Eigenvectors and eigenvalues are properties of operators, not of Hilbert spaces. Hence, answering the first part of your question, if $| n \rangle$ is an orthonormal basis of $\mathcal{H}$ (the Hilbert space), it is not correct to say they are the eigenvalues of the Hilbert space. We simply say they are an orthonormal basis.
Suppose now you are given an operator $A \colon \mathcal{H} \to \mathcal{H}$ on the Hilbert space. Loosely speaking, a matrix. A vector $| \Psi \rangle \in \mathcal{H}$ is said to be an eigenvector of $A$ associated to the eigenvalue $a$ when $A | \Psi \rangle = a | \Psi \rangle$. Notice that I need the operator to define what I mean by eigenvector: eigenvectors are the vectors in which $A$ acts as if it was simply a number, this number being called an eigenvalue.
Now if $A$ is hermitian, there are two cool results:
- the eigenvalues of $A$ are all real;
- the eigenvectors of $A$ provide an orthonormal basis of the Hilbert space.
Hence, if you are given a hermitian operator on the Hilbert space, you can use it to obtain a basis. We usually pick the Hamiltonian, for example, because then each state in the basis has a simple time-evolution in terms of its energy (more specifically, in terms of the eigenvalue of the Hamiltonian to which it is associated). By the very definition of what a basis is, we can then write any vector in the Hilbert space in terms of eigenvectors of the Hamiltonian, i.e., in terms of states of definite energy. This provides a nice way to write down the time evolution of the state, since the Hamiltonian eigenstates have simple time-evolution rules.
We could pick other operators, if we so desired. Instead of the Hamiltonian, you could pick some other hermitian operator to obtain a basis, but it would probably be less convenient to work with.
Finally, it is worth mentioning that sometimes a few linearly independent states might be associated to the same eigenvalue. We call this degeneracy. In this case, we often need a few more operators (such as angular momentum squared and angular momentum in the $\vec{z}$ direction) to properly label all the states in the basis in a unique way. This is what happens when we solve for the hydrogen atom and need three labels to define uniquely which state in the basis is which, because a few of them have the same energy (as a consequence, we distinguish them by using their angular momentum properties).
I set $\hbar = 1$
Another way to come at this is to consider the Hamiltonian to be the generator of unitary time evolution. That is, say we have an initial state $|\psi(0)\rangle$. We can take it as a postulate of quantum mechanics that the state at a later time is given by
$$
|\psi(t)\rangle = U(t) |\psi(0)\rangle
$$
Where $U$ is a unitary operator. By taking the time derivative of this expression we arrive at the Schrodinger equation
$$
\frac{d}{dt}|\psi(t)\rangle = -i H(t) |\psi(t)\rangle
$$
Where I've defined the Hamiltonian
$$
H(t) = i \frac{dU}{dt} U^{\dagger}
$$
From this definition and unitary of $U(t)$ it can be proven that
$$
H(t) = H^{\dagger}(t)
$$
and, if $H(t)$ happens to be time-independent:
$$
U(t) = e^{-i H t}
$$
What you seem to be gesturing at in your question is that the Hermiticity of $H$ seems to intuitively follow from the reversibility of dynamics/interactions. This reversibility is exactly encoded in the unitarity of the time evolution operator, so deriving Hermiticity of $H$ from this unitarity should then, maybe, be satisfying to you.
Best Answer
I believe that it is true as long as there does not exist a non-trivial unitary operator $U$ that commutes with the Hamiltonian ($[H,U] = 0$) in the subspace of ground states. If such an operator exists then for a ground state $|\phi_0\rangle$ with energy $E_0$ we have $$HU|\phi_0\rangle = UH|\phi_0\rangle = E_0\left(U|\phi_0\rangle\right)$$ and so $U|\phi_0\rangle$ also has the lowest possible energy $E_0$ and it thus also a ground state. Note that the statement of non-triviality of $U$ is important. It needs to be non-trivial in the subspace of ground states, that is $U|\phi_0\rangle \neq e^{i\theta}|\phi_0\rangle$ for any phase $\theta$, otherwise there is no degeneracy. (Unitarity is needed so that $U|\phi_0\rangle$ is a state with norm 1)
More succinctly, if there exists a unitary operator $U$ such that $[H,U]=0$ and $U|\phi_0\rangle\neq e^{i\theta}|\phi_0\rangle$ for any phase $\theta$ then we have ground state degeneracy.
In the example you have given we see that the matrix elements in the basis given $\{|a\rangle,|b\rangle,|c\rangle\}$ is $$H = \begin{pmatrix}1&0&0\\0&1&0\\0&0&2\end{pmatrix}$$ from which we see there exists a unitary operator, with matrix elements $$U = \begin{pmatrix}0&1&0\\1&0&0\\0&0&1\end{pmatrix}$$ which commutes with $H$ and is non-trivial in the ground-state space.
Proof that non-existence of $U$ implies non-degenerate ground state:
Assume $\nexists U$ s.t. $\{[H,U]=0 ~~\mbox{and}~~ U|\phi_0\rangle \neq e^{i\theta}|\phi_0\rangle\}$
Now, for every state $|a\rangle$ and $|b\rangle$, $\exists U_{ab}$ which is unitary that takes us from $|a\rangle\rightarrow|b\rangle$. We are interested in the operator that take us from $|\phi_0\rangle$ to any $|a\rangle$ in our Hilbert space (which obviously includes all possible ground states), which we denote by $U_{a0}$. This means that any state $|a\rangle$ can be written as $|a\rangle = U_{a0}|\phi_0\rangle$. By our starting assumption $U_{a0}$ either satisfies $$(1)~~~~~~ [H,U_{a0}]\neq 0,~~~~~~~~\mbox{or}~~~~~~~~(2)~~~~~U_{a0}|\phi_0\rangle = e^{i\theta}|\phi_0\rangle$$ If (1), then we have $$H|a\rangle = H U_{a0}|\phi_0\rangle \neq U_{a0}H|\phi_0\rangle = E_0|a\rangle~~~\implies~~~H|a\rangle \neq E_0|a\rangle$$ and so $|a\rangle \neq |\phi_0\rangle$ is not a ground state.
If (2), then $|a\rangle = e^{i\theta}|\phi_0\rangle$ and so $|a\rangle$ and $|\phi_0\rangle$ represent the same state.
Thus the non-existence of $U$ implies the non-existence of a second ground state and thus non-degeneracy.