What determines a bonding/antibonding molecular orbital? Citing from this site:

Bonding molecular orbitals are formed by in-phase combinations of atomic wave functions, and electrons in these orbitals stabilize a molecule. Antibonding molecular orbitals result from out-of-phase combinations of atomic wave functions and electrons in these orbitals make a molecule less stable.

This made me think about two problems:

*

*A quantum state, as my professor once said, should not distinguish a specific phase from any other, i.e., what really matters is not ${| \psi \rangle }$ but rather ${| \psi \rangle }\langle \psi |$, and this last is invariant under the transformation  ${| \psi \rangle } \mapsto  e^{i\phi}{| \psi \rangle }$, which precisely corresponds to a particular phase choice.


*Let's assume that phase matters in describing orbitals. Then why are there just two possibilities for the overlapping: in-phase or out-of-phase? The phase is not discreet: $\phi \in [0, 2\pi)$, so I would expect a continuous.
 A: First of all, quantum states are invariant under global phaseshifts $\psi \rightarrow e^{i\phi} \psi$. If you shift the phase between two contributions in a superposition, you get a different state.
You are right that you can build a continuous number of states $\psi_1 + e^{i\phi} \psi_2$, and they will all be different. However, there are only two choices of $\phi$ which diagonalize the hamiltonian in the subspace {$\psi_1,\psi_2$} (remember that the $\psi_{1,2}$ are the hydrogen groundstates at the first or second proton), and those are $\phi = \pi$ or $\phi=2 \pi$.
Just to clarify, the Hamiltoniam I'm talking about is
$$ H = \frac{p^2}{2m}+\frac{Ze}{|r-R_1|}+\frac{Ze}{|r-R_2|} = \begin{pmatrix}
\epsilon & V\\
 V & \epsilon
\end{pmatrix}$$
The last equal sign is in the 2 state subspace.
Usually, if we want to describe actual systems in nature, it is helpful to know the eigenstates of the Hamiltonian (for several reasons). This is why people always talk about those two states and ignore the others.
