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?


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.

Particles with different energy tunneling to the negative kinetic energy region.

  • $\begingroup$ But when you actually solve the TISE for $l=0$ you arrive at a 1D problem for the radial part, with the restriction that there must be a root at $r=0$. If I'm not totally mistaken this yields an equation that gives the exact lower bound to $\hbar^2\pi^2/8ma^2$. Doesn't this contradict your statement? $\endgroup$ – Pascal Engeler Jan 10 '15 at 0:19
  • $\begingroup$ Do you mean the limiting case where there exists no potential well ($l = 0$)? In this case, the wavefunction is nonnormalizable anyway, because there exists no potential well. However, for any $l > 0$, the total energy can be made arbitrarily close to the potential, making the exponential coefficients for the outside region as close to zero as needed to make the normalization condition true. Note that in this case, the uncertainty of the position does not decrease (at least not much) when decreasing $l$, as reducing $l$ just makes the lowest energy state tunnel more outside. $\endgroup$ – kristjan Jan 10 '15 at 11:46

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