I would like to present a sketch of the approach via the unitary transformation; please have a look.
We start with the scale transformation U generated by G:
U[\lambda] = e^{iG\lambda}
G = (px+xp)/2
with time-dep. lambda, such that
x^\prime = UxU^\dagger = x\,e^\lambda
p^\prime = UpU^\dagger = p\,e^{-\lambda}
results in a fixed interval
x^\prime \in [0,L_0]
The problem is equivalent to a Hamiltonian
H_1 = \frac{p^2}{2me^{2\lambda}}
with time-dep. mass plus an additional term
H_2 = - iU\partial_tU^\dagger = -G\dot{\lambda}
The ansatz for a solution of the time-dep. problem is
|n,t\rangle = e^{-i\int_0^t d\tau \, E_n(\tau)}\,|n\rangle
|\psi,t\rangle = \sum_n a_n(t) \, |n,t\rangle
with momentum eigenstates
p|n\rangle = k_n|n\rangle
E_n = \frac{k_n^2}{2me^{2\lambda}}
We have to solve the following coupled differential equations:
i\dot{a}_m + \dot{\lambda}\,\sum_n\langle m|G|n\rangle\,a_n = 0
which is formally
i\dot{A} = -\dot{\lambda}gA
with
A = (a_n)
g = \langle m|G|n\rangle = (k_m + k_n) \, \langle m|x|n \rangle \, / \, 2
The formal solution is
A(t) = e^{i\int_0^t d\tau \,\dot{\lambda} g}\,A_0 = e^{i(\lambda - \lambda_0) g}\,A_0
and therefore
|\psi,t\rangle = \sum_n e^{-i\int_0^t d\tau \, E_n(\tau)} \left[ e^{i(\lambda - \lambda_0) g}\,A_0 \right]_n |n\rangle
This basically corresponds to a kind of interaction picture with "free, time-dep." H_1 plus interaction term H_2 ~ G. The first term does not cause transitions between "instantaneous eigenstates".
The "instantaneous eigenstates" of the full, time-dep. Hamiltonian are constructed as follows
(H-E)|u,t\rangle = 0
|u,t\rangle = \sum_n u_n(t)|n,t\rangle
with eigenvalue equation for E
(E_m - E)u_m - \dot{\lambda}\sum_n e^{i\int_0^t d\tau\,(E_m-E_n)} \, \langle m|G|n\rangle \, u_n = 0