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From the Heisenberg's Uncertainty Principle for position and momentum we know that, $x$ and $p$ of a particle cannot be measured simultaneously with arbitrary accuracy.

$$\Delta x \Delta p \geq \hbar/2 $$

How does the uncertainty principle work in the following experiment?

Consider an electron hitting a fluorescent screen which is mounted on a very sensitive (hypothetical) spring. When the electron hits the screen, we know its position at the instant of it hitting the screen. Besides this, it imparts momentum at the instant of hitting the sunscreen which can be calculated from the compression of the spring. Thus getting both position and momentum simultaneously.

What part of the above experiment am I confused about that gives me this wrong conclusion?

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    $\begingroup$ You only know its position if you know the position of the screen. How did you measure that? $\endgroup$
    – chepner
    Jan 6, 2022 at 19:38
  • $\begingroup$ I can in principle measure only the position of screen to arbitrary accuracy. Right? $\endgroup$
    – Lost
    Jan 7, 2022 at 10:21
  • $\begingroup$ Sure, but the more accurately you measure the position, the less accurately you measure the momentum of the screen, which decrease your ability to accurately measure the momentum of the electron upon impact. (The compression of the spring is basically a proxy for the change in momentum of the screen.) $\endgroup$
    – chepner
    Jan 7, 2022 at 12:58

4 Answers 4

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If I understand you correctly you have an experimental setup like this.

enter image description here
(image partially taken from The Physics of Springs)

You measure two things:

  • the $x$ position given by the position where the fluorescent spot appears on the screen
  • $p_y$, the $y$-component of momentum given by the compression of the spring when hit by the particle

You claim there should be no restriction between these two measurements. And you are right. Let us see how this is no contradiction to the uncertainty relation.

Heisenberg's uncertainty relation says $$\Delta x\ \Delta p_x \geq \frac{\hbar}{2}$$

However, this relation doesn't apply to your experiment because you measure $p_y$, but not $p_x$.

There is the more general Robertson uncertainty relation between two arbitrary observables $A$ and $B$: $$\Delta A\ \Delta B \geq \frac{1}{2} \left|\left<[A,B]\right>\right|$$ where $[A,B]$ is the commutator between $A$ and $B$. (Heisenberg's uncertainty relation obviously is a special case of this because $[x,p_x]=i\hbar$.)

Now the commutator between your two measured observables is $$[x,p_y]=0$$ and therefore you have the uncertainty relation $$\Delta x\ \Delta p_y \geq 0.$$

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    $\begingroup$ @ThomasFritsch well, we also know where the screen is in y-direction, in this case $\Delta y p_y \geq \frac{\hbar}{2}$ applies, no ? $\endgroup$ Jan 5, 2022 at 16:22
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    $\begingroup$ @Lost : Why do you believe you know the $y$-coordinate of the electron? Phosphors are not geometric planes -- they have thickness (of several tens to hundreds of nanometers, if they are to have a high probability of interacting with the electron). $\endgroup$ Jan 5, 2022 at 18:00
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    $\begingroup$ @EricTowers Correct me if I'm wrong: In principle, there is nothing in nature preventing a perfectly flat atom-thick phosphor screen which we use in this experiment. In the unlikely event that an electron is absorbed by one of these atoms, we would know its y position only to within the width of the screen (i.e. 1 atom) so the HBU is recovered. $\endgroup$
    – Poo2uhaha
    Jan 5, 2022 at 22:59
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    $\begingroup$ @Lost Eric Towers is right. You definitely cannot know the $y$ position of your screen with enough precision if you are to measure the impact $p_y$ of the electron. You'd have to think up an experiment for that and there will perforce be an impossibility. The setup by Thomas Fritsch does give unlimited precision for $x$ and $p_y$ allowed by Heisenberg. Try another setup for $y$ and $p_y$ and you'll see it won't work. $\endgroup$
    – Alfred
    Jan 5, 2022 at 23:03
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    $\begingroup$ @Poo2uhaha No, again Eric Towers is right. To determine with enough precision the $p_y$ of the electron, you'd need to have determined already the $p_y$ of your screen as well as its position in $y$ . But for that you have to imagine a way to do so. And there is no way to do so, because of Heisenberg principle. It is a circular proposal. Try to find a way to determine the $y$ and the $p_y$ of the screen before interaction with the electron. You'll find it is imposssible. $\endgroup$
    – Alfred
    Jan 5, 2022 at 23:09
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Thomas Fritsch's answer covers the fact that your experiment doesn't probe the essential problem. So what is the essential problem?

Your measurement system can record whatever it records to whatever precision it's capable of. The HUP constrains your ability to predict what it will record. If you prepare many electrons in exactly the same way, so their positions and momenta should be the same for this purpose, your measurement system will record a distribution of positions and momenta whose width is at least as much as the HUP demands.

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When an electron hits the screen, it will give us a spot of diameter $\Delta x$, whereas the compression of the spring is also measured with a certain precision, which will limit the precision of measuring momentum to $\Delta p$. The Heisenberg uncertainty principle then says that $$\Delta x \Delta p \geq \frac{\hbar}{2}.$$

There is nothing special about this type of measurement as compared to other kinds of measurements that one could envisage, however it is worth adding a few remarks:

  • The material of the screen and the spring are also made of atoms. Having precision of measurement sufficiently high to try to break the HUP would require treating these as quantum, which would make the survival of HUP in this experiment far less mysterious.
  • Experiments of this type have been actively tried (and corresponding calculations have been performed) in the last decade in the context of nanomechanics - see the references in this Wikipedia article. There is a lot of interesting stuff going there, but, unsurprisingly, no breaking of the fundamental QM principles.
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You are assuming that the screen is a classical object with a perfectly well-defined position, a perfectly well defined distribution of detection points, and that the spring has a perfectly well defined length. In reality, the screen and the spring are made of electrons and ions, all of which have imperfectly defined positions and momenta in accordance with the HUP. So the HUP imposes a limit on how well you can define the position and momentum of your detecting equipment, which in turn prevents you from pinning down the exact position and momentum of the electron, even supposing it had an exact position and momentum, which, of course, it doesn't.

The HUP not only means that the electron's position and momentum can't be measured exactly together, it goes further and says that the electron cannot have an exact momentum and position at the same time.

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