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Is possible a non-deterministic propagation of the wave function in the QM?

Qmechanic
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peterh
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    Yes, in the measurement process. – Trimok Aug 08 '14 at 09:39
  • Comment to the question (v1). Given an initial wave function $\psi(x,0)$, the Schr. eq. determines uniquely the wave function $\psi(x,t)$ itself in the future (under mild regularity assumptions). Are you thinking about bifurcation/non-uniqueness of solutions to Schr. eq., etc? The non-deterministic aspect of QM usually only enters via the measurement process, as Trimok and John Rennie write. – Qmechanic Aug 08 '14 at 11:31
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    You might be interested in this, this and this question – ACuriousMind Aug 08 '14 at 13:15

2 Answers2

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In the Copenhagen interpretation the collapse of the wavefunction is non-deterministic. Following the collapse the wavefunction evolves in a deterministic manner until the next collapse.

However the Copenhagen interpretation gives no account of what the collapse process is and by what mechanism it occurs, and this has led to dissatisfaction with it in some quarters. There are other interpretations, such as Many Worlds, that avoid the collapse. In the Many Worlds interpretation the evolution of the wavefunction is completely deterministic.

John Rennie
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Schrödinger equation is deterministic , so if a system is in a state A it will evolve into a unique and perfectly determined state B.

Some people say that a measurement is not deterministic, but imagine we consider the state of the system we want to measure plus the measurement apparatus, or even the whole universe : the universe wavefunction at time 0 is $\psi(0)$, and at time t the wavefunction $\psi(t)$ is deterministically given by Schrödinger equation.

So either the measurement is not deterministic and Quantum theory is not consistent with itself, since it gives different results according to the way we choose the system.

Or the wavefunction collapse comes from the Schrödinger equation and results from the interaction of the system we want to measure with another system (the measurement apparatus) which contains much more degrees of freedom.

I think precise measurement on the way the wavefunction collapse happens favours the last solution (I'll try to find some papers if I have the time)

agemO
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