# Why can't this be a post measurement state for position measurement?

We know that the post measurement states after a projective measurement has the form

$$\frac{\hat{\Pi}|\Psi(t)\rangle}{\sqrt{\langle\psi(t)|\hat{\Pi}|\psi(t)\rangle}} \qquad (1)$$

When a position measurement is made, with finite accuracy $$\delta$$, we can define $$\hat{\Pi}_{x_i}= \int_{x_i }^{x_i+\delta}{|x\rangle\langle x|dx}$$ as the projectors.

Why can't we define $$\hat{\Pi}_x=|x\rangle\langle x|$$ as the projectors instead,assuming that there is no theoretical limit to the precision? That is, I have come across statements saying that

$$\frac{|x\rangle \langle x|\Psi(t)\rangle}{\sqrt{\langle\psi(t)|\hat{\Pi}|\psi(t)\rangle}} \qquad (2)$$

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 $$|x\rangle$$, we have $$\int_{-\infty }^{\infty}\langle x'|x\rangle\langle x|x'\rangle dx'=\infty$$, 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?

• On what hilbert space do your proposed projectors act? Commented May 23, 2020 at 21:59
• @WillO, on the infinite dimensional Hilbert representing the system under consideration? Commented May 23, 2020 at 22:05
• Hi Varun, I've edited your question to make the MathJax a bit more readable, and I added subscripts to your projectors. Feel free to revert the edits if you would like. Commented May 23, 2020 at 22:24
• Varun: And what Hilbert space would that be, exactly? Commented May 23, 2020 at 22:59

The simple answer is that your proposed projector takes you out of the Hilbert space $$\mathcal H$$, because the norm of the state (2) is found to be $$\langle x | x\rangle = \delta(0) \rightarrow \infty$$.
"Non-normalizable states" are useful devices from a computational standpoint, but they are purely mathematical tools; they do not lie in $$\mathcal H$$, and do not correspond to physical states in which the system can exist. A genuine physical state can only be a continuous superposition of position eigenstates, i.e. $$|\psi\rangle = \int f(x) |x\rangle\ dx$$ for some function $$f\in L^2(\mathbb R)$$.
Put differently, the position eigenstates are formal devices which only make real sense when they appear in an integral like the one above. Bare position eigenstates $$|x\rangle$$ must be used with care, and with the understanding that there is an implicit integral around whatever expression you find them in.