Phy5645/Particle bouncing on the floor in Earth's Gravitational field

We need to solve the Schrodinger's equations with the boundary condtion $\psi \left(0\right)=0\;\;and\;\;\psi \left(+\infty \right)=0$

$-{\frac {\hbar ^{2}}{2m}}{\frac {\partial ^{2}\psi \left(z\right)}{\partial z^{2}}}+mgz\psi \left(z\right)=E\psi \left(z\right)$

Now, with the change of variable $x=({\frac {\hbar ^{2}}{2m^{2}g}})^{\frac {2}{3}}{\frac {2m}{\hbar ^{2}}}(mgz-E)$ we can reduce the above equation to

${\frac {\partial ^{2}\phi \left(x\right)}{\partial x^{2}}}-x\phi (x)=0$

This is a standard differential equation whose solution is given by

$\phi (x)=B{\textit {A}}_{i}(x)$ where ${\textit {A}}_{i}(x)={\frac {1}{\pi }}\int _{0}^{\infty }cos\;({\frac {t^{3}}{3}}+xt)\;dt$ are called the Airy Function.

When z=0, we have $x=({\frac {2}{mg^{2}\hbar ^{2}}})^{\frac {1}{3}}$

The boundary condition $\psi \left(0\right)=0$ yields $\phi [-({\frac {2}{mg^{2}\hbar ^{2}}})^{\frac {1}{3}}{\textit {E}}\;\;=0]$

which means ${\textit {A}}_{i}[-({\frac {2}{mg^{2}\hbar ^{2}}})^{\frac {1}{3}}{\textit {E}}\;\;=0]$.

The Airy function has zeros only at certain values$R_{n}$ such that ${\textit {A}}_{i}(R_{n})=0withn=0,1,2,3...$.

The first few roots of the Airy function are:

$R_{0}=-2.338,\;R_{1}=-4.088,\;R_{2}=-5.521,\;R_{3}=-6.787$ The boundary condition $\psi \left(0\right)=0$ therefore gives the discrete set of energy levels which can be expressed in terms of the roots of the Airy functions

${\textit {A}}_{i}[-({\frac {2}{mg^{2}\hbar ^{2}}})^{\frac {1}{3}}{\textit {E}}\;\;=0]\;\Rightarrow \;-({\frac {2}{mg^{2}\hbar ^{2}}})^{\frac {1}{3}}{\textit {E}}_{n}\;=\;R_{n}$

Thus the discrete energy levels of the particle are given by $E_{n}=-({\frac {1}{2}}mg^{2}\hbar ^{2})^{\frac {1}{3}}R_{n}$

The wavefunction of the particle is $\psi _{n}(z)=B_{n}{\textit {A}}_{i}[(-{\frac {2m^{2}g^{2}}{\hbar ^{2}}})^{\frac {1}{3}}z-R_{n}]$

The first few energy levels of the particle are:

$E_{0}=2.338({\frac {1}{2}}mg^{2}\hbar ^{2})^{\frac {1}{3}}$

$E_{1}=4.088({\frac {1}{2}}mg^{2}\hbar ^{2})^{\frac {1}{3}}$

$E_{2}=5.521({\frac {1}{2}}mg^{2}\hbar ^{2})^{\frac {1}{3}}$

$E_{3}=6.787({\frac {1}{2}}mg^{2}\hbar ^{2})^{\frac {1}{3}}$