what does it mean that "f(x) has a delta function"?

Question

I've found some lecture on quantum mechanics and the professor is talking about functions being continuous, having jumps, or having delta functions. What does it mean to have a delta function? I know what Dirac delta function is, I would understand someone saying that "f(x) is a delta function" as "f(x) is a Dirac delta function". But what does it mean to say that f(x) has a delta function?

Answer 1

What it roughly means is that in addition to discontinuities of the function which are finite jumps, there are also discontinuities of the form of dirac deltas, i.e. $"f(x)=f_1(x) + \sum_{i=1}^n\ \alpha_i \delta(x-x_i)"$, where $f$ is an e.g. piecewise continuous function and the sum is a sum of (scaled) Dirac deltas at points $x_1, x_2 ,\dots, x_n.$ But then, this $f$ is not a function anymore.

It's rather tricky to answer this question rigorously, because the usual mathematical frameworks where you would have an object like a Dirac Delta, the set of objects considered are either so-called distributions (which means functionals on function spaces), or measures (i.e. things you integrate with respect to). Both classes of objects are generalizations of functions - so not every distribution is a function - the Dirac Delta is such an example of a non-function.

Since each measure defines an integral, and every integral is a linear functional, these notions are closely connected.

To see how discontinuities behave, let's consider a continuously differentiable function $F:\mathbb R \rightarrow \mathbb R$ and let's define an integral w.r.t. $F$ by defining for any continously differentiable (as often as you like) function $g$ with compact support:

$$\int_{\mathbb R}g(x)d(F(x))= \int_{\mathbb R}g(x)F'(x)d x.$$

(If you know some probability theory, then you'll recognize $F$ as a cumulative distribution function and $F'$ as probability density function).

But what if $F$ has a discontinuity? Consider the Heaviside function $\theta(x)=\begin{cases}0, x < 0 \\ 1, x \geq 0\end{cases}$: Then we can't use the formula above, as $\theta'$ is not well-defined. On the other hand, if we look at the right hand side, what we can do in the case of a cont. differentiable function $F$, is integration by parts and we end up with $$\int g(x)F'(x)dx = -\int g'(x)F(x)dx$$ (no boundary terms since $g$ has compact support). Now the right-hand- side is well-defined for functions $F$ which are not continuously differentable (piecewise continuous is enough), and for the choice above, we get:

$-\int g'(x) \theta(x) dx = -\int g('x) \theta(x) dx = -\int_{[0,\infty)}g'(x)dx = -\left[g(x)\right]_0^{\infty}= g(0)$

So, be defining $\int g(x)d(F(x)) = -\int g'(x) F(x) dx$, we have:

$$\int g(x) d(\theta(x)) = g(0)$$ In the sense of the definition above, the delta distibution which sends $g$ to $g(0)$ is the derivative (so.called distributional derivative) of the Heavyside function and we could write $g(0) = \delta(g)$.

If we relax our rigor, we might write also formally (!) $"g(0) = \int g(x)\delta(x) dx"$. (Don't do this!)

Instead of the heaviside function, we might also consider functions which are either equal to $0$ or $1$ and have a finite number of jumppoints $x_1,x_2,\dots, x_n$. Then, we can do the same trick above (split the integral at $x_1,x_2,\dots, x_n$ not just at $0$) and get that the distributional derivative is the sum of dirac deltas at the points $x_1,x_2,\dots,x_n$.

Finally, if the function is not constant between jump points, then we write $F=F_1 + \sum_{i=1}^n \alpha_i \theta(x-x_i)$, where $\alpha_i$ accounts for the jump height at $x_i$ and $F_1$ is continuous. Then, we can calculate

$$\int g(x) d(F(x))= - \int g'(x)F_1(x) dx + \sum_{i=1}^n \alpha_i g(x_i),$$

thus the distributional derivative of $F$ is a continuous function plus some added dirac deltas - if $F_1$ is continuously differentiable.

Answer 2

Here is an example of an interesting distribution:

$$\Phi[\varphi] = \frac{1}{2} \varphi(0) + \int_{-\infty}^{\infty} \left( \frac{1}{2 \sqrt{\pi}} e^{-x^2} \right) \varphi(x) \, \mathrm{d} x $$

The distribution $\Phi$ (which, incidentally, is a probability distribution) is interesting, because at most places it is given by a density function: the smoothly varying quantity $e^{-x^2}/(2\sqrt{\pi})$. However, at the origin, $\Phi$ is given by (half of) a delta distribution.

In fact, we might make the common abuse of notation to write a "density function" for $\Phi$:

$$ \Phi(x) = \frac{1}{2 \sqrt{\pi}} e^{-x^2} + \frac{1}{2} \delta(x) $$

It would be reasonable to describe the shape of $\Phi$ has being continuous everywhere except the origin, where it has a singularity of dirac delta type.