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Many explanations about the uncertainty principle and its related EPR paradox state that it is impossible to measure opposite complementary variables on different entangled particles; for example, measuring the position of one particle and the momentum of the other at the exact time.

I fail to understand what it means for this to be "impossible" in this case. What would happen if such an attempt was made?

I can only think of 4 ways of interpreting this impossibility:

  1. For every known pair of complementary variables, we only know how to measure one of the two variables for any given particle; for example, we only know how to measure position, not momentum (or vice versa). This is obviously incorrect.
  2. It is impossible to measure any of the complementary variables of a particle at the exact time we measure any other complementary variable of another particle. For example, given any two particles, we cannot measure both of their positions at the same time. I'm guessing this is also incorrect.
  3. Same as #2, but exclusive for entangled particles, implying that there is something in the process of the creation of such a pair that prevents them from being measured at the same exact time.
  4. In the case of an entangled pair being measured at the same time for opposite complementary variables, these measurements would "somehow" no longer correlate between the two, thus rendering the particles entangled no more.

Furthermore, if #4 is the case, was it verified? Was there an experiment showing the measured particles indeed were no longer entangled, perhaps by proof of contradiction; i.e, observing a later measurement that could only occur if these values were different than the values assumed based on the original measurements and the correlation between the particles?

I would love it if the answer would be simple as possible, as I do not have a strong background in neither physics nor mathematics.

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    $\begingroup$ What would happen is that you'd observe some eigenstate of the combined observation, with probabilities given by the Born rule, just as with any other measurement in quantum mechanics. $\endgroup$ – WillO Apr 15 at 16:33
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    $\begingroup$ In particular, the impossibility assertion that you attribute to "many explanations" is simply false. $\endgroup$ – WillO Apr 15 at 16:35
  • $\begingroup$ @WillO So that means it is possible to measure / know (not predict) the "incompatible observables" of a particle at a given time -- say its position and its momentum? $\endgroup$ – Eyal Roth Apr 15 at 17:13
  • $\begingroup$ Yes, of course it's possible. What could conceivably prevent it? Alice gets out her measuring apparatus and uses it in the usual way; Bob, across town, does likewise. Alice 's choice about what to measure obviously can't constrain Bob. $\endgroup$ – WillO Apr 15 at 17:25
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    $\begingroup$ No. Of course not. $\endgroup$ – WillO Apr 15 at 17:43
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This has really been fully answered in the comments, but here is more detail:

You want the particles to be in an entangled state that guarantees that momentum measurements on both particles are certain to agree. So take the entangled state $\xi=A\otimes A+B\otimes B$, where $A$ and $B$ are eigenstates of momentum. ("Momentum" here can stand for any observable of your choice.)

Then $X=A+B$ and $Y=A-B$ are eigenstates of the complementary observable (which we can call "position"). Check that (up to multiplication by a constant)

$$\xi=X\otimes X+Y\otimes Y$$

so that position measurements of both particles are also certain to agree.

But

$$\xi=X\otimes A+X\otimes B+Y\otimes A-Y\otimes B$$ so that if you measure the position of the first particle and the momentum of the second, all four outcomes are equally likely.

Moreover, if you measure the position of the first particle and it comes out to be, say, $Y$, then the state collapses to $Y\otimes A-Y\otimes B$, leaving you with maximum uncertainty about the outcome of a momentum measurement on the second.

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  • $\begingroup$ Is there a quick way to check that $X,Y$ are eigenstates of a complimentary (in the sense of commuting to a constant) variable? $\endgroup$ – jacob1729 Apr 16 at 15:28
  • $\begingroup$ @jacob1729 : I'm not sure I fully understand your question, but if you take $A$ and $B$ to be the column vectors $(1,0)^T$ and $(0,1)^T$, then you can explicitly write down matrices representing momentum and position and can explicitly compute their commutator. $\endgroup$ – WillO Apr 16 at 16:31
  • $\begingroup$ yes that's what I meant. Is there a smarter way of seeing that (eg X,Y are a 45 degree rotation of A,B). $\endgroup$ – jacob1729 Apr 16 at 16:37
  • $\begingroup$ "if you measure the position of the first particle ... leaving you with maximum uncertainty about the outcome of a momentum measurement on the second" - but what does that uncertainty mean in practice? As far as I understand the term, "uncertainty" in practice would mean that predictions made using said uncertain measurement will have a weaker correlation with actual results than it would with a more "certain" measurement. For example, predicting the (later) position of the second particle according to its momentum (comparing a certain and uncertain measurements of momentum). $\endgroup$ – Eyal Roth Apr 19 at 8:35
  • $\begingroup$ "maximum uncertainty" here means there's a 50% probability of each outcome. $\endgroup$ – WillO Apr 19 at 12:07

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