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?
2 Answers
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).
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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.
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