# Simple property of Dirac's $\delta$-function.

I'm on Page 63 of R. Shankar's "Principles of Quantum Mechanics".

I'm trying to do Exercise 1.10.1 by proving that $\displaystyle{\delta(ax) = \frac{\delta(x)}{|a|}}$, where $a \in \mathbb R \backslash\{0\}$.

We know that $\displaystyle{\int_{-\infty}^{+\infty}\delta(t)~\mathrm dt = 1}$.

Making the substitution $t=ax$, where $a>0$ gives $$\int_{-\infty}^{+\infty} \delta(ax)~\mathrm dx = \frac{1}{a}$$ Making the substitution $t=-ax$, where $a<0$ gives $$\int_{-\infty}^{+\infty}\delta(-ax)~\mathrm dx = -\frac{1}{a}$$ Since $\delta(t) \equiv \delta(-t)$ we can re-write this as $$\int_{-\infty}^{+\infty}\delta(ax)~\mathrm dx = -\frac{1}{a}$$ Hence: $$\int_{-\infty}^{+\infty}\delta(ax)~\mathrm dx = \left\{ \begin{array}{ccc} 1/a & : & a>0 \\ -1/a &:& a < 0\end{array}\right.$$ $$\implies \ \ \int_{-\infty}^{+\infty}\delta(ax)~\mathrm dx = \frac{1}{|a|}$$ I can see why this might suggest the result, since we can conclude that $$\int_{-\infty}^{+\infty}\delta(ax)~\mathrm dx = \int_{-\infty}^{+\infty}\frac{\delta(x)}{|a|}~\mathrm dx$$ However, I can't see why it proves that $\displaystyle{\delta(ax) = \frac{\delta(x)}{|a|}}$. After all, there are many different functions whose integrals over $\mathbb R$ are all equal.

• Comment to the question formulation (v1): To make it rigorous, you need to insert test functions in various places. – Qmechanic Jun 10 '17 at 17:43
• The Diract delta function is an even function. For that reason the absolute value is required. The Dirac delta is defined as follows: $$f(x_0) = \int_{\mathbb{R}}{\delta(x-x_0)f(x)\,dx}$$ – HBR Jun 10 '17 at 17:44
• @Qmechanic The changes were purely cosmetic, e.g. inserting line breaks, etc. What do you mean by inserting test-functions? – Fly by Night Jun 10 '17 at 17:49
• @HBR Thank you, but I already know that: as I said $\delta(t) \equiv -\delta(t)$. I used that fact in my post... – Fly by Night Jun 10 '17 at 17:50
• It is completely true. My fault. I did not notice that. – HBR Jun 10 '17 at 18:09

You could make an even stronger argument:

For all $A \subset \mathbb{R},$ we have $$\int_A \delta(ax)dx = \int_A \frac{\delta(x)}{|a|} dx$$

Now you can use this type of result.

• How would I prove the more general result? Do you think the question meant to ask me to prove the integrals were the same? It's a book for physicists and it haven't used (or even mentioned) any measure theory whatsoever. – Fly by Night Jun 10 '17 at 17:55
• Just note that both integrals are $1/|a|$ if $0 \in A$ and zero otherwise. The fact that it's a physics book helps you here: you won't ever need a rigorous proof for a physics problem. In fact, modern (theory of) physics often uses math which isn't really rigorously defined! The question most likely meant for you to indeed show what you have shown in the question you posted. – Piotr Benedysiuk Jun 10 '17 at 18:03

Dirac delta is not a true function. Functions in the integral sign can be understood as differentials $f(x)\,dx$, which describe a small quantity of size $f(x)\,dx$ sitting on a small interval $[x,x+dx]$. For the Dirac delta, however, the 'differential' $\delta(x)\,dx$ describes the unit mass concentrated near $x = 0$. No true function can cope with this situation.

So, if $\delta$ is not a function then how can we deal with them? You can understand them by examining how they react to various test functions. It is like you wear an eye-mask and try to understand an object by touching it. If you carefully touch the object in every possible way you can think of, then you may identify the shape of it. The same analogy works here. If you have an object $?(x)$ over the space, then you can test it by evaluating the integral

$$\int \varphi(x) \, ?(x) \, dx$$

with every sensible choices of function $\varphi$. This allows you to pinpoint what $?(x)$ is. For instance, $\delta(x)$ is an object which is characterized by the following property

$$\int \varphi(x) \delta(x) \, dx = \varphi(0)$$

for all nice functions $\varphi$. For your question, notice that

$$\int \varphi(x) \delta(ax) \, dx \stackrel{y = ax}{=} \int \varphi\left(\frac{y}{a}\right) \delta(y) \, \frac{dy}{|a|} = \frac{1}{|a|}\varphi(0).$$

We have tested how $?(x) = \delta(ax)$ behaves when it is acted upon test functions $\varphi$ and now see that the result is just proportional to what $\delta$ yields. This does indeed tell you that $\delta(ax) = \frac{1}{|a|}\delta(x)$.