Estimate for minimum potential depth required for bound states in 3D attractive potential well

Question

Consider a 3D spherically symmetric potential well,

$$H = \frac{p^2}{2m} + V(r)$$

with $V(r) = - V_0$ for $r < a/2$ and $0$ else, for some $V_0 > 0$.

Now, it is well known that $V_0$ needs to be a minimum value for the well to be able to bind a state. A quick estimate with HUP and $T + V < 0$ yields

$$V_0 > \frac{\hbar^2}{2ma^2}.$$

However, the same argument also works for the 1D symmetric well, but in 1D such a well can bind at least one state for any $V_0 > 0$. The same is true for 2D, where any such well can at least bind one state marginally.

I know that a precise calculation yields the desired result, but why does this estimate not work?

Answer 1

The estimate only applies, when the particle is really pressed into the well ($V_0 = \infty$). However making the potential close to zero, the particle's wavefunction expands outside due to tunneling. In your estimation, $a$ should not be the diameter of the well, but rather the characteristic diameter of the wavefunction. This goes arbitrarily large (when potential goes arbitrarily small), thus there are no lower limits to the kinetic energy.