Ordering of eigenstates in the quantum adiabatic theorem Suppose we have an 'initial' Hamiltonian $H_{i}$ whose eigenvalues are all non-degenerate, which we order as follows:
$ E^{0}_{i} < E^{1}_{i} < \dots < E^{N-2}_{i} < E^{N-1}_{i}$
Suppose also that we have a 'final' Hamiltonian $H_{f}$, again with non-degenerate eigenvalues which we can order as before:
$ E^{0}_{f} < E^{1}_{f} < \dots < E^{N-2}_{f} < E^{N-1}_{f}$
Now, if we have some time-dependent Hamiltonian $H(t)$, such that $H(t=0)=H_{i}$ and $H(t=T)=H_{f}$, then the quantum adiabatic theorem, as defined in the 'Proof' section of its Wikipedia page, states that if the state of the system at $t=0$ is an eigenstate of $H_{i}$, so that $|\psi(t=0)\rangle = |E_{i}^{j}\rangle$, then for sufficiently large $T$, $|\psi(t=T)\rangle$ will be an eigenstate of $H_{f}$.
One important application of the quantum adiabatic theorem is in the field of adiabatic quantum computing, where one encodes the solution to a problem in some problem Hamiltonian $H_{p}$, and evolves the system Hamiltonian from some easy-to-prepare initial Hamiltonian $H_{0}$ to $H_{p}$. The main idea is that if one uses some $H_{p}$ with a non-degenerate ground state that encodes the solution to your problem (e.g., to factor a number $N$, use a Hamiltonian $H_{p}=(N-\hat{x}\hat{y})^{2}$, where $\hat{x}$ and $\hat{y}$ are operators whose eigenvalues are natural numbers), and starts the system off in the ground state of $H_{0}$ (presumed non-degenerate), then the final state vector will be the ground state of $H_{p}$. In the case of the example I gave, this solves the problem because the ground state of $H_{p}$ has an energy of 0, and so calculating $\langle \hat{x} \rangle$ and $\langle \hat{y} \rangle$ gives us the natural numbers $x$ and $y$ such that $N = xy$, assuming that $\hat{x}$ and $\hat{y}$ commute with the Hamiltonian.
My question is this: how do we know that the system at time $T$ will be in the ground state of $H_{p}$ if it was prepared in the ground state of $H_{0}$, and not simply some arbitrary eigenstate of $H_{p}$? Put another way, does adiabatic quantum evolution preserve the ordering of eigenvalues? If the answer is no, what is special about the ground state that means that its time-evolved vector is always the ground state of the resultant Hamiltonian?
 A: The answer is given in the comment by Martin. The adiabatic theorem makes a stronger assertion than the one proposed in the question. It asserts that, if the conditions of the theorem hold, then the final state is not just any eigenstate of $H_f$, but the particular one whose energy-level quantum number is the same as that of the starting state. The conditions of the theorem include that
$$
\tau \gg \frac{\hbar}{\Delta E}
$$
where $\Delta E$ is energy gap between the starting state and nearest other state during the change and $\tau$ is the timescale of the change. In order that $\tau$ should be finite while obeying this condition, it follows that $\Delta E$ must not be zero. This means the starting state must not be degenerate, and there must be no level-crossing of its energy eigenvalue with that of another state during the evolution.
It is characteristic of ground states that they tend to be non-degenerate. For such a case $\Delta E$ is the smallest value of the energy gap between ground and first excited state during the adiabatic evolution.
