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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?

J. Murray
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Varun Premkumar
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1 Answers1

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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.

J. Murray
  • 69,036