91Ó°ÊÓ

This is still a harmonic oscillator, so its energy spectrum is given as${\mathbf{E}}_{\mathbf{n}}^{\mathbf{,}}\mathbf{=}{\mathbf{Ä\S Ó\neg }}^{\mathbf{\text{'}}}\left(\frac{1}{2}+n\right)$

where${\mathit{Ó\neg }}^{\mathbf{\text{'}}}\sqrt{\frac{{\mathbf{k}}^{\mathbf{\text{'}}}}{\mathbf{m}}}\mathbf{=}\sqrt{\frac{\mathbf{k}\left(1+\stackrel{Ë™}{\mathrm{ο}}\right)}{\mathbf{m}}\mathbf{=}\mathbf{Ó\neg }\sqrt{\mathbf{1}\mathbf{+}\stackrel{\mathbf{Ë™}}{\mathbf{ο}}}}$and$\mathit{Ó\neg }$is the frequency of a harmonic oscillator with spring constant$\mathbf{k}$. Therefore, the energy is given as${\mathbf{E}}_{\mathbf{n}}^{\mathbf{\text{'}}}\mathbf{=}{\mathbf{E}}_{\mathbf{n}}\sqrt{\mathbf{1}\mathbf{+}\stackrel{\mathbf{Ë™}}{\mathbf{ο}}}$

a)

In this problem solve the case of a harmonic oscillator whose spring constant changes slightly as

$\mathrm{k}→{\mathrm{k}}^{\text{'}}=\left(1+\stackrel{˙}{\mathrm{ο}}\right)\mathrm{k}$

Use the Taylor expansion for the square root if is very small, which is given as

$$\begin{gathered} \mathrm{f}\left(\stackrel{˙}{\mathrm{ο}}\right)=\sqrt{1+\stackrel{˙}{\mathrm{ο}}} \\ =\mathrm{f}\left(0\right)+\stackrel{˙}{\mathrm{ο}}\frac{\mathrm{df}}{\mathrm{d}\stackrel{˙}{\mathrm{ο}}}{\left|{}_0+\frac{1}{2}\stackrel{˙}{\mathrm{ο}}{}^2\frac{{\mathrm{d}}^2\mathrm{f}}{\mathrm{d}{\stackrel{˙}{\mathrm{ο}}}^2}\right|}_0+...\mathrm{n} \\ ≈1+\stackrel{˙}{\mathrm{ο}}\frac{1}{2}\frac{1}{\sqrt{1+\stackrel{˙}{\mathrm{ο}}}}\overline{)\underset{0}{\overset{}{}}+\frac{1}{2}}{\stackrel{˙}{\mathrm{ο}}}^2\left(-\frac{1}{2}.\frac{1}{2}\right)\frac{1}{{\left(1+\stackrel{˙}{\mathrm{ο}}\right)}^{3/2}}+... \\ ≈1+\frac{1}{2}\stackrel{˙}{\mathrm{ο}}-\frac{1}{8}{\stackrel{˙}{\mathrm{ο}}}^2+... \end{gathered}$$

The expanded energy is given as

${\mathrm{E}}_{\mathrm{n}}^{\text{'}}=\mathrm{Ä\S Ó\neg }\left(\frac{1}{2}\mathrm{n}\right)\left(1+\frac{1}{2}\stackrel{Ë™}{\mathrm{ο}}-\frac{1}{8}{\stackrel{Ë™}{\mathrm{ο}}}^2+...\right)$

The first term in the expansion is the same as the regular harmonic oscillator with spring constant$\mathrm{k}$.

b)

Calculate the first-order perturbation in the energy using

the formula

The perturbed Hamiltonian is obtained as the total Hamiltonian minus the unperturbed Hamiltonian${\mathrm{H}}_0$

${\mathrm{H}}^{\text{'}}=\mathrm{H}-{\mathrm{H}}_0$

The unperturbed Hamiltonian has a potential

${\mathrm{V}}_0=\frac{1}{2}{\mathrm{kx}}^2$

and the total Hamiltonian with the perturbation has the potential

$\mathrm{V}=\frac{1}{2}{\mathrm{k}}^{\text{'}}{\mathrm{x}}^2=\frac{1}{2}{\mathrm{k}}^{\text{'}}{\mathrm{x}}^2\left(1+\stackrel{,}{\mathrm{ο}}\right)$

The perturbed Hamiltonian is therefore

${\mathrm{H}}^{\text{'}}=\frac{1}{2}{\mathrm{k}}^{\text{'}}{\mathrm{x}}^2-\frac{1}{2}{\mathrm{kx}}^2=\frac{1}{2}{\mathrm{kx}}^2\stackrel{˙}{\mathrm{ο}}=\stackrel{˙}{\mathrm{ο}}\mathrm{V}$

Therefore, the first-order correction to the energy is obtained as

Calculate this using the virial theorem. Since both the kinetic and potential energy for a harmonic oscillator is squared, hence,

and also

Therefore, the result obtained is,

and the first-order correction becomes${\mathrm{E}}_{\mathrm{n}}^1=\frac{\stackrel{Ë™}{\mathrm{ο}}}{2}\mathrm{Ä\S Ó\neg }\left(\frac{1}{2}+\mathrm{n}\right)$

which is the second term in the energy expanded with respect to obtained above.