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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.

Nameless Paladin
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  • You can find some justification for this if you look up "test functions", "distributions", and "Schwartz spaces". – Tom Collinge Dec 09 '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 09 '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 09 '16 at 11:37
  • Why? Because it’s fruitful. – A rural reader Sep 17 '24 at 16:19

2 Answers2

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Almost always, when dealing with Physics and Engineering, the answer to "How do they get away with doing that?" is simple: They get correct answers.

The delta function really started with Heaviside, who is considered to be the Father of modern Electrical Engineering. His version of the delta function was the derivative of the step function. The step function described what happened when you flipped a switch. Heaviside developed abstract methods of solving differential equations of Engineering by consider evolution operators before this was a well-formed idea in Mathematics, recognizing that evolution operators associated with time invariant systems such as circuits had an exponential property. Both the delta function as well as the Laplace transform directly evolved out of Heaviside's methods. (It appears that Mathematicians named the transform after Laplace, however, because Heaviside was a prick, and an integral resembling the transform had been looked at by Laplace.) Mathematicians were very troubled by his methods, especially when it came to expanding operators in a binomial series and applying that to an impulse function. Most people at the time would have dismissed Heaviside as a crackpot were it not for one important fact: he got correct answers, and he verified those answers. When Mathematicians would try to insist that Heaviside be more rigorous, he told them that it was "their job" to clean it up, which didn't make the Mathematicians any friendlier.

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.

Are Physicists lucky that everything works out? Not really. There was great intuition behind the methods leading to answers, and people checked their answers. You can well imagine that many wrong ideas were discarded along the way, "wrong" meaning they gave wrong answers.

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.

Disintegrating By Parts
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  • "It appears that Mathematicians named the transform after Laplace, however, because Heaviside was a prick..." LOL! – Viktor Vaughn Dec 10 '16 at 13:01
  • It is possible that people were pricks with him too. – Red Banana Jan 03 '17 at 04:29
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    This is an exceptionally good answer imho. As a physicist who has always valued and enjoyed learning the rigor and beauty of pure mathematics, it took me a long time to come to terms with the statements in your last paragraph. As much as it pains me to admit this, I'd actually go even further and remark that mathematical rigor in physics can sometimes be a distraction, or worse, a hindrance to making progress. – joshphysics May 13 '21 at 04:08
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    @joshphysics : I agree. Whatever your major, you need to keep your eye on the ball and not veer too far off course. – Disintegrating By Parts May 13 '21 at 05:58
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    @joshphysics : "Am I to refuse to eat because I do not fully understand the mechanism of digestion?" - Heaviside – Disintegrating By Parts Apr 16 '22 at 04:51
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Disintegrating By Parts gave a nice answer already, but here is something useful worth adding.

It is often said that Quantum Mechanics works in a Hilbert space. A Hilbert space is essentially an inner product space where the "vectors" can now be infinite dimensional. Vectors in a Hilbert space can therefore be functions, and an inner product can be defined: $$\int_{-\infty}^{+\infty} f(x) g(x) dx.$$ Of course, we have to assume here that the vectors, which are functions $f:\mathbb{R} \to \mathbb{R},$ behave nicely enough so that the above inner product above is always defined (they go to zero fast enough; are integrable in the first place, etc.) Also, in Quantum Mechanics, we use complex numbers.

Now, it would be tempting to be able to think of a function $f(x)$ as being the sum of the basis functions which are just defined at a point. Let's call these functions $\Delta_r:$ $$\Delta_r(x) = 0 \text{ for } x \neq r$$ $$\Delta_r(x) = 1 \text{ for } x = r$$

Then we could write $$f(x) = c_1 \Delta_{r_1}(x) + c_2 \Delta_{r_2}(x) + c_3 \Delta_{r_3}(x) + ...$$ In analogy to finite dimensional vector spaces $$\vec{v} = c_1 \hat{e}_1 + c_2 \hat{e}_2 + c_3 \hat{e}_3 + ... + c_N \hat{e}_N.$$ or to countably-infinite dimesional vector spaces: $$\vec{v} = \sum_{n}^{\infty} c_n \hat{e}_n$$

Of course, such a sum does not actually make sense in an uncountably infinite space. We would instead use the integral: $$f(x) = \int c_r \Delta_r(x) dr.$$

Unfortunately, the $\Delta_r$ we have defined have a "length" of zero by the inner product we have defined, and no matter how you scale them they are always zero. But one of the axioms of an inner product space is if the length of a vector is zero, then that vector must just be the zero vector. So the idea is to define new "functions" $$\delta(x-r) = \infty \cdot \Delta_{r}(x)$$ where the infinity is of just the right size to make the length of the "vector" $\delta(x-r)$ to be 1 (or $2 \pi)$.

That is the intuition and it should hopefully convince you that the physicists are not just "getting lucky." After all, if you wanted to, instead of looking at the uncountably infinite and continous case, you could just instead look at an extremely large finite system with a countably infinite number of basis vectors. Then, at the end, you could take the limit as the size of the finite system becomes infinite.

But there is still a contradiction here: these $\delta$ functions are not "functions" and so should not be allowed to be in the Hilbert space. The answer to the question of how to make this rigorous then, is that Quantum Mechanics does not actually happen in a Hilbert space, it instead happens in what is called a "rigged Hilbert space."

Historically though, the formalisms and rigorous definitions of rigged Hilbert spaces came after physicists had already been using Dirac-Delta functions. This is not unique in the history of mathematics, calculus was used for many years before a rigorous footing for it was found.

Jbag1212
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