I'm trying to understand the renormalization group (RG), and in particular, how it is used to study entanglement entropy (EE) and c-theorems in quantum field theory (QFT). But I'm having trouble with some conceptual aspects that I haven't been able to get around by reading things.

My basic understanding of how RG works in QFT, which I've cobbled together from reading various textbooks and notes on the internet, is as follows: You have a theory specified by a Lagrangian $\mathcal{L}$ with coupling constants $\{g_i\}$. The coupling constants $\{g_i\}$ that appear in the Lagrangian are called "bare" coupling constants. You can also have a "UV-cutoff" $\epsilon$ that specifies some energy scale $1/\epsilon$ much larger than any scale we care about, and regulates infinities in calculations. Bare couplings $\{g_i\}$ are not things that you can physically measure, but they are related to things that we can measure, which we call "physical" couplings $\{g_i^{phys}\}$. These physical couplings, however, depend on the energy scale $\Lambda$ at which we measure them. The physical couplings are thus functions of the bare couplings and of the energy scale $\Lambda$. When doing calculations of what we expect to measure for these physical couplings, the UV-cutoff $\epsilon$ comes into the calculation. But $\epsilon$ is an artificial parameter, so in order to get the correct calculation of the $\{g_i^{phys}\}$, we need to allow the bare couplings $\{g_i\}$ to be able to vary with $\epsilon$ so that the $\epsilon$ dependence cancels out of the calculations of $\{g_i^{phys}\}$. My understanding is that the "Renormalization group flow" is a description of how the physical couplings $\{g_i^{phys}\}$ change with the energy scale at which you measure them, $\Lambda$.

Okay, I think the above is correct, but here's where I start to get confused: people often talk about the RG flow as a "trajectory in theory space" where at each point along the flow you imagine that you have a different Lagrangian $\mathcal{L}$ whose coupling constants are the $\{g_i^{phys}\}$ which change as you move along the flow (by changing $\Lambda$), thus changing your theory. But that's not right, is it? The Langrangian's coupling constants are by definition the BARE ones $\{g_i\}$ which are invariant under change of the scale $\Lambda$, right? But people seem to insist on talking about RG flows this way, and say things like "fixed points (sometimes) correspond to Conformal Field Theories (CFTs)" and stuff like that. Is it that they are just talking about a different Lagrangian whose bare couplings ARE the physical couplings (which isn't really the Lagrangian of the theory that you are studying)?

I know that in one particular impelementation of the RG, the Wilsonian approach, you ARE actually changing your action by changing a so-called cutoff $\Lambda$ (note: as seen in comments below, I am not clear on whether to think of the $\Lambda$ in the Wilsonian approach as being a UV-cutoff like $1/\epsilon$ above, as a physical measurement scale as in $\Lambda$ in the above, or neither/both, this is probably the crux of my confusion). Lowering $\Lambda$ is accomplished by functionally integrating over the field modes corresponding to momenta higher than the $\Lambda$ you want to set, which gives you an action that looks like it's based on the same Lagrangian as your old one but with different couplings. Are the couplings in your action in Wilsonian RG actually the physical coupling constants?? Why?? What happened to the bare ones??? Is this method of implementing the RG incompatible with my above understanding of what the RG flow is? Or is there perhaps some sort of duality between [1. fixing your theory/Lagrangian and watching your physical quantities change with $\Lambda$] and [2. fixing a physical quantity and watching your theory/Lagrangian change with $\Lambda$]??

I think that my confusion about these details are contributing to my confusion about how the RG is actually implemented in situations where one uses EE to study c-theorems for QFT, but perhaps I'll describe that later if answers to the above don't solve my problems...

  • $\begingroup$ I think you have it backwards. The physical couplings do not depend on the regularization scale. The bare couplings do. In Wilson language, the regularization scale is an arbitrary choice you make, splitting high energy and low energy modes, so no physical quantity should depend on this split. However, making a good choice may have significant practical, computational advantages, like making perturbation theory converge faster. $\endgroup$
    – Andrew
    Dec 17, 2016 at 2:32
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    $\begingroup$ So, the way I explained things in the second paragraph, the physical couplings are not depending on the regularization scale. The regularization scale is $\epsilon$, and this is distinct from the energy scale $\Lambda$ at which measurements are made. It is natural for the physical measurements to depend on $\Lambda$, for example, you'll get a different measured value for electric charge depending on the CM momentum of the electrons you scatter in your experiment. $\endgroup$
    – user9999
    Dec 17, 2016 at 22:00
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    $\begingroup$ Part of my confusion is about what the Wilsonian cutoff represents. Is it a regularization scale? Is it a measurement scale? Is it neither? Both? How do I explain how the Wilsonian picture works in terms of the setup I have in my second paragraph? $\endgroup$
    – user9999
    Dec 17, 2016 at 22:05
  • $\begingroup$ Normally I think of the renormalization group as the running of the bare couplings with the Wilson scale ($1/\epsilon$ in your language). One reason it is confusing to learn this material is because there is more than one way to think about it. The way I think about, the Wilson scale $1/\epsilon$ is an arbitrary scale you choose when doing the calculation. Then, it is frequently convenient to chose this scale to be of order $\Lambda$, the measurements you are making. For example, 'large logs' in perturbation theory can be resummed in this way. $\endgroup$
    – Andrew
    Dec 17, 2016 at 22:38
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    $\begingroup$ I think an excellent reference of this subject is the review by Burgess, arxiv.org/abs/hep-th/0701053. For example, one of the subtle points Burgess discusses is that the 'conceptual picture' of renormalization is often described in terms of a Wilson cutoff, but in practical calculations we often use dimensional regularization, in which the momentum integrals go to arbitrarily large scales. $\endgroup$
    – Andrew
    Dec 17, 2016 at 22:41


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