[Physics] The delta function as an eigenfunction of the position operator explanation

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$\delta (\textbf{r})$ can be interpreted as a wavefunction. […] It is non-vanishing only for $\textbf{r}=0$. […] $\delta(\textbf{r})$ is an eigenfunction of the position operator with eigenvalue zero.

How is this last statement true?

Best Answer

For that to be true you will have to go beyond the Hilbert space and consider the rigged Hilbert space, that is the Hilbert space plus distributions. Hence this can be made formal if you identify Dirac's bras and kets with the (anti)linear rigged Hilbert space.

In more intuitive terms, the delta function $\delta(\mathbf r)$ can be realised as limits of a normalised gaussian centred at $\mathbf r = 0$, where you assume that the width is shrinking on to that point (indeed this would give you an approximate eigenvector of the position operator with eigenvalue 0). Observe that the $\delta$ function as it is is already normalised in the so-called $\delta$-normalisation.

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