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I am just reading the "Statistical Interpretation of Quantum Mechanics" from L. Ballentine. In section 3.2. he discusses the difference between the observer effect and the statistical Schrödinger-Robertson uncertainty relation. Then he shows a classical example used in introductory physics texts to "derive" the uncertainty relation and disusses why this is misleading and that one might violate the uncertainty relation if it is interpreted as an observer effect.

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It is clear that you can make the error $\delta y$ of $y$ arbitrarily small by just making the detector smaller. However why can the error $\delta p_y$ of $p_y$ also become arbitrarily small? He writes that this is done by just making $L$ large enough. But I don't understand this point.

My second question is about the reinterpretation of this Gedankenexperiment as preparation procedure. Ballentine removes one of the detectors leaving a hole. If it passes trough this hole we call this particle "prepared".

Then consider an ensemle of such identically prepared particles. Now you can measure for the first half of the ensemble the momentum $p_y$ and for the second half the position $y$. Then you get a statistical variation $\Delta p_y$ of your momentum and $\Delta y$ of your position.

From the experimental setup and the de-Broglie relation he infers that $\Delta y = \delta y$ and $\Delta p_y = \frac{\delta y}{h}$.

I don't understand this because I am not sure where and how the measurement of the prepared ensemble is done (to determine experimentally $\Delta y$ and $\Delta p_y$).

Is it measured just by placing a second array of detectors in a line parallel (and right to) $x = x_2$?

Why not just consider the slit in $x_1$ as preparation procedure and the array in $x_2$ for measuring the ensemble?

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  1. However why can the error $\delta p_y$ of $p_y$ also become arbitrarily small?

The discussion refers to the y component of the momentum, $p_y = p\sin\theta = p\frac{y}{\sqrt{y^2 + L^2}}$, when the initial momentum $p$ is known. In terms of uncertainties in $y$ and $L$, the uncertainty in $p_y$ is $$ \delta p_y = \frac{p}{\left(y^2 + L^2\right)^{3/2}}\left[ L^2\delta y - yL\delta L \right] $$ The factor $\left(y^2 + L^2\right)^{-3/2}$ ensures that it decreases as $L$ increases. There is a tendency to confuse this with the angular uncertainty corresponding to the angle subtended by the detector, but it is not the same.

  1. Where and how the measurement of the prepared ensemble is done (to determine experimentally $\Delta y$ and $\Delta p_y$). Is it measured just by placing a second array of detectors in a line parallel (and right of) $x = x_2$?

It is a position measurement done at the location of the removed detector. $\Delta y$ would be just the $y$ span of the detector, that is, $\delta y$. $\Delta p_y$ is given by the $y$ spread of the particle beam after passing the slot of the removed detector, so yes, it can be done with a 2nd screen to the right of $x_2$. The problem is though that the formula cited in the text, $\Delta p_y \approx \delta y/\hbar$, is dimensionally incorrect, so you do have reason to be concerned. I don't know if there was an Errata for this paper, but the correct relation should be $\Delta p_y \approx \hbar/\delta y$, which follows from an argument similar to the one for the first microscope example in the preceding paragraph. In this case it follows indeed that $\Delta p_y \Delta y \approx \hbar$.

Why not just consider the slit in $x_1$ as preparation procedure and the array in $x_2$ for measuring the ensemble?

Because removing a detector is more fun! :D Just kidding. Physically the same, but probably preferred for economy/continuity of examples.

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