Angular Momentum of “He-d”
The quantized angular momentum states are a manifestation of the central force angular components of a wave function. This is discussed in all quantum physics books in the development of the hydrogen wave function, and not reproduced here. The result of the quantized angular momentum is that the angular momentum can take on only discrete values, namely;
Angular momentum, L ={ l(l +1)}1/2 ħ l = 1,2,3,……. eq 4-3
figure 4-5a
Figure 4-5 pictures the transition from the channeled “free helium” atom approaching deuterium and then the moment of transition to attraction of the two particles. This attraction (“He-d” formation) can take place only when the angular momentum criteria above are met. The angular momentum arises because the wave function of the helium has 0.084eV of energy in the direction exactly orthogonal (perpendicular) to the radius vector between the two particles. This transverse energy must be incorporated into the “He-d” molecule as angular momentum. The deuterium does not contribute to the angular momentum because its wave function is symmetrical in all three dimensions. In terms of the “zero point” transverse energy of the helium, the angular momentum will be:
Angular momentum = L = ro(movt) = ro(2mo Ezero)1/2 eq 4-4
ro= nuclear separation
vt= transverse velocity
mo =reduced mass =mHe md/( mHe+ md)
Ezero= helium transverse channel energy
This means that the transition to the “He-d” molecule can only happen at very specific nuclear separations. Putting the two equations 4-3 and 4-4 together gives;
ro(2mo Ezero)1/2 = { l(l+1)}1/2 ħ l = 1,2,3,…….
which rearranges to:
ro = { l(l+1)}1/2 ħ / (2mo Ezero)1/2
With the helium transverse “zero point” energy of 0.084eV, the allowed nuclear separations of the reaction “activated complex” will then be:
| l = | 1 | 2 | 3 | 4 |
| ro (angstroms)= | 0.19 | 0.33 | 0.47 | 0.61 |
What this means, is that in the dynamic “activated complex” the transition will only take place while the nuclear wave functions are at these separations. For l = 3& 4 , the nuclear separation is far outside the average “1s” orbital diameter of 0.23 angstroms for helium, and can not form the localized linear momentum states required for the “He-d” bond. At the proper kinetic energy, as the particles approach the l=1 or 2 separation criteria, the nuclear repulsion slows the particles down. The slowing does two things to aid the transition;
1) the lower relative velocity gives more time for the reaction and,
2) the lower velocity (momentum) makes the wave function expand and spend more time traversing the critical separation (see figure 4-6)
figure 4-6
The probability for a transition from the standard state, to the “He-d” state has two components;
1) probability with time = P(time) ~ e-t(E2-E1)/ħ ;
where E2=energy of ‘He-d’
E1=initial energy
2) overlap of the initial(1) and final(2) wave functions;
Overlap probability=P(Ψ1èΨ2)= {Ψ1 (He) Ψ1(d) Ψ1(elec)}*{ Ψ2(He) Ψ2(d) Ψ2(elec)}
The transition probability will then be:
P(transition) ~F( P(time):P(Ψ1èΨ2))
The exponential time function comes from the solution to the time dependence of two states using the Schrodinger equation, and can be found in most quantum texts. The coefficients of the two states (the localized and standard Linear Momentum state electron) oscillate with time as the sine and cosine functions. The oscillation cycle goes as:
sin(tΔE/ħ)=2π
Assuming each cycle delivers ~50% of the probability, the half life would be ;
t1/2 ~ 2π ħ /ΔE
Time becomes an issue because of the relative (“free particle”) motion of the helium and deuterium. The analysis for the initial and final (activation complex) states is done in the center of mass coordinates. The center of mass angular momentum and center of mass relative velocity determines the requisite wave function overlap of the three particles, as well as the reaction time.
The kinetic energy of the nuclear states is small compared to the electron (10’s of eV for electron and ~0.1 eV for the nuclei), thus the difference in energy, (E1-E2), will be determined by the difference in the initial and final electron wave function energies. This will determine the time required for the transition.
With a wide separation of the deuterium and helium(>2 angstroms) there will be little interaction of the two atoms electrons. As they approach, to ~1 angstrom, there will be substantial interaction. The wave function for atoms, such as the helium and deuterium system, is generally modeled by combining the wave functions of the individual atoms. The wave functions can be combined in two ways, adding and subtracting. The addition of the wave functions will give the familiar “1s” bonding molecular orbital which is similar to the hydrogen (H2) molecular bond. In the hydrogen molecule, there are only two electrons, thus the “1s” bonding orbital is filled. The orbital formed by subtracting the two wave functions gives the “1s” anti-bonding orbital. It is called an anti-bonding orbital because it does not have as low an energy level as the “1s” bonding orbital. Physically this is manifested as a lower electron concentration between the nuclei and additional electron concentration mostly around the helium. The decrease in energy can be thought of as the energy of repulsion of the higher electron concentration around helium, without the benefit of higher electron concentration between the atoms.
As the nuclear separation decreases, the anti-bonding orbital will look more and more like a “2p” orbital of lithium(Density Function calculations would be needed for accuracy). The consequence of these changes is a lower energy , as the electron is taken from the anti-bonding state close to the He nucleus, out into a metallic bonding state, with more electron probability outside the “1s” state.
The transition of the anti-bonding electron to a ~“2p” metallic bond allows for a more stable “1s” bonding state between the helium and deuterium, since the electron-electron interaction from the anti-bonding electron is removed. This is important because it lowers the energy required to bring the two particles together. In the literature it has been calculated that with the removal of the anti-bonding electron, an (He-d)+ ion, actually has a stable bond, with a bond energy of ~-2.9eV. The equilibrium separation is about 0.74 angstroms. From this, we are guessing at the rest of the energy versus separation curve, as in figure 4-7 (Again, a great need for Density Function Calculations)
figure 4-7
Intuitively, as the particles approach each other, the helium “1s” electrons can be pictured as assuming the “1s” bonding orbital, while the free electron charge around the deuterium assumes the “1s” anti-bonding and then the “2p” orbital. This is easy to comprehend if you look at the extreme results of the two nuclei coming together. This would give a lithium atom with two “1s” electrons and a metallic “free electron”, in the “2s” orbital. As the two atoms come together, the two “1s” bonding electrons will decrease in energy; going from the combined energy in helium of -77 eV, to the “1s” electrons of lithium with -199 eV. This electron energy (199-77=122eV) plays a role in the “activated complex”, as it allows the helium and deuterium nuclei to approach closer than they otherwise could. Instead of the energy being radiated away (as photons) as the electron drops to a lower energy state (as in free space), the energy is the “work” used to overcome the nuclear repulsion. Next, we will show how the ~“2p” orbital overlaps with the localized “linear momentum state” electron wave function.
figure 4-8
figure 4-9In chapter 2 we showed some crude results for the n=1 and 2 linear momentum state wave functions for the He-d molecule. In chapter 3 the ground state, n=1, wave function was discussed. Figure 4-8 shows the n=2 wave functions at various He-d nuclear separations. Similarly, Figure 4-9 and 4-10 show the n=3 and 4 wave functions respectively. There are no other higher n, localized linear momentum state wave functions. The higher “n”, energy levels, would have wave functions that span the crystal, just as in chapter 2. It is the linear combination of these 4 states (n=1, 2, 3, 4) that allows for the smooth energy transition from a few 10’s of electron volts, to the over 100,000 eV of the ground state “He-d” bond. Figure 4-11 shows how the linear momentum states (n=1, 2, 3 & 4) energy changes with nuclear separation.
figure 4-10
figure 4-11a
Figure 4-12a shows how the n=3 and 4 wave functions compare (overlap) with a simulate “2p” orbital for lithium in the solid state. Note that the“2p” wave function does not go to zero as the radial distance increases, but rather joins the adjacent atom with a boundary condition of the first derivative equal to zero. This is the essence of a metallic bond. The overlap of the “2p” with the n=3 and 4 linear momentum wave functions is very good. Figure 4-13 shows a cartoon of how the two types of states overlap. The wave function for the linear momentum state will decay into a much lower energy state very quickly, where as the lithium ~“2p” state is relatively static. This means that the probability will quickly drain away to the linear momentum state. There is not room here to explain how this works. The chapters on time dependence of quantum states and probability flow should be read in any quantum physics text to understand the details.