# Nuclear fusion: what causes this “resonance” peak?

Can anyone explain why the $^{11}\mathrm{B}\mathrm{H}$ fusion cross-section has a peak near 150 keV, and why $\mathrm{D}\mathrm{D}$ and $\mathrm{D}\mathrm{T}$ have no such sharp peaks?

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This is a very difficult question to answer. There are (at least) two reasons. First, we have detailed, numerically exact wave functions for stable, light nuclei only up to, just recently, $A=12$ (like $^{12}C$). The Argonne-Los Alamos-Urbana collaboration uses quantum Monte Carlo (QMC) techniques to evaluate the ground and excited states of bound nucleons (ie. nuclear states that are $L^2$ normalized). And the fact that the QMC methods look at only bound states indicates the second reason: we're really interested in the eigenstates of the nuclear system in the continuum -- that is, the scattering states. This is a much tougher problem than evaluating the energy of the bound states, whose nucleons range over effectively bounded regions, since we have to do an integral over and infinite region numerically; or be clever and figure out an equivalent, finite-region problem. (There's been some recent work in this direction by Bob Wiringa and Ken Nollett, building on earlier work by Kievsky and collaborators -- check the preprint archive for their recent work.) So although we know a little about the wave function for $^{12}C$, the $A=12$ scattering problem is something we're just starting to learn more about.

Before talking about an alternative to bound QMC for describing (ie., parametrizing, not solving) the scattering states, let me digress on the issue of the meaning of ab initio solutions of many-body quantum mechanical problems. Basically, even if you solve the problem exactly, unless you're very lucky (and smart) and identify a single (or very few) physical mechanism(s) (usually a collective phenomenon like GDR, pairing, etc.) that's particularly relevant for the experimental observation you're trying to describe, you're probably not going to have a great one-line, 'take-home' message that says, "The reason that $DD$ and $DT$ don't have as narrow a resonance as $p^{11}B$ is XXX." The answer to your question would require: 1) very good wave functions and 2) a concomitant study of the 2-,3-,...,?-body correlation functions in the $A=4$, $5$, and $12$ problems (with, of course, the right quantum numbers). Even then you might not identify an 'smoking gun' mechanism that says, "Here, look, that's why $A=4$ and $5$ don't show the narrow peak that $A=12$ has." But you might...

One way, alternative to QMC, that we have of studying/describing/parametrizing the reactions of light nuclei, that doesn't assume that the states are bound, is Wigner's $R$ matrix. (There are ab initio methods like the resonating group method and no-core shell model, too.) You can find a lot of literature through google scholar. But the basic idea is that one (artificially, if you will) separates the scattering problem into 'internal' and 'external' regions. The internal region is hard to solve -- all the dynamics of the interacting nucleons, when they're all close together (the "compound nucleus"), are at play. The external region is easy to solve: one ignores 'polarizing' (ie., non-Coulomb) forces (because they're small). The complicated hypersurface in the $3A-3$ dimensional space that separates the internal from the external is called the channel surface. We generally assume a sharp, simple form for this surface that roughly corresponds to (though is usually significantly less than) the distance between the target and projectile (or daughter particles) and parametrized by the "channel radii", $R_c$. (We only consider two-body channels -- a limitation of the method.)

Now, the wave functions are known in the external region (just appropriate sums of regular and irregular spherical Bessel functions modified by Coulomb phases if it's a charged channel reaction). Inside, however, we describe the system by Wigner's $R$ matrix: \begin{align} R_{c'c} &= \sum_{\lambda=1}^\infty \frac{\gamma_{\lambda c'}\gamma_{\lambda c}}{E_\lambda-E}, \end{align} which you might recognize as the Green's function in the presence of some boundary conditions (Wigner's insight gave a particularly useful, simple condition) at the channel radii. The $R$ matrix is a meromorphic function of the energy, $E$ and depends on an infinite number of levels, $E_\lambda$ corresponding to the eigenstates of the Schr\"{o}dinger equation in the finite cavity (with given, Wigner-type BC's). The reduced widths (basically the fractional parentage coefficients of the bound, compound system as it "decays" to particular channels $c,c'$) are related (in a very complicated way) to the partial widths of the compound nucleus. In sum, the $R$ matrix makes an almost intractable problem a little easier.

So, what's my point? You can calculate or parametrize the $R$ matrix, then derive the $T$ (transition) (or $S$, scattering) matrix from it and find its poles near the physical region. This will tell you where the resonances are. And this procedure will give you insight into why a particular compound nucleus has resonance at a particular energy. If there is a "strong" (ie., small reduced width) $R$-matrix level at a particular energy, you can learn what the relevant (LS) quantum numbers of that level are.

The next step in the program is to ask: what type of 2-, 3-, ..., ?-body correlations/forces give rise to strong interaction in this LS channel-state? This, unfortunately, is a much more difficult question to answer and, incidentally, occupies a good part of my "free-time" as this is the type of research that I'm currently working on.

And I'm pretty sure that we have a ways to go to answer it.

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mass of a proton 1.00728 AMU, mass of C12 12 AMU, while the mass of B11 is 11.00931 AMU, so that the difference in mass p+B11-C12 is .00728+.00931 = .01659 AMU, or multiplying by 931 MeV/AMU, the energy is 15.445 MeV, and this means that the C12 is formed from B11 and a proton with 15.445 MeV more energy than the ground state.

There are lots of narrow energy levels of C12 around this region (taken from Energy Levels of Light Nuclei A = 12 by F. Ajzenberg-Selove a and T. Lauritsen):

Energy spin/parity (Isospin)

15.62 MeV
16.11 MeV 2/+ (1) (6KeV)

16.58 MeV 2/- (1) (wrong parity) (295 KeV)
17.23 MeV 1/- (1) (wrong parity)(1160)
17.77 MeV 0/+ (wrong spin)

18.37 MeV 2/+ (280 KeV)
18.40 MeV (46 KeV)
18.85 MeV (90 KeV)
19.26 MeV (450 KeV)
19.42 MeV (45 KeV)
19.67 MeV (180 KeV)
19.88 MeV (90 KeV)
20.27 MeV (180 KeV)
20.49 MeV (180 KeV)

Subtracting the fusion energy, this gives resonances at collision energies (KeV)

(15.62- 15.445) = 175 KeV
(16.11 - 15.445) = 665 KeV (width 6)

1135 (not there)
1785 (not there)
2325 (not there)

2925 (280)
2955 (46)
3505 (90)
3546 (450)
3975 (45)
4225 (180)
4435 (90)
4825 (180)
5045 (180)

The first two resonances match the position of the first and second peaks precisely (although the second resonace is broader than one expects). The next three resonances are inaccessible by conservation laws. The last peak is a broad combination of whichever of the other resonances can be excited, the ones that are accessible from the collision of the proton and the B11 nucleus. You can't tell because their spin/isospin/parity is not given.

The accessible resonances are found by using the conserved quantities for the fusion, the spin, isospin, and parity. For such low momenta, the nuclei collide with zero orbital angular momentum (the scale of spatial wavefunction variations is much larger than the scale of the collision), so that the spin of the proton and the Be nucleus have to add up to the spin of the resonance. So you are restricted to resonances of spin 1,2 (since B11 has spin 3/2). Likewise, the isospin must add up, and the Be nucleus has isospin 1/2, like the proton, so you need isospin 0,1. The parity of the state must be + (these are the states that show up).

The narrow resonances are due to the dynamics of the nucleus which have a hard time falling apart, due to the difficulty of concentrating collective energy of many nucleons into one proton or neutron, or one alpha, to eject, or due to the long time to emit a gamma. For d-d or d-T fusion, all the resonances are extremely broad, they fall apart easily because there are so few particles, so it is easy to eject a p or n, or two deuterons again, and you don't get narrow peaks at all, and it's all background. For d-T, there is a real peak, which is the resonance for the unstable alpha-neutron He5 nucleus.

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Nice answer, but the symbol of boron is $\mathrm{B}$, $\mathrm{Be}$ is beryllium. –  mmc Dec 4 '12 at 17:16

Nobody knows the law relating the nucleus content and it's binding energy.

Your peak means that at this energy some additional states are forming. The width of the peak means it's lifetime (the narrower the peak, the longer lifetime).

According to this document: http://www.oecd-nea.org/janis/book/book-proton.pdf

your picture occurs when He4 and 2 alpha particles are forming:

This picture is on page 25.

I can't say yet what does this peak mean.

UPDATE

Original data source: http://www.nndc.bnl.gov/exfor/servlet/X4sGetSubent?reqx=12111&subID=130017002

Measured by G.M.Hale in 1979 in Los Alamos National Laboratory

May be this is some ionization peak? The energy of 150 kev looks like non-nuclear by nature.

UPDATE2

Also neither Boron-11 nor alpha particles have hundred-kevs excitation levels:

http://www.nndc.bnl.gov/chart/getdataset.jsp?nucleus=4HE&unc=nds

http://www.nndc.bnl.gov/chart/getdataset.jsp?nucleus=11B&unc=nds

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Gerry Hale didn't measure this reaction -- he performed the multichannel $R$ matrix fit to produce this curve. –  MarkWayne Dec 6 '12 at 3:50
But why the curve is not continuous then? –  Suzan Cioc Dec 6 '12 at 8:04
yeah, that's Gerry's analysis, all right. It's just evaluated on a grid. Experimentally observed data should be -- and nearly always is -- presented with error bars. (Theory should be too, and we're doing that now.) –  MarkWayne Dec 6 '12 at 16:58
I found data by title "Experimental data sets are identified by their EXFOR entry number". So EXFOR entry number was D0017.002, and "EXFOR" stands for "Experimental Nuclear Reaction Data". How do you know this is a fit? –  Suzan Cioc Dec 6 '12 at 17:57
If you look at the EXFOR entry carefully, in the field "METHOD" it says. "Evaluation, from the measurements..." Gerry isn't an experimentalist. He's a theorist. This is an analysis via the unitary, multichannel $R$ matrix method. A note of caution here: not all the entries in EXFOR are observed data! Some are analyses. And this is a little unfortunate because it causes this kind of confusion. –  MarkWayne Dec 6 '12 at 18:08