Consider a particle of mass m that is bouncing vertically and elastically on a smooth reflecting floor in the Earth’s gravitational field
V(z)=\begin{cases} mgz, & z>0, \\ +\infty , & z\leq 0,\end{cases},
where g is a constant (the acceleration due to gravity). Find the energy levels and wave function of this particle.
We need to solve the Schrödinger equation with the boundary conditions \psi (0)=0 and \psi (+\infty )=0 :
-\frac{\hbar ^{2} }{2m} \frac{d^{2}\psi (z) }{dz^{2} } +mgz\psi (z)=E\psi (z)\Longrightarrow \frac{d^{2}\psi (z) }{dz^{2} }-\frac{2m}{\hbar ^{2}} (mgz-E)\psi (z)=0. (4.338)
With the change of variable x=(\hbar ^{2}/(2m^{2}g))^{2/3} (2m/\hbar ^{2})(mgz-E), we can reduce this equation to
\frac{d^{2}\phi (x) }{dx^{2}}-x\phi (x)=0. (4.339)
This is a standard differential equation; its solution (which vanishes at x\longrightarrow =+\infty , i.e., \phi (+\infty )=0) is given by
\phi (x)=BAi(x) where Ai(x)=\frac{1}{\pi } \int_{0}^{\infty }\cos \left(\frac{1}{3}t^{3}+xt \right) dt, (4.340)
where Ai(x) is called the Airy function.
When z = 0 we have x=-(2/(mg^{2}\hbar ^{2} ))^{1/3} E. The boundary condition \psi(0)=0 yields \phi [-(2/ (mg^{2}\hbar ^{2} ))^{1/3}E ]=0 or Ai[-(2/(mg^{2}\hbar ^{2} ))^{1/3}E ]=0. The Airy function has zeros only at certain values of R_{n}:Ai(R_{n}) =0 with n = 0, 1, 2, 3, …. The roots R_{n} of the Airy function can be found in standard tables. For instance, the first few roots are R_{0}=-2.338,R_{1} =-4.088,R_{2}=-5.521,R_{3} =-6.787.
The boundary condition \psi(0)=0 therefore gives a discrete set of energy levels which can be expressed in terms of the roots of the Airy function:
Ai\left[-\left(\frac{2}{mg^{2}\hbar ^{2}} \right) ^{1/3}E \right] =0\Longrightarrow -\left(\frac{2}{mg^{2}\hbar ^{2}} \right) ^{1/3}E_{n}= R_{n}; (4.341)
hence
E_{n}=-\left(\frac{1}{2}mg^{2}\hbar ^{2} \right) ^{1/3} R_{n}, \psi _{n}(z)=B_{n}Ai\left[-\left(\frac{2m^{2}g^{2} }{\hbar ^{2}} \right) ^{1/3}z-R_{n} \right]. (4.342)
The first few energy levels are
E_{0}=2.338\left(\frac{1}{2}mg^{2}\hbar ^{2} \right) ^{1/3}, E_{1}=4.088\left(\frac{1}{2}mg^{2}\hbar ^{2} \right) ^{1/3}, (4.343)
E_{2}=5.521\left(\frac{1}{2}mg^{2}\hbar ^{2} \right) ^{1/3}, E_{3}=6.787\left(\frac{1}{2}mg^{2}\hbar ^{2} \right) ^{1/3}. (4.344)