# How do you calculate the mass of diproton helium-2 nucleus?

I understand the diproton form of helium, helium-2 has a half-life of about a nanosecond. It seems safe to assume it's mass cannot be measured experimentally, yet I understand there is a way to calculate what it's mass should be. Is this true?

• It is certainly true that Wikipedia reports a mass for the diproton, which seems odd since it is not a bound state. I cannot find any reference for this claimed mass. Commented Jul 23 at 7:36
• I think I have tracked down the argument used to estimate the proton mass, though I don't think this is rigorous enough to post as an answer. For two protons to come close enough to interact and convert to p + n + e⁺ + ν they have to get through a potential barrier of about 1.4MeV. So at the top of this barrier the total energy is 2mₚ + 1.4MeV/c². If you add this up it comes to the mass that Wikipedia quotes for the ²He state. Commented Jul 23 at 7:44
• I've seen that number thrown around but without anything close to a rigorous definition. I have no idea how Wikipedia can claim such a high precision. Commented Jul 23 at 8:07
• Note that many of Wikipedia's anonymous editors have a young student's misunderstanding of significant figures. The distance from "this reputable source says something surprising" to "this reputable source contains a mistake" is shorter for Wikipedia than it is for other high-quality sources.
– rob
Commented Jul 23 at 14:57
• I think the lifetime of $^2$He is MUCH shorter than a nanosecond. If you assume it is unbound then a brief consideration of the uncertainty pronciple suggests a lifetime of $\sim 10^{-21}$ s. This is also what you get from a consideration of p-p fusion in the Sun and the branching ratio between decay back to protons or a deuteron plus positron. Commented Jul 23 at 15:53

If two particles can almost form a bound state, there will be what is called a [resonance in the scattering amplitude ] as a function of the center-of-mass energy. This is sometimes also called a quasi-bound state. One can easily see this with certain square-well potentials or other simple examples, using the Schrodinger equation for the relative position. Note that in $$3$$ dimensions, looking just at the partial wave for the angular momentum that gives the quasi-bound state, will reduce the problem to the [one-dimensional case].

Basically this will then allow computation of the [differential cross-section], which can be measured, or computed, and will show the resonance peak. Clearly, the energy where the resonance is peaking is above the combined mass of the two particles (since the incoming particles in scattering have positive kinetic energy) so it is not a bound state, although going to complex energy in the analysis will show some similarity. So we can use this to assign a mass to the quasi-bound state based on the peak position of the resonance, or (more or less equivalently) the position of a pole in the complex plane for the scattering amplitude.

If the width of the resonance is small, this mass is quite well determined. For p-p scattering this is not so clear, see for instance Fig. 3 of [arxiv 2404.06318v1], which shows:

Where we can see a bump which is indeed somewhat above $$1$$Mev. And if we describe this with a complex pole in scattering theory (or perhaps with some other form of "fitting") then the value of $$1.4$$MeV can be found.

• Is this a calculation or just a measurement? Isn't the 1.4 MeV figure just the Coulomb barrier for the protons? Commented Jul 23 at 15:49
• It's good to have an expert on hand! But I don't entirely follow. The combined (nuclear + Coulomb) potential can't be very deep or the diproton would be stable. Isn't there some sort of local minimum inside a local maximum of about 1.4 MeV (which is the Coulomb energy at about 1.75 fermis where the nuclear potential is ~zero)? In which case, isn't the "diproton mass" the sum of two proton masses and the height of the local minimum? Commented Jul 24 at 9:19
• @ProfRob At these low energies, experiment and theory are very much in agreement (when I worked on it, it was already within about 1 percent, see inspirehep.net/authors/2592190 ). Bound state formation is of course governed by the repulsive Coulomb tail and the short range attractive potential well, which looks like this: i.sstatic.net/QsgmgvPn.png . But it is not true that the highest point of the potential (the barrier) is directly related to bound state energy or to resonance position. For this case, the height is about 0.3 MeV, which is not the 1.4 MeV. Commented Jul 24 at 9:58
• @ProfRob Note that I added the separate Coulomb potential to the plot in the comment. It is indeed about 1.4 MeV at the point where the total potential $V_\text{tot}=0$ (which is around 1 fm). But that is accidental because the value of the Coulomb potential alone is not what matters for the position of bound states. (Also, if you change the strength of the short-range attraction a little bit you can shift the resonance a lot, or even make it a truly bound state, while the zero crossing remains close to 1 fm and consequently the value of $V_\text{coul}$ at that point remains close to 1.4 MeV.) Commented Jul 24 at 10:00
• @ProfRob To have a bound state for a 3D wave function you need to have a quarter sine wave in the negative potential well. If the depth and/or width of the well are not sufficient to get that, there is no bound state. This differs from the 1D case, even with a spherically symmetric wave function $\psi(r)$, because the Laplace operator becomes: $$\nabla^2\psi=\frac1r\frac{d^2}{dr^2}(r\psi)$$ so the substitution $u(r) = r\psi(r)$ reduces it to the 1D case for the new function $u(r)$ but then with the extra requirement $u(0)=0$, which is not present in 1D. Commented Jul 25 at 9:04