What would the implication of orthogonal states being mapped to nonorthogonal be? Would information be "lost" in this process? Let's say we have 2 states, $| a \rangle$ and $| b \rangle$, and we suppose them to be orthogonal, and eigenstates with values $a$ and $b$ to some Operator $\hat{O}$. 
What would happen if they were to evolve into 2 states $| \tilde{a} \rangle$ and $| \tilde{b} \rangle$, which are no longer orthogonal anymore. 
What are the implications of this? Although the mapping is still bijective, I am under the impression that we lost some information in the process, because the 2 states lost a special property which they had before. Is there a way to make this statement mathematically rigorous? Or am I wrong, and there isn't any information lost in the process?
 A: I think there's no loss in an absolute sense, but the measurement process tells you less than it otherwise would. 
When you measure an observable, the result is going to be an eigenvalue of the associated operator. If your eigenvectors are orthogonal and non-degenerate, then there is a one to one mapping of your measurements to the associated eigenstate. By virtue of making the measurement, you know the state as well as possible. For a given eigenstate, the same measurement produces the same result. 
If two states are non-orthogonal, they are necessarily mixed states composed of multiple eigenstates. If you measure a particle in one state, you'll get different measurements at different times corresponding to the combination of eigenstates of which the mixed state is composed. So by performing a single measurement on a mixed state, you do not know you are in a specific eigenstate. Further, mixed states of the Hamiltonian evolve over time further complicating the measurement process. 
A given state can be represented as a sum of multiple states of one observable, or multiple states of a different observable. For example, a particle in the ground state of the infinite potential well is in two different states of the momentum but only one state of the energy.  The ground state of the Harmonic Oscillator is a combination of infinitely many states of the momentum. So in this sense, you have the same state, the same information, represented in different ways. 
Given two non-orthogonal states, the Graham-Schmidt orthonormalization process can be used to break them up into orthonormal states. 
