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Final edit: I think I pretty much understand now (touch wood)! But there's one thing I don't get. What's the physical reason for expecting the correlation functions to be independent of the cutoff? I.e. why couldn't we just plump for one "master Lagrangian" at the Planck scale and only do our integration up to that point?

  • Perhaps it has something to do with low energy experiments not being influenced by Planck scale physics.
  • Maybe it's because there isn't any fundamental scale, i.e. that $\Lambda$ must be arbitrary in a QFT approximation, for some reason.

I'll award the bounty to anyone who can explain this final conundrum! Cheers!

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Apologies if this question is too philosophical and vague! I've been thinking about QFTs and continuum mechanics, and reading about their interpretation as effective theories. In these theories we have natural cutoffs at high momentum (small scales). We make the assumption ($\star$) that the large scale physics is decoupled from the small-scale. Therefore we hope that our predictions are independent of the cutoff (after some renormalization if necessary).

Why is the assumption ($\star$) so reasonable? I guess it seems observationally correct, which is powerful empirical evidence. But could it not be the case that the small scale physics had ramifications for larger scale observations? In other words, would it be reasonable to expect that the predictions of a TOE might depend on some (Planck scale) cutoff?

This question may be completely trivial, or simply ridiculous. Sorry if so! I'm just trying to get a real feel for the landscape.

Edit: I'd like to understand this physically from the purely QFT perspective, without resorting to analogy with statistical physics. It might help if I rephrase my question as follows.

In the Wilsonian treatment of renormalization we get a flow of Lagrangians as the energy scale $\Lambda$ changes. For a renormalizable theory we assume that there's a bare Lagrangian independent of $\Lambda$ in the limit $\Lambda \to \infty$. We calculate with this quantity, by splitting it into physical terms and counterterms. I think these counterterms come from moving down the group flow, but I'm not quite sure...

But why do we care about (and calculate with) the bare Lagrangian, rather than one at some prescribed (high) energy scale (say the Planck scale)? I don't really understand the point of there existing a $\Lambda\to \infty$ limit.

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Related: physics.stackexchange.com/q/57965/2451 –  Qmechanic Aug 6 '13 at 22:50
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Hi Edward, do you have some understanding of RG flows (especially the "Wilsonian picture")? This should help people (maybe me, tomorrow) write a proper answer. –  Vibert Aug 6 '13 at 22:54
    
@Vibert: I'm just starting to read up on the renormalization group at the moment. I don't know that much about it atm, but will certainly focus on it more if it helps me to understand your (potential) answer! Thanks in advance. –  Edward Hughes Aug 6 '13 at 23:29
    
Physics at different scales are related. See for instance running of coupling constants in this general paper –  Trimok Aug 7 '13 at 16:43
    
Just to clarify - I now understand RG flows. –  Edward Hughes Sep 29 '13 at 17:02
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2 Answers

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This is a very interesting question which is usually overlooked. First of all, saying that "large scale physics is decoupled from the small-scale" is somewhat misleading, as indeed the renormalization group (RG) [in the Wilsonian sense, the only one I will use] tells us how to relate the small scale to the large scale ! But usually what people mean by that is that if there exists a fixed-point in the RG flow, then some infrared (IR) [large scale] physics is independent of the details at small scale [ultraviolet (UV)], that is it is universal. For instance, the behavior of the correlation functions at long distance is independent of the bare parameters (to fix the setting, say a scalar field with bare parameters $r_\Lambda, g_\Lambda$ for the quadratic and quartic interaction and $\Lambda$ is the (for now) finite UV cut-off).

But one should not forget that a lot of physical quantities are non-universal. For example, the critical value of $r_\Lambda$ (at fixed $g_\Lambda$ and $\Lambda$) to be at the critical point is not universal. And this is a physical quantity in condensed-matter/stat-phys, the same way that $\Lambda$ also has a physical meaning.

The point of view of the old-school RG (with conterterms and all that) is useful for practical calculations (beyond one-loop), but make everything much less clear. In the spirit of high-energy physics with a QFT of everything (i.e. not an effective theory), one does not want a cut-off, because it has no meaning, the theory is supposed to work at arbitrary high-energy. This mean that we should send $\Lambda$ to infinity. And here comes another non-trivial question : what do we mean by $\Lambda\to\infty$ ?

The perturbative answer to that is : being able to send $\Lambda\to\infty$ order by order in perturbation in $g$. But is it the whole answer to the question ? Not really. When we say that we want $\Lambda\to\infty$, it means that we want to define a QFT, at a non-perturbative level, which is valid at all distance, and we want this QFT to be well-defined, that is defined by a finite number of parameters (say two or three). And in fact, this non-perturbative infinite cut-off limit (that I will call the continuum limit) is much more difficult to take. Indeed, having a theory described in the limit $\Lambda\to\infty$ by a finite number of parameter means that the RG flows in the UV to a fixed point. In the same way, the RG has to flow in the IR to another fixed point in order to be well controlled. This implies that very few QFTs in fact exist in the continuum limit, and that some QFTs which are perturbatively renormalizable ($\Lambda\to\infty$ order by order in perturbation in $g$) are not necessarily well defined in the continuum limit !

For instance, some well known QFTs in dimension four (such as scalar theories or QED) do not exist in the continuum limit ! The reason is that even if these theories are controlled by a fixed point in the IR (at "criticality", which for QED means at least electrons with zero masses), it is not the case in the UV, as the interaction grows with the cut-off. Therefore one has to specify the value an infinite number of coupling constants (even "non-renormalizable") to precisely select one RG trajectory.

One of the QFTs which exists in the continuum limit is the scalar theory in dimension less that four (say three). In that case, at criticality, there exists one trajectory which is controlled by a fixed point in the UV (the gaussian fixed point) and in the IR (the Wilson-Fisher fixed point). All (!) the other trajectories are either not well defined in the UV (critical theories but with otherwise arbitrary coupling constants) or in the IR (not a critical theory). One then sees why this $\Lambda\to\infty$ limit is less and less seen as important in the modern approach to (effective) QFTs. Unless one wants to describe the physics at all scale by a QFT, without using a fancy up-to-now-unknown theory at energies above $\Lambda$. Nevertheless, this idea of controlling a QFT both in the IR and the UV is important if you want to prove that General relativity is (non-perturbatively) renormalizable (i.e. can be described at all scales by few parameters) in the asymptotic safety scenario : if there is a non trivial UV fixed point, then there exists a trajectory from this fixed point to the gaussian fixed point (which is, I think, Einstein gravity), and you can take the continuum limit, even though the perturbative $\Lambda\to\infty$ does not exists.

Reference : Most of this is inspired by my reading of the very nice introduction to the non-perturbative RG given in arXiv 0702.365, and especially by the section 2.6 "Perturbative renormalizability, RG flows, continuum limit, asymptotic freedom and all that".

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Many thanks for your detailed answer! So am I right to think that the usual renormalization procedure is pointless then? Rather one should just calculate with an effective Lagrangian with a cutoff at some experimental scale $\mu$. Then the answers will automatically depend on $\mu$ but that's okay because the coupling constants do? I feel like there's still something wrong with my reasoning there. I don't understand what exactly you're meant to calculate with in the Wilsonian picture. Any ideas? –  Edward Hughes Sep 30 '13 at 8:16
    
If by usual you mean "old-school", no, it's not useless. It does not help on the conceptual level, but it is really useful for calculations. The reason is that in this approach, you don't need to take care of the infinite number of coupling constant that "exist" in the Wilsonian skim. (The old-school scheme corresponds to projecting all RG trajectories that start close enough to the gaussian fixed point to the only trajectory that relates gaussian and Wilson-Fisher fixed point, see the discussion in the ref I gave.) –  Adam Oct 1 '13 at 3:12
    
Also, in the Wilsonian RG, the action (or lagrangian) at scale $\mu$ is not physical by itself. Only some quantities that can be extracted from the RG (such as critical exponents) are physical. That one of the main shortcoming of this approach (but see the non-perturbative RG, wilsonian in spirit, which allows to compute physical quantities such that thermodynamics or correlation functions). On the other hand, the "old-school" RG computes physical quantities : for instance $g(\mu)$ is a vertex function at some specified momentum being equal to $\mu$. This is measurable and physical. –  Adam Oct 1 '13 at 3:15
    
Right - so is the following reasoning correct? If you could just confirm this for me I'll happily award the bounty! We want our physical quantities (e.g. amplitudes) to be independent of cutoffs. Why? Because otherwise we could get information about small scale physics by doing large scale experiments. Is this correct? I don't quite see exactly why that would be the case. I'm pretty sure that this is just the definition of an EFT now, but I'd like to have a physical feeling about why cutoff independence is good. Thanks in advance! –  Edward Hughes Oct 3 '13 at 13:33
    
People use to want everything independent of the cut-off, and sadly it is still presented that way in most text books. You only want that if you think that your theory is the ultimate theory that will describe all phenomena at all energies. But usually, you don't want that (all cond-mat/stat-phys, low energy QCD, Fermi theory of weak interactions...), and you're happy there is a (physical) cut-off. But of course, having a 'renormalizable theory' (keeping only the most relevant interactions) is very useful technically, as it makes everything simpler (cut-offs make calculations complicated). –  Adam Oct 3 '13 at 22:27
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At every stage of renormalization, the Hamiltonian changes $\mathcal{H} \rightarrow \mathcal{H}_{\textrm{ren.1}}\rightarrow \mathcal{H}_{\textrm{ren.2}} \rightarrow \ldots$; in the process, energy modes and length scales are excluded as you say. But the point is that every $\mathcal{H}, \mathcal{H}_{\textrm{ren.1}}, \mathcal{H}_{\textrm{ren.2}}, \ldots$ (including the 'original' $\mathcal{H}$) is an effective or emergent theory applicable only within its domain $\Omega, \Omega_{\textrm{ren.1}}, \Omega_{\textrm{ren.2}}, \ldots$. That is, there being no fundamental theories even in particle physics was a key point stressed by K. G. Wilson. Therefore, for instance in field theories, the bare electron mass $m$ becomes simply a mathematical construct; the true one as measured and measurable is the renomalized value $m^*$.

As regards the decoupling, I'll take this from the point of view of critical phenomena. At this critical point where there are correlations across the entire system, the lattice spacing does not matter as we well know; therefore, its the long wavelength modes that stretch across the system that contribute most. Clearly, the decoupling of length scales is justified in such a situation; because QFT and statistical mechanics are essentially equivalent via Feynman's path integral notation, the decoupling is justified in renormalizable field theories. If anyone can make this mathematically rigorous, please feel free ...

As an analogy, think of a classical system with many configurations $i$ with energies $\epsilon_i$; depending on the temperature $T$, the contribution of a configuration will be largely decided by its Boltzmann weights $e^{-\epsilon_i/k_BT}$. In which case, we may discard all other contributions or modes that have negligible weights.

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