When solving one dimensional problems in quantum mechanics it is often assumed that the first derivative of the wave function is continuous. What justifies this assumption?
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$\begingroup$ The question (v2) is essentially a duplicate of physics.stackexchange.com/q/19667/2451 and links therein. $\endgroup$– Qmechanic ♦Commented Feb 11, 2015 at 10:00
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I am assumming you're solving the "Chemist's" Schrödinger equation, i.e. expressing the quantum state in position co-ordinates to find the shape of orbitals and probability density to find a particle at a particular position.
In this case, the reason for the first derivative's continuity is the conservation of probability: we can define a probability flux whose divergence must vanish for steady state solutions. A nonzero divergence at a point means that the particle is "gathering around" or "spreading from" that point: it is either becoming more or less likely to be found there as time advances, which cannot happen at steady state, by definition.
As well as zero divergence, the probability current must be continuous at interfaces: otherwise, we should be describing a particle that is not at steady state and which is showing the "gathering / spreading" behaviour I describe above at the interface.
answered Feb 11, 2015 at 10:08