# The dirac delta function

aaaa202
What's the reason that you write δ(x-x') rather than just δ(x') both indicating that the function is infinite at x=x' and 0 everywhere else? For me that notation just confuses me, and in my opinion the other notation is easier.

## Answers and Replies

Staff Emeritus
Gold Member
What's the reason that you write δ(x-x') rather than just δ(x') both indicating that the function is infinite at x=x' and 0 everywhere else? For me that notation just confuses me, and in my opinion the other notation is easier.

I don't understand exactly what you want to do, but it doesn't seem to make sense. Consider the equality $$\int f(x')\delta(x'-x)dx'=f(x).$$ Are you suggesting that we write the left-hand side as
$$\int f(x')\delta(x')dx'$$ or as $$\int f(x')\delta(x)dx'\ ?$$ The former is independent of x, so it can't be equal to f(x). The latter at least contains an x, but since the "δ(something)" doesn't contain an x', it's not involved in the integration, so we should be able to take it outside the integral.

It would however make sense to define one "delta function" ##\delta_x## for each real number x: ##\delta_x(y)=\delta(y-x)##. Now x indicates where the "function" has its peak.

Homework Helper
$\delta(x')$ is not a function, it is a number. Specifically, it is 0 if x' is not 0, undefined if x= 0.

alan2
No, it is a continuous linear functional, or distribution. Physicists just call it a function but it is not a number.

You really need to think before you speak. $\delta(x)$ and $\delta(x- x')$, for fixed x', are "functionals" or "generalized functions". $\delta(x')$ is a number, just as $(x- x')^2$, for fixed x', is a function of x, but $x'^2$ is a number.