- #26

- 481

- 55

Thanks$$\frac{\delta(x-x')}{\sqrt{\delta(0)}}=\sqrt{\delta(x-x')}$$

- Thread starter GIM
- Start date

- #26

- 481

- 55

Thanks$$\frac{\delta(x-x')}{\sqrt{\delta(0)}}=\sqrt{\delta(x-x')}$$

- #27

- 11,220

- 3,884

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.

$$\delta_{\epsilon}(x)=1/\epsilon \;\;{\rm for}\;\; |x|<\epsilon/2$$

$$\delta_{\epsilon}(x)=0 \;\;{\rm for}\;\; |x|>\epsilon/2$$

$$\delta_{\epsilon}(x)=1/2\epsilon \;\;{\rm for}\;\; |x|=\epsilon/2$$

It satisfies

$$\int_{-\infty}^{\infty} dx \, \delta_{\epsilon}(x)=1$$

By Taylor expansion ##f(x)=f(0)+xf'(0)+x^2f''(0)/2+...## one obtains

$$\int_{-\infty}^{\infty} dx \, f(x)\delta_{\epsilon}(x)=f(0) +\epsilon^2\frac{f''(0)}{24}+...=f(0) +{\cal O}(\epsilon^2)$$

Repeating my previous renormalization procedure by a replacement ##\delta\rightarrow\delta_{\epsilon}## at the right places, one obtains

$$\psi_{ren}(x)=\frac{\delta_{\epsilon}(x-x')}{\sqrt{\delta_{\epsilon}(0)}}

=\sqrt{\delta_{\epsilon}(x-x')}$$

which is well defined for an arbitrarily small positive ##\epsilon##. Putting also the limit ##\epsilon\rightarrow 0## in the right places, one covers also the ##\delta## "function" case. But such pedantry makes all equations more cumbersome, so for the sake of practical calculus I would prefer not to take care of all these details. For me, it's sufficient to know that I

Last edited:

- #28

- 11,220

- 3,884

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:

$$\frac{\delta(0)}{\sqrt{\delta(0)}}=\sqrt{\delta(0)}$$

which equals the right-hand side.

If you want a more rigorous argument, see the hint in the post above.

- #29

- 11,220

- 3,884

$$\delta^4(0)=\frac{TV}{(2\pi)^4}$$

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.

- #30

- 16,750

- 7,998

$$\langle x | x' \rangle=\delta(x-x'),$$

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.

- #31

- 16,750

- 7,998

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.

$$\delta^4(0)=\frac{TV}{(2\pi)^4}$$

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.

- #32

- 11,220

- 3,884

The square root in #27 certainly does make sense.

$$\langle x | x' \rangle=\delta(x-x'),$$

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.

- #33

- 11,220

- 3,884

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.

- #34

samalkhaiat

Science Advisor

- 1,721

- 1,000

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?

[Mentor's note: This text had been edited to fix the Latex formatting. everyone is reminded that there's section explaining how to make Latex work with the Physics Forums software on our help page: https://www.physicsforums.com/help/] [Broken]

[tex]\Delta[A] = \det \left| \frac{\delta G}{\delta \theta}\right|_{G = 0} ,[/tex]

into the path integral so that we can factor out the infinite group volume [itex]\int \mathcal{D} g(\theta)[/itex] which causes the infinite over-counting, i.e. summing over equivalent gauge field configurations. Plus, in actual calculations, you never need to interchange the order of integrations because once you choose the gauge-fixing surface [itex]G^{a}[A] = 0[/itex] so that it intersects every group-orbit exactly once, all pieces of the integrand will become independent of the group coordinates [itex]\theta^{a}(x)[/itex], and you can safely pull out the group volume [itex]\int \mathcal{D} g(\theta)[/itex]. For example in the covariant gauge [itex]G^{a} = \partial^{\mu}A^{a}_{\mu}[/itex], the path integral becomes

[tex]\int \mathcal{D}A_{\mu} \ e^{i S[A]} = \int \mathcal{D}A_{\mu} \int \mathcal{D}g(\theta) \ \prod_{x} \delta [\partial^{\mu}A_{\mu}] \det \left| \partial^{\nu}D^{ab}_{\nu}\right| \ e^{i S[A]} .[/tex] Clearly, nothing is there to prevent you from factoring out the infinite volume of the gauge group [itex]\mbox{Vol}(\mathcal{G}) = \int \mathcal{D} g[/itex]. The only important point to observe is this, your choice of gauge fixing [itex]G^{a}[A][/itex] which defines the local cross section on the principal fibre bundle (gauge slice transverse to every [itex]\mathcal{G}[/itex]-orbit)

Last edited by a moderator:

- #35

- 8

- 0

- #36

samalkhaiat

Science Advisor

- 1,721

- 1,000

Schwartz said it isOh... my... Gawd!

I'd love to see you try and make that rigorous.

So, in order to define a product of two distributions [itex]f[/itex] and [itex]g[/itex], it is necessary that they have the following properties: [itex]f[/itex] must be just as

Last edited:

- #37

samalkhaiat

Science Advisor

- 1,721

- 1,000

If I know how much you know about the subject, I might be able to explain it better for you. Feel free to ask.

- Replies
- 6

- Views
- 926

- Replies
- 14

- Views
- 1K

- Replies
- 4

- Views
- 3K

- Replies
- 1

- Views
- 13K

- Last Post

- Replies
- 8

- Views
- 3K

- Replies
- 6

- Views
- 4K

- Last Post

- Replies
- 6

- Views
- 3K

- Last Post

- Replies
- 8

- Views
- 4K

- Last Post

- Replies
- 24

- Views
- 4K

- Replies
- 19

- Views
- 6K