Integration of Dirac delta w/ different dimensionalities

[FredMadison]

Hi!

The Dirac delta satisfies

$$\int dx f(x) \delta(x-a) = f(a)$$

But how about

$$\int d^3x f(x) \delta^{(4)}(x-a)$$

Here, x and a are four-momenta, and the integral is over the regular 3-dim momentum.

How does the delta behave here?

 

[CompuChip]

The notation you used usually means
$$\delta^{(4)}(x - a) = \delta(x^0 - a^0) \delta(x^1 - a^1) \delta(x^2 - a^2) \delta(x^3 - a^3)$$

So in that case
$$\int d^3x f(x) \delta^{(4)}(x-a) = f(\vec a) \delta(x^0 - a^0)$$
where $a = (a^0, \vec a)$.

 

[FredMadison]

Ok, thanks!

 

[CompuChip]

By the way, note that this is what you might call "physicist's notation".
Of course, technically the remaining delta makes no sense whatsoever, unless you put it under an integral sign again.

However, we usually interpret it as 'this quantity has the value f(a), but it is only well-defined* if x⁰ = a⁰' or something along those lines.

* well-defined being: having a physical meaning

 

[FredMadison]

So, if I end up with something like

$$\int d^3p f(p) \delta(p^0-a^0)$$

where the argument in the delta is zero from energy conservation, you're saying that I should regard this as

$$\int d^3p f(a)$$

?