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What prohibits, mathematically, that a particle cannot be found outside the box ? Here, I am referring to particle in a box problem (infinite potential on both ends & zero potential along the length of the box).

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    $\begingroup$ well, infinite potential along the boundary of the box is a mathematical way of saying the particle cannot cross it with any finite energy (which any fnite particle has) $\endgroup$ – Nikos M. Sep 30 '14 at 11:17
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To explicitly verify this, one solves the problem for a box of finite depth $V_0$. If you additionally assume the wavefunction and its first derivative to be continuous across the potential step, the solution becomes a matter of

  1. Solve the Schrödinger equation in the distinct regions in- and outside of the box.
  2. Match $\phi$ and $\phi'$ at the potential steps, say at $x=\pm L$

Essentially, the wavefunction behaves in the classically forbidden regions, where $V_0\gt E$ like $$ \phi(x) \sim \exp[-\sqrt{2m(V_0-E)}\,x] $$ Thus sending $V_0 \rightarrow \infty$, you'll find $\phi$ vanishing in those regions.

Keep in mind, that we demanded $\phi$ and its derivative to be continuous everywhere. It's not immediately clear, at least for the second condition, why this should hold. I believe, demanding a continuous derivative is necessitated by the fact, that our potential is discontinuous. If one considers a smooth potential, I think one can drop this condition.

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  • $\begingroup$ In a smooth potential the continuity of the derivative is guaranteed anyways - this can be seen e.g. from the WKB approximation which always returns a smooth function if a smooth potential is used as input. $\endgroup$ – Neuneck Sep 30 '14 at 10:00
  • $\begingroup$ The wave function will have a discontinuous derivative if the potential contains a Dirac delta function. You can see this by integrating the Schrödinger equation (see en.wikipedia.org/wiki/…). $\endgroup$ – Eric Angle Sep 30 '14 at 13:08

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