Derivation of the 3 Body, Solid State Bonding, to form "He-d" [March 3, 2007]
As a ~200eV helium atom, channeled in the palladium lattice, approaches a deuterium atom in the tetrahedral position, the particles intimately involved with the collision process will be:
1) helium nucleus (alpha particle)
2) Deuterium nucleus(deuteron)
3) 2 angular momentum state electrons (1s state) from helium
4) every free electron in the crystal; all free electron wave functions contribute to charge on deuterium
The helium will have 2 electrons in the 1s state, as it moves through the lattice. The deuteron in the tetrahedral site of the crystal is surrounded by one electron charge worth of the wave functions from every free electron in the crystal (as we have discussed in chapter 2). All the linear momentum state electrons contribute a fraction of probability around the deuteron. As the helium approaches within about 1 angstrom of the deuterium, the helium and deuterium will momentarily bond using the 1s state from the helium, plus the 1s state of deuterium. The addition of the two 1s states gives a bonding orbital (1s(He) +1s(d)) and subtraction, an anti-bonding orbital (1s(He)-1s(d)). This same bonding in free space holds the hydrogen molecule, H2, together. The two electrons of the hydrogen molecule both reside in the bonding orbital. In the "He-d" bond, one electron must go into the anti-bonding orbital. The overall "He-d" angular momentum bond is only a few electron volts worth of energy, not enough to overcome the repulsion of the nuclei (or the ~70eV center of mass energy). The anti-bonding charge density is provided by the free electrons (linear momentum state electrons). There is no localized electron in the anti-bonding orbital; the charge distribution is somewhat the same as would be in free space, but is provided by the sum of free elecron probabilities.
As the helium and deuterium nuclei approach much closer, to say 0.2 angstrom, the charge distribution will start to look like that of lithium (a +3 nucleus). Lithium is a classic metal with two core electrons in the 1s state (from the helium), and one charge worth of free electron contributions to the wave function around the ~central +3 charge. The anti-bonding 1s state will morph into a ~2s state as the nuclei converge. This 2s state is filled with linear momentum state electron charge density. There can be no angular momentum state for “free electrons” in the palladium crystal environment. In the next chapter, we will discuss the activated complex alluded to here, which leads to the localization of a linear momentum state electron. It is sufficient for now to understand that there are no angular momentum states for the “3rd” electron (outside the 1s state) in the momentarily formed “He-d molecule”.
Since the two 1s electrons , from helium, are well understood and do not contribute to the high energy bond, they are not part of the analysis. Later in the activated complex calculations, the 1s electrons will be accounted for by using a screening factor. A screening factor must be used for any state with appriciable electron probability more than ~0.1 angstrom from the helium nucleus. For the ground state of the He-d molecule this screening is included, but has little affect on the results. This is because the average radius of the ground state linear momentum state electron is less than 0.01 angstroms, well inside most of the 1s electron probability of helium (average= ~0.26 angstroms).
Derivation of the 3-particle wave equation
Since the helium's 1s electrons are to be ignored, the three particles to be analyzed using the Schrodinger equation, are:
1) helium nucleus
2) deuterium nucleus
3) a localized linear momentum state electron
The starting point for this derivation is the time independent multiple particle Schrodinger equation (equation 3-1).
Equation 3-1 is the Schrodinger equation for an arbitrary number of particles, with an arbitrary potential between each particle, Vi,j. Recall that the symbol, (del)2 represents the second spatial derivative. The subscript “i” refers to the ith particle.
Now we write the time independent Schrodinger equation for a 3-particle system consisting of an alpha particle (helium nucleus), a deuteron and a free electron using equation 3-1 to give equation 3-2. The three particles are labeled 1, 2 & 3 respectively.
The Born-Oppenheimer approximation allows us to hold the two nuclei stationary, while calculating the wave function of the electron (This derivation is similar to others given in many texts). This approximation works well because of the much higher velocity of the electron compared to the nuclei at comparable energies. Essentially the electron wave function is able to adjust, or explore its probability space, in the time that the nuclei move very little. This is a universally applied assumption used for Density Function Theory calculations (one of the main “ab initio” molecular modeling techniques). As a further simplification, the electron mass (being less than 0.01% of the nuclei masses) will be neglected so that the reduced mass of the two nuclei may be used to solve the wave function. The reduced mass is an exact analogy to how the proton-electron reduced mass is used to solve the hydrogen wave equation.
Now we replace ψ with “AB”, where “A” is the reduced mass wave function for the alpha-deuteron pair, and “B”, the electron wave function. We then write equation 3-2 as equation 3-3;
Equation 3-3 can be rearranged by moving wave function “B” out of the derivative of the first term. This can be done because, although “B” is a function of ro, we will see later that “B” does not change much in the ground state with changes in ro (= the distance between the alpha and deuteron particles). Changes in “B” will be insignificant with small changes of ro.
The Born-Oppenheimer approximation lets us hold everything associated with the nuclei constant, while we calculate the electron wave function, “B”. This allows us to pull “A” from the derivative of “B”. Rearranging equation 3-3 to put all the “constant” (relative to “B”) terms associated with the nuclei, on the right gives equation 3-4:
The deffinition of E*, equation 3-5, can be inserted into equation 3-4 to give equation 3-6. Equation 3-6 is the familiar one particle Schrodinger equation with the potential (V1,3+V2,3), the attraction of the electron to the two nuclei. We will numerically solve equation 3-6 for E* as a function of ro, using a spreadsheet calculation.
Once E* is known as a function of ro, equation 3-5 can be rearranged to give equation 3-7. Again, this is a fairly standard looking Schrodinger equation, which will be solved using the pseudo potential, (V1,2 + E*). The E* is an attractive potential due to the energy state (kinetic energy) of the electron at the given spacing of the nuclei, ro. V1,2 is the repulsive potential of the positively charged helium and deuterium nuclei.