2
$\begingroup$

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 $|\Psi_{S_\lambda} \rangle$.

Improve this question
$\endgroup$
2
  • 1
    $\begingroup$ Isn't this basically just another formulation of the measurement problem? $\endgroup$
    – Daniel
    Commented Dec 15, 2015 at 20:17
  • $\begingroup$ 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. $\endgroup$
    – Ján Lalinský
    Commented Nov 11, 2018 at 13:43

1 Answer 1

Reset to default
2
$\begingroup$

Question in the title seems to be different from the question in the body. Born rule is not consistent with unitary evolution because collapse is not supposed to be a part of unitary evolution, it is an orthogonal projection onto an eigenspace (instantaneous one under the OP stipulation), not a unitary transformation.

There is indeed no guarantee that after the collapse the state is still factorizable, the system can become entangled with the measuring apparatus, and strictly speaking it does. However, the measuring apparatus is usually "classical", that is it involves miriads of quantum objects whose interactions quickly destroy any entanglement for all practical purposes (albeit not quite instantaneously). In fact, measuring apparatus is specifically designed to produce approximately factorizable states, or it wouldn't be a very good measuring apparatus.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Doesn't quite answer the OP concern: what the maths shows is an entanglement between system and apparatus which ends up with a density matrix which is for all practical purposes the same as a mixed state density matrix. Each state in this mixture is a factorizable state. Then you have to make some meta-argument (which depends on your interpretation) why it as acceptable to forget all the states in the mixture except the one you observed. This is the measurement problem. $\endgroup$
    – isometry
    Commented Oct 30, 2016 at 9:27

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.