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

Question

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?

Answer 1

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.

Answer 2

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 $\alpha\,\delta(x)$.

Answer 3

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

Answer 4

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?