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This question already has an answer here:

In theory, one could create entanglement for two particles for any observable by taking two normalized eigenvectors $\psi_1$, $\psi_2$ of such observable and considering: $$\psi=\frac{1}{\sqrt{2}}(\psi_1\psi_2+\psi_2\psi_1)$$

Imagine that you could create quantum entanglement for two particles for both position and momentum. This is, a two-particle state in which a measure of $\psi_1$ in position (or momentum) for particle A would imply $\psi_2$ for particle B and vice versa. Furthermore, you are able to measure at exactly the same time position (with arbitrarily good accuracy) for A and momentum (with arbitrarily good accuracy) for B (and therefore for A). However, Heisenberg principle states: $$\Delta x\Delta p\geq\frac{\hbar}{2}$$

Why would the uncertainty principle not be violated? Is such experiment even possible to create?

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marked as duplicate by Norbert Schuch, Community Aug 24 '17 at 15:21

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  • $\begingroup$ Why do you think the uncertainty principle would apply in any other fashion than usual? Are you aware that an exact measurement of position/momentum is impossible? Are you aware that the standard version of the uncertainty principle is not about how good the measurement device is? What does it mean to create "quantum entanglement[...]for both position and momentum" - can you write down the actual entangled states you are thinking about, please? $\endgroup$ – ACuriousMind Aug 23 '17 at 11:59
  • $\begingroup$ Hi @ACuriousMind, thanks for your comment. There is no discussion on the accuracy of the measuring device on my question. Let me rephrase please. $\endgroup$ – Diego F Medina Aug 23 '17 at 12:31
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    $\begingroup$ If this is not about the accuracy of the devices, I do not understand why you would expect the uncertainty principle to be violated. Also, I do not see how you can have a state that would entangle position and momentum as you claim you can - since position states have wholly uncertain momentum and vice versa, any entanglement like $\lvert x\rangle \lvert y\rangle + \lvert y\rangle\lvert x\rangle$ will not look like that w.r.t. momentum basis. $\endgroup$ – ACuriousMind Aug 23 '17 at 12:50
  • $\begingroup$ "Exactly the same time" I'm not aware of any way you would be able to actually prove that. Since the entangled state is broken upon measurement and quantum states lack a stopwatch, you would lack information as to whether the measurement was simultaneous or not since it would be impossible to really verify afaik. So you'd end up getting no useful information in trying to do both at the same time $\endgroup$ – user95137 Aug 23 '17 at 12:55
  • $\begingroup$ @ACuriousMind Of course such a state exists -- $\delta(x_1-x_2)\delta(p_1+p_2)$. AFAIK this is the example from the original EPR paper. But you can raise the same question with a 2-qubit singlet state, where the measurements in all bases are perfectly anti-correlated. $\endgroup$ – Norbert Schuch Aug 24 '17 at 15:05
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Suppose the two particles are entangled in position. When you measure, say, the position of the particle A, the wavefunction of the two-particle system collapses to $$\phi_{x_1}\phi_{x_2}$$ where $\phi_{x_1}$ and $\phi_{x_2}$ are two eigenfunctions of the position operators. Now the momentums are utterly unknown.

You cannot have two wavefunctions for the two-particle system at the same time (one entangled in position, the other entangled in momentum).

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  • $\begingroup$ I upvoted this but it's not quite responsive to the question, because the OP has one observer measuring position and the other measuring momentum, whereas this answer has both measuring position. To be consistent with the question as asked, one should say that the two-particle system collapses to a state $A\otimes B$ where $A$ is an eigenstate of position and $B$ is an eigenstate of momentum, with the probabilities for different pairs of eigenstates calculable from the original entangled state. $\endgroup$ – WillO Aug 23 '17 at 21:52

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