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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.

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  • $\begingroup$ Yes, a self-adjoint operator must have a dense set of analytic vectors, but if $|0\rangle$ is not analytic it does not imply that $U_k$ is not unitary. It would require a long answer to explain rigorously about unbounded Hermitian operators. John Baez gives some intuition for the inverse cube force law, where a classical particle spirals in to its doom in a finite time (scroll down to "Quantum aspects"). I don't think Braunstein and McLachlan prove that the Pade converges, much less what it converges to. $\endgroup$ Dec 20, 2021 at 23:25
  • $\begingroup$ @KeithMcClary I appreciate this comment, but I don't understand the first part: does $|0\rangle$ not being an analytic vector of $U=\exp iA$ imply that it is also not an analytic vector of $A$? One can clearly show that $A|0\rangle$ is normalizable here, so I don't see the problem with $A$ having a dense set of analytic vectors including $|0\rangle$. Also B&M indeed plot what the matrix element converges to. $\endgroup$ Dec 21, 2021 at 14:02
  • $\begingroup$ $\psi$ is an analytic vector for $A$ if the formal series expansion of $U$ converges (in the specified sense). AFAIK we do not speak of analytic vectors for $U$. $\endgroup$ Dec 21, 2021 at 18:39
  • $\begingroup$ When we are talking about self-adjointness we need to be mathematically rigorous. B&M have shown numerically that it appears to converge. Even if that can be made rigorous somehow, it also needs to be proven that it converges to the same thing as you would get from Stone-von Neumann. $\endgroup$ Dec 21, 2021 at 18:50
  • $\begingroup$ That makes sense. So then would you disregard B&M? A recent paper On the derivation of exact eigenstates of the generalized squeezing operator seemed to say the problems with $U$ can be ignored because they found exact eigenstates for it and because B&M solved all the problems, but your comments here make me think one should be more suspicious of the new papers and less suspicious of the original by Fisher, Nieto, & Sandberg. $\endgroup$ Dec 21, 2021 at 20:03

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