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Heisenberg's uncertainty principle states that it is impossible to measure two properties of a particle (like $S_z$ and $S_x$) with certainty at the same time. Consider the following experiment (from this page):

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The middle part is a radioactive substance (of total spin of zero) which emits electron pairs in opposite directions. The right filter is oriented at 0 degrees (measuring $S_z$) and the left one oriented at 90 degrees (measuring $S_x$).

Here is the part which I don't understand: if an electron passes through the left filter, it means that $S_x$ = +h/2. Therefore, the corresponding right electron has $S_x$ = -h/2. Now, the right electron is also passing through the $S_z$ filter at the same time, and hence we can measure $S_z$.

Isnt this against the uncertainty principle and what does the principles of relativity have to say about this? Sure it will take time to communicate the $S_x$ result, but we are measuring $S_x$ and $S_z$ at the same time. Therefore, aren't we able to measure both of the properties at the same time, even though the result of one measurement is communicated at a later time?

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  • $\begingroup$ Welcome to physics.SE! You can put dollar signs before and after your math to make it be displayed properly. $\endgroup$
    – user4552
    Commented Jan 1, 2018 at 14:45
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    $\begingroup$ "Therefore, aren't we able to measure both of the properties at the same time" - But there is no electron state with definite values for $S_x$ and $S_z$. $\endgroup$
    – Alfred Centauri
    Commented Jan 1, 2018 at 14:59

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If you observe the left-hand electron to get through the filter, this is equivalent to preparing the right-hand electron in the state $S_x=-\hbar/2$. Actually, you could have done that without entanglement. Just take an unpolarized electron beam that's going to the right, and pass it through a filter that requires $S_x=-\hbar/2$. Then pass it through a second filter that requires $S_z=+\hbar/2$. This doesn't mean that the electron has definite values of both $S_x$ and $S_z$ simultaneously. After the $S_z$ measurement, the electron no longer has a definite value of $S_x$.

BTW, there is no uncertainty principle of the standard form for $S_x$ and $S_z$, although they are not compatible observables, just like $p$ and $x$ are not compatible observables. But you can make this paradox into an equivalent one by having the electrons be emitted from a particle at rest in the lab frame, so they're in entangled states $p_1=-p_2$. Then you can put them through $x$ and $p$ filters. The resolution of the paradox is the same.

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  • $\begingroup$ Hi! I agree with your response but here's my argument: we are not actually measuring $S_x$ of the right electron directly. We are deriving it from the left electron's result. But at the same time, we measure $S_z$. Since it is at the same time, its not like measuring $S_x$ first and then $S_z$. $\endgroup$
    – Ravi Sankar
    Commented Jan 1, 2018 at 16:00
  • $\begingroup$ @RaviSankar: When A and B are entangled, there is no distinction between measuring A and measuring B. Measuring A also means measuring B. $\endgroup$
    – user4552
    Commented Jan 1, 2018 at 16:48
  • $\begingroup$ Yes! That's what I am curious about. When we measure the left electron, we get the corresponding property of the right electron and at the EXACT time, we also measure a different property (and presumably, get the "right" result since the measurement operations are not sequential; they are done parallel) $\endgroup$
    – Ravi Sankar
    Commented Jan 1, 2018 at 17:15
  • $\begingroup$ @RaviSankar, (1) after the measurements, the left electron has definite $S_x$ (with indefinite $S_z$) and the right electron has definite $S_z$ (with indefinite $S_x$). This must be the case since there is no electron state with definite values of $S_x$ and $S_z$, (2) you insist that the measurements occur at the same "EXACT" time but the time of measurement is itself uncertain. $\endgroup$
    – Alfred Centauri
    Commented Jan 1, 2018 at 18:28
  • $\begingroup$ @AlfredCentauri Can you please help me out on the (2) point? Why is the time of measurement uncertain $\endgroup$
    – Ravi Sankar
    Commented Jan 2, 2018 at 4:29
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This seems to correspond to the Einstein-Podolsky-Rosen (EPR) paradox (in the Bohm formulation) which has since been resolved.

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