The Two-Minima Problem and the Ammonia Molecule

Abstract

The system of a particle moving in a potential field containing two equal minima is treated by the Wentzel-Kramers-Brillouin method of approximation. The energy levels are grouped in pairs and the object of the computation is to find the separation between two levels forming a pair. This is accomplished by connecting the oscillatory and exponential approximate solutions of the wave equation by means of the Kramers connection formulae. If $\ensuremath{\Delta}$ is the separation of a pair and $h\ensuremath{\nu}$ the distance between two pairs $\frac{\ensuremath{\Delta}}{h\ensuremath{\nu}}=\frac{1}{\ensuremath{\pi}{A}^{2}}$ where $A=\mathrm{exp}[(\frac{2\ensuremath{\pi}}{h})\ensuremath{\int}{0}^{{x}_{1}}{[2m(V\ensuremath{-}E)]}^{\frac{1}{2}}\mathrm{dx}]$. A particular potential curve is chosen consisting of two equal parabolae connected by a straight line. The expression for $\ensuremath{\Delta}$ may then be evaluated explicitly as a function of the length of the joining line, $2({x}_{0}\ensuremath{-}\ensuremath{\alpha})$ and the distance between two minima, $2{x}_{0}$. These formulae may be applied to determining the form of the ammonia molecule. Substituting the experimental values for ${\ensuremath{\Delta}}_{0}$ and ${\ensuremath{\Delta}}_{1}$, it is found that ${x}_{0}=3.161$ and $\ensuremath{\alpha}=1.916$. An exact solution for this particular potential curve may be found by joining Weber's function ${D}_{n}(x\ensuremath{-}{x}_{0})$ and ${D}_{n}(x+{x}_{0})$ to a hyperbolic sine or cosine. This process also leads to expressions for $\ensuremath{\Delta}$ which may be equated to the experimental values yielding ${x}_{0}=3.182$ and $\ensuremath{\alpha}=1.930$, in good agreement with the earlier determination. Finally ${x}_{0}$ is used to compute $2{q}_{0}=0.760\ifmmode\times\else\texttimes\fi{}{10}^{\ensuremath{-}8}$ cm, the distance between the two potential minima, and the following dimensions of the ammonia molecule, H - H = 1.64\ifmmode\times\else\texttimes\fi{}${10}^{\ensuremath{-}8}$, N - H = 1.02\ifmmode\times\else\texttimes\fi{}${10}^{\ensuremath{-}8}$ cm.