Fall 2007 6 Qm

A particle of mass m is placed above a rigid horizontal surface. In the presence of a gravitational field g its vertical motion is quantized. Find the asymptotic expression for the nth energy level for n » 1, with n = 0 being the ground state.

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To get the requested accuracy, it is sufficient to use the standard WKB approximation,

(1)
\begin{align} \psi_n(z)=\frac{\cos[\phi(z)-\pi/4]}{\sqrt{k(z)}},\;\;\phi(z)=\int_z^Hk(\zeta) d\zeta,\;\;k(z)=\sqrt{\frac{2m}{\hbar^2}(E_n-mgz)} \end{align}

where En is the WKB value of the energy and H = En/mg is the classical turning point. The phase shift $\pi/4$ in the argument of the cosine is important: it ensures that (n matches onto an exponentially decaying solution at z > H. The boundary condition $\phi_n(0)=0$ is met if $\phi(0)=(n+3/4) \pi$, i.e.,

(2)
\begin{align} \int_0^H d\zeta\sqrt{\frac{2m}{\hbar^2}(E_n-mg\zeta)}=\frac{2}{3}\frac{\sqrt{2m}}{mg\hbar}E_n^{3/2}=(n+3/4)\pi, \;\; n=0,1,2,... \end{align}

so that

(3)
\begin{align} E_n \simeq \left[ \frac{9}{8}\left(n+\frac{3}{4}\right)^2 \pi^2\hbar^2g^2m\right]^{1/3}, \;\; n \gg 1 \end{align}

This expression is accurate to $\mathcal{O}(1/n)$, and so keeping the term 3/4 on the background of n is still legitimate.

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