Talk:Dirac delta function

Does this abuse of notation used in the text an accurate treatment

In the section "Resolutions of the identity" (https://en.wikipedia.org/wiki/Dirac_delta_function#Resolutions_of_the_identity) it is said: "Nevertheless, it is common to abuse notation and write:

$f(x)=\int \,\delta (x-\xi )f(\xi )\,d\xi ,\qquad$"

when the accurate result is:

$\int \,\delta (x-\xi )f(\xi )\,d\xi =f(x)H(\xi -x)+\mathbf {C}$

with $H(x)$ the Heaviside step function and $\mathbf {C}$ it is the integration constant.

I don't have a clue about what it is the topic about, but it could be nice to include the explanation about why in the specific used described both formulations could be used equivalently. 191.125.179.92 (talk) 19:57, 22 March 2023 (UTC)

It is an abuse of notation not for the reason you state, but because this is not an integral of a function. 74.111.98.156 (talk) 12:56, 4 July 2023 (UTC)

Proposal: change name from Dirac delta function to Dirac delta distribution

I suggest to rename the mathematical object and the page "Dirac delta distribution". Although using the word function is common, it is also common to call it distribution, which is more appropriate mathematically. Skater00 (talk) 16:35, 21 March 2024 (UTC)

I think WP:COMMONNAME favors the current "Dirac delta function". I will add another reason for keeping things as they are: prospective readers of the article will all have heard of "function", but not know "distribution", and may as a result be uncertain whether they have arrived at the correct article. Thus the current naming is the least likely to cause confusion. Tito Omburo (talk) 09:10, 22 March 2024 (UTC)

complex analysis

This also comes up in complex harmonic analysis, right? Is there a corresponding theory of generalized functions in C? It doesn't like it can be done the same way as in the reals. 03:29, 1 May 2024 (UTC) 2601:644:8501:AAF0:0:0:0:6CE6 (talk) 03:29, 1 May 2024 (UTC)

Dirac delta in quantum mechanics

@Tito Omburo you've reverted part of my edit. Could you please give an example of a wave function that cannot be expressed as a linear combination of an orthonormal basis and complex coefficients? As that is what is being suggested by the current phrasing.

This suggestion doesn't make a lot of sense to me as according to the postulates of quantum mechanics the wave function belongs, by definition, to a separable complex Hilbert space. Roffaduft (talk) 07:39, 2 May 2024 (UTC)

A set of orthonormal functions need not be a basis. That is what is being defined by the sentence in question. Tito Omburo (talk) 09:01, 2 May 2024 (UTC)

That doesn't alter the fact that we are talking about wave functions in $L^{2}$ here, they span the "state space" (i.e. Hilbert space).

Lets turn the phrasing around for a second:

If no wave function (in $L^{2}$ can be expressed as a linear combination [...], then a set of orthonormal wave functions is "incomplete" in $L^{2}$

Which makes even less sense.

If the wave functions are normalizable, they belong to $L^{2}$ which, in this case, is a separable Hilbert space. Therefore both $\psi$ and $\varphi$ can be expressed as linear combinations of ONBs with complex coefficients. Roffaduft (talk) 12:35, 2 May 2024 (UTC)

Look, the sentence is just defining what it means for an orthonormal set to be complete. Just being orthonormal is not enough, because they might not span the space. Tito Omburo (talk) 12:42, 2 May 2024 (UTC)

I get that, from a purely mathematical point of view. Which would be true were it not that the sentence is not “just” defining what it means for “an” orthonormal set to be complete. We are talking about wave functions here, not arbitrary functions.

For something to be a wave function, it satisfies additional conditions. The “if” statement falsely ignores these conditions. Roffaduft (talk) 12:50, 2 May 2024 (UTC)

Not sure what you mean. Please clarify. Tito Omburo (talk) 13:02, 2 May 2024 (UTC)

I just gave an explanation on why (normalizable) wave functions always belong to $L^{2}$. I also showed why the “if” statement didn’t make sense by giving the opposite implication as an example. You’re the one who claims the “if” statement holds for wave functions, so I think you should be the one clarifying your claim.

Again, we’re not talking about arbitrary functions here. Roffaduft (talk) 13:18, 2 May 2024 (UTC)

So every orthonormal set in L^2 is complete in L^2? (I did not engage with your contrapositive, but it is not the correct contrapositive.) Tito Omburo (talk) 13:23, 2 May 2024 (UTC)

Where did I say “every” orthonormal set? For the third time: we’re talking about normalizable wave functions here Roffaduft (talk) 13:26, 2 May 2024 (UTC)

Does "normalizable wave functiom" mean something different from "element of complex Hilbert space L^2"? Tito Omburo (talk) 13:30, 2 May 2024 (UTC)

Maybe you want to lookup the meaning of the normalization condition in quantum mechanics first. Also, you forgot the “separable” part; it’s not just a complex Hilbert space. I provided two hyperlinks in my initial reaction that may be of use to you. Roffaduft (talk) 13:40, 2 May 2024 (UTC)

Does "normalizable wave function" mean something different from "element of a complex separable Hilbert space L^2"? Tito Omburo (talk) 13:44, 2 May 2024 (UTC)

Can any element of a separable complex hilbert space be expressed as the linear combination of an ONB with a (complex) coefficient? Roffaduft (talk) 13:50, 2 May 2024 (UTC)

Yes. Is every orthonormal set in a separable Hilbert space an orthonormal basis? Tito Omburo (talk) 13:56, 2 May 2024 (UTC)

I see what you mean, but that doesn't alter the fact that the statement implies normalizable wave functions can exist (in $L^{2}$) that cannot be expressed as a linear combination of the normalized wave function (in $L^{2}$) with a complex coefficient. So can we at least agree that the "if" statement should be an "if and only if"? Roffaduft (talk) 14:24, 2 May 2024 (UTC)