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On page 74 of Timo Weigand's QFT notes, right at the top, the following equality is used:

$$\left[(2\pi)^4\delta^{(4)}(p_f-p_i)\right]^2=(2\pi)^4\delta^{(4)}(p_f-p_i)(2\pi)^4\delta^{(4)}(0) \tag{2.167},$$

however I cannot figure out why this should be the case. I can't even see where to begin, could someone more enlightened please explain this step?

Qmechanic
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Charlie
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    $\delta\left(0\right)$ is not defined normally. You would need non-standard analysis to square a delta. Or it is a sloppy notation for a limit of test functions that is later repaired by "dividing by $\delta\left(0\right)$". See, for example, (2.149), looking equally dodgy to me without further context - which I'm sure there must be. And there is: See e.g. chapter 1.8. – kricheli Jul 25 '22 at 18:32
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    For regular functions $f$ there is the rule $\delta\left(x\right) f\left(x\right) = \delta\left(x\right) f\left(0\right)$, which is easily proven by applying this distribution to a test function and seeing that the delta fixes the argument of $f$ to zero. This makes the formula at least seem plausible in analogy, but again: it looks dodgy. – kricheli Jul 25 '22 at 18:45
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    These are all idealized limits. You may regularize everything by suitable spacetime volume normalizations. This is what the explanatory Volume pointer in the complete formula means. – Cosmas Zachos Jul 25 '22 at 21:27

1 Answers1

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  1. The square of the Dirac delta distribution does not make mathematical sense, cf. e.g. this Phys.SE post.

  2. However, physicists often implicitly imply that the spacetime is a finite (large) box. Then momentum eigenvalues become discrete, integrals are replaced by sums, Dirac delta distributions are replaced by Kronecker delta functions, etc., cf. e.g. this Phys.SE post. In this regularization, one can make sense of $\delta(0)$ and eq. (2.167).

Qmechanic
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