Deuterium Nuclei Repulsive Energy
Above we estimated the energy to bring two electrons into helium against their mutual repulsion. A more accurate method to do this is to add up all the possible combinations of positions of the two electrons and their contributions to the overall energy. This is calculated using the double volume integral equation 5-30.
This works well for particles in “s” states since there is no relative movement. In an angular momentum state there are two ways the angular momentum can add together, either parallel or anti-parallel. The states with parallel orbital angular momentum have lower energy, which can be understood with a classical model. The states with parallel angular momentum have particles traveling in the same orbital rotation. In this state, the particles tend to stay at opposite sides of the molecule as they speed around at the same velocity. In the anti-parallel state, the particles move in opposite directions and must cross paths with each other twice for every revolution; a high energy condition. Because of this, we will assume that the two deuterium will always fall into the lower “parallel” orbital angular momentum state. This is similar to electrons in multi-electron atoms, which in general, seek to align their angular momentum to minimize energy. The three states are sketched in figure 5-3 showing the respective spatial configurations. The reason for the lower energy is obvious from the separate regions of probability in the angular momentum states. These separate regions are much farther apart, allowing the nuclei to spend much of their time at greater separations than in the spherically symmetrical “1s” state.
Figure 5-3
Since the electrons in the same angular momentum state “correlate” in a way to minimize energy, we will use the simplistic model as we did with helium to capture this affect when the deuterium nuclei are in the same angular momentum state. This essentially puts one particle in apposing lobes of the distributions. When the deuterium nuclei are in a “1s” or different states, we will use equation 5-30 to calculate the repulsive energy.
Figure 5-2
Note in figure 5-2 that when the d-d repulsion energy is included, the even states become the “stable” low energy states. These states are unable to decay via photon emission as there is no state, with L=±1, with a lower energy to decay to. Since the half-life to fusion of these “He-d2” molecules is greater than 10-5 seconds, and most photon emission half-lives much shorter, this is critical to the mechanism. If these states where able to decay to the “1s” state, the tritium and neutron producing reactions would prevail. Since the even angular momentum states are stable to photon emission, they will remain in their respective states until fusion takes place.
Nuclear Spin
In the energy level diagram, figure 5-2, the spin states are noted as well as the energy level, n, and angular momentum, L. The spin states are critical to the fusion mechanism, as they force the “He-d2” molecule into specific states preventing the tritium and neutron producing fusion reactions. There are 9 equally probable spin states, only one of which is an anti-symmetric state. The other 8 spin states are symmetric (or parallel) spin states. When the deuterium nuclei are in the same state, the spins are parallel. When the original state forms with anti-parallel spin in the activation complex, the “He-d2” molecule slides down the energy levels, but must stop at a higher energy state, since the spins must be parallel when in the same state. Looking at the energy levels in figure 5-2 under L= 2, we see that the state [n=1,L=1:n=3,L=1], is the lowest state with anti-parallel spin. This state will immediately decay into the [n=2,L=1: n=2,L=1] state by flipping the spin of a deuterium nuclei.
Normally a “spin flip” is not very probable because of the small energy difference involved and a radiative mechanism. The existence of the electrons from the outer helium “1s” state facilitate this “spin flip”. The deuterium nuclei in the [n=3,L=1] and [n=2,L=1] wave functions substantially overlap, allowing the spin flip to occur in conjunction with a drop to the lower energy state. This provides a driving energy difference of ~two thousand electron volts. The spin angular momentum of 2ħ is then transferred to a “1s” electron moving it to a “3d” state with angular momentum of 2ħ. The motive force for the excitation of the “1s” electron is the oscillating charge density caused by the superposition of the initial and final states of the “He-d2”, just as in photon emission dynamics. The charge density oscillation for this transition does not have to be anti-symmetric as in photon emission, since the action is on an individual electron; not a dipole interaction. The electron reaction time will only be a little longer than the oscillation frequency of ~10-18 seconds, instead of many more orders of magnitude required for photon emission.
We will see in the next chapter how this reaction time, half-life, for transition of spins and angular momentum determines the amount of tritium formed. The fusion reaction to tritium will be shown to require the deuterium nuclei to be in an anti-parallel spin state. The half-life for the tritium producing reaction is calculated to be ~10-17 seconds; which with the ~10-18 second transition time to parallel spins, gives a ratio of tritium to Helium4 on the order of 10-5, consistent with experiments. The neutron reaction will be shown to be completely suppressed, even with anti-parallel spins in “He-d2”.