# Heisenbergs uncertainity principle clashing with momentum conservation

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?

• Don't forget about the uncertainties of the original particle. – PM 2Ring Jun 11 '19 at 11:52
• Ohh i really didnt look at that... Can you tell me more what does it imply? – SOSXX Jun 12 '19 at 18:01
• $m_1v_1 = -m_2v_2$ 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. – PM 2Ring Jun 13 '19 at 2:34

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.