I don't understand exactly what is your question:
If you start in an eigenstate of $H_p$, say $\varphi_j$ and then your Hamiltonian changes like $H(t) = H_u + f(t) (H_p - H_u)$ where $f(0) = 1$ and $f(T) = 0$ -$T$ being the final time of the perturbation- then in order to follow the evolution of $\psi(t)$ all you need to do is to solve the time dependent Schroedinger equation
$$i\hbar \partial \psi(t) / \partial t = H(t) \psi(t)$$
where $\psi(0) = \varphi_j$. There are many numerical procedures to do so. For instance if you are interested, as it seems, in following the dynamics in the basis of the eigenstates of $H_u$, then you expand $\psi(t) = \sum_k a_k(t)
\phi_k$, where $\phi_k$ are eigenstates of $H_u$, and you solve a set of ordinary differential equations.
If $f(t)$ is very fast ($T << (\Delta E)^{-1}$, where $\Delta E$ is a characteristic energy of the system, say the energy difference between adjacent eigenstates) then you can use the sudden approximation; if $f(t)$ is very slow you can use the adiabatic approximation, if $H_p - H_u$ is small you can use first order perturbation theory, etc.
In any case it is important to realise that in Quantum Mechanics there is no relaxation unless you couple your initial state to a continuum and there is no return to equilibrium unless you do the same (couple to continuum) and work with the proper formalism that allows you to define the (thermal) equilibrium, which are the density matrices, unless by equilibrium you are referring to the initial state. Again, in ordinary Quantum Mechanics $\psi(t)$ will oscillate and will not relax to the initial state $\varphi_j$ (or $\phi_0$); at most, it will return to the initial state as part of an oscillation, called revival.
