A harmonic oscillator has its minimum possible energy, what is the probability of finding it in the classically forbidden region? (Note: At some point, a calculator able to do numerical integration will be needed.)

Potential energy of simple harmonic oscillator is given by,

$U=\frac{1}{2}K{x}^2$

Here, k represents spring constant and x represents displacement

The lowest energy is given by,

$\text{E=}\frac{\text{h}}{\text{2}}\sqrt{\frac{\text{k}}{\text{m}}}$

Here, h represents Planck’s constant and m represents mass of particle.

When the potential energy is more than the total energy, the region becomes classically forbidden,

$\begin{array}{l}\frac{1}{2}k{x}^2=\frac{h}{2}\sqrt{\frac{k}{m}}\\ x=\pm \frac{\sqrt{h}}{{\left(km\right)}^{1/4}}\end{array}$

But, $b={\left(\frac{mk}{h^2}\right)}^{1/4}$

The classically forbidden region lies from$-\frac{\sqrt{h}}{{\left(km\right)}^{1/4}}$ to$\frac{\sqrt{h}}{{\left(km\right)}^{1/4}}$ .

The wave function is given by:

${\psi }_0\left(x\right)={\left(\frac{b}{\sqrt{\pi }}\right)}^{1/2}{e}^{-b^2{x}^2/2}$

The probability to find the particle in the forbidden region,

${\int }_{\frac{\sqrt{\text{A}}}{{\text{(km)}}^{\text{1/4}}}}^{\frac{\sqrt{\text{h}}}{{\text{(km)}}^{\text{1/4}}}}{\text{ψ(x)ψ}}^{\text{0}}\text{(x)dx}$

By symmetry the above function is changed into,

$\begin{array}{l}{\int }_{\frac{\sqrt{\text{h}}}{{\text{(km)}}^{\text{14}}}}^{{\text{(km)}}^{\text{1/4}}}{\text{ψ(x)ψ}}^{\text{0}}\text{(x)dx}\\ \text{=2}{\int }_{\frac{\sqrt{\text{h}}}{{\text{(km)}}^{\text{14}}}}^{\text{¥}}{\left(\sqrt{\frac{\text{b}}{\sqrt{\text{π}}}}{\text{e}}^{{\text{-b}}^{\text{2}}{\text{x}}^{\text{2}}\text{/2}}\right)}^{\text{2}}\text{dx}\\ \text{=2}\frac{\text{b}}{\sqrt{\text{π}}}{\int }_{\frac{\sqrt{\text{h}}}{{\text{(km)}}^{\text{1/4}}}}^{\text{¥}}{\text{e}}^{{\text{-b}}^{\text{2}}{\text{x}}^{\text{2}}}\text{dx}\\ \text{=}\frac{\text{2}}{\sqrt{\text{π}}}\underset{\frac{\sqrt{\text{h}}}{{\text{(km)}}^{\text{3/4}}}}{\overset{\text{¥}}{\int }}{\text{e}}^{{\text{-b}}^{\text{2}}{\text{x}}^{\text{2}}}\text{bdx}\end{array}$

By change of variable method, substitute$y=bx$and$dy=bdx$.

$\begin{array}{l}\frac{2}{\sqrt{\pi }}{\int }_{\frac{\sqrt{h}}{{\left(km\right)}^{1/4}}}^{\infty }{e}^{-b^2{x}^2}bdx\\ =\frac{2}{\sqrt{\pi }}{\int }_{\frac{b\sqrt{\hslash }}{{\left(km\right)}^{1/4}}}^{\infty }{e}^{-y^2}dy\end{array}$

Since, $b={\left(\frac{mk}{h^2}\right)}^{1/4}$above integral becomes,

$\begin{array}{l}\frac{2}{\sqrt{\pi }}\underset{1}{\overset{\infty }{\int }}{e}^{-y^2}dy=\frac{2}{\sqrt{\pi }}\left(0.1394\right)\\ =0.157299\end{array}$

Therefore,

The required probability of finding the particle is 0.1573