What is the problem with quantizing GR in the Effective Field Theory approach?

Question

In the modern view due to Wilson, the cut-off $\Lambda$ is an intrinsic property of a theory and renormalization just means that the theory is invariant under scale transformations below $\Lambda$. The divergences are now absent due to the cut-off. In this case, what would be the problem in performing the Feynmann path integral using the Einstein-Hilbert (EH) action and then get a quantized theory for gravity?

I understand that what is done is to add higher order terms to the Lagrangian, but why the EH action is to be seen as a low energy theory? I think my confusion is related to how we determine $\Lambda$ itself. For example, what is the value for $\Lambda$ that makes EH action be a low energy theory?

By adding higher order terms to the Lagrangian, I can see the lack of predictivity that we would get if we worked in arbitrarily high energies, since we would get an infinite number of coupling constants to be fixed by experiments. But this would not be a problem if we could use only the EH term, so seems that quantizing EH action gives us a finite and predictable theory. So what is the problem with all of these?

Answer 1

There is nothing wrong with effective quantum general relativity as a QFT. It works fine and we can get effective answers for calculations. That said, most are not interested in calculating scattering cross sections and amplitudes with quantum GR in a relatively low energy limit, they are interested in the character of space time and Plank scale physics which effective GR isn't all that useful for.

Most would argue that we need a background independent, UV complete theory to answer big questions about the nature of spacetime at the Plank scale.

Edit:

We can work with just the EH action in an effective approach but the scattering amplitudes will depend on our energy scale. If we go up to too high an energy scale, our answers won't make any sense unless we add a counter term of higher order to $\mathcal{L}$ multiplied by some a priori unknown coupling constant which has to be determined by experiment. If we start asking about even higher energies we'll need to add an even higher order term to $\mathcal{L}$.

If we only use the term in the Lagrangian we know from the classical theory (ie. The Einstein Hilbert Lagrangian), we can learn a little about quantum gravity, but not all that much. What to do about this situation is largely up to personal taste.

If you ask someone who takes seriously the insights from classical gravity they will tell you that the problem is not the Lagrangian, but perturbation theory. Perturbation theory destroys the background independence of General Relativity and we shouLdnt be too surprised that perturbation theory isn't a straightforward and easy way to learn everything we know about quantum gravity. Imagine in classical gravity trying to find the Schwarzchild solution pertubatively (ie, linearized gravity), that's clearly going to be extremely troublesome.

Question: What is an appropriate cutoff for effective gravity?

In units where $c=\hbar=1$, when one calculates the scattering amplitude for gravitons they find that the energy scale dependence goes as $$1+GE^2+(GE^2)^2+...$$ Where G is Newtons constant.

So, we need to start worrying when GE^2 gets to of order 1 because then that series is divergent. This tells us that $\Lambda \approx \sqrt{\frac{1}{G}}$, which is just the Plank energy in our unit system! This tells us that an appropriate cutoff for quantum gravity is the Plank scale.

This answer is adopted from Quantum Field Theory in a Nutshell by Zee, page 172