physically as "going to small energy scale"? I thought the opposite (and, as you pointed out, which inevitably leads to the wrong conclusion). To me, if one fixes the coupling at a given scale $M$, and measure (or calculate) the coupling at a $p<M$ (determined by Callan-Symanzik equation) is "going to a smaller scale". But, as discussed above, $\bar{\lambda}$ as a function of $p$ just does the opposite. Again, many thanks for the explanation. $\endgroup$
    – gamebm
    Oct 2, 2017 at 0:38
  • $\begingroup$ @gamebm $p \to 0$ is considered going to small energies because energy $E$ and momentum $p$ are related by $E^2=p^2 + m^2$, so lower momentum means lower energy and higher momentum means higher energy. $\endgroup$ Oct 2, 2017 at 1:29
  • $\begingroup$ Sorry I did not make myself clear. My doubts originates from the following thoughts. Generically, how do we study the coupling at a small scale $p$? Intuitively, I will carry out a collision between few (four) $\phi$s with energy roughly $E\sim p$, and try to find the effective coupling by measuring the cross section of the process, right? So I think the couping we talk about ($\bar{\lambda}$) should be paired to the corresponding momentum (not $p$ in this context). So why $p\rightarrow 0$ is interpreted as "going to small scale"? $\endgroup$
    – gamebm
    Oct 2, 2017 at 1:48
  • $\begingroup$ I edited my question to clarify my doubts, so that it might be easier to spot my misunderstanding. $\endgroup$
    – gamebm
    Oct 3, 2017 at 2:22
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I think that I figured it out.

From the ``apparent'' physical interpretation, it seems $\lambda$ carries the physical content of running coupling, not $\bar{\lambda}$. In fact, the coupling $\lambda$ runs according to the definition of the $\beta$ function (the first equation of (12.39)), and in this sense, we do not even need the Callan-Symanzik equation for this. So while I understood part of the arguments given by Moths, I cannot see why.

Now I realized two things. First according to (12.73), $\bar{\lambda}(p;\lambda)$ indeed satisfies the mathematical ``defintion'' of the running coupling, namely, $$\frac{d}{d\log(p/M)}\bar\lambda(p;{\lambda})=\beta(\bar\lambda) .$$ If, instead, one tries to similarly write down the equation satisfied by $\lambda(p;\bar{\lambda})$, one finds $$\frac{d}{d\log(p/M)}\lambda(p;\bar{\lambda})=-\beta(\lambda) ,$$ which is simply not the proper equation (12.39) one would have expected.

Secondly, the reason for this weird outcome is the following. (12.73) is a direct consequence of (12.70) $$\frac{\partial}{\partial t}\bar x(t';x)=-v(\bar x) ,$$ together with the mapping $$\log(p/M) \leftrightarrow t,$$ $$\lambda \leftrightarrow x,$$ $$-\beta(\lambda) \leftrightarrow v(x).$$ One notes there is an extra minus sign on the right-hand side of the equation. This counter-intuitive minus sign is due to the fact that the physical interpretation of the derivative is not to calculate velocity, but the ratio of displacement of the initial position of a bacteria to the time variation when it is observed later at position $x$ and time $t$. It is noted that two infinitesimal quantities, namely, the displacement and time variation, are not associated with the same time instant, either the same location. The above subtly, namely, the deviation from the ordinary definition of velocity, gives rise to this additional minus sign, and explains why it is $\bar\lambda$ but not $\lambda$ satisfies the definition of running coupling, when we are talking in terms of the variables involving the formal solution of the Callan-Symanzik equation. In short, regarding (12.82), one may simply treat $\lambda$ as an irrelevant constant, and think of $\bar\lambda$ as $\lambda$.

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I agree with Brian Moths that $p \rightarrow 0$ means going to small energies.

However, I do not understand the following statement.

as you integrate out the wavenumbers between $𝑀$ and $p$, the coupling constant $\bar{\lambda}$ gets multiplied by a large factor to give the renormalized coupling $\lambda$.

In the example given in the book, we have a positive $\beta = \frac{3\lambda^2}{16\pi^2}$. Therefore the factor is in fact smaller than 1 as we move to smaller energies ($p \rightarrow 0$).

This means that if we fix $\bar{\lambda}$ at $M$ and move to smaller energies ($p \rightarrow 0$), the effective $\lambda$ should become smaller and smaller. Now if we instead fix $\lambda$ as we move to smaller energies ($p \rightarrow 0$), we would have to make $\bar{\lambda}$ bigger and bigger to make the equation hold.

For reference, see the sentence immediately following (12.28) on Page 404.

[Update]

After dreaming about this thing the whole night, I believe I have a more satisfactory answer.

I am afraid that Brian Moths was wrong with the following statement, for the objection that I posted originally.

My interpretation is that you are right when you say...

Below is what I believe should be the correct interpretation.

The bacteriological analogy is a formal analogy, in the sense that Equation (12.68) has the same form as Equation (12.66) with the replacements in (12.69). It does not mean that we should interpret $\bar{\lambda}$, which is something that we just make up to make it formally look like $\bar{x}$ defined by (12.70), in the same way as we interpret $\bar{x}$.

In fact, the definition of $\lambda$ has never changed since (12.30). It is simply the effective coupling at $p^2=-M^2$ (or equivalently at $p=M$).

Later we "figure out" that $\bar{\lambda}$ is in fact the effective coupling at any $p$, which can be different to $M$.

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  • $\begingroup$ Thanks for the answer. In the case $\beta=\frac{3\lambda^2}{16\pi^2}$, since $\beta>0$, it is the normal screen effect so the coupling increase with energy. This can also be seen by observing (12.61), where the $\beta$ function for QED reads $\beta(e)=\frac{e^3}{12\pi^2}$, but QED gives the normal screening effect. For QCD, the $\beta$ function is possibly negative as shown in (17.13), so one might get the famous anti-screening effect, where the coupling decreases with increasing energy scale. Again, thanks for the comments. $\endgroup$
    – gamebm
    May 23, 2021 at 9:59

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