# Why is $R^2$ gravity not unitary?

I have often heard that $R^2$ gravity (as studied by Stelle) is renormalisable but not unitary. My question is: what is it that causes the theory to suffer from problems with unitarity?

My naive understanding is that if the the Hamiltonian is hermitian then the $S$-matrix $$\langle \text{out} \mid S \mid \text{in} \rangle = \lim_{T\to\infty} \langle \text{out} \mid e^{-iH(2T)} \mid \text{in} \rangle$$ must be unitary by definition. So why is this not the case for $R^2$ gravity?

I see that Luboš Motl has a nice discussion related to such things here, but I am not sure which, if any, of the reasons he mentions relate to $R^2$ gravity.

Are there other well known theories that have similar problems?

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The problem is either the theory is unstable (the Hamiltonian is not bounded from below) or contains states with negative norm (the former is the real problem, and the latter is why it is said that theory is not unitary). It is a general problem of theories with higher order timer-derivatives. –  drake Apr 12 at 2:23
The point is that $R^2$ gravity theories allow for non-hermitean Hamiltonians. Another sign of non-unitarity are fourth-order equations of motions, which show up in precisely such theories (like conformal gravity). Resulting propagators allow for negative energy modes propagating forward in time. There are attempts which circumvent this problem by making use of PT-symmetry, but I am not sure to which extent this issue can be regarded as resolved.
Thanks for your reply. But I don't see that a simple fourth-order Hamiltonian for a real scalar field (for example) is not Hermitian. If it is Hermitian how can the $S$-matrix not be unitary? –  Mistake Ink Apr 11 at 13:54