# Baptiste Ravina

PhD student - high energy physics
CERN/ATLAS/SUSY/TOP

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 $1/\lvert\alpha\rvert$. Even though the graph might look the same, it is important to repeat that $\delta(\alpha x)\neq\delta(x)$! The normalization is not the same ($1$ vs. $1/\lvert\alpha\rvert$), 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 $\mathcal{F}\left\{G(\sigma)\right\}\sim G(1/\sigma)$. Taking the Gaussian representation of the delta-function we mentioned earlier, we see that in the limit $\sigma\to 0$ the transform $\mathcal{F}\left\{\delta(x)\right\}$ 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 :