Consider a particle of massm in the n th stationary state of the harmonic oscillator (angular frequency $\mathit{\omega }$ ).

(a) Find the turning point, x2 .
(b) How far (d) could you go above the turning point before the error in the linearized potential reaches 1%? That is, if $\frac{\mathbf{V}\left(x_2+d\right)\mathbf{-}{\mathbf{V}}_{\mathbf{Iin}}\left(x_2+d\right)}{\mathbf{V}\left(x_2\right)}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{01}$ what is ?
(c) The asymptotic form of Ai(z) is accurate to 1% as long as localid="1656047781997" $\mathbf{z}\mathbf{\ge }\mathbf{5}$. For the din part (b), determine the smallest nsuch that$\mathit{\alpha }\mathit{d}\mathbf{\ge }\mathbf{5}$ . (For any n larger than this there exists an overlap region in which the liberalized potential is good to 1% and the large-z form of the Airy function is good to 1% .)

Theexact potential is given by:

$V\left(x\right)=\frac{1}{2}m{\omega }^2{x}^2$ …………………(1)

Let $x_2$be the turning point, the values of the potential and the energy function must equal, therefore we can find $x_2$by setting the exact potential equal to the energy at $x_2$and then solve for $x_2$, as:

$$\begin{gathered} E_n=\frac{1}{2}m{\omega }^2{x}_2^2 \\ {x}_2=\frac{1}{\omega }\sqrt{\frac{2E_n}{m}} \\ \mathrm{The}\mathrm{energy}\mathrm{of}\mathrm{the}\mathrm{harmonic}\mathrm{oscillator}\mathrm{is}\mathrm{given}\mathrm{by}: \\ {\mathrm{E}}_{\mathrm{n}}=\left(\mathrm{n}-\frac{1}{2}\right)\mathrm{ħ\omega } \\ \left(\mathrm{n}=1,2,3,....\right) \\ \mathrm{Thus}, \\ {x}_2=\sqrt{\frac{\left(2\mathrm{n}-1\right)\mathrm{ħ}}{\mathrm{m\omega }}} \end{gathered}$$

The liberalized potential at is the energy at turning point (or the exact potential, since both have the same value at the turning point) plus the derivative of the potential at turning point multiplied by the$\mathbf{x}\mathbf{-}{\mathbf{x}}_{\mathbf{2}}$where is the point, so we can write the liberalized potential at as:

$V_{\mathrm{lin}}\left(x\right)=v\left(x_2\right)+{\overline{)\frac{dv}{dx}}}_{x-x_2}\left(x-x_2\right)$ ……………..(2)

from equation (1) we can write the two term of the liberalized potential as:

$V\left(x_2\right)=\frac{1}{2}m{\omega }^2{x}_2^2{\overline{)\frac{dv}{dx}}}_{x-x_2}=m{\omega }^2{x}_2$

substitute into equation (2), so we get:

$V_{\mathrm{lin}}\left(x_2+d\right)=\frac{1}{2}m{\omega }^2{x}_2^2+m{\omega }^2{x}_{2}d$

the liberalized potential at point$x=x_2+d$is therefore:

$V_{\mathrm{lin}}\left(x_2+d\right)=\frac{1}{2}m{\omega }^2{x}_2^2+m{\omega }^2{x}_{2}d$

we need to find the point, which is the distance that we can go above the turning point before the error in the liberalized potential reaches 1% . Mathematically:

$$\begin{gathered} \frac{V\left(x_2+d\right)-V_{\mathrm{lin}}\left(x_2+d\right)}{V\left(x_2\right)}=0.01 \\ \mathrm{substitute}\mathrm{with}\mathrm{the}\mathrm{values}\mathrm{of}{\mathrm{V}}_{\mathrm{lin}}\left({\mathrm{x}}_2+\mathrm{d}\right),\mathrm{V}\left({\mathrm{x}}_2+\mathrm{d}\right)\mathrm{and}\mathrm{V}\left({\mathrm{x}}_2\right),\mathrm{so}\mathrm{we}\mathrm{get}: \\ \frac{\mathrm{V}\left({\mathrm{x}}_2+\mathrm{d}\right)-{\mathrm{V}}_{\mathrm{lin}}\left({\mathrm{x}}_2+\mathrm{d}\right)}{\mathrm{V}\left({\mathrm{x}}_2\right)}=\frac{{\displaystyle \frac{1}{2}}{\mathrm{m\omega }}^2{\left({\mathrm{x}}_2+\mathrm{d}\right)}^2-{\displaystyle \frac{1}{2}}{\mathrm{m\omega }}^2{\mathrm{x}}_2^2-{\mathrm{m\omega }}^2{\mathrm{x}}_2\mathrm{d}}{\frac{1}{2}{\mathrm{m\omega }}^2{\mathrm{x}}_2^2} \\ \frac{\mathrm{V}\left({\mathrm{x}}_2+\mathrm{d}\right)-{\mathrm{V}}_{\mathrm{lin}}\left({\mathrm{x}}_2+\mathrm{d}\right)}{\mathrm{V}\left({\mathrm{x}}_2\right)}=\frac{{\mathrm{x}}_2^2+2{\mathrm{x}}_2\mathrm{d}+{\mathrm{d}}^2-{\mathrm{x}}_2^2-2{\mathrm{x}}_2\mathrm{d}}{{\mathrm{x}}_2^2} \\ \frac{\mathrm{V}\left({\mathrm{x}}_2+\mathrm{d}\right)-{\mathrm{V}}_{\mathrm{lin}}\left({\mathrm{x}}_2+\mathrm{d}\right)}{\mathrm{V}\left({\mathrm{x}}_2\right)}={\left(\frac{\mathrm{d}}{{\mathrm{x}}_2}\right)}^2=0.01 \\ \mathrm{Consequently}: \\ \mathrm{d}=0.1{\mathrm{x}}_2=0.1\sqrt{\frac{\left(2\mathrm{n}-1\right)\mathrm{ħ}}{\mathrm{m\omega }}} \\ \mathrm{d}=0.1\sqrt{\frac{\left(2\mathrm{n}-1\right)\mathrm{ħ}}{\mathrm{m\omega }}} \end{gathered}$$

The patching wave function in this region is given by the Airy function, that is:

$$\begin{gathered} Ai\left(z\right)=Ai\left(\alpha \left(x-x_2\right)\right) \\ Ai\left(z\right)=Ai\left(\alpha d\right) \end{gathered}$$

The asymptotic form of Ai(z) is accurate to 1% as long as $z=\alpha d\ge 5$. From equation 8.34 , we have:

$$\begin{gathered} \alpha ={\left(\frac{2m}{{ħ}^2}{\overline{)\frac{dV}{dx}}}_{x-x_2}\right)}^{1/3} \\ \mathrm{Thus}, \\ \mathrm{\alpha }={\left(\frac{2\mathrm{m}}{{\mathrm{ħ}}^2}\sqrt{\left(2\mathrm{n}-1\right){\mathrm{ħm\omega }}^3}\right)}^{1/3} \\ \mathrm{Therefore}, \\ \mathrm{\alpha d}=0.1{\left(\frac{2\mathrm{m}}{{\mathrm{ħ}}^2}\sqrt{\left(2\mathrm{n}-1\right){\mathrm{ħm\omega }}^3}\right)}^{1/3}\sqrt{\frac{\left(2\mathrm{n}-1\right)\mathrm{ħ}}{\mathrm{m\omega }}} \\ \mathrm{\alpha d}=0.1{\left(2\right)}^{1/3}\left(2\mathrm{n}-1\right) \\ \mathrm{The}\mathrm{condition}\mathrm{require}\mathrm{\alpha d}\ge 5,\mathrm{so}: \\ 0.1{\left(2\right)}^{1/3}{\left(2\mathrm{n}-1\right)}^{2/3}\ge 5 \\ 2\mathrm{n}-1\ge 250 \\ \mathrm{n}\ge 125.5 \\ \mathrm{Thus},\mathrm{the}\mathrm{minimum}\mathrm{n}\mathrm{such}\mathrm{that}\mathrm{the}\mathrm{asymptotic}\mathrm{form}\mathrm{of}\mathrm{Ai}\left(\mathrm{z}\right)\mathrm{is}\mathrm{accurate}\mathrm{to}1\%\mathrm{is}: \\ {\mathrm{n}}_{\min }=126 \end{gathered}$$