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.

  • $\begingroup$ There are already some simplications here. In the CM frame you propose, including both $\Theta[ (p_3)_0 + q_0 ] = \Theta[ M + q_0]$ and $\Theta(q_0)$ is redundant. $\endgroup$ – Vibert Mar 1 '13 at 11:14
  • $\begingroup$ OK but if we forget about the Thetas, how would you proceed in doing the q_0 integrals? The deltas are coupled and I really don*t know what to do. $\endgroup$ – Loser Mar 1 '13 at 11:25
  • $\begingroup$ The second delta function tells you that $q_0 = \pm |\mathbf{q}|$, and due to the Heaviside, only the + sign is possible. Due think of the chain rule: $\delta(f(x)) = \ldots$ when working this out. $\endgroup$ – Vibert Mar 1 '13 at 11:30
  • $\begingroup$ Yes this is exactly how I do...and then when I plug this positive q_0 into the argument of the second delta I get the strange result delta(M). Maybe I shouldn't just plug this in. Too bad these operations are really hand wavy and I've never seen them properly defiined, unfortunately. For instance in Srednicki Chapter 10 (If i recall correctly) he uses $\delta(x)^2 = \delta(x)\cdot\delta(0)$ etc... $\endgroup$ – Loser Mar 1 '13 at 11:34
  • $\begingroup$ Srednicki eq. 11.13 for instance: ' web.physics.ucsb.edu/~mark/ms-qft-DRAFT.pdf $\endgroup$ – Loser Mar 1 '13 at 11:41

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


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

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.

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.

  • $\begingroup$ My mistake was that i did NOT replace the q_0 immediately but instead replacing AFTER I had written the second delta in a the composition-form...this is wrong one cannot do that. One should immediately replace, just as Michael Brown did. $\endgroup$ – Loser Mar 1 '13 at 15:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.