Dirac Delta function with a complex argument

Question

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.

Answer 1

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 $c$).

Answer 2

This is a comment but I do not belong to the happy few who have this privilege. The exponential function (even with real arguments) is a Schwartzian distribution. The purpose of this comment is to document the fact that Schwartz showed that the Fourier transform can be extended to the space of ALL distributions (not just tempered ones as is in his seminal text) in 1952 (consult Wikipedia under Paley-Wiener theorem, in particular, Schwartzian Paley-Wiener theorem): His method establishes an isomorphism between the space of distributions on the line and a certain space of entire functions subject to suitable growth conditions. (The precise details can be found in accessible form in Strichartz' book on Fourier transforms and distributions). This allows a completely rigorous derivation of the above formula for the FT of such functions. The easiest way to do this is to use the usual trick of first calculating the FT of the Dirac function (with complex singularity---there is no mystery about this---the Dirac "function" is a measure and so can be defined at any point even in a topological space) which follows immediately from the latter's filtering property. One then uses the fact (which is valid also in this context) that the FT is its own inverse (modulo the usual games with constants).

I don't have the resources to check whether the Schwartz approach predates that of Gelfand and Silov mentioned above.

Answer 3

An integral of $\delta(s)$ for $s\in\mathbb{C}$ makes sense if-and-only-if it can be mapped to an integral of $\delta(x)$ for $x\in\mathbb{R}$. This can be done for integrals along lines both parallel and perpendicular to the real axis as illustrated below.

Below I refer to the real part of $z$ as $\Re_z$ and the imaginary part of  $z$ as $\Im_z$, in other words $z=\Re_z+i\,\Im_z$.

The integral along a line parallel to the real axis illustrated in (1) below is evaluated with the variable substitution $s=t+i\,\Im_z$.

(1) $\quad\int\limits_{-\infty+i\,\Im_z}^{\infty+i\,\Im_z}\delta(s-z)\,f(s)\,ds=\int\limits_{-\infty}^{\infty}\delta(t-\Re_z)\,f(t+i\,\Im_z)\,dt=f(z)$

The simple integral along the imaginary axis illustrated in (2) below is evaluated with the variable substitution $s=i\,t$.

(2) $\quad\int\limits_{-i\,\infty}^{i\,\infty}\delta(i\,s)\,dt=\int\limits_{-\infty}^{\infty}\delta(-t)\,i\,dt=\int\limits_{-\infty}^{\infty}\delta(t)\,i\,dt=i$

The slightly more complicated integral along the imaginary axis illustrated in (3) below is again evaluated with the variable substitution $s=i\,t$.

(3) $\quad\int\limits_{-i\,\infty}^{i\,\infty}\delta(i\,s)\,f(s)\,dt=\int\limits_{-\infty}^{\infty}\delta(t)\,f(i\,t)\,i\,dt=i\,f(0)$

The more general (and even more complicated) integral along a line perpendicular to the real axis illustrated in (4) below is evaluated with the variable substitution $s=i\,t+\Re_z$ which leads to $ds=i\,dt$. Since $t=-i\,(s-\Re_z)$ the lower integration limit becomes $-i\,((\Re_z-i\,\infty)-\Re_z)=-\infty$ and the upper integration limit becomes $-i\,((\Re_z+i\,\infty)-\Re_z)=\infty$. Since $\delta(-x)=\delta(x)$, $\delta(i\,(s-z))=\delta(i\,((i\,t+\Re_z)-(\Re_z+i\,\Im_z)))=\delta(-t+\Im_z)=\delta(t-\Im_z)$.

(4) $\quad\int\limits_{\Re_z-i\,\infty}^{\Re_z+i\,\infty}\delta(i\,(s-z))\,f(s)\, ds=\int\limits_{-\infty}^{\infty}\delta(t-\Im_z)\,f(\Re_z+i\,t)\,i\,dt=i\,f(z)$

Integral along lines perpendicular to the real axis have application in the theory of Mellin transforms. For example, consider the evaluation of the inverse Mellin transform in (5) below which illustrates the Mellin transform of $1$ is $2\,\pi\,\delta(i\,s)$. The integral in (5) below is evaluated with the variable substitution $s=i\,t$.

(5) $\quad\mathcal{M}_s^{-1}[2\,\pi\,\delta(i\,s)](x)=\frac{1}{2\,\pi\,i}\int\limits_{-i\,\infty}^{i\,\infty}2\,\pi\,\delta(i\,s)\,x^{-s}\,ds=\int\limits_{-\infty}^{\infty}\delta(t)\,x^{-i\,t}\,dt=x^0=1$

More generally, consider the evaluation of the inverse Mellin transform in (6) below which illustrates the Mellin transform of $x^z$ is $2\,\pi\,\delta(i\,(s+z))$. The integral in (6) below is evaluated with the variable substitution $s=i\,t-\Re_z$ which leads to $ds=i\,dt$. Since $t=-i\,(s+\Re_z)$, the lower integration limit becomes $-i\,((-\Re_z-i\,\infty)+\Re_z)=-\infty$ and the upper integration limit becomes $-i\,((-\Re_z+i\,\infty)+\Re_z)=\infty$. Note $x^{-s}=x^{-(i\,t-\Re_z)}=x^{\Re_z-i\,t}$, and since $\delta(-x)=\delta(x)$, $\delta(i\,(s+z))=\delta(i\,((i\,t-\Re_z)+(\Re_z+i\,\Im_z)))=\delta(-t-\Im_z)=\delta(t+\Im_z)$.

(6) $\quad\mathcal{M}_s^{-1}[2\,\pi\,\delta(i\,(s+z))](x)=\frac{1}{2\,\pi\,i}\int\limits_{-\Re_z-i\,\infty}^{-\Re_z+i\,\infty}2\,\pi\,\delta(i\,(s+z))\,x^{-s}\,ds$ $\qquad\qquad\qquad=\frac{1}{2\,\pi\,i}\int\limits_{-\infty}^{\infty}2\,\pi\,\delta(t+\Im_z)\,x^{\Re_z-i\,t}\,i\,dt=x^{\Re_z-i\,(-\Im_z)}=x^z$

Answer 4

A generalization of the Dirac delta function to complex arguments, under certain circumstances, is described in the following references: [1] R. A. Brewster and J. D. Franson. Generalized Delta Functions and Their Use in Quasi-Probability arXiv: 1605.04321, May 13, 2016. [2] R. A. Brewster and J. D. Franson. Generalized delta functions and their use in quantum optics. Journal of Mathematical Physics, 59(1):012102, January 2018. [3] Ismo V. Lindell. Delta function expansions, complex delta functions and the steepest descent method. American Journal of Physics, 61(5):438, 1993.

Answer 5

In "On the Mellin Transforms of Dirac’S Delta Function, The Hausdorff Dimension Function, and The Theorem by Mellin" by N Südland & G Baumann (http://hdl.handle.net/10525/1235), it is proven that the Mellin transform of 2πδ(s) is unity.