The standard Vesman model

As mentioned earlier, resonant molecular formation refers to the process:

$$(\mu t)_F + [D_2]_{\nu _{i}, K_{i}} \rightarrow [(d\mu t)^{S}_{Jv}dee]_{\nu _{f},K_{f}},$$ (38)

where the kinetic energy of $\mu t$ and the energy released upon formation of $d\mu t$ is absorbed in the rotational (K) and vibrational ($\nu$) excitation of the molecular complex $[(d\mu t)dee]^{*}$, a hydrogen like molecule with $d\mu t$ playing the role of one of the nuclei. The process is resonant in nature because of the discrete spectra in the final states, corresponding mainly to the ro-vibrational levels ( $\nu _{f},K_{f})$, and the collision energy has to match the resonant condition. The resonance condition can be written, when considering purely two body collisions (i.e., neglecting three-body or phonon effects):

$$\epsilon _{res} [\mu t + D_2]= \epsilon ^{FS}_{11}[d\mu t] + E_{\nu _f K_f}[(d\mu t)dee] - E_{\nu _iK_i}[D_2].$$ (39)

Figure 2.3 gives a schematic energy level diagram for the resonant $d\mu t$ formation process. Plotted on the left is the potential curve of $d\mu t$ showing the ``shallow'' (in the muonic scale) bound state, while on the right the molecular complex energy levels are plotted. Note that due to the reduced mass difference, the energy levels of D₂ and [ $(d\mu t)dee$] are different (by 33.7 meV for the ground states according to Faifman et al. [138]). In fact it is this difference in reduced mass which introduces the dependence of the $d\mu t$ formation rates on the target molecular species, i.e., $\mu t + D_2$ versus $\mu t+HD$ or $\mu t+DT$.

 

Under ordinary target conditions, only $\nu _i = 0$ is populated. On the other hand, the initial rotational population depends on the target preparation procedure as well as on temperature. At 3 K, an equilibrated target would have nearly 100% K_i=0 population (ortho deuterium). Our targets, however, were prepared by rapidly freezing the statistically populated deuterium (67% K_i = 0 and 33% K_f = 1) from a hot palladium filter, and because the rotational relaxation is very slow in the absence of a catalyst, we expect this population will last for the entire measurement time. The initial ortho-para population is relevant, because angular momentum conservation requires:

$${\bf L} + {\bf K_i} = {\bf J} + {\bf K_f},$$ (40)

where L is the relative angular momentum in the reaction (2.24). Since the $d\mu t$ rotational angular momentum J is 1, we have $K_f = K_i \pm 1$ for low energy collisions with L=0 [139]. However, at epithermal energies L>0 becomes important, allowing various K_f states. This is one of the reasons why we expect such a high rate at epithermal energies.