# Semiclassical quantization of bouncing ball

Consider an elastically bouncing ball of mass $m$ and energy $E$. This has a triangular potential $$V(x)~=~\left\{\begin{array}{ll} mgx & \text{if } x>0, \\ \infty & \text{if } x<0, \end{array}\right.$$ where the $x$-axis points upwards. Let $\hbar = m = g = 1$, so that the maximum height reached is $E$.

The classical frequency of motion is $$\omega = \frac{\pi}{\sqrt{2 E}}.$$ I can recall there is a quantization rule that says that the spacing between energy levels equals the classical frequency (in the semiclassical regime). This would imply that $$E_{n+1} - E_n \propto E_n^ {-1/2}$$ so $$E_n \propto n^{-2}.$$ However, the area in phase space enclosed by the orbit is proportional to $E^{3/2}$. This area must be an integer times $2\pi$, which gives a different quantization condition $$E_n \propto n^{2/3}.$$ What went wrong? I am pretty sure the second result is correct, but why is the first result wrong?

• Why do you expect several ad hoc "quantization" procedures to give the same result at all, let alone the correct one? – ACuriousMind Nov 10 '14 at 13:46

## 3 Answers

Unfortunately I cannot tell you what went wrong on your first try, since I don't exactly know what you did. However, I sat down and tried to solve the system you describe:

We are looking for Eigenstates of the Hamiltonian $$H = \frac{p^2}{2} + V(x),\qquad \text{where }V(x)=\left\{\begin{array}{ll}\infty & \text{if } x<0 \\ x & \text{otherwise}\end{array}\right.$$

For $x<0$ we have $\psi(x)=0$, because the potential is infinite there.

For $x\geq 0$, we have a little more work to do. With the definition of $p = -i\frac{d}{dx}$, we can write $H\psi(x) = E_n\psi(x)$ as:

$$-\frac{1}{2}\psi'' + x\psi = E_n\psi$$

I don't think this equation has an analytic solution. Wolphram Alpha says the solution is called the Airy function $\text{Ai}(x)$, which looks like this: The solution is\

$$\psi(x) = c \cdot\text{Ai}\left(2^\frac{1}{3}(x-E_n)\right)$$

To be able to stitch together both parts of the solution, they must coincide at $x=0$, which means we need $\psi(0) = 0$. So we have the condition

$$E_n = -2^{-\frac{1}{3}}\cdot (n\text{th zero of Ai})$$

The first few values are 0.826645, 1.44531, 1.95181, 2.39946, 2.80867 etc. (consult Wolphram Alpha for more values). These values closely resemble $E_n = \left(n+\frac{1}{\sqrt{2}}\right)^{\frac{2}{3}}$, but they (probably) diverge ultimately. For the last statement I have no prove, however, only a hunch. So, with your second try, you got the exponent somewhat right, but there is no direct proportionality.

I) Ignoring the metaplectic correction/Maslov index, the Bohr-Sommerfeld quantization rule reads

$$\tag{1} N ~\approx~\int_a^b \!\frac{\mathrm{d}x}{\pi\hbar} p(x) ~=~ \int_a^b \! \frac{\mathrm{d}x}{\pi\hbar} \sqrt{2m(E-V(x))},$$

so that

$$\tag{2} \frac{dN}{dE} ~\approx~\int_a^b \! \frac{\mathrm{d}x}{\pi\hbar} {\sqrt{\frac{m}{2(E-V(x))}}}~=~\int_a^b \! \frac{\mathrm{d}x}{\pi\hbar v(x)} ~=~\frac{T}{2\pi\hbar }~=~\frac{1}{\hbar\omega },$$

or

$$\tag{3} \frac{dE}{dN} ~=~\hbar\omega,$$

in agreement with OP's recollection.

Example: If $$\tag{4} E~\propto~N^{\frac{2}{3}}$$ then $$\tag{5}\frac{dE}{dN}~\stackrel{(4)}{\propto}~N^{-\frac{1}{3}}~\stackrel{(4)}{\propto}~E^{-2},$$ and vice-versa.

II) Now consider the potential

$$\tag{6}V(x)~=~\left\{\begin{array}{ll} mgx & \text{if } x>0, \\ \infty & \text{if } x<0, \end{array}\right.$$

with $m=g=\hbar=1$. Next let us include the metaplectic correction/Maslov index. The turning point at an infinitely hard wall and an inclined potential wall have Maslov index $2$ and $1$, respectively, cf. e.g. this Phys.SE post. In total $3$. We should then adjust the Bohr-Sommerfeld quantization rule with a fraction $\frac{3}{4}$.

$$\tag{7}n+\frac{3}{4}~\approx~\int_a^b \!\frac{\mathrm{d}x}{\pi} k(x) ~=~ \int_0^{E_n} \!\frac{\mathrm{d}x}{\pi} \sqrt{2(E_n-x)}~=~\frac{(2E_n)^{\frac{3}{2}}}{3\pi},$$ where $n\in\mathbb{N}_0$. Inverting eq. (7) yields

$$\tag{8} E_n~\approx~ \frac{1}{2}\left(3\pi( n+\frac{3}{4})\right)^{\frac{2}{3}}.$$

I think your error is in assuming that $E_{n+1} - E_{n}$ is proportional to $n$. At least, I assume you assumed it; it's the only way I can see that you could go from the statement $$E_{n+1} - E_n \propto E_n^{-1/2}$$ to the statement $$E_n \propto n^{-2}.$$ Really, what the first proportionality above implies is that $$\frac{dE}{dn} \propto \frac{1}{\sqrt{E}}$$ in the limit of large $n$; and if you solve this equation, you get $$\sqrt{E} \, dE \propto dn \quad \Rightarrow \quad E^{3/2} \propto n \quad \Rightarrow \quad E \propto n^{2/3}$$ as expected.