Calculate the energy levels of an anharmonic oscillator with potential of the form V(x) = κ/2 x² + ξ x³ ћω where κ is the spring constant for a harmonic potential. Show that the difference between two adjacent perturbed levels is En − En−1 = ћ ω (1 − 15ξ² (ћ/ mω)³ n/2). A heterodi-atomic molecule

Question

Calculate the energy levels of an anharmonic oscillator with potential of the form

[latex]V(x)={\frac{\kappa}{2}}x^{2}+\xi x^{3}\hbar \omega [/latex]

where κ is the spring constant for a harmonic potential. Show that the difference between two adjacent perturbed levels is [latex]E_{n}-E_{n-1}{\bf}=\hbar{}\omega(1-15\xi^{2}(\hbar {}/m\omega)^{3}n/2).[/latex] A heterodi-atomic 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?

Accepted Answer

The Hamiltonian for the one-dimensional harmonic oscillator subject to perturbation W is

[latex]H={\frac{p^{2}}{2m}}+{\frac{\kappa}{2}}x^{2}+ W [/latex]

where [latex]\kappa=m\omega^{2}.[/latex] From Section 10.2.5 we have for a perturbation [latex]W =\xi x^{ 3 }\hbar\omega(m\omega/\hbar)^{3/2}[/latex]

[latex]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[/latex]

which, expanding the square, may be written

[latex]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\omega[/latex]

Hence,

[latex] 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\omega[/latex]

so that

[latex]E_{n}-E_{n-1}=\hbar\omega-{\frac{15}{2}}\xi^{2} n[/latex]

Because, in this exercise, W did not explicitly contain the factor [latex](m\omega/\hbar)^{3/2}[/latex], we need to put this back in. Since [latex]ξ[/latex] appears as a squared term, the inverse of the factor [latex](m\omega/\hbar)^{ 3/2}[/latex] is also squared to give

[latex]E_{n}-E_{n-1}=\hbar\omega\biggl(1-\frac{15}{2}\xi^{2}\biggl(\frac{\hbar}{m\omega}\biggr)^{3}n \biggr)[/latex]

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

[latex]E_{n}-E_{0}=n\hbar\omega-{\frac{15}{4}}\xi^{2}{\bigg(}n^{2}+n+{\frac{1}{4}}{\bigg)}\hbar\omega-{\frac {1}{16}}\xi^{2}\hbar\omega[/latex]

and the wavelength is given by [latex]\lambda=\hbar c/( E_{n}-E_{0}).[/latex]

Question 10.2: Calculate the energy levels of an anharmonic oscillator with......

This Question has been Answered.

Already have an account?

Login

Related Answered Questions

Verified Answer:

Verified Answer:

In this chapter we solved for the first excited state of a two-dimensional harmonic oscillator subject to perturbation W = κ′ x y. How do the three-fold degenerate energy E = 3ћω and the four-fold degenerate energy E = 4ћω separate due to the same perturbation? ...

Verified Answer:

Verified Answer:

Verified Answer:

Verified Answer:

Verified Answer: