Unless the wavefunction collapses to an eigenstate of the Hamiltonian, the subsequent time-evolution will produce a superposition.
The postulates clearly state that, if you measure the observable $\Lambda$ and obtain the outcome $\lambda$ (assumed non-degenerate for simplicity), then the state collapses to the eigenstate $\vert\psi_{\lambda}\rangle$ of $\hat \Lambda$, and the subsequent evolution is given by $$ \sum_{k}e^{-iE_k t}\vert \Psi_{E_k}\rangle\langle \Psi_{E_k}\vert\psi_{\lambda}\rangle $$$$ \sum_{k}e^{-iE_k t/\hbar}\vert \Psi_{E_k}\rangle\langle \Psi_{E_k}\vert\psi_{\lambda}\rangle $$ where $\vert \Psi_{E_k}\rangle$ is an eigenstate of $H$ with eigenvalue $E_k$. Thus, unless $\langle \Psi_{E_k}\vert\psi_{\lambda}\rangle=\delta_{E_{k}\lambda}$, the system will revert to a superposition.
Edit: after the measurement the state $\vert \psi_{\lambda}\rangle$ functions as an initial state and its time development is obtained in the usual manner by expanding over a complete set of eigenstates of $H$ using $$ \hat 1=\sum_k\vert\Psi_{E_k}\rangle\langle \Psi_{E_k}\vert $$ so that $$ \vert\Psi(0)\rangle=\vert\psi_\lambda\rangle= \sum_k\vert\Psi_{E_k}\rangle\langle \Psi_{E_k}\vert\psi_\lambda\rangle $$ and evolving the $H$-eigenstates $$ \vert\Psi(t)\rangle= \sum_k\,e^{-iE_{k}t/\hbar }\vert\Psi_{E_k}\rangle\langle \Psi_{E_k}\vert\psi_\lambda\rangle\, . $$