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In this Wikipedia article there are interesting statements:

A quantum field theory is said to be trivial when the renormalized coupling, computed through its beta function, goes to zero when the ultraviolet cutoff is removed. Consequently, the propagator becomes that of a free particle and the field is no longer interacting.

For a $φ^4$ interaction, Michael Aizenman proved that the theory is indeed trivial, for space-time dimension $D ≥ 5$.

For $D = 4$, the triviality has yet to be proven rigorously, but lattice computations have provided strong evidence for this. This fact is important as quantum triviality can be used to bound or even predict parameters such as the Higgs boson mass. This can also lead to a predictable Higgs mass in asymptotic safety scenarios.

These statements are entirely counter-intuitive and strange for me.

Could somebody explain, how theory with initial nontrivial 4-particle interaction becomes trivial?

Maybe there are some toy examples of such phenomena?

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    $\begingroup$ By the way, a proof has been announced that the 4d theory is trivial. arxiv.org/abs/1912.07973 $\endgroup$ – user1504 Jun 5 at 20:17
  • $\begingroup$ I am confused. Does this apply to the Higgs field? $\endgroup$ – lalala Jun 6 at 19:38
  • $\begingroup$ @lalala The Higgs field has also many other interactions, so it's not immediately obvious. $\endgroup$ – MannyC Jun 7 at 10:36
  • $\begingroup$ @lalala That would be a good question to ask in a separate post. $\endgroup$ – user1504 Jun 7 at 11:10
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One can get a physical sense of the theory might be trivial in more than four dimensions by thinking of the trajectories of the $\phi$-field particles. In $d$ dimensions two geometric objects of the same dimension $k$ typically intersect in sets of dimension $2k-d$ For examples two curves ina plane typically intersect in $2-2=0$ dimensional pbejcts -- i.e point. Two $k=2$ surfaces in $3$ dimensions typically intersect in $4-3=1$ dimensional curves. Now a $\lambda \phi^4$ interaction means that particles only interact if their spacetime trajectories touch. The particle trajectories in a path integral are random walks that have Hausdorf dimension $2$, so a random walk in three dimensions will typically self intersect in a set of dimension $1$ -- lots of interactions therefore. In four dimensions the particles only intersect in isolated points -- not so much interactions therefore. In more than four dimensions the trajectories of randomly walking particles typically do not self intersect, and so no matter how strong the interactions, nothing happens -- the theory is free.

This reasoning might sound overly simplistic, but the real triviality proof is a version of this one, only with rigorous definitions and estimates. I think that the original idea is due to Giorgio Parisi: See G Parisi "Hausdorff dimensions and gauge theories" Physics Letters B 81 (1979) 357-360.

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    $\begingroup$ That’s a very interesting point. Can you please provide a reference to the triviality proof that uses this idea? $\endgroup$ – Prof. Legolasov Jun 6 at 1:19
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    $\begingroup$ The proof of the statement about the intersection of random paths was given in the following paper link.springer.com/article/10.1007/BF02020942 $\endgroup$ – spiridon_the_sun_rotator Jun 6 at 7:28
  • $\begingroup$ Then it turns out that the particles do not interact at all in 4D. And there are not any interacting theories... How does your logic works with YM theory? $\endgroup$ – Nikita Jun 6 at 18:43
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    $\begingroup$ The argument only applies to point contact ineractions i.e. to $\lambda \phi^4$. When set up on a lattice YM is theory of intersecting surfaces. The 4d case was always marginal which is why the $d=4$ proof took so long while $d>4$ was quicker in coming. $\endgroup$ – mike stone Jun 6 at 18:48
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    $\begingroup$ @Nikita Look at G Parisi "Hausdorff dimensions and gauge theories" Physics Letters B 81 (1979) 357-360. $\endgroup$ – mike stone Jun 6 at 20:55
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Suppose you prove, non perturbatively, that the $\beta$ function stays larger than a positive constant. This implies that the coupling grows. You can do physical renormalization in which you define the coupling as $$ \lambda(\mu) \equiv \Gamma_{p_1p_2p_3p_4}\big|_{|\mathbf{p}_i|^2 = \mu^2}\,,\tag{1}\label{rc} $$ where $\Gamma$ is the four-point amplitude. Let's define a reference IR scale $\mu_0$ and the associated coupling $\lambda_{\mathrm{IR}}$ as $\lambda(\mu_0)$. As a consequence, since the derivative of the coupling is strictly positive, you have $$ \lambda_{\mathrm{UV}} \equiv \lim_{\mu\to\infty}\lambda(\mu) = \begin{cases} \infty&\lambda_{\mathrm{IR}} \neq0\,,\\ 0 &\lambda_{\mathrm{IR}} =0\,. \end{cases} $$ Notice that here there are no cutoff issues. We can, for instance, renormalize using dim-reg, impose the renormalization condition \eqref{rc}, and then send $\varepsilon\to0$.

So if we removed the cutoff and obtained a non trivial theory all the way to the UV, what on earth is going on? The problem is that the theory that we obtained is garbage. If we try to compute the S-matrix for $\phi+\phi \to \phi+\phi$ we obtain a divergent answer $$ T_{12\to34} \sim \Gamma_{p_1p_2p_3p_4} \underset{\mu\to\infty}{\longrightarrow} \infty\,. $$ But $|T_{12\to34}|^2$ is a probability, it has to be less than $1$. So the only consistent value for the coupling is $\lambda_{\mathrm{UV}} = \lambda_{\mathrm{IR}} = 0$. If you instead introduce a cutoff, the UV coupling is the value at the cutoff $$ \lambda_{\mathrm{UV}} \equiv \lambda(\Lambda)\,. $$ This may be huge with respect to $\lambda_{\mathrm{IR}}$ but not necessarily infinite. So you can just tune $\lambda_{\mathrm{IR}}$ to be small enough in order to respect unitarity at high energy.

In conclusion

If you want to prove quantum triviality you have to show that the $\beta$ function is eventually larger than a positive constant (so that the coupling grows to infinity from any initial condition).


Some comments:

The source of (at least my) confusion was the following: the statement that "when the cutoff is removed, the theory becomes trivial," may be misunderstood as the fact that the coupling goes to zero at the UV. The situation is precisely the opposite, the coupling grows! (Asymptotic freedom is when the coupling goes to zero, and that requires the $\beta$ function to be negative instead.)

Here is not the RG flow that sets your coupling to zero in the UV. You decide to put it to zero because that's the only option you have to preserve the S-matrix unitarity.

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    $\begingroup$ Thanks for a great answer, +1. What you mentioned in comments on @Stratiev's post has been confusing me since a long time, this clears it up to a nice extent. $\endgroup$ – Dvij D.C. Jun 7 at 3:05
  • $\begingroup$ Thanks for the clarification, this is nice. $\endgroup$ – Stratiev Jun 7 at 8:17
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    $\begingroup$ Thanks a lot, now I've got the point. This means, that the only way to make a consistent UV theory with such interaction is a theory without such interaction $\endgroup$ – spiridon_the_sun_rotator Jun 7 at 9:56
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    $\begingroup$ Unitarity is a bit of a red herring. The rigorous proofs of triviality make no use of it. Triviality is purely a renormalization phenomenon. It can happen in statistical field theories, too, even if they have no Wick rotation. $\endgroup$ – user1504 Jun 7 at 10:55
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    $\begingroup$ The correct answer is the one I explained in our discussion below: Triviality occurs simply because, for any regularization, the interactions disappear too rapidly as you head to the IR. When you try to remove the cutoff, you're always left with a theory whose observables have trivial correlations. $\endgroup$ – user1504 Jun 7 at 11:02
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The 1d Ising Model is a good toy example for this phenomenon.
$$ H = K_a \sum_{i\in a\mathbb{N}} \sigma_i\sigma_{i+1} $$ For this model, you can write down the block spin renormalization transformations exactly, integrating out the variables on odd-numbered sites, and so passing from the lattice $a \mathbb{N}$ to $2 a \mathbb{N}$.

Nice notes on this here.

What you find is that the renormalization flow reduces the coupling as you flow out to longer distances, via $$ K_{2a} = \frac{1}{2} \ln(\frac{e^{2K_a} + e^{-2K_a}}{2}). $$ This always shrinks $K$. (Proof: Flip the sign in the second exponent. $K_{2a} < \frac{1}{2} \ln(\frac{e^{2K_a} + e^{+2K_a}}{2}) = K_a$.)

Once it's small enough, this becomes $$ K_{2a} \simeq \frac{1}{2} \ln(1 + K_a^2) \simeq \frac{1}{2} K_a^2 $$ So the renormalization flow rapidly scales the interaction to zero. Thus, the long distance behavior of 1d Ising is trivial.

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    $\begingroup$ It is a good example, however, of other phenomena - applying the block-spin transformation you strengthen the cutoff, or in other words drive the model to IR limit. In the case conserned the question is whether the coupling vanishes in the UV limit at some point. $\endgroup$ – spiridon_the_sun_rotator Jun 5 at 21:04
  • $\begingroup$ @spiridon_the_sun_rotator I've given an example of the concept the OP was asking about. In this situation, the renormalized coupling going to zero at fixed energy as the UV regularization is removed is the same thing as coupling flowing to zero as we look at long distances on a fixed lattice. If you read Aizenman's papers, you'll see that the statements he proves are analogous to the one I outlined. $\endgroup$ – user1504 Jun 5 at 21:12
  • $\begingroup$ OP's question is about the UV limit, your example is the IR limit. Long distance = Coarse lattice spacing = IR. $\endgroup$ – MannyC Jun 5 at 22:25
  • $\begingroup$ @MannyC Spiridon said the same thing above. It's not true. OP's question is about the renormalized coupling (which is defined at some finite distance scale $A$) and how it changes as the UV scale $a \to 0$. This is equivalent to keeping $a$ fixed and taking $A$ to infinity; the answer only depends on the dimensionless ratio $A/a$. $\endgroup$ – user1504 Jun 5 at 22:45
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    $\begingroup$ I am inclined to agree with @user1504. In momentum-shell RG of (Euclidean) $\phi^4$ theory with fixed momentum cutoff $\Lambda$, repeatedly integrating out degrees of freedom within shells $[\Lambda_0, \Lambda]$, with $\Lambda_0/\Lambda = b$, & rescaling momenta $k \rightarrow k/b$ will drive the quartic interaction to $0$ in $d > 4$ as $b \rightarrow 0$. Is this not equivalent to the QFT approach of taking $\Lambda \rightarrow \infty$ (which also sends $b \rightarrow 0$)? I don't think the issue is about running the RG flow back to the UV, for which the quartic coupling grows in $d > 4$. $\endgroup$ – bbrink Jun 6 at 14:53
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I have taken and slightly rephrased this from Srednicki's QFT book.

Consider the renormalization group equation

\begin{equation} \frac{d\lambda}{d \ln \Lambda} = \beta(\lambda), \tag{1} \end{equation}

for $\phi^4$ theory, where $\lambda$ is the quartic coupling, and $\Lambda$ is an energy scale. Now we integrate between the physical scale $\Lambda = m_{\text{phys}}$ up to the cutoff scale $\Lambda = \Lambda_0$ we have \begin{equation} \int_{\lambda(m_{\text{phys}})}^{\lambda(\Lambda_0)} \frac{d \lambda} {\beta(\lambda)}= \ln \frac{\Lambda_0}{m_{\text{phys}}}. \tag{2} \end{equation}

Now if we approximate the beta function by its leading order term $\beta(\lambda)= \frac{3\lambda^2}{16\pi^2}$ and we try to take the limit of the cut off to infinity $\Lambda_0 \rightarrow \infty$, since we would like to have a theory that is consistent at all energy scales. If we assume that the beta function is monotonic, we get that the coupling should grow with energy and hence $\lambda(\Lambda_0) \rightarrow \infty$. But if that is the case, the LHS of (2) becomes \begin{equation} \lim_{\lim \Lambda_0 \rightarrow \infty}\int_{\lambda(m_{\text{phys}})}^{\lambda(\Lambda_0)} \frac{d \lambda} {\beta(\lambda)}=\int_{\lambda(m_{\text{phys}})}^{\infty} \frac{d \lambda} {\beta(\lambda)} =\frac{16 \pi^2}{3\ \lambda(m_{\text{phys}})}. \end{equation}

This is clearly not infinte if $\lambda(m_{\text{phys}}) \neq 0$, so the RHS of equation (2) could not be infinite either, which means that $\Lambda_0$ cannot be taken to infinity. This tells us that there is a maximum value of the cutoff $\Lambda_0$ that we can take. Namely, \begin{equation} \Lambda_{\text{max}} = m_{\text{phys}} e^{\frac{16 \pi^2}{3 \lambda(m_{\text{phys}})}}. \end{equation} If we wish to actually take the cut off to infinity, we need $\lambda(m_{\text{phys}})=0$. But that's just a non-interacting theory, which is trivial.

So in a way, in QFT, "trivial" means that you can't both take a UV limit and have this theory be interacting.

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    $\begingroup$ I have two confusions: 1. $\beta(\lambda)$ cannot be replaced by its 1-loop value for $\lambda$ in the range $[O(1),\infty]$, so the integral after (2) needs some justifications. 2. By using the renormalization scheme of my choice I can compute any quantity at infinite cutoff, for any $\lambda$. But since I cannot physically take the cutoff to infinity, something must go wrong somewhere. I can't figure out what. $\endgroup$ – MannyC Jun 5 at 20:11
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    $\begingroup$ By the way, I'm not criticizing your answer (+1) because it's the standard explanation. I just wanted to start a discussion. $\endgroup$ – MannyC Jun 5 at 20:12
  • $\begingroup$ @MannyC Sorry, for the slow response. Thought about it for a while and can't really come up with a good answer to those. I'd be happy to hear if you do have one. $\endgroup$ – Stratiev Jun 6 at 19:18
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    $\begingroup$ The point (I believe) is that if your coupling is infinite in the UV, then you violate unitarity. And if the $\beta$ function is positive, the coupling in the UV is either zero or infinite. If it's zero, it's zero all the way. Have a read at my answer and let me know if it makes sense. $\endgroup$ – MannyC Jun 7 at 2:16

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