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Are the proton, electron and anti-neutrino entangled after beta decay? If you measured the spin of the neutron before beta decay and it was $-1/2$ and you measure the spin of the electron afterwards and it is $+1/2$ then you would instantly know the spin of the proton and the anti neutrino. But if the electron's spin is $-1/2$ you wouldn't know the spin of the proton and the anti neutrino. So are they entangled or not?

Harish Chandra Rajpoot
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user142370
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    Due to conservation in momentum and energy there should also be entanglement in the spatiotemporal degrees of freedom – flippiefanus Jun 06 '20 at 03:04
  • Unentangled states have positive codimension in the state space, so absent a good reason to believe otherwise, you should expect entanglement. – WillO Jul 07 '20 at 20:23

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The entanglement is with spin and it a tripartite entanglement. Suppose we have a neutron with the spin state $$ |\psi\rangle~=~\frac{1}{\sqrt{2}}|n\rangle\left(|+\rangle~+~|-\rangle\right). $$ The $|\pm\rangle$ states correspond to the spin up and down. The neutron state decays into anti-neutrino and and proton states. The state is then $$ |\psi\rangle~\rightarrow~\frac{1}{2}|p\rangle|\bar\nu\rangle|e\rangle\Big[|+\rangle_p(|+\rangle_e|-\rangle_\bar\nu~+~|-\rangle_e|+\rangle_\bar\nu)~+~|-\rangle_p(|+\rangle_e|-\rangle_\bar\nu~+~|-\rangle_e|+\rangle_\bar\nu)\Big] $$ $$ =~\frac{1}{2}|p\rangle|\bar\nu\rangle|e\rangle\Big[\big(|+\rangle_p|+\rangle_e|-\rangle_\bar\nu~+~|+\rangle_p|-\rangle_e|+\rangle_\bar\nu\big)~+~\big(|-\rangle_p|+\rangle_e|-\rangle_\bar\nu~+~|-\rangle_p|-\rangle_e|+\rangle_\bar\nu\big)\Big]. $$ This is a superposition of two entangled states. Here we have that the particle states $|p\rangle$, $|e\rangle$ and $|\bar\nu\rangle$ correspond to other quantum numbers, such as baryon, lepton and antilepton numbers and these do not enter into the entanglement.

CDCM
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Lawrence B. Crowell
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  • It is not entirely clear from the way you write it, but it seems that you say that a $|+\rangle_n$ always gives rise to a $|+\rangle_p$. For sure, in the state you write the proton is unentangled with the electron and neutrino. Is there a reason for that? – Norbert Schuch Sep 21 '17 at 13:36
  • Why did you not choose a spin eigenstate as the initial wave function? – my2cts Feb 01 '20 at 09:41