Double delta function value

Question

I have a question from a past paper of a university physics course.

"Calculate $\int_{-\infty}^{\infty}\delta(y-x)\delta(y-z)dy$"

We believe the answer is $1$ only if $x=z$, otherwise the function evaluates to $0$, but is this true? How do we deal with double delta functions in a single dimension?

Answer 1

A simple dimensional analysis tells you that your result cannot be right. The dimension of the delta function is the inverse of the argument. The proper result would have to have the dimension $1/x$.,

Don't let yourself get misled by the double delta. Just treat one of them as the usual function $f(x)$ in the defining relation of the delta function: $$\int_{-\infty}^{\infty} \mathrm dx \ f(x) \delta(x-x_0) = f(x_0)$$ You'll find the result to be $$\delta(z-x)$$ which also has the proper dimension.

Answer 2

Let $$\hat f(x,z) = \int \mathrm dy \ \delta(y-x) \delta(y-z)$$

Given some arbitrary smooth, compactly-supported test function $G(x)$, we can see how $\hat f$ behaves under an integral sign:

$$\int \mathrm dx \ G(x) \hat f(x,z) = \int \mathrm dx \int \mathrm dy\ G(x) \delta(y-x)\delta(y-z)= \int \mathrm dy \ G(y) \delta(y-z) = G(z)$$

So in summary, $\int \mathrm dx \ G(x) \hat f(x,z) = G(z)$ for every suitably well-behaved function $G$. Can you think of a simpler way to write the object $\hat f(x,z)$ which has this property?