📂Functions

The History of the Delta Function and Why Dirac Used the Delta Function

The History of the Delta Function^{1 2 3}

The delta function started to appear in the works of scholars such as Poisson (1815), Fourier (1822), and Cauchy (1823, 1827) who made significant contributions to mathematics and physics in the early 19th century. However, at that time, there was not a focus on rigorously defining the delta function as we do today. Later, Kirchhoff (1882, 1891) and Heaviside (1893, 1899) were the first to propose a mathematical definition of the delta function. Of course, from a modern perspective, even these were not mathematically rigorous.

Heaviside described the delta function as the derivative of the unit step function. Since then, the delta function began to be freely used especially in electrical engineering, associated with the Laplace transform. Then, Paul Dirac (1927) introduced the delta function in his works on the theory of quantum mechanics, and it became famous as it came to be widely used thereafter. Because of this historical background, the delta function is referred to as the Dirac delta function. As of that time, however, no one had yet provided a mathematically rigorous definition for the delta function.

The person who achieved this was the French mathematician Schwartz. Schwartz, after first encountering the delta function in 1935, rigorously explained it in 1950 through the book “Theorie des Distribution”. By defining things like the delta function as functionals within function spaces, he legitimized what was being rather loosely used previously.

Why Dirac Needed the Delta Function

To put it simply, Dirac needed something that could play a role similar to the Kronecker delta for continuous variables, and that something was the delta function. It seems this is why Dirac chose $\delta$ as the symbol for the delta function.

Let’s suppose a sequence $\left\{ a_{1}, a_{2}, a_{3}, \cdots \right\}$ is given. How might one select a particular element $a_{i}$ in some manner? ‘Something other than $0$ from the multiplication with a particular element’ multiplied by all elements and then added together could select a specific $a_{i}$. This can be defined as below and is called the Kronecker delta.

$$ \begin{equation} \delta_{ij}=\begin{cases} 1,&i=j \\ 0,& i\ne j\end{cases} \end{equation} $$

Then, $a_{i}$ can be represented as follows.

$$ \begin{equation} a_{i}=\sum \limits_{j=1}\delta_{ij}a_{j} \end{equation} $$

Additionally, the following formula holds true for the Kronecker delta.

$$ \begin{equation} \sum \limits_{j=1}\delta_{ij}=1 \end{equation} $$

Let’s now move on to quantum mechanics. Assume a wave function $\phi \in L^2(\mathbb{R})$ is given. If $\phi$ is represented in $n$ states, the amplitude of each state can be shown as below using the Kronecker delta.

$$ \begin{align*} && | \phi \rangle &= a_{1}|\phi_{1}\rangle +a_{2}|\phi_{2}\rangle+\cdots+a_{n}|\phi_{n}\rangle=\sum \limits _{j=1} ^{n}a_{j}|\phi_{j}\rangle \\ \implies && a_{i} &= \langle\phi_{i} | \phi \rangle=\sum_{j=1}^{n}a_{j}\langle \phi_{i}|\phi_{j}\rangle =\sum \limits _{j=1} ^{n}a_{j}\delta_{ij} \end{align*} \tag{4} $$

However, observable physical quantities⁴ might not be divided discretely like this. Thinking in terms of location, we aren’t standing at intervals of 1 meter; we can stand anywhere. Therefore, the eigenfunctions of the position operator $|x\rangle$ can be non-countable, and instead of being represented by the Kronecker delta and $\sum$, it must be expressed as an integral. If the amplitude at each location is $f(x)$, then

$$ | \phi \rangle = \int f(x) |\phi \rangle dx $$

To find the amplitude $f(x_{0})$ of the eigenfunction $|x_{0}\rangle$ at location $x_{0}$,

$$ f(x_{0})=\langle x_{0} |\phi \rangle = \int f(x) \langle x_{0}|x \rangle dx $$

What we need to do now is to ensure the right-hand side integral becomes $f(x_{0})$. This can be solved by considering $\langle x_{0} | x\rangle$ as some function $\delta$ with the following properties.

$$ \begin{equation} \begin{aligned} \delta (x-x_{0})=0,\quad x\ne x_{0} \\ f(x_{0}) = \int f(x)\delta (x-x_{0})dx \\ \int \delta (x-x_{0})dx =1 \end{aligned} \end{equation} $$

Comparing $(5)$ with $(1)$, $(2)$, and $(3)$, it can be seen that Dirac’s delta function is an extension of the Kronecker delta for continuous variables. In his book⁵, Dirac explained that ‘inserting $x_{0}$ into $f(x)$ and integrating $f(x)$ multiplied by $\delta (x-x_{0})$ across the entire region are the same’. He also mentioned that it is not exactly a function in the mathematical sense and thus should only be used in simple expressions to avoid inconsistencies.