Consider an entangled pair described by the wavefunction $$\lvert1,0\rangle = \frac{1}{\sqrt{2}}(\lvert\uparrow_1\downarrow_2\rangle-\lvert\downarrow_1\uparrow_2\rangle)$$ in in the $S_z$-basis. If the first measurement finds Alice's particle to be in the $\lvert\uparrow\rangle$ state, then Bob's particle is found to be in the $\lvert\downarrow\rangle$ state i.e. the entangled state above collapses to $\lvert\uparrow_1\downarrow_2\rangle$ which is not entangled. Since it is not entangled in the $S_z$ basis, it will also be unentangled in the $S_x$ basis.

Am I correct to conclude from this that an entangled pair becomes unentangled after the first measurement? Here, I am assuming that making a measurement of $S_z$ by Alice or Bob doesn't affect the space part of the wavefunction, which I am not sure though.

  • 2
    $\begingroup$ A comment about your last line - Two particles can be doubly entangled i.e. entangled in spin and some other property like position. In such a case, measuring the spin will not destroy the entanglement in the other property. $\endgroup$
    – rnva
    Commented Oct 28, 2020 at 12:28

1 Answer 1


It depends on the kind of measurement.

If you do a projective measurement with respect to a basis made of separable states, then yes, every post-measurement result will be separable. In your example, you are essentially considering a measurement wrt the basis $\{|s,s'\rangle : \, s,s'\in\{\uparrow,\downarrow\}\}$ (that is, the basis of eigenstates of $S_z\otimes S_z$), therefore the measurement breaks the entanglement.

If you were to instead measure wrt a Bell basis, then the post-measurement results would also be entangled.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.