# Homework Help: The dirac delta function

1. Feb 29, 2012

### 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.

2. Feb 29, 2012

### Fredrik

Staff Emeritus
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.

3. Feb 29, 2012

### HallsofIvy

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

4. Feb 29, 2012

### alan2

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

5. Feb 29, 2012

### HallsofIvy

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.

6. Feb 29, 2012

### alan2

I didn't say for fixed x prime. The delta distribution is defined in terms of the independent variable. Look at the second integral in Fredrik's post above. Clearly well defined and equal to f(0). I can call my variables anything that I want and it doesn't change the meaning of the distribution. I still think that calling it a number is rather loose.