Welcome to MathOverflow

Question

3

2

According to:

Dirac, P. A. M. (1927). "The physical interpretation of the quantum dynamics." Proceedings of the Royal Society of London. Series A, Containing Papers of a Mathematical and Physical Character 113(765), pp.621–641.

For any $y \in \mathbb{C}$, if $f$ is analytic, then $\int_{-\infty}^\infty f(x) \delta(y-x)dx = f(y)$.

And, according to Wolfram Mathematica, the Fourier transform: $$ \int_{-\infty}^\infty \frac{e^{i \lambda x}}{\sqrt{2\pi}} e^{cx} dx = \sqrt{2\pi} \delta(\lambda - i c). $$ which essentially means that for any $\lambda \in \mathbb{C}$ we have $$ \int_{-\infty}^\infty \frac{e^{i \lambda x}}{2\pi}dx = \delta(\lambda). $$ This all seems consistent since $$ e^{cx} = \int_{-\infty}^\infty \frac{e^{-i \lambda x}}{\sqrt{2\pi}} \sqrt{2\pi} \delta(\lambda - i c) d\lambda = e^{-i (i c)x} = e^{cx} . $$ But, I am wondering if anybody knows where I can find some sort of justification for these formal manipulations. I'd like to have a better understanding of what is going on here. And neither Mathematica nor Dirac provide any sort of justification for the above results.

flag	asked Jan 5 at 1:11 psyduck 261●5

1

The theory of distributions was invented precisely to make rigorous sense of the Dirac delta function. Most functional analysis textbooks (for instance Rudin's book or Folland's book on real analysis) discuss these issues in detail, and because of this your question may be closed. – Paul Siegel Jan 5 at 1:32

4

Functions like $e^{cx}$ are distributions, but not tempered distributions. Hence the theory of the Fourier transform as expounded in most textbooks does not apply to them. The Fourier transforms of distributions are a class of objects known as analytic functionals. An exposition of the theory can be found in Gelfand and Shilov, Generalized Functions. – Michael Renardy Jan 5 at 2:03

1

I agree with Michael that Gelfand-Shilov is an excellent source. Another good source for hyperfunctions (a.k.a. analytic functionals) is Hormander, Analysis of linear differential operators..., vol. I Chap 9. A very readable book is also Kaneko, Introduction to hyperfunctions. – Alexandre Eremenko Jan 5 at 6:46

It might be useful to point out that the first equation in the question has two different interpretations. If $y$ is real, the equation holds in the theory of distributions for any $f$ in the appropriate Schwartz space, not necessarily analytic. But if $y$ isn't real, the equation expresses $f(y)$ in terms of values of $f(x)$ on the real axis; for those values to determine $f(y)$, we need $f$ to be analytic, and therefore we need to work in the theory of analytic functionals. – Andreas Blass Jan 5 at 16:23

Carlo Beenakker · Accepted Answer · 2013-01-05 13:37:21Z UTC

I am afraid this is due to a misunderstanding of what Dirac meant. He does not write "for any $y\in \mathbb{C}$" but he refers to a "c-number". The c stands for "classical" as opposed to quantum, and what Dirac means is that $y$ is a real number and not a Hermitian operator. Dirac never considered the delta function of a complex argument, only of real numbers.

When working with a complex number $z$, you can introduce the product of the delta function of the real and imaginary parts of $z$, and if you wish you can call that $\delta(z)\equiv \delta(\Re z)\delta(\Im z)$. So ultimately the fundamental object remains the delta function of a real number.

Concerning Mathematica: I am not able to reproduce your finding that Mathematica would return the delta function of a complex number. if I input

Integrate[Exp[I*lambda*x]*Exp[c*x]/Sqrt[2*Pi],{x,-Infinity,Infinity}]

into the online Mathematica interface at Wolfram Alpha it returns "integral does not converge", which seems to me to be the only sensible answer (without further information on $\lambda$ and $c$).

Dirac Delta function with a complex argument

Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

1 Answer

You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.