We know that the post measurement states after a projective measurement has the form $\frac{\hat{\Pi}|\Psi(t)>}{\sqrt{<\psi(t)|\hat{\Pi}|\psi(t)>}}$ -(1)
.$$\frac{\hat{\Pi}|\Psi(t)\rangle}{\sqrt{\langle\psi(t)|\hat{\Pi}|\psi(t)\rangle}} \qquad (1) $$
When a position measurement is made, with finite accuracy $\delta$, we can define $\hat{\Pi}= \int_{x_i }^{x_i+\delta}{|x><x|dx}$$\hat{\Pi}_{x_i}= \int_{x_i }^{x_i+\delta}{|x\rangle\langle x|dx}$ as the projectors.
Why can't we define $\hat{\Pi}=|x><x|$$\hat{\Pi}_x=|x\rangle\langle x|$ as the projectors instead,assuming that there is no theoretical limit to the precision? That is, I have come across statements saying that $\frac{|x><x|\Psi(t)>}{\sqrt{<\psi(t)|\hat{\Pi}|\psi(t)>}}$ -(2)
is
$$\frac{|x\rangle \langle x|\Psi(t)\rangle}{\sqrt{\langle\psi(t)|\hat{\Pi}|\psi(t)\rangle}} \qquad (2)$$
is not a valid post measurement state for position measurements. Why is that so? I have heard that it is related to the fact that for position states $|x>$$|x\rangle$, we have $\int_{-\infty }^{\infty}{<x'|x><x|x'>dx'=\infty}$$\int_{-\infty }^{\infty}\langle x'|x\rangle\langle x|x'\rangle dx'=\infty$, i.e unnormalizable. Although I might be missing something basic, I am not able to see why this fact prohibits equation (2) as a post measurement state?
