Problem with section in Schwartz's QFT text

  • #1
692
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Beginning of page 303 of Schawrtz's QFT text (section 16.1.1), there's a part on the renormalization of the scalar propagator in ##\phi^3## theory to second order in ##g## the bare coupling. He writes (quote begins):

$$M(Q)=M^0 (Q)+M^1 (Q)=\frac{g^2}{Q^2}(1-\frac{1}{32 \pi^2} \frac{g^2}{Q^2}\text{ln}{ \frac{Q^2}{\Lambda^2}}+...) $$

Note that ##g## is not a number in ##\phi^3## theory but has dimensions of mass (...) Let us substitute for ##g## a new Q-dependent variable ##\tilde{g}^2=\frac{g^2}{Q^2}##, which is dimensionless. Then,

$$M(Q)=\tilde{g}^2-\frac{1}{32 \pi^2} \tilde{g}^4\text{ln}{ \frac{Q^2}{\Lambda^2}}+... $$

Then we can define a renormalized coupling ##\tilde{g}_R^2=M(Q_0)## (...)
It follows that ##\tilde{g}_R^2## is a formal power series in ##\tilde{g}##:

$$\tilde{g}_R^2=\tilde{g}^2-\frac{1}{32 \pi^2} \tilde{g}^4\text{ln}{ \frac{Q_0^2}{\Lambda^2}}+... $$
(...)

Substituting into ... produces a prediction fo rthe matrix element at the scale Q in terms of the matrix element at the scale ##Q_0##

$$M(Q)=\tilde{g}_R^2-\frac{1}{32 \pi^2} \tilde{g}_R^4\text{ln}{ \frac{Q^2}{Q_0^2}}+... $$

(End quote)

So I've got a real issue with this derivation, it looks like he's treating the ##\tilde{g}##s in the 2nd and 3rd equations as being the same when this is clearly not so. Shouldn't the two be related by
$$\tilde{g}^2(Q)=\frac{Q_0^2}{Q^2}\tilde{g}^2(Q_0)?$$ I'm very confused.
 

Answers and Replies

  • #2
king vitamin
Science Advisor
Gold Member
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I don't follow the last equation you give. There are three different couplings being defined here: [itex]g[/itex], [itex]\tilde{g}[/itex], and [itex]\tilde{g}_R[/itex]. The latter two are related by the third equation you've given:
[tex]
\tilde{g}_R^2=\tilde{g}^2-\frac{1}{32 \pi^2} \tilde{g}^4\text{ln}{ \frac{Q_0^2}{\Lambda^2}} + \cdots
[/tex]
where there are higher-order contributions on the right-hand side due to [itex]\tilde{g}^n[/itex] for [itex]n>4[/itex]. So they are clearly not the same. Where are you getting your last equation from?
 
  • #3
692
142
First note that because ##\tilde{g}^2=\frac{g^2}{Q^2}##, ##\tilde{g}## is a function of the scattering momentum and hence not a constant. Because the author defined $$\tilde{g}_R^2=M(Q_0),$$ the ##\tilde{g}## in equation 3 has to be ##\frac{g^2}{Q_0^2}## and not ##\frac{g^2}{Q^2}##.

I guess the crux of my question is, since $$\tilde{g}_R^2=M(Q_0)$$

why does he then write

$$\tilde{g}_R^2=\tilde{g}^2-\frac{1}{32 \pi^2} \tilde{g}^4\text{ln}{ \frac{Q_0^2}{\Lambda^2}}+... $$

which is essentially

$$\tilde{g}_R^2=\frac{g^2}{Q^2}(1-\frac{1}{32 \pi^2} \frac{g^2}{Q^2}\text{ln}{ \frac{Q_0^2}{\Lambda^2}}+...) $$

Which is clearly not ##M(Q_0)##. Why is he ignoring the momentum dependence in ##\tilde{g}##?
 
Last edited:
  • #4
king vitamin
Science Advisor
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Ok, I think I see the confusion. The coupling [itex]\tilde{g} = \tilde{g}(Q)[/itex] is not being fixed at a particular value of [itex]Q[/itex], its implicit [itex]Q[/itex]-dependence is being included in the subsequent definition of [itex]\tilde{g}_R[/itex]. So really
[tex]
\tilde{g}_R^2=\tilde{g}(Q_0)^2-\frac{1}{32 \pi^2} \tilde{g}(Q_0)^4\text{ln}{ \frac{Q_0^2}{\Lambda^2}}+\cdots
[/tex]
where [itex]\tilde{g}(Q_0) = g/Q_0[/itex].

You may just want to skip defining [itex]\tilde{g}[/itex] in the first place. Just define
[tex]
\tilde{g}_R^2 = M(Q_0) = \frac{g^2}{Q^2}(1-\frac{1}{32 \pi^2} \frac{g^2}{Q^2}\text{ln}{ \frac{Q^2}{\Lambda^2}}+\cdots)
[/tex]
directly in terms of the bare coupling [itex]g[/itex], and you get the correct final result:
[tex]
M(Q)=\tilde{g}_R^2-\frac{1}{32 \pi^2} \tilde{g}_R^4\text{ln}{ \frac{Q^2}{Q_0^2}}+\cdots
[/tex]
 
  • #5
692
142
I've been trying what you suggested for the past hour but I can't get it to work. Inverting,

$$\tilde{g}_R^2=\frac{g^2}{Q_0^2}(1-\frac{1}{32\pi^2}\frac{g^2}{Q_0^2}\text{ln}\frac{Q_0^2}{\Lambda^2})$$ I get
$$\frac{g^2}{Q_0^2}=\tilde{g}_R^2+\frac{1}{32\pi^2}\tilde{g}_R^4\text{ln}\frac{Q_0^2}{\Lambda^2}$$
But when I plug this back into ##M(Q)##, it works out like this:
$$M(Q)=\frac{g^2}{Q^2}(1-\frac{1}{32\pi^2}\frac{g^2}{Q^2}\text{ln}\frac{Q^2}{\Lambda^2})$$
$$=\frac{Q_0^2}{Q^2}\frac{g^2}{Q_0^2}(1-\frac{1}{32\pi^2}\frac{Q_0^2}{Q^2}\frac{g^2}{Q_0^2}\text{ln}\frac{Q^2}{\Lambda^2})$$
$$=\frac{Q_0^2}{Q^2}(\tilde{g}_R^2+\frac{1}{32\pi^2}\tilde{g}_R^4\text{ln}\frac{Q_0^2}{\Lambda^2})[1-\frac{1}{32\pi^2}\frac{Q_0^2}{Q^2}(\tilde{g}_R^2+\frac{1}{32\pi^2}\tilde{g}_R^4\text{ln}\frac{Q_0^2}{\Lambda^2})\text{ln}\frac{Q^2}{\Lambda^2}],$$ up to ##\tilde{g}_R^4## this is
$$=\frac{Q_0^2}{Q^2}[\tilde{g}_R^2-\frac{1}{32\pi^2}\frac{Q_0^2}{Q^2}\tilde{g}_R^4\text{ln}\frac{Q^2}{\Lambda^2}+\frac{1}{32\pi^2}\tilde{g}_R^4\text{ln}\frac{Q_0^2}{\Lambda^2}]$$

Have I made another mistake somewhere?

Also, if you go through Schwartz's derivation, you will see that it only works if ##\tilde{g}## is fixed at ##Q## in all expressions.
 
  • #6
king vitamin
Science Advisor
Gold Member
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I'm sorry! It appears that I made a mistake, and in addition Schwartz made a mistake.

The real definition of [itex]\tilde{g}[/itex] should be
[tex]
\tilde{g}^2 = \frac{g^2}{\Lambda^2}
[/tex]
It really doesn't make sense to scale [itex]g[/itex] by [itex]Q[/itex], as I think we both understood, but I was wrong above in that I missed that a scaling was still needed. The only other scale we can use here is [itex]\Lambda[/itex].

Now,
[tex]
M(Q)=\frac{\Lambda^2}{Q^2}\tilde{g}^2-\frac{1}{32 \pi^2} \frac{\Lambda^4}{Q^4} \tilde{g}^4\text{ln}{ \frac{Q^2}{\Lambda^2}}+\cdots
[/tex]
and then we define
[tex]
g_R \equiv M(Q_0) = \frac{\Lambda^2}{Q_0^2}\tilde{g}^2-\frac{1}{32 \pi^2} \frac{\Lambda^4}{Q_0^4} \tilde{g}^4\text{ln}{ \frac{Q_0^2}{\Lambda^2}}+\cdots
[/tex]
With this definition we get
[tex]
M(Q)=\left( \frac{Q_0^2}{Q^2} \right)\tilde{g}_R^2-\frac{1}{32 \pi^2} \left( \frac{Q_0^4}{Q^4} \right) \tilde{g}_R^4\text{ln}{ \frac{Q^2}{Q_0^2}}+\cdots
[/tex]

Schwartz's answer must be wrong, because it is well-known that a massless super-renormalizable theory (which is the theory this amplitude comes from) is perturbatively sick in the IR (that is, for small [itex]Q[/itex]). We need some manifestation of this, and you see it here: the proper renormalized perturbation series includes increasingly higher factors of [itex](Q_0/Q)^2[/itex], signaling a breakdown of perturbation theory in the infrared.

Sorry again about leading you down the wrong track!
 
  • #7
692
142
Thanks for that, would've bothered me for a good while otherwise.
 

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