Coherent state evolution - nonlinear Hamiltonian

[CptXray]

TL;DR Summary

I have a weird Hamiltonian and I can't find the evolution of coherent state

Given the hamiltonian:
$$\hat{H} = \hbar \omega_{0} \hat{a}^{+}\hat{a} + \chi (\hat{a}^{+}\hat{a})^2,$$
where $\hat{a}^{+}$, $\hat{a}$ are creation and annihilation operators.
Find evolution of the state $|\psi(t) \rangle$, knowing that initial state $|\psi(0)\rangle = |\alpha\rangle$, where $|\alpha\rangle$ is a coherent state.
So, in Schrodinger picture:
$$|\psi(t)\rangle = \hat{U}|\psi(0)\rangle = \hat{U}|\alpha\rangle$$
Evolution operator:
$$\hat{U} = e^{-i\hat{H}t} = e^{-i\omega_{0}t\hat{n} - i\frac{\chi}{\hbar} \hat{n}^2} = e^{-i\omega_{0}t\hat{n}}e^{- i\frac{\chi}{\hbar} \hat{n}^2},$$
in the last equality I used BCH formula.
Expanding coherent state:
$$|\alpha\rangle = e^{-|\alpha|^2/2}\sum_{n}\frac{\alpha^n}{\sqrt{n!}}|n\rangle$$
So, the evolution of the initial state is:
$$|\psi(t)\rangle = e^{-i\omega_{0}t\hat{n}}e^{- i\frac{\chi}{\hbar} \hat{n}^2} e^{-|\alpha|^2/2}\sum_{n}\frac{\alpha^n}{\sqrt{n!}}|n\rangle.$$
By expanding the exponents (from right to left) i get that nasty $n^2$ term and I can't recreate the state.
$$e^{-|\alpha|^2/2}\sum_{n}e^{-i\chi t / \hbar n^2}e^{-i\omega_{0}tn}\frac{\alpha^n}{\sqrt{n!}}|n\rangle.$$

 

[Demystifier]

It seems that you tacitly assume that if the state is initially a coherent state, then it must be a coherent state at all times. But that assumption is wrong. It's true only for the harmonic oscillator Hamiltonian, which corresponds to $\chi=0$.