Crazy Dirac Deltas

Question

I'm not expecting any rigor in the following and the answers...since we're dealing with Dirac deltas in the context of QFT.

Consider the integral

$$ \int d^4q\ \Theta(q_0)\Theta(p_{3,0}+q_0)\ \delta((p_3+q)^2)\delta((q^2))\frac{1}{(q+p_2)^2-m^2+i\epsilon} $$

The $p$'s and $q$'s are all Minkowski 4-vectors and $\ p_3=p_1+p_2$. For simplicity one can work in the $p_3$ CM-Frame so that $\ p_3 =(M, \vec{0})$.

I have problem with the deltas (since they are coupled), I was thinking about the $q_0$ integration first, how would you guys proceed doing the $q_0$-integration?. I use the composition formula for the Dirac delta Dirac Delta Comp. Wikipedia, but when I replace $q_0$ in the argument of the other delta I get something like $\delta(M)$ where $M$ is the mass of $p_3$. This delta makes no sense to me since we're not even integrating over $M$? Anyone know how to do this integral from the begining? Any hint appreciated, thanks.

Answer 1

I'm getting zero.

Using $p_{3,0}=M$ and assuming $M>0$ the $\Theta(p_{3,0}+q_0)$ factor is redundant and you have: $$ \int d^4q\ \Theta(q_0)\ \delta((p_3+q)^2)\delta(q^2)\frac{1}{(q+p_2)^2-m^2+i\epsilon} $$

Now

$$ \begin{array}{lcl} \delta((p_3+q)^2)\delta(q^2) &=& \delta((M+q_0)^2 - \vec{q}^2)\delta(q_0^2 - \vec{q}^2)\\ &=& \delta((M+|\vec{q}|)^2 - \vec{q}^2)\delta(q_0^2 - \vec{q}^2) \\ &=& \delta(M^2+2M|\vec{q}|)\delta(q_0^2 - \vec{q}^2)\\ &=& \frac{1}{2M}\delta(|\vec{q}|+M/2)\delta(q_0^2 - \vec{q}^2)\\ \end{array}$$

The first line just subs in. The second line uses $q_0 = +|\vec{q}|$ because of the second delta function and the theta function. Third line just expands and simplifies in the first delta. Last line is the chain rule identity for the delta function.

Now the $\delta(|\vec{q}|+M/2)$ only has support for negative $|\vec{q}|$ which is clear nonsense, so the integral is zero. If you had $p_3 - q$ in the original integral instead of $p_3 + q$ you would get a nonzero answer.

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In general you can use $ \delta(x - a) \delta(x - b) = \delta(x-a) \delta(a-b)$ since the first delta kills the integral everywhere except $x=a$. If you happen to have $a=b$ you get $\delta(x-a)^2=\delta(x-a) \delta(0)$ and the $\delta(0)$ needs to be defined by some prescription, such as putting the system in the box.