pp chain gamma decay / intermediate state Based on the first equation at http://en.wikipedia.org/wiki/Proton%E2%80%93proton_chain_reaction
$$_1^{1}H+_1^{1}H \to _2^{2}He+\gamma$$
Is it correct to say that in the P-P chain two hydrogen atoms fuse to form a helium-2 isotope in an excited state which decays by gamma emission to a helium-2 isotope in a non-excited state, emitting a photon in the process?
$$_1^{1}H+_1^{1}H \to _2^{2}He^* \to _2^{2}He+\gamma$$
Or is it just a one-step process without excited helium-2 intermediate?
 A: In a very general sense a lot of reaction that are written in one step can also be written in two. I.e. alpha capture on carbon-131 is often written 
$$\alpha + ^{13}\!\mathrm{C} \to ^{16}\!\mathrm{O} + \text{various photons and leptons} \,,$$ 
but may be written as one of
$$\begin{align} 
\alpha + ^{13}\!\mathrm{C} &\to ^{16}\!\mathrm{O} \\
\alpha + ^{13}\!\mathrm{C} &\to ^{16}\!\mathrm{O}^* \to ^{16}\!\mathrm{O} + e^+ + e^-\\
\alpha + ^{13}\!\mathrm{C} &\to ^{16}\!\mathrm{O}^{**} \to ^{16}\!\mathrm{O} + \gamma\,(\text{6.05 MeV}) \\
\dots
\end{align}$$
However, in order for that to be reasonable, there must be a bound intermediate state to talk about.
In the case of PP fusion there is no bound $^2\mathrm{He}$ state (much less a bound excited state).2 Fusion only results if there is a weak tranformation at a time when both protons are very close to one another.
Nor can we talk about one proton turning into a neutron and then finding the other proton, because that intermediate state is energetically forbidden.
So, long story short, I don't think that you should write the proton-proton fusion process with an intermediate state. It's all or nothing.

1 A reaction selected entirely because I know it well.
2 Contrast this with the oxygen system where the excited states are real and have been studied in detail by neutron knock-out reaction on Oxygen-17.
