The paper Impossibility of naively generalizing squeezed coherent states proves that the generalized squeezing operators $$U_k(z)=\exp(z a^{\dagger k}-z^* a^k)$$ have some of their matrix elements diverge for integers $k>2$ and bosonic operators $a$, in that $\langle 0|U_k(z)|0\rangle$ diverges for all $z$, where $a|0\rangle=0$. This leads to the assertion that $U_k(z)$ is not unitary and that the Hamiltonian that generates it, $$A_k(z)=\frac{z a^{\dagger k}-z^* a^k}{-i},$$ is not self-adjoint, because $|0\rangle$ is not an analytic vector of $U_z$. Analytic vectors $|\psi\rangle$ are defined as having a finite radius of convergence for $\sum_n ||A_k^n |\psi\rangle|| t^n/n!$ for some finite $t$.
Then, later papers such as Generalized squeezing claim that the divergence of $\langle 0|U_k(z)|0\rangle$ does not affect the unitarity of $U_k(z)$, because one can numerically calculate matrix elements such as $\langle 0|U_k(z)|0\rangle$ using Padé approximants to avoid the pole on the imaginary time axis of $\exp[-i A_k(z) t]$. The latter paper even claims that the former paper does not question the self-adjointness of $A_k$, which seems like a strong contradiction to me.
Who is "correct?" Does the lack of an analytic expression for $U_k|0\rangle$ imply that $U_k$ is not unitary? Does it ever make sense to talk about unitary operators whose matrix elements are undefined in a standard basis?
In terms of motivation, these Hamiltonians arise in considerations of high-order nonlinear optics. One might contend that an experimental demonstration of spontaneous parametric down conversion with a pump photon splitting into three lower-energy photons would demonstrate the existence of operators such as $U_k$, but this is not immediately apparent: $U_k$ can be thought of as a semiclassical approximation of some other operator $$V_k=\exp(b a^{\dagger k}-b^\dagger a^k)$$ that does not have any divergence problems according to Photon number divergence in the quantum theory of n-photon down conversion. I was led to the first paper by this useful answer.
