91影视

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

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

a)

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

$\mathrm{k}鈫�{\mathrm{k}}^{\text{'}}=\left(1+\stackrel{藱}{\mathrm{慰}}\right)\mathrm{k}$

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

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

The expanded energy is given as

${\mathrm{E}}_{\mathrm{n}}^{\text{'}}=\mathrm{魔蝇}\left(\frac{1}{2}\mathrm{n}\right)\left(1+\frac{1}{2}\stackrel{藱}{\mathrm{慰}}-\frac{1}{8}{\stackrel{藱}{\mathrm{慰}}}^2+...\right)$

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

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

The perturbed Hamiltonian is therefore

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

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${\mathrm{E}}_{\mathrm{n}}^1=\frac{\stackrel{藱}{慰}}{2}\mathrm{魔蝇}\left(\frac{1}{2}+\mathrm{n}\right)$

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