Timeline for Can quantum information conservation verified in this specific situation?
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| Sep 2, 2022 at 13:53 | comment | added | don't train ai on me | @Davius Another argument that shows that the projections cannot be modeled as a unitary operation: Every unitary transformation has an inverse, because by definition a unitary operator is invertible with $U^{-1} = U^{\dagger}$. With any invertible operation, given the final state you can find the initial state. But, given the outcome "spin up", there are many different initial states that you could have started with. So the evolution into the spin up state cannot be described unitarily. | |
| Aug 25, 2022 at 22:05 | comment | added | don't train ai on me | So, there is no possibility for $|\chi_0 \rangle \neq |\bar{\chi}_0 \rangle$, because we just define them to start with the detector in the same state. | |
| Aug 25, 2022 at 22:04 | comment | added | don't train ai on me | Also, looking back at the definition of $|\chi_0 \rangle$ vs $\bar{\chi}_0 \rangle$, we only need to say that $|\chi_0 \rangle$ is one possible initial state of the detector. Then, we could consider a system measuring states other than $|\Psi_1 \rangle$, which start with the detector in the same state $|\chi_0 \rangle$. Physically, we have detectors which are capable of detecting many different results... e.g. the same stern-gerlach apparatus can detect either spin up or spin down. So, by choosing one reference starting state for the detector, the other ones are just defined to be the same. | |
| Aug 25, 2022 at 22:00 | comment | added | don't train ai on me | How do you define $|\chi_1 \rangle$ vs $|\bar{\chi}_1\rangle$? | |
| Aug 25, 2022 at 21:38 | comment | added | Davius | I am not sure, assume that initially we have $|\Psi_1\rangle \otimes |\chi_0\rangle$, maybe, then, the unitary evolution leads to $|\Psi_1\rangle \otimes |\chi_0\rangle \to |\Psi_1\rangle \otimes |\bar{\chi}_1\rangle$ with $|\bar{\chi}_1\rangle \neq |\chi_1\rangle$. The problem is that we do not know the details of an hypothetical unitary evolution (under the assumption of an interpretation without "objective collapse" of wave function, which underlie "quantum information paradox": en.wikipedia.org/w/…). | |
| Aug 25, 2022 at 16:01 | comment | added | don't train ai on me | Because starting from the initial pre-measurement state of the detector $|\chi_0 \rangle$, we could choose to measure either a particle in superposition or not. The state of the detector can be chosen independently of the state of the particle since before the measurement happens, they are just completely separate systems. | |
| Aug 25, 2022 at 15:57 | comment | added | Davius | But what if the state $|\bar{\chi}_0\rangle$ in the evolution $|\Psi_1\rangle\otimes|\bar{\chi}_0\rangle \to |\Psi_1\rangle\otimes|\chi_1\rangle$, and the state $|\chi_0\rangle$ in the evolution $(\alpha|\Psi_1\rangle+\beta\Psi_2)\otimes|\chi_0\rangle \to |\Psi_1\rangle\otimes|\chi_1\rangle$ are not equal? I am not sure why you are not assuming that $|\bar{\chi}_0\rangle \neq |\chi_0\rangle$? | |
| Aug 25, 2022 at 15:46 | history | answered | don't train ai on me | CC BY-SA 4.0 |