# How is the Born rule consistent with unitary evolution?

Consider a system $|\Psi_T \rangle_{t = 0} = |\Psi_E \rangle \otimes |\Psi_S \rangle$ where $|\Psi_S \rangle$ is a system that collapses into an eigenstate upon measurement. $|\Psi_E \rangle$ is the system that performs the measurement.

So at time $t = 0$, we have $|\Psi_T \rangle_{t = 0} = |\Psi_E \rangle \otimes |\Psi_S \rangle$. At some time $t'$ after measurement on $|\Psi_S \rangle$, we have

$$|\Psi_T \rangle_{t = t'} = e^{-iHt} \left( |\Psi_E \rangle \otimes |\Psi_S \rangle \right)$$ $|\Psi_S \rangle$ has "turned into" eigenstate $|\Psi_{S_\lambda} \rangle$ via the Born rule. I'm not exactly sure what this entails for the overall state though. Perhaps something like $|\Psi_T \rangle_{t = t'} = |\Psi_{E'} \rangle \otimes |\Psi_{S_\lambda} \rangle$ for some unknown environment state $|\Psi_{E'} \rangle$?

However, this makes no sense to me, because at this point in time ($t = t')$ there is no guarantee that $|\Psi_T \rangle_{t = t'}$ is still separable!

How is this inconsistency resolved?

Assume the Born rule is exact — i.e., there is a discontinuity in the evolution of $|\Psi_S \rangle$ whereupon it instantly becomes $|\Psi_{S_\lambda} \rangle$.

• Isn't this basically just another formulation of the measurement problem? Dec 15 '15 at 20:17
• Some remarks: 1) Born rule only gives probability of results of measurement. It does not imply that there is a collapse - that is merely one possible interpretation. 2) In your expressions, you assume $H$ is for the whole system $E+S$. Evolution of tensor product from E,S will in general lead to a state that is not a tensor product from E,S. You only can talk about state of $E+S$, E or S do not have individual states. Nov 11 '18 at 13:43