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On the derivative of a Heaviside step function being proportional to the Dirac delta function

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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?

This post imported from StackExchange Mathematics at 2014-06-09 19:12 (UCT), posted by SE-user Joebevo
asked Nov 15, 2012 in Mathematics by Joebevo (35 points) [ no revision ]

4 Answers

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Use the fundamental theorem of calculus ... If $\Theta'(x) = \delta(x)$, then we ought to have

$$\int_{-1}^{+1} \delta(x) = \int_{-1}^{+1} \Theta'(x) = \Theta(1) - \Theta(-1)$$

The left-hand-side is 1: One of the defining properties of the delta function is that it has area 1 under the peak at 0. The right-hand-side is also 1 because the theta function steps up by exactly 1 unit at 0.

Do you see the connection here?

This post imported from StackExchange Mathematics at 2014-06-09 19:12 (UCT), posted by SE-user Steve B
answered Nov 15, 2012 by Steve B (125 points) [ no revision ]
It should be noted that this works for arbitrary small integration limits $\pm \epsilon$, which excludes most/all functions but $\delta(x)$ from fulfilling the requirement.

This post imported from StackExchange Mathematics at 2014-06-09 19:12 (UCT), posted by SE-user Claudius
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Suppose $H(x)$ is the unit step function, and it's derivative is the Delta function,

$$\frac{d}{dx} H(x) = \delta(x)$$

Then multiplying by some factor $\alpha$ gives,

$$\frac{d}{dx} \left(\alpha H(x)\right) = \alpha \frac{d}{dx} H(x) = \alpha \delta(x)$$

So the statement really follows from the fact that differentiation is a linear operation.

This post imported from StackExchange Mathematics at 2014-06-09 19:12 (UCT), posted by SE-user Olaf
answered Nov 15, 2012 by Olaf (320 points) [ no revision ]
I understood it mathematically. But isn't the delta function supposed to be of infinite height? In that case, what exactly does it mean to say that something is $\alpha$ times infinity?

This post imported from StackExchange Mathematics at 2014-06-09 19:12 (UCT), posted by SE-user Joebevo
The delta 'function' isn't really a function at all. Technically, it is something called a distribution; in down-to-Earth terms, it is only defined under an integral: $\int_{a}^{b}dx\, f(x)\delta(x) = f(0)$ if $a<0<b$, and $0$ otherwise. Then it's clear how to interpret $\alpha\delta(x)$.

This post imported from StackExchange Mathematics at 2014-06-09 19:12 (UCT), posted by SE-user Rhys
@Joebevo you have to take into account that the area "under" the delta function is also normalized. So you always have $\int_{-\infty}^\infty \delta(x) = 1$. Therefore $\delta(x)$ and $\alpha\delta(x)$ are not the same function, because if you integrate the latter you get $\alpha$.

This post imported from StackExchange Mathematics at 2014-06-09 19:12 (UCT), posted by SE-user Olaf
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Here is a more direct way of seeing it. Instead of working with the Heaviside function, consider the function $$\frac{1}{2}\,\left(1+\tanh{\frac{x}{\epsilon}}\right)~.$$ Now, this is a nice, smooth function for finite $\epsilon$. But if you start to make $\epsilon$ smaller (try plotting it for values like $\epsilon=1$ and $\epsilon=0.1$) you'll see that it looks more and more like the Heaviside step function. Taking the derivative with respect to $x$ gives $$\frac{1}{2\epsilon}\,\text{sech}^{2}\frac{x}{\epsilon}~,$$ which, as expected, looks more and more like the delta function for smaller and smaller values of $\epsilon$. In fact, these functions give nice representations of the step and delta functions, respectively, in the $\epsilon \to 0$ limit.

This representation of the step function has a "jump" of 1 at $x=0$, so to get a jump of $\alpha$ you could start with $$\frac{\alpha}{2}\,\left(1+\tanh{\frac{x}{\epsilon}}\right)~.$$ Taking the derivative with respect to $x$ explains why you get $\alpha\,\delta(x)$.

This post imported from StackExchange Mathematics at 2014-06-09 19:12 (UCT), posted by SE-user Robert McNees
answered Nov 15, 2012 by Robert McNees (50 points) [ no revision ]
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A more formal proof:

We're treating the Heaviside function $\Theta$ and the delta function $\delta_0$ with support at $0$ both as distributions. This means they're defined by their action on functions $f$ which vanish sufficiently rapidly at $\infty$.

$\delta_0(f) = f(0)$.

$\Theta(f) = \int_{-\infty}^\infty \Theta(x) f(x) dx$

The derivative of a distribution $d$ is defined by duality: $d'(f) = -d(f')$. (Intuition: This is integration by parts, with $f$ vanishing at infinity.)

So $\Theta'(f) = -\Theta(f') = -\int_{-\infty}^\infty \Theta(x) f'(x)dx = - \int_0^\infty f'(x) dx = f(0) - f(\infty) = f(0)$.

Likewise, by linearity, $(\alpha\Theta)'(f) = \alpha\Theta'(f) = \alpha \delta_0(f)$.

This post imported from StackExchange Mathematics at 2014-06-09 19:12 (UCT), posted by SE-user user1504
answered Nov 15, 2012 by user1504 (1,100 points) [ no revision ]

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