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?
 A: 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.
A: I) Ignoring the metaplectic correction/Maslov index, the Bohr-Sommerfeld quantization rule reads
$$\begin{align} N ~\approx~&\int_a^b \!\frac{\mathrm{d}x}{\pi\hbar} p(x)\cr 
~=~& \int_a^b \! \frac{\mathrm{d}x}{\pi\hbar} \sqrt{2m(E-V(x))},\end{align} \tag{1} $$
so that
$$\begin{align}  \frac{dN}{dE} ~\stackrel{(1)}{\approx}~&\int_a^b \! \frac{\mathrm{d}x}{\pi\hbar} 
{\sqrt{\frac{m}{2(E-V(x))}}}\cr
~=~&\int_a^b \! \frac{\mathrm{d}x}{\pi\hbar v(x)}\cr 
~=~&\frac{T}{2\pi\hbar }\cr
~=~&\frac{1}{\hbar\omega }, \end{align} \tag{2}$$
or
$$ \frac{dE}{dN} ~\stackrel{(2)}{\approx}~\hbar\omega, \tag{3} $$
in agreement with OP's recollection.
Example: If
$$ E~\propto~N^{\frac{2}{3}}\tag{4}$$
then
$$\frac{dE}{dN}~\stackrel{(4)}{\propto}~N^{-\frac{1}{3}}~\stackrel{(4)}{\propto}~E^{-2},\tag{5}
$$
and vice-versa.
II) Now consider the potential
$$ V(x)~=~\left\{\begin{array}{ll} mgx & \text{if } x>0, \\ 
\infty & \text{if } x<0, \end{array}\right. \tag{6}$$
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 $\frac{3}{4}$.
$$\begin{align} n+\frac{3}{4}~\approx~&\int_a^b \!\frac{\mathrm{d}x}{\pi} k(x)\cr
~=~& \int_0^{E_n} \!\frac{\mathrm{d}x}{\pi} \sqrt{2(E_n-x)}\cr
~=~&\frac{(2E_n)^{\frac{3}{2}}}{3\pi}, \end{align} \tag{7}$$
where $n\in\mathbb{N}_0$. Inverting eq. (7) yields
$$  E_n~\stackrel{(7)}{\approx}~ \frac{1}{2}\left(3\pi( n+\frac{3}{4})\right)^{\frac{2}{3}}. \tag{8} $$
A: 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.
