Question 10.2: Calculate the energy levels of an anharmonic oscillator with......
Calculate the energy levels of an anharmonic oscillator with potential of the form
V(x)=\frac{\kappa }{2} x^{2} +\xi x^{3} \hbar \omegawhere \kappa is the spring constant for a harmonic potential. Show that the difference between two adjacent perturbed levels is E_{n}-E_{n-1}=\hbar \omega (1-15\xi^{2}(\hbar /m\omega )^{3}n/2 ). A heterodiatomic molecule can absorb or emit electromagnetic waves the frequency of which coincides with the vibrational frequency of anharmonic oscillations of the molecule about its equilibrium position. For a molecule initially in the ground state, what do you expect to observe in the absorption spectrum of the molecule?
The Hamiltonian for the one-dimensional harmonic oscillator subject to perturbation \hat{W} is
\hat{H} =\frac{\hat{p}^{2} }{2m} +\frac{\kappa }{2} \hat{x}^{2} +\hat{W}where \kappa =m\omega^{2}. From Section 10.2.5 we have for a perturbation
\hat{W}=\xi x^{3} \hbar \omega (m\omega /\hbar )^{3/2}E_{n} =\left(n+\frac{1}{2} \right) \hbar \omega -\frac{15}{4} \xi ^{2} \left(n+\frac{1}{2} \right)^{2} \hbar \omega -\frac{7}{16}\xi^{2} \hbar \omega
which, expanding the square, may be written
E_{n} =\left(n+\frac{1}{2} \right) \hbar \omega -\frac{15}{4} \xi ^{2} \left(n^{2}+ n+\frac{1}{4} \right) \hbar \omega -\frac{7}{16}\xi^{2} \hbar \omegaHence,
E_{n-1} =\left(n-\frac{1}{2} \right) \hbar \omega -\frac{15}{4} \xi ^{2} \left(n^{2}- n+\frac{1}{4} \right) \hbar \omega -\frac{7}{16}\xi^{2} \hbar \omegaso that
E_{n}-E_{n-1}=\hbar \omega -\frac{15}{2} \xi ^{2} nBecause, in this exercise, \hat{W} did not explicitly contain the factor (m\omega /\hbar )^{3/2}, we need to put this back in. Since \xi appears as a squared term, the inverse of the factor (m\omega /\hbar )^{3/2} is also squared to give
E_{n} -E_{n-1} =\hbar \omega \left(1-\frac{15}{2}\xi^{2} \left(\frac{\hbar }{m\omega } \right)^{3} n \right)The absorption spectrum of an anharmonic diatomic molecule initially in the ground state will consist of a series of absorption lines with energy separation between adjacent lines that decreases with increasing energy. The absorption lines will be at energy
E_{n} -E_{0} =n\hbar \omega -\frac{15}{4} \xi ^{2} \left(n^{2} +n+\frac{1}{4} \right) \hbar \omega -\frac{1}{16} \xi ^{2} \hbar \omegaand the wavelength is given by \lambda =\hbar c/(E_{n}-E_{0} )