Talk:Dirac delta function

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)

I will agree with Tito because we start by clarifying that there is no function having this property. As long as the scare quotes remain, our opening paragraph immediately corrects laypeople new to the topic. Skater is of course right, and that is why the clarification belongs in the introduction, and why my support is conditional on that.

It might still be best to improve the rest of the article, though K Smeltz (talk) 21:50, 25 August 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)

For Banach spaces of holomorphic functions, it is usually the case that evaluation at a point is a continuous linear functional, that is, an ordinary element of the dual space. For example, Hilbert spaces of holomorphic functions are reproducing kernel Hilbert space, the most basic example of which is the Bergman kernel, which in some sense represents the "Dirac delta" in this situation. Tito Omburo (talk) 00:55, 27 August 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)

Now that I read back my remarks I understand that I wasn't clear in addressing the issue I have, which is that the sentence suggests that normalized wave functions (can) form an orthonormal basis.

For example, the wave function (in position representation) is denoted $\Psi (x)$ and defined in terms of the state vector and position basis

$\Psi (x)=\langle x|\Psi \rangle$

A set of vectors $\{|\psi _{i}\rangle \}$ is complete if every state of the quantum system can be represented as:

$|\Psi \rangle =\sum _{i=1}^{n}c_{i}|\psi _{i}\rangle$

I hope this example elucidates the issue I have with the current ambiguous phrasing and mathematical notation. Roffaduft (talk) 08:43, 3 May 2024 (UTC)

Time-delayed Dirac delta

Dirac_delta_function#As_a_measure and Dirac_delta_function#Resolutions_of_the_identity

appear to disagree with Dirac_delta_function#Translation

The result is used at Uncertainty_principle#Proof_of_the_Kennard_inequality_using_wave_mechanics ;ones 7->8 Darcourse (talk) 16:58, 26 December 2024 (UTC)