There are already probably too many articles on the Dirac delta-function out there, but see if I care… More seriously, the following notes will just point out a few aspects of the famous function that might come in handy to the budding particle physicist.

## Not a function!

You may have seen the following definition of the Dirac delta-function : *it is zero everywhere, except at the origin where it’s infinite*. This corresponds to the graph of the function :

It also satisfies the integral relation

But there is no real function that satisfies these two properties together! Indeed, the “spike” has no width, and so it’s integral **must** be zero, from a traditional point of view. To preserve a semblance of mathematical rigour, we’ll refrain from writing such nonsense as $\delta(0)=\infty$ and instead only consider the delta-function as a function *when integrated*.

### One possible approximation

Any function $\phi$ that satisfies $\phi(x)=0$ if $x\neq 0$ and $\int\phi(x)\mathrm{d}x=1$ is a possible analytic representation of the delta-function. The most commonly used one in physics is an *infinitesimally narrow Gaussian distribution* :

*(In the above animation, $a$ is in fact $\sigma$.)*

## Properties

### Translation

One of the most common uses of the delta-function is its ability to “pick out” a specific value of a given function. To do so, we simply need to multiply them together. Indeed, since $\delta(x-a)$ is only non-zero at $x=a$, the following relation holds :

and so, because we are normalizing to unity, for any function $f(x)$ :

As you’ve seen in your particle physics course, and as it is commonly used in QFT, this sifting property enables us to enforce e.g. energy conservation in a $a\to 1+2$ decay :

### Scaling

Because it is not really a function (or because the integral definition loses uniqueness), we can define the following **scaling** property :

An outline of a proof involves a change of variables to $y=\alpha x$ ($\alpha>0$) :

Repeating this calculation for introduces a minus sign upon inverting the integral limits, leading to the final overall factor of . Even though the graph might look the same, it is important to repeat that ! The normalization is not the same ($1$ vs. ), and this has direct consequences in Fourier space.

### Fourier transform

The Fourier transform of the delta-function is a uniform distribution of height the scaling factor (so $1$ in the usual case) :

In general, for a Gaussian distribution of width $\sigma$, $G(\sigma)$, its Fourier transform is . Taking the Gaussian representation of the delta-function we mentioned earlier, we see that in the limit $\sigma\to 0$ the transform behaves like $G(\infty)$ - a uniform distribution.

An other important result that turns up **everywhere in QFT** comes from the so-called Fourier integral theorem, which can be stated (in the usual particle physics convention) as :

Changing the order of the integrals (which is fine) :

where we’ve rewritten the integral in brackets as some function $\phi$ of the *constant* difference $x-y$ (as far as *this* integral is concerned). So we’re integrating some function $f(y)$ over all possible values of $y$, and multiplying it by some function $\phi(x-y)$ picks out the value $f(x)$… It’s not too hard at this stage to guess that

and the **incredibly useful** (I cannot stress how important it is in QFT) result we were after is then

### Delta-function of a function

Technically, we’ve already considered the delta-function of $x-a$, but what about a general function $f(x)$? We expect it to be proportional to $\delta(x-x_0)$, with $x_0$ the zero of $f$ on the interval considered, up to a normalization factor. Let’s go back to the definition of the delta-function :

Now change variables to $y=f(x)$ and the measure to $\mathrm{d}x$ :

On the LHS, we can pull out the derivative, which will be evaluated at $x=x_0$. We can also rewrite the RHS as the integral of $\delta(x-x_0)$ (by definition). Getting rid of the integral notation :