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Suppose I momentarily, with hundred percent certainty, assure that a particle is in range $(x,x+a)$, by using some sort of detector. If initial state was ket $\alpha$, what is the final state?

Is it some ket $\beta$ with $\psi_\beta(x) = 0$, for all $x \notin (x,x+a)$, with $\psi_\beta\neq\psi_\alpha$ for $x\in (x,x+a)$?

If yes, then can you explain the following, taken from Sakurai's textbook, where he talks of this experiment? Specifically in (1.6.5), he says that measurement reduces the state to RHS. He further suggests that such an action 'may' not affect the wavefunction..

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The initial wavefunction is $\psi_\alpha(x)\equiv\langle x|\alpha\rangle$. The RHS in (1.6.5) corresponds to a wavefunction $\psi_\beta$ which is $0$ outside the interval and equals to the old one $\psi_\beta=\psi_\alpha$ within the interval. This is a usual projection measurement that simply cuts everything that doesn't correspond to the measured value. Of course for the following measurements you need to normalize wavefunction to $1$.

You may also read this answer of mine.

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  • $\begingroup$ So, by that you mean, the final normalized wavefunction would be none other than some $c\psi_\alpha(x')$ for $x' \in (x,x+a)$, for constant c? Is that physically necessary? $\endgroup$ – physicophilic Jul 15 '17 at 17:21
  • $\begingroup$ @physicophilic What's necessary? That wavefunction doesn't change within the interval? It comes from our definition of the coordinate representation of wavefunction and what constitutes idealized coordinate measurement. Think what is $|x\rangle\langle x|$ $\endgroup$ – OON Jul 15 '17 at 17:37
  • $\begingroup$ Although I think I got the overall idea, can you explain what point you are talking about with this outer product $|x⟩⟨x|$? $\endgroup$ – physicophilic Jul 15 '17 at 18:12
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I would model the situation as follows. Let $$ \hat \Pi= \int_{x-\Delta/2}^{x+\Delta/2}\,dx\,\vert x\rangle\langle x\vert \tag{1} $$ denote your "measuring apparatus". Basically $\hat \Pi$ is a projection to the interval where your apparatus produces a click. It generalizes to continuous spectrum the usual projection $$ \hat \Pi=\sum_{\lambda=\lambda_0}^{\lambda_1} \vert \lambda\rangle\langle \lambda\vert \tag{2} $$ where $\lambda$ is an eigenvalue of $\hat \Lambda$ and $\vert\lambda\rangle$ is and eigenvector of the operator $\hat \Lambda$.
You would use (2) to compute the total probability of getting $\lambda_0\le \lambda\le \lambda_1$ if you measured the observable $\Lambda$.

Note that (1) assumes all positions in the interval have equal probability of detection, i.e. the detector has uniform sensitivity over the interval. If not then (1) would be modified by multiplying the uniform probability $dx$ by $dx\, \phi(x)$, where $\phi(x)$ accounts for the sensitivity of the detector at detecting at location $x$ in the interval.

The probability you are looking for is then $$ \langle \alpha\vert\hat \Pi\vert\alpha \rangle =\int_{x-\Delta/2}^{x+\Delta/2}\,dx\, \langle\alpha\vert x\rangle\langle x\vert\alpha\rangle = \int_{x-\Delta/2}^{x+\Delta/2}\,dx\, \vert\langle x\vert\alpha\rangle\vert^2\, . $$ If you assume as suggested that $\vert\langle x\vert\alpha\rangle\vert^2$ is constant over the interval then you can pull it out of the integral and you obtain the probability as $\vert\langle x\vert\alpha\rangle\vert^2 dx$ where $dx=\Delta$ is the width over which your detector will click.

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  • $\begingroup$ Excellent way to look at it. However, my main question is regarding (1.6.5). What I want to know is, what is $\hat{\Pi}(ket \alpha)$. (how to use braket notation?). Another thing, the eigenkets x in the range $(x-\Delta/2, x+\Delta/2)$ don't correspond to same eigenvalues, so its not completely analogous.. right? $\endgroup$ – physicophilic Jul 15 '17 at 17:45
  • $\begingroup$ The eigenvalues do not enter in the computation of probabilities. $\endgroup$ – ZeroTheHero Jul 15 '17 at 17:49

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