Problem with section in Schwartz's QFT text

[HomogenousCow]

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.

 

[king vitamin]

I don't follow the last equation you give. There are three different couplings being defined here: $g$, $\tilde{g}$, and $\tilde{g}_R$. The latter two are related by the third equation you've given:
$$\tilde{g}_R^2=\tilde{g}^2-\frac{1}{32 \pi^2} \tilde{g}^4\text{ln}{ \frac{Q_0^2}{\Lambda^2}} + \cdots$$
where there are higher-order contributions on the right-hand side due to $\tilde{g}^n$ for $n>4$. So they are clearly not the same. Where are you getting your last equation from?

 

[HomogenousCow]

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 $\tilde{g}$?

 

[king vitamin]

Ok, I think I see the confusion. The coupling $\tilde{g} = \tilde{g}(Q)$ is not being fixed at a particular value of $Q$, its implicit $Q$-dependence is being included in the subsequent definition of $\tilde{g}_R$. So really
$$\tilde{g}_R^2=\tilde{g}(Q_0)^2-\frac{1}{32 \pi^2} \tilde{g}(Q_0)^4\text{ln}{ \frac{Q_0^2}{\Lambda^2}}+\cdots$$
where $\tilde{g}(Q_0) = g/Q_0$.

You may just want to skip defining $\tilde{g}$ in the first place. Just define
$$\tilde{g}_R^2 = M(Q_0) = \frac{g^2}{Q^2}(1-\frac{1}{32 \pi^2} \frac{g^2}{Q^2}\text{ln}{ \frac{Q^2}{\Lambda^2}}+\cdots)$$
directly in terms of the bare coupling $g$, and you get the correct final result:
$$M(Q)=\tilde{g}_R^2-\frac{1}{32 \pi^2} \tilde{g}_R^4\text{ln}{ \frac{Q^2}{Q_0^2}}+\cdots$$

 

[HomogenousCow]

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.

 

[king vitamin]

I'm sorry! It appears that I made a mistake, and in addition Schwartz made a mistake.

The real definition of $\tilde{g}$ should be
$$\tilde{g}^2 = \frac{g^2}{\Lambda^2}$$
It really doesn't make sense to scale $g$ by $Q$, 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 $\Lambda$.

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

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 $Q$). We need some manifestation of this, and you see it here: the proper renormalized perturbation series includes increasingly higher factors of $(Q_0/Q)^2$, signaling a breakdown of perturbation theory in the infrared.

Sorry again about leading you down the wrong track!

 

[HomogenousCow]

Thanks for that, would've bothered me for a good while otherwise.