The “He-d2” Schrodinger Equation
In solving the Schrodinger equation for the “He-d2” molecule we will simplify the procedure by skipping the electron part of the derivation. We will assume that the linear momentum state electron bond, as discussed in chapter 3, mediates the potential between the helium and the deuterium atoms. There is a substantial interaction of the electrons with the potential of the “other” deuteron, but it is fairly constant for the different states and we will ignore it. This leaves us with the three particle (one helium and two deuterium nuclei) Schrodinger equation 5-1.
Equation 5-1 can not be solved as shown. If we assume that the two deuterium atoms are in the same state, we can split the helium differential operator (particle 3) into two parts as in Equation 5-2. We will see later that this is a reasonable assumption. This is a rigorous assumption if the deuterium are in the same state (If the two states are orthogonal the normalization factor would be (1/2)1/2 , instead of 1/2). We can then associate one-half of the operator with each deuterium nucleus (particles 1 and 2). This has the affect of looking like a doubling of the helium mass in the reduced mass representation in Equation 5-4. The terms in each set of brackets have independent variables (the distance between the helium and the respective deuterium nuclei) so they can be solved independently. The positive repulsive potential between the two deuterium nuclei, V1,2 , must be ignored for now. This is perfectly analogous to ignoring the electron interaction in the solution of the wave equation for the “1s” electrons in helium. We will add back this energy after solving for the Helium-deuterium nuclei wave functions, just as is done with the helium electrons. We will see that the affect on the overall energy is relatively much less in the “He-d2” molecule compared to the electrons in helium, but still critical to the analysis.
The energies calculated from the Schrodinger equation for the individual deuteron-helium states are shown in table 5-1.
Table 5-1
| state | Energy of state- eV |
| 1s | -156,705 |
| 2p | -155,643 |
| 3d | -154,160 |
Recall from chapter 4 that only the two angular momentum states are formed out of the “activated complex”. The energy for each deuteron-helium pair is calculated using the Schrodinger equation above (this time we include the angular momentum term). There are 3 basic states arranged in 5 combinations. The three states are; 1s, 2p, 3d (using the electron orbital notation). Only the 2p and 3d states are formed from the activated complex seen last chapter. The “1s” state is the zero angular momentum state, or ground state (n=1). The “1s” state is considered here since it appears to be a lower energy than the angular momentum states and photon emission decay to this state would, at first glance, be expected. The 2p state is the n=2, l=1 state, with angular momentum of (1*(1+1))1/2ħ=21/2ħ. The 3d state has n=3, l=2 with angular momentum of (2*(2+1))1/2 ħ =61/2 ħ. These individual states are then put together to form “He-d2” molecules of different states as shown below;
1s,1s => (n=1, l=0)(n=1, l=0) ground state
1s,2p => (n=1, l=0)(n=2, l=1)
2p,2p => (n=2, l=1)(n=2, l=1)
2p,3d => (n=2, l=1)(n=3, l=2)
3d,3d => (n=3, l=2)(n=3, l=2)
The energy of the combination of states for the “He-d2” molecule (shown above) is shown in Table 5-2. The energies follow what one would expect, with the ground state (1s,1s) being the lowest energy. Note that the separation in the energies is a small percentage of the total energy; less than 1%. This is a manifestation of the “shallow” and wide potential created by the linear momentum state electron bond. With the small difference between the energy levels, other factors can control the relative stability of the “He-d2” states. This is where the deuterium repulsion plays a critical role. Figure 5-2 shows the energy levels graphically with corrections for the deuterium-deuterium repulsion energy.
Table 5-2
| “He-d2” state | Energy of state without d-d repulsion-eV |
| n=1, L=0 : n=1, L=0 | -313,410 |
| n=1, L=0 : n=2, L=1 | -312,348 |
| n=2, L=1 : n=2, L=1 | -311,286 |
| n=2, L=1: n=3, L=2 | -309,803 |
| n=3, L=2 : n=3, L=2 | -308,320 |
Figure 5-2
The deuterium nuclei spins are also shown in figure 5-2. The nuclear spins become important because deuterium nuclei are bosons (electrons are fermions) and bosons must have a symmetric total wave function. What this means is that if the spatial wave function is anti-symmetric, then the spin state must also be anti-symmetric. If the spatial wave function is symmetric then the spin state of the two deuterium nuclei must also be symmetric. In the symmetric spin states the spin angular momentum, from the two nuclei, are added together instead of cancelling each other as in anti-symmetric spin states. When two identical boson particles are in the same state, they are necessarily in a symmetric spatial state. This means that in order to be in this state together, the spins must be parallel.