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

We need to solve the Schrödinger equation for this problem,

$-{\frac {\hbar ^{2}}{2m}}{\frac {d^{2}\psi (z)}{dz^{2}}}+mgz\psi (z)=E\psi (z),$

subject to the boundary condtions, $\psi (0)=0$ and $\psi (z\to \infty )=0.$

If we make the change of variable, $x=\left({\frac {2}{mg^{2}\hbar ^{2}}}\right)^{\frac {1}{3}}(mgz-E),$ then we can reduce the above equation to

${\frac {d^{2}\phi (x)}{dx^{2}}}-x\phi (x)=0.$

This is a standard differential equation, known as the Airy equation. The only physical solution to this equation is

$\phi (x)=B{\text{Ai}}(x),\!$

where

${\text{Ai}}(x)={\frac {1}{\pi }}\int _{0}^{\infty }\cos \left({\frac {t^{3}}{3}}+xt\right)\,dt.$

At $z=0,\,x=-\left({\frac {2}{mg^{2}\hbar ^{2}}}\right)^{\frac {1}{3}}E.$

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

or

${\text{Ai}}\left[-\left({\frac {2}{mg^{2}\hbar ^{2}}}\right)^{\frac {1}{3}}E\right]=0.$

The Airy function has zeros only at certain values $R_{n}$ such that

${\text{Ai}}(R_{n})=0\!$ for $n=0,1,2,3\ldots$

The first few roots of the Airy function are:

$R_{0}=-2.338,\,R_{1}=-4.088,\,R_{2}=-5.521,\,R_{3}=-6.787,\,\ldots$

The boundary condition $\psi (0)=0$ therefore results in a discrete set of energy levels that can be expressed in terms of the roots of the Airy function; i.e.,

${\text{Ai}}\left[-\left({\frac {2}{mg^{2}\hbar ^{2}}}\right)^{\frac {1}{3}}E\right]=0\;\Rightarrow \;-\left({\frac {2}{mg^{2}\hbar ^{2}}}\right)^{\frac {1}{3}}E_{n}=R_{n}.$

Thus the discrete energy levels of the particle are given by

$E_{n}=-\left({\frac {mg^{2}\hbar ^{2}}{2}}\right)^{\frac {1}{3}}R_{n},$

and the associated wave functions are $\psi _{n}(z)=B_{n}{\text{Ai}}\left[-\left({\frac {2m^{2}g^{2}}{\hbar ^{2}}}\right)^{\frac {1}{3}}z-R_{n}\right].$

The first few energy levels of the particle are:

$E_{0}=2.338\left({\frac {mg^{2}\hbar ^{2}}{2}}\right)^{\frac {1}{3}},\,E_{1}=4.088\left({\frac {mg^{2}\hbar ^{2}}{2}}\right)^{\frac {1}{3}},\,E_{2}=5.521\left({\frac {mg^{2}\hbar ^{2}}{2}}\right)^{\frac {1}{3}},\,E_{3}=6.787\left({\frac {mg^{2}\hbar ^{2}}{2}}\right)^{\frac {1}{3}}$

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