I 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:
<br /> \tilde{g}_R^2=\tilde{g}^2-\frac{1}{32 \pi^2} \tilde{g}^4\text{ln}{ \frac{Q_0^2}{\Lambda^2}} + \cdots<br />
where there are higher-order contributions on the right-hand side due to \tilde{g}^n for n&gt;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
<br /> \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<br />
where \tilde{g}(Q_0) = g/Q_0.

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

 

[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
<br /> \tilde{g}^2 = \frac{g^2}{\Lambda^2}<br />
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,
<br /> 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<br />
and then we define
<br /> 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<br />
With this definition we get
<br /> 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<br />

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.