Is the derivative of a discontinuity a delta function?

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In these notes, https://ocw.mit.edu/courses/physics...-2016/lecture-notes/MIT8_04S16_LecNotes10.pdf, at the end of page 4, it is mentioned:

(3) V(x) contains delta functions. In this case ψ'' also contains delta functions: it is proportional to the product of a continuous ψ and a delta function in V. Thus ψ' has finite discontinuities.

If ψ' has finite discontinuities and ψ'' has delta functions, does that mean that the derivative of a discontinuity is a delta function?
 

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  • #2
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Using informal mathematics, you can think of the "derivative" of a step function being the Dirac delta function times the step amount (including sign). A function with a simple step at a point ("simple" meaning that otherwise it would be differentiable) can be thought to have a derivative with a Dirac delta term at that point. There are other kinds of discontinuities that do not work that way.

This can be made mathematically rigorous by defining "distributions".
 
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