How can you approximate the number of bound states in a potential well of depth $-V_0$ and width $-a$ to $+a$ using uncertainty principle? How can you approximate the number of bound states in a potential well of depth $-V_0$ and width $-a$ to $+a$ using uncertainty principle?
 A: The $n$-th state wave function has $n-1$ zeros. In a well of $2a$ width, the uncertainty of coordinate between two zeros is
$$
\Delta x \sim \frac{a}{n}.
$$
Hence the uncertainty of momentum is
$$
\Delta p \sim \frac{\hbar}{\Delta x} \sim \frac{\hbar n}{a}.
$$
Estimate of the $n$-th's state energy is
$$
E_n \sim -V_0 +\frac{(\Delta p)^2}{2m} \sim -V_0 + \frac{\hbar^2n^2}{2ma^2}
$$
For bound states, $E_n <0$ and we obtain an estimation of the number of bound states
$$
n_b \sim \sqrt{\frac{ma^2V_0}{\hbar^2}}.\quad (1)
$$
Update. Let's show the applicability of (1) in a simple example. Consider a truncated harmonic oscillator well with the potential
$$
U(x) = \left\{
\begin{array}{ccc}
\frac{m\omega^2 x^2}{2}-V_0 & \mbox{if} & |x| < a\\
0 & \mbox{if} & |x| > a
\end{array}
\right.
$$
From the $E_n = \hbar\omega(n+1/2)$ formula, which is valid for the untruncated harmonic oscillator, one expects a number of bound states
$$
n_b\approx \frac{E}{\hbar\omega}.\quad (2)
$$
If the vertical walls at $|x|=a$ are absent, then the parameters $a$ and $V_0$ are related to each other in the following way:
$$
\frac{m\omega^2a^2}{2} = V_0\quad\longrightarrow\quad a\sim\sqrt{\frac{V_0}{m\omega^2}}.
$$
Substitution of the last relation into (1) leads to
$$
n_b \sim \frac{E}{\hbar\omega}.
$$
This expression is equivalent to (2).
A: *

*The HUP is usually only used to give a (often crude) estimate for the ground state energy, cf. e.g. my Phys.SE answer here.


*To find the number of bound states, short of solving the TISE exactly, one may rely on the WKB estimate that there's 1 bound state per $h$-volume in phase space, i.e.
$$ N ~\sim~\frac{2p_{\max}\cdot 2x_{\max}}{h} ~=~\frac{2\sqrt{2m|V_0|}\cdot 2a}{h},$$
cf. e.g. my Phys.SE answer here.
