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I am learning Quantum Mechanics, and came across this fact that the derivative of a Heaviside unit step function is Dirac delta function. I understand this intuitively, since the Heaviside unit step function is flat on either side of the discontinuity, and hence its derivative is zero, except at the point where it jumps to 1, where it is infinite. However, what I don't understand intuitively is that when the discontinuity is, say $\alpha$, then the derivative is $\alpha \delta(x)$, while one would naively expect it to remain $\delta(x)$. After all, whether the discontinuity is 1 unit or 10 units, the slope still remains infinite. The intuitive reasoning that worked for the unit step function seems to break down here. It seems like voodoo at this point. Can someone throw some light on this, at the level of someone fairly new to the Dirac delta function?

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    Possible duplicate of [How to prove that the derivative of Heaviside's unit step function is the Dirac delta?](http://math.stackexchange.com/questions/13898/how-to-prove-that-the-derivative-of-heavisides-unit-step-function-is-the-dirac)2015-11-26

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