# Squaring the delta function in QFT(Srendicki ch11)

1. May 3, 2010

### LAHLH

Hi,

In Srednicki's chapter on cross sections, when he calculating the probability of a particular process from the overlap $$\langle f\mid i\rangle$$ he comes across:

$$[(2\pi)^4\delta^4(k_{in}-k_{out})]^2$$

He states this is can be equated as follows:

$$[(2\pi)^4\delta^4(k_{in}-k_{out})]^2= (2\pi)^4\delta^4(k_{in}-k_{out})\times (2\pi)^4\delta^4(0)$$

I presume from the rest of the text that he is evaluating this as if it was being integrated over $k$, but I still can't see where this comes from. I've tried google but all I seem to find is alot of discussion about people not being sure if the square of the delta function is even well defined.

Thanks

2. May 3, 2010

### alxm

Could it be the delta-function of a 4-vector rather than the delta-function raised to the 4th?

3. May 3, 2010

### ansgar

Use the delta function as the function f(x)

integral f(x) delta(x_o) dx = f(x_o)

$$(2\pi)^3\delta^{(3)}(\mathbf{0}) = \int d^3x\, e^{i\mathbf{0}\cdot\mathbf{x}} = V$$

thus

$$\delta(x)^2 = \delta(0)\delta(x)$$

(which is "sloppy")

4. May 3, 2010

### clem

The V that comes from the integral of \delta(0) should get canceled by a 1/V in the normalization.

5. May 3, 2010

### ansgar

notice that I have no such normalization, and neither do Srecnicki where these things comes from...

6. May 4, 2010

### LAHLH

I take from this that you mean

$$\int (2\pi)^4\delta^4(k_{in}-k_{out})\times (2\pi)^4\delta^4(k_{in}-k_{out}) d^4k$$
Now treat one of the delta as $$f(x)$$, in the usual $$\int f(x)\delta(x-x_0)=f(x_0)$$ . Which sets $$k_{out}=k_{in}$$ in one the delta's, so you have $$\delta^4(0)$$

and can make the replacement

$$[(2\pi)^4\delta^4(k_{in}-k_{out})]^2= (2\pi)^4\delta^4(k_{in}-k_{out})\times (2\pi)^4\delta^4(0)$$

Which seems to make sense to me, I don't know if it's mathematically correct/rigourous but it's probably what Srednicki was doing I think.

Thanks

7. May 4, 2010

### ansgar

yes that is what srednicki is doing, but it is not mathematically rigour

8. May 6, 2010

### DrFaustus

Unless you study QFT from Glimm&Jaffe, say, you might as well forget mathematical rigour. ALL the most popular books on QFT (Srednicki, Weinberg, Peskins&Schroeder, Zee...) are essentially dealing with fields by formal manipulations. In other words, they use them as physicists do, half way between a classical object and a non-commuting quantum operator. The reason is that this sort of formal mainuplations is justified to some degree by rigorous maths, which is WAAAAAAAY too complicated to be of general physical interest (functional analysis, distribution theory and operator algebras mainly).

Now to answer your question. The Dirac delta is a distribution and in general, you cannot multiply distributions pointwise. And the delta is one such example - the square oof the Dirac delta is formally infinite (you can see this for yourself by multipying two sequences of Gaussians, each of which gives you the delta in the limit, and then take the limit.) The way Srednicki's manipulations are to be understood is essentally by pretending to live on a 3D torus, where the dirac Delta is a Kronecker delta multiplied by V, the volume of the torus, consider the density of whatever quantity you're computing so V gets canceled, and then take the infinite volume limit. The result is what Srednicki has. And the reason why he says things like "$$\delta(o)$$ is just the volume". Thank you Mark, too bad it's infinite!

Again, this sort of sloppiness is typical in QFT and unless you wanna go into some serious maths, you just have to live with it.