Feynman's ammonia molecule I have been reading Feynman's description of the quantum behaviour of an ammonia molecule. He assumes that the $\rm N$-atom is either pointing up or down as a two-states basis. He then says there is a little probability that the state pointing UP becomes DOWN and vice-versa. But in the same chapter Feynman says that basis states are orthogonal. So how the UP state can become a DOWN one?
Can you please help me understand this?
 A: Let us say I make a basis $\{|\uparrow\rangle,|\downarrow\rangle\}$ such that the two states are orthogonal $\langle \uparrow|\downarrow\rangle=0$. I can use this basis two write all posible states of a two level system, such that $|\psi\rangle=a_\uparrow|\uparrow\rangle+a_\downarrow|\uparrow\rangle$ with $|a_\uparrow|^2+|a_\downarrow|^2=1$.
That is all great to write the dynamics and any state. However, this does not tell us anything about the Hamiltonian $H$. The Hamiltonian describes the dynamics of the system. And in particular the basis I have just written has not necessary to be the eigenbasis of $H$. Maybe it looks something like this in the basis that I have written:
$$H=\begin{pmatrix} h_{\uparrow\uparrow} & h_{\uparrow\downarrow} \\ h^*_{\uparrow\downarrow}  & h_{\downarrow\downarrow}\end{pmatrix}$$
In a toy model of ammonia you could have a Hamiltonian where the two states of the basis are approximately the eigenstates but there is a small coupling term $h_{\uparrow\downarrow}\neq0$  in $H$ between the two basis states. If that is the case, for a initial state  $|\uparrow\rangle$, after enough time it would evolve into $|\downarrow\rangle$ and so on.
In Feynman's notation you can see that he clearly writes a non-diagonal Hamiltonian:

we set $H_{11}=H_{22}=E_0$, and $H_{12}=H_{21}=−A$

