Triviality of interacting QFT 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?
 A: 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.
A: 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.
A: 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.
A: 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.
