# Why do physicists get away with thinking of the Dirac Delta functional as a function?

For instance they use it for finding solutions to things like Poisson's Equation, i.e. the method of Green's functions.

Moreover in Quantum Mechanics, it's common practise to think of the delta functions $\delta_x$ as being a sort of standard basis for the vector space of square integrable functions but $\delta$ is obviously not a square integrable function itself so how can it form a basis for something that it's not a even an element of.

(On a slightly separate issue, it also doesn't make sense to me why you can think of the functions $e^{ikx}$ (Fourier Basis) as a basis square integrable functions since $e^{ikx}$ is also not square integrable. )

Are physicists just really lucky that these things work out or is there some deeper underlying reason why it's okay to think in these terms? I've come across a lot of great books by a lot of great physicists who've simply assumed this to be the case. It's hard to believe that they were all naive enough to dismiss the mathematical fallacy in their arguments. Clearly I 'm missing something. My question is what is it?

An elaborate answer would be much appreciated.

• You can find some justification for this if you look up "test functions", "distributions", and "Schwartz spaces". – Tom Collinge Dec 9 '16 at 11:28
• It is not because something works for finite sums $\sum_k f(k) \delta_{x_k}$ and $\sum_k F(k) e^{i \omega_k x}$ that it works for integrals $\int_{-\infty}^\infty f(y) \delta_{x-y} dy$ and $\int_{-\infty}^\infty F(\omega) e^{i \omega x}d\omega$. Now those physicists have heard that the mathematicians proved it works "the same" with finite sums and integrals in those case, so they trust us and do as if it was allowed. And in general, of course it is a good idea to look at finite sums for getting some ideas about integrals. – reuns Dec 9 '16 at 11:28
• And most non-mathematicians define $\delta(x)$ by $\int_{-\infty}^\infty g(x) \delta(x)dx = \lim_{\epsilon \to 0} \int_{-\infty}^\infty g(x) \frac{1_{|x| < \epsilon}}{2 \epsilon}dx$ which is perfectly valid and enough for being convinced of how we use it in many cases. – reuns Dec 9 '16 at 11:37

It appears that Dirac picked up the formalism from Heaviside's work that came several decades before Dirac's work. He too got correct answers. Mathematicians eventually came up with a rigorous formalism (actually more than one) for studying the delta function. But this work is not simple. However, it turned out to be incredibly useful on a theoretical level when it came to studying PDEs, especially elliptic operators. So Mathematicians did their job, and much (but not all) that the Physicists do with $\delta$ makes sense in that advanced context.
Why do Physicists teach an intuitive approach using $\delta$ functions? The short answer is this: there's not enough time for Physicists to become Mathematicians before they start Physics. And who would tolerate many years of tedious Mathematics before being allowed to start Physics? Life is short, and people have to get things done. And in getting things done, maybe we'll advance Science to eventually end up living long enough to do more of that sort of thing, or we'll become experienced enough to compact the Math to be taught at a lower level.