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I'd be content just having this passage explained to me:
Thanks$$\frac{\delta(x-x')}{\sqrt{\delta(0)}}=\sqrt{\delta(x-x')}$$
Thanks$$\frac{\delta(x-x')}{\sqrt{\delta(0)}}=\sqrt{\delta(x-x')}$$
It's very simple indeed to make it more rigorous. One can always replace the ##\delta## "function" with a true function, such as a very narrow Gaussian parameterized with a small width ##\epsilon##. Or instead of a Gaussian, an even better choice is the narrow "wall" function ##\delta_{\epsilon}(x)## defined byOh... my... Gawd!
I'd love to see you try and make that rigorous.
Here is an explanation at the level of practical calculus. When ##x\neq x'##, both sides are zero, so there is a match. When ##x=x'##, the left-hand side isI'd be content just having this passage explained to me:
That's the usual short cut often used to define the square of S-matrix elements, but it's only a hand-waving shortcut. The correct way is to use true states rather than distributions (plane waves are distributions, namely momementum-eigenfunctions in position representation) for the asymptotic free states. Then there's no problem in squaring the S-matrix element and taking the weak limit to generalized momentum-eigenfunctions for the asymptotic free states. See Peskin/Schroeder for the details.Note also that physics textbooks often do some similar "illegitimate" manipulations. For example, for the Dirac ##\delta## in the momentum space ##\delta^4(k)## physicists often write
$$\delta^4(0)=\frac{TV}{(2\pi)^4}$$
where ##V## is the volume of the "laboratory" and ##T## is the time duration of the experiment. Without carrying about rigor, in this way they obtain results which agree with experiments.
The square root in #27 certainly does make sense.Come on, it doesnt' make sense to take a square or a square root of a ##\delta## distribution (NOT FUNCTION). In QT you must have
$$\langle x | x' \rangle=\delta(x-x'),$$
because otherwise the entire Dirac formalism of bras and kets breaks down, and it's well defined in the sense of distributions. You can take various weak limits to define this properly.
Yes, just like #23 is a hand-waving shortcut which can be justified by more rigorous procedure sketched in #27.That's the usual short cut often used to define the square of S-matrix elements, but it's only a hand-waving shortcut. The correct way is to use true states rather than distributions (plane waves are distributions, namely momementum-eigenfunctions in position representation) for the asymptotic free states. Then there's no problem in squaring the S-matrix element and taking the weak limit to generalized momentum-eigenfunctions for the asymptotic free states. See Peskin/Schroeder for the details.
There is no problem with such formal manipulations at all. The whole purpose of the Faddeev-Popov procedure is to introduce the correct integration measure,Can somebody help me? I am studying Faddeev-Popov trick, following the Peskin and Schroeder's QFT book, but I can't understand one thing. After they inserted the Faddeev-Popov identity,
$$I = \int {{\cal D}\alpha \left( x \right)\delta \left( {G\left( {{A^\alpha }} \right)} \right)\det \left( {\frac{{\delta G\left( {{A^\alpha }} \right)}}{{\delta \alpha }}} \right)}$$
they exchanged the order of integration, but to my knowledge, since it includes delta function, there's no guarantee to exchange the order. Where can I find the reasoning for this?
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Schwartz said it is impossible. In fact he showed that it is impossible to define (associative and commutative) multiplication in the class of generalized functions (distributions). Indeed, it is easy to show that such multiplication leads to contradictions: Consider the two most common distributions, the Dirac delta function [itex]\delta (x)[/itex] and the Principal value function [itex]\mathscr{P}(1/x)[/itex]. We can rigorously prove the following relations [tex]x \ \delta (x) = \delta (x) \ x = 0, \ \ \ \ x \ \mathscr{P}(\frac{1}{x}) = \mathscr{P}(\frac{1}{x}) \ x = 1 .[/tex] If a product existed, then, using these relations, we would have the following contradictory chain of equalities [tex]0 = 0 \ \mathscr{P}(\frac{1}{x}) = \left( \delta (x) \ x \right) \mathscr{P}(\frac{1}{x}) = \delta (x) \left( x \mathscr{P}(\frac{1}{x}) \right) = \delta (x) .[/tex]Oh... my... Gawd!
I'd love to see you try and make that rigorous.
If I know how much you know about the subject, I might be able to explain it better for you. Feel free to ask.Thank you, @samalkhaiat! I think this would be the best correct answer, even though I didn't understand it well. I will try to understand it along this direction. Thank you again.