Cutting off our Ignorance:renormalization

[haushofer]

Hi,

I have a question about the "QFT in a nutshell"-book by A. Zee, chapter 3.1 (page 148-149). It's about renormalization and regularization, and I still don't get the exact point.

Zee looks at meson-meson scattering in $\lambda^{4}$ theory. The $\lambda^{2}$-term is a diverging integral, as can easily be seen. Now, the introduction of a cutt-off $\Lambda$ is clear to me; you don't expect theories to be valid for all energies, so you regularize your integral. After some rewriting, the scattering amplitude M up to second order in $\lambda$ becomes

$$M = -i\lambda + iC\lambda^{2}[\log{\Lambda^{2}/s}]+ O(\lambda^{3})$$

Here the $\log{\Lambda^{2}/s}$-term is actually the sum of 3 terms with kinematic variables in them, but their exact form doesn't concern us; we focus on one kinematic variable s.

Now, the question in this chapter is: what does $\lambda$ exactly mean? Zee introduces $\lambda_{P}$, a physical coupling constant as measured by an actual experiment. Then, after formula (3) he states that "according to our theory",

$$-i\lambda_{P} = -i\lambda + iC\lambda^{2}[\log{\Lambda^{2}/s_{0}}]+ O(\lambda^{3})$$

where $s_{0}$ is the value found of the kinematic variable s of the experiment. Why is this the case? Why is $M = -i\lambda_{P}$? Does this physical coupling include ALL orders of $\lambda$ and so gives directly the physical scattering amplitude M because an experiment concears all of these lambda-orders?

He also states that $\lambda$ is a function of $\Lambda$ in order that the actual scattering amplitude M doesn't depend on $\Lambda$. But for my feeling, $\lambda$ and $\Lambda$ are 2 different things, and I don't see intuitively why they should be related besides the invariance-argument.

Can anyone clarify things up to me? I've read quite some QFT-stuff and I'm also quite familiar with the idea of renormalization and regularization, but these pages keep troubling me. Thanks!

 

[Demystifier]

Why is $M = -i\lambda_{P}$?

Think on it from the point of view of an experimentalist. He can measure $M$, but he does not know much about quantum field theory. He knows only how to calculate the lowest (tree level) contribution of QFT to $M$, and this contribution is given by $M = -i\lambda$. Thus, from his point of view, it is natural to DEFINE $\lambda_{P}$ through the identity
$M = -i\lambda_{P}$
In other words, this equation should be understood as a definition of $\lambda_{P}$ natural from the point of view of an experimentalist.

So yes, this physical coupling includes ALL orders of $\lambda$ and so gives directly the physical scattering amplitude M because an experiment concerns all of these lambda-orders.

 

[Demystifier]

He also states that $\lambda$ is a function of $\Lambda$ in order that the actual scattering amplitude M doesn't depend on $\Lambda$. But for my feeling, $\lambda$ and $\Lambda$ are 2 different things, and I don't see intuitively why they should be related besides the invariance-argument.

It is best to think on this stuff in a solid-state physics style. In a true fundamental theory, which we do not know, M depends both on $\lambda$ and $\Lambda$. They are independent parameters. M also depends on some other parameters about which we can say nothing because we do not know the true fundamental theory. However, since we do not know the true theory, we want to parameterize physics in terms of quantities in the effective theory we do know, which is QFT. But pure QFT (i.e., QFT without a cutoff) does not have any free parameter $\Lambda$. Therefore, we want to reformulate QFT with a cutoff such that all dependence on $\Lambda$ is eliminated. The only meaningful way to do it is to absorb the dependence on $\Lambda$ into a redefinition of the allowed parameter $\lambda$.

 

[haushofer]

Think on it from the point of view of an experimentalist. He can measure $M$, but he does not know much about quantum field theory. He knows only how to calculate the lowest (tree level) contribution of QFT to $M$, and this contribution is given by $M = -i\lambda$. Thus, from his point of view, it is natural to DEFINE $\lambda_{P}$ through the identity
$M = -i\lambda_{P}$
In other words, this equation should be understood as a definition of $\lambda_{P}$ natural from the point of view of an experimentalist.

So yes, this physical coupling includes ALL orders of $\lambda$ and so gives directly the physical scattering amplitude M because an experiment concerns all of these lambda-orders.

Ok, I see the point. Renormalization ofcourse in general only pops up at "loop-level" and beyond, not at tree-level.

It is best to think on this stuff in a solid-state physics style. In a true fundamental theory, which we do not know, M depends both on $\lambda$ and $\Lambda$. They are independent parameters. M also depends on some other parameters about which we can say nothing because we do not know the true fundamental theory. However, since we do not know the true theory, we want to parameterize physics in terms of quantities in the effective theory we do know, which is QFT. But pure QFT (i.e., QFT without a cutoff) does not have any free parameter $\Lambda$. Therefore, we want to reformulate QFT with a cutoff such that all dependence on $\Lambda$ is eliminated. The only meaningful way to do it is to absorb the dependence on $\Lambda$ into a redefinition of the allowed parameter $\lambda$.

Ok, but this is more or less the "invariance argument" I mentioned. I can see this is the "only meaningful way", but as I understand from your text this is more or less the only argument (so I shouldn't try to find an intuitive reason why the cutt-off and the coupling parameter should be related)?

Thanks for your clear answer! It's very nice to exchange thoughts about this kind of stuff; it can clearify things in minutes which normally would take days or perhaps weeks :D

 

[Demystifier]

but as I understand from your text this is more or less the only argument (so I shouldn't try to find an intuitive reason why the cutt-off and the coupling parameter should be related)?

As far as I know, there is no other (more intuitive) reason for this. If you find some, let me know!

 

[haushofer]

As far as I know, there is no other (more intuitive) reason for this. If you find some, let me know!

Yeah sure, I will ;) But somehow this bothers me a little, so the quest is not over yet :D Thanks!

 

[Demystifier]

Maybe this can help:
Theories with different values of $$\lambda$$ and $$\Lambda$$ can be thought of as theories with different measurable predictions. However, we cannot determine $$\Lambda$$ from experiments. Why not? Because for each value of $$\Lambda$$ there is a corresponding value of $$\lambda$$ such that measurable predictions (that is, M) will stay unchanged. Thus, the function $$\lambda(\Lambda)$$ can be thought of as a parameterization of a class of different theories that lead to the same measurable predictions.

 

[RedX]

It is best to think on this stuff in a solid-state physics style. In a true fundamental theory, which we do not know, M depends both on $\lambda$ and $\Lambda$. They are independent parameters. M also depends on some other parameters about which we can say nothing because we do not know the true fundamental theory. However, since we do not know the true theory, we want to parameterize physics in terms of quantities in the effective theory we do know, which is QFT. But pure QFT (i.e., QFT without a cutoff) does not have any free parameter $\Lambda$. Therefore, we want to reformulate QFT with a cutoff such that all dependence on $\Lambda$ is eliminated. The only meaningful way to do it is to absorb the dependence on $\Lambda$ into a redefinition of the allowed parameter $\lambda$.

In this case, don't we know the true fundamental theory? It's a scalar $$\phi^4$$ theory. So $$\lambda$$ is a function of $$\Lambda$$.

If you start out with a cutoff however, then $$\lambda$$ and $$\Lambda$$ are independent, for one value only - the value of $$\lambda$$ at the cutoff. When you change your cutoff however, then $$\lambda$$ is a function of the cutoff in order to retain the same results.