In Griffiths' Intro to QM [1] he gives the eigenfunctions of the Hermitian operator $\hat{x}=x$ as being
$$g_{\lambda}\left(x\right)~=~B_{\lambda}\delta\left(x-\lambda\right)$$
(cf. last formula on p. 101). He then says that these eigenfunctions are not square integrable because
$$\int_{-\infty}^{\infty}g_{\lambda}\left(x\right)^{*}g_{\lambda}\left(x\right)dx ~=~\left|B_{\lambda}\right|^{2}\int_{-\infty}^{\infty}\delta\left(x-\lambda\right)\delta\left(x-\lambda\right)dx ~=~\left|B_{\lambda}\right|^{2}\delta\left(\lambda-\lambda\right) ~\rightarrow~\infty$$
(cf. second formula on p. 102). 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
$$\tag{2.95}\int_{-\infty}^{\infty}f\left(x\right)\delta\left(x-a\right)dx~=~f\left(a\right)$$
(cf. second formula on p. 53).
References:
- D.J. Griffiths, Introduction to Quantum Mechanics, (1995) p. 101-102.
Answer
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.
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