Don't understand the integral over the square of the Dirac delta function

Question

In Griffiths' Introduction to Quantum Mechanics he gives the eigenfunctions of the Hermitian operator $\hat{x}=x$ as being

$$g_{\lambda}\left(x\right)~=~B_{\lambda}\delta\left(x-\lambda\right).$$

He then says that these eigenfunctions are not square integrable because

$$\int_{-\infty}^{\infty}g_{\lambda}\left(x\right)^{*}g_{\lambda}\left(x\right)dx ~=~\left|B_{\lambda}\right|^{2}\int_{-\infty}^{\infty}\delta\left(x-\lambda\right)\delta\left(x-\lambda\right)dx ~=~\left|B_{\lambda}\right|^{2}\delta\left(\lambda-\lambda\right) ~\rightarrow~\infty.$$

My question is, how does he arrive at the final term, more specifically, where does the $\delta\left(\lambda-\lambda\right)$ bit come from?

My total knowledge of the Dirac delta function was gleaned earlier on in Griffiths and extends to just about understanding

$$\int_{-\infty}^{\infty}f\left(x\right)\delta\left(x-a\right)dx~=~f\left(a\right).$$

Answer 1

You need nothing more than your understanding of $$ \int_{-\infty}^\infty f(x)\delta(x-a)dx=f(a) $$ Just treat one of the delta functions as $f(x)\equiv\delta(x-\lambda)$ in your problem. So it would be something like this: $$ \int\delta(x-\lambda)\delta(x-\lambda)dx=\int f(x)\delta(x-\lambda)dx=f(\lambda)=\delta(\lambda-\lambda) $$ So there you go.

Answer 2

Well, the Dirac delta function $\delta(x)$ is a distribution, also known as a generalized function.

One can e.g. represent it as a limit of a rectangular peak with unit area, width $\epsilon$, and height $1/\epsilon$; i.e.

$$\tag{1} \delta(x) ~=~ \lim_{\epsilon\to 0^+}\delta_{\epsilon}(x), $$ $$\delta_{\epsilon}(x)~:=~\frac{1}{\epsilon} \theta(\frac{\epsilon}{2}-|x|) ~=~\left\{ \begin{array}{ccc} \frac{1}{\epsilon}&\text{for}& |x|<\frac{\epsilon}{2}, \\ \frac{1}{2\epsilon}&\text{for}& |x|=\frac{\epsilon}{2}, \\ 0&\text{for} & |x|>\frac{\epsilon}{2}, \end{array} \right. $$

where $\theta$ denotes the Heaviside step function.

The product $\delta(x)^2$ of the two Dirac delta distributions does strictly speaking not$^1$ make mathematical sense, but for physical purposes, let us try to evaluate the integral of the square of the regularized delta function

$$\tag{2} \int_{\mathbb{R}}\! dx ~\delta_{\epsilon}(x)^2 ~=~\epsilon\cdot\frac{1}{\epsilon}\cdot\frac{1}{\epsilon} ~=~\frac{1}{\epsilon} ~\to~ \infty \quad \text{for} \quad \epsilon~\to~ 0^+. $$

The limit is infinite as Griffiths claims.

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$^1$ We ignore Colombeau theory. See also this mathoverflow post.

Answer 3

Suppose I want to show $$\int \delta(x-a)\delta(x-b) dx = \delta(a-b) $$ To do that , I need to show $$\int g(a)\int \delta(x-a)\delta(x-b) dx da = \int g(a)\delta(a-b) da$$ for any function $g(a)$. $$LHS = \int \int g(a) \delta(x-a)da \ \delta(x-b) dx $$ $$=\int g(x)\delta(x-b)dx $$ $$=g(b) $$ But RHS clearly = $g(b)$ too.

The result follows putting $a=b=\lambda$