In answer to the edit, any transitions due to single-graviton exchange will involve energies that are just impossibly small. To convince yourself of this, remember that the energy levels of the hydrogen atom are given by:
$$E = \frac{\mu k^{2}e^{4}}{2\hbar^{2}n^{2}} = \frac{13.6\,\,{\rm eV}}{n^{2}}$$
If you do the same for two solar mass neutron stars bound by gravity, you get (up to a fudge factor of order 1 to account for the reduced mass of the system):
$$E = \frac{ G^{2}M^{5}}{2\hbar^{2}n^{2}}$$
Now, this tells us that the energy for the first transition is:
$$\frac{G^{2}M^{5}}{2\hbar^{2}}\left(1 - \frac{1}{4}\right) = \frac{3G^{2}M^{5}}{8\hbar^{2}}$$
Now, let's dump this energy into kinetic energy in one of the stars:
$$\begin{align}
\frac{1}{2}Mv^{2} &= \frac{3G^{2}M^{5}}{8\hbar^{2}}\\
v&=\frac{\sqrt{3}GM^{2}}{2\hbar}
\end{align}$$
Comparing this to escape velocity from radius $r$, we have:
$$\begin{align}
\frac{v}{v_{e}} &= \frac{\sqrt{3}GM^{2}}{2\hbar}\sqrt{\frac{r}{2GM}}\\
&=\frac{\sqrt{3GM^{3}r}}{2^{3/2}\hbar}\\
&=5.15\times 10^{79}\left(\frac{M}{M_{\bigodot}}\right)^{3/2}\left(\frac{r}{\rm 1\,\, AU}\right)^{1/2}
\end{align}$$
So, it should be clear that you're not going to observe any quantum transitions in bulk astronomical orbits, which makes single-graviton effects nigh unobservable (and justifies me ignoring the reduced-mass effects).