If a particle of mass $M$ spontaneously explodes breaking up into two pieces we know by energy conservation $m_1 v_1 = -m_2v_2$, so if i measure the position of $m_1$ accurately and measure the velocity $v_2$ then by this equation i could also accurately get the velocity of $m_1$ along with its position which violates the uncertainty principle. What is going on here? Is momentum conservation not applicable in this case?
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$\begingroup$ Don't forget about the uncertainties of the original particle. $\endgroup$– PM 2RingCommented Jun 11, 2019 at 11:52
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$\begingroup$ Ohh i really didnt look at that... Can you tell me more what does it imply? $\endgroup$– SOSXXCommented Jun 12, 2019 at 18:01
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$\begingroup$ $m_1v_1 = -m_2v_2$ in the rest frame of the original particle, where the momentum of that original particle is zero. But that means the uncertainty of that original momentum is small and so the uncertainty in its position is large, and so there's some uncertainty in the positions of the final particles relative to the original particle. $\endgroup$– PM 2RingCommented Jun 13, 2019 at 2:34
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Conservation means that $m_1v_1$ is exactly equal and opposite to $m_2v_2$. But how can that be true if $m_1v_1$ is uncertain?
Here’s where the idea of correlated states enters in. The final state has both equal & opposite momenta and uncertain moments because the two sides are correlated. If, due to uncertainty, $m_1v_1$ is a bit small, so will $m_2v_2$ be. Ditto is they’re a bit larger, they’re both larger.
answered Jun 11, 2019 at 4:45
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$\begingroup$ But i am doing my best to keep m1v1 the least uncertain(can measure velocity and the mass doesnt change here) but it seems to me the positions are also connected how is that possible? $\endgroup$– SOSXXCommented Jun 12, 2019 at 2:49
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$\begingroup$ If you measure v1 accurately, you also know v2 from conservation. $\endgroup$ Commented Jun 12, 2019 at 3:00
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$\begingroup$ But i measure only v1 and i focus on the other particles position only so does it mean i cant study it because of maybe... Quantum entanglement? $\endgroup$– SOSXXCommented Jun 12, 2019 at 18:00
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1$\begingroup$ Right. Entanglement is how both particles together obey the conservation laws. $\endgroup$ Commented Jun 12, 2019 at 18:02