Question

For the harmonic oscillator$\left[V\left(x\right)=\left(1/2\right){\mathrm{kx}}^2\right]$, the allowed energies are${\mathbf{E}}_{\mathbf{N}}\mathbf{=}\left(n+1/2\right)\mathbf{ħ\omega }\mathbf{,}\mathbf{}\mathbf{}\left(n=0.1.2,..\right)\mathbf{,}$whererole="math" localid="1656044150836" $\mathit{\omega }\mathbf{=}\sqrt{\mathbf{k}\mathbf{/}\mathbf{m}}$is the classical frequency. Now suppose the spring constant increases slightly:$\mathbf{k}\mathbf{\to }\left(1+\stackrel{\text{'}}{o}\right)\mathbf{k}$(Perhaps we cool the spring, so it becomes less flexible.)

(a) Find the exact new energies (trivial, in this case). Expand your formula as a power series in$\stackrel{\mathbf{,}}{\mathbf{o}}$, up to second order.

(b) Now calculate the first-order perturbation in the energy, using Equation 6.9. What ishere? Compare your result with part (a).

Hint: It is not necessary - in fact, it is not permitted - to calculate a single integral in doing this problem.

Short answer

The exact new energies${\mathrm{E}}_{\mathrm{n}}=\mathrm{ħ\omega }\left(\frac{1}{2}\mathrm{n}\right)\left(1+\frac{1}{2}\stackrel{,}{\mathrm{o}}-\frac{1}{8}\stackrel{,}{{\mathrm{o}}^2+...}\right)$

The first-order perturbation in the energy${\mathrm{E}}_{\mathrm{n}}^1=\frac{\stackrel{,}{\mathrm{o}}}{2}\mathrm{ħ\omega }\left(\frac{1}{2}+\mathrm{n}\right)$

Step-by-step solution

Energy of still Harmonic oscillator.

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

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

Step 2: Find the exact new energies (trivial, in this case) and expand the formula as a power series in , up to second order.

a)

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

$\mathrm{k}\to {\mathrm{k}}^{\text{'}}=\left(1+\dot{\mathrm{o}}\right)\mathrm{k}$

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

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

The expanded energy is given as

${\mathrm{E}}_{\mathrm{n}}^{\text{'}}=\mathrm{ħ\omega }\left(\frac{1}{2}\mathrm{n}\right)\left(1+\frac{1}{2}\dot{\mathrm{o}}-\frac{1}{8}{\dot{\mathrm{o}}}^2+...\right)$

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

Calculate the first-order perturbation in the energy.

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

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

The perturbed Hamiltonian is therefore

${\mathrm{H}}^{\text{'}}=\frac{1}{2}{\mathrm{k}}^{\text{'}}{\mathrm{x}}^2-\frac{1}{2}{\mathrm{kx}}^2=\frac{1}{2}{\mathrm{kx}}^2\dot{\mathrm{o}}=\dot{\mathrm{o}}\mathrm{V}$

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${\mathrm{E}}_{\mathrm{n}}^1=\frac{\dot{o}}{2}\mathrm{ħ\omega }\left(\frac{1}{2}+\mathrm{n}\right)$

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