Propagator in normal modes

I started with the Hamiltonian of coupled oscillators in a circular lattice(with $m=hbar=1$ and $x_{a+N}=x_{a}$)
$$H=frac{1}{2}sum_{a=0}^{N-1}left[p_a^2+omega^2 x_a^2+Omega^2left(x_a-a_{a+1}right)right]$$
Then I used the normal-modes
$$tilde{x}_kequivfrac{1}{sqrt{N}}sum_{a=0}^{N-1}expleft(-frac{2pi i k}{N}aright)x_aquad tilde{p}_kequivfrac{1}{sqrt{N}}sum_{a=0}^{N-1}expleft(frac{2pi ik}{N}aright)p_k$$ to ‘decouple’ the oscillators:
$$H=frac{1}{2} sum_{k=0}^{N-1}left(|tilde{p_k}|^2+tilde{omega_k}^2 |tilde{x_k}|^2 right)$$
where$$tilde{omega}_k=omega^2+4Omega^2sin^2left(frac{pi k}{N}right)$$
In terms of the normal-modes the wavefunction is
$$psi_0left(tilde{x_0},tilde{x_1},..right)=prod_{k=0}^{N-1}left(frac{tilde{omega}_k}{pi}right)^frac{1}{4}expleft(-frac{1}{2}tilde{omega}_k|tilde{x}_k|^2right)$$
Now, I want to time evolve this state by using the product of propagators of free-oscillators. If $tilde{x}_k$ were real, then I would proceed with the propagator as
$$Kleft(tilde{x}_0,tilde{x}_1,..;tilde{x}’_0,tilde{x}’_1;tright)=prod_{k=0}^{N-1}sqrt{frac{tilde{omega}_k}{2pi i sinleft(tilde{omega}_k tright)}}expleft[frac{itilde{omega}_k}{2 sinleft(tilde{omega}_ktright)}{left(tilde{x_k}^2+tilde{x_k}’^2right)cosleft(tilde{omega}_ktright)-2tilde{x}_ktilde{x}’_k}right]$$
And I would time evolve the state $psi_0$ as
$$psi_1 left(tilde{x_0},tilde{x_1},..;tright) =int dtilde{x}’_0 dtilde{x}’_1.. Kleft(tilde{x}_0,tilde{x}_1,..;tilde{x}’_0,tilde{x}’_1…;tright) psi_0left(tilde{x_0},tilde{x_1},..right) $$
How can I find the propagator knowing that $tilde{x}_k$ is not real and then find the time evolved state?