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The theory is the equations. Take the Schrödinger equation. It is often misunderstood in discussions of time-symmetry and determinism. The Schrödinger equation

by mablap 10y ago

The theory is the equations. Take the Schrödinger equation. It is often misunderstood in discussions of time-symmetry and determinism. The Schrödinger equation describes the evolution of a wavefunction, deterministically, and symmetrically in time. The equations are "indeterministic" in another sense, that of Heisenberg: can't know position and momentum with infinite accuracy. But that's a subtly different point. The result of a measurement is not dealt with by the equation. Schrödinger only lets you calculate the probabilities of measurement outcomes, but not which one it will be. That's why Einstein proposed we just didn't knew the whole picture. He thought we needed more information.

marcosdumay 10y ago
The theory is whatever tells you what you will observe in the real world. On the case of QM, that's the set of equations, observable operators, and the instructions of hoe to interpret them. None of those predict anything alone.
- mablap 10y ago
  The Schrödinger equation alone predicts how a state evolves in time. Just not what the result of a measurement will be.
  - marcosdumay 10y ago
    And that state is a completely made up mathematical artifact. The only thing real on QM are measurements. Besides, the Schrodinger equation does not predict how the state will evolve after a measurement is done. That's the entire problem.
    - mablap 10y ago
      And that state is a completely made up mathematical artifact. The wavefunction is not measurable, but its absolute value squared is. This does not mean it's not "real". I understand what you mean, but you need to provide us with another better answer that supports all observations before saying it's "completely made up". And yes, the Schrödinger equation predicts exactly how a state evolves, even after a measurement. It does not tell you what the outcome of a measurement is - you give it the outcome, and it gives you the time evolution. There is a problem, and it lies in what happens during measurement - or what is the exact nature of measurement. But given any state, the SE tells you how it evolves.

by mablap 10y ago

marcosdumay 10y ago

The theory is whatever tells you what you will observe in the real world. On the case of QM, that's the set of equations, observable operators, and the instructions of hoe to interpret them. None of those predict anything alone.

mablap 10y ago
The Schrödinger equation alone predicts how a state evolves in time. Just not what the result of a measurement will be.
- marcosdumay 10y ago
  And that state is a completely made up mathematical artifact. The only thing real on QM are measurements. Besides, the Schrodinger equation does not predict how the state will evolve after a measurement is done. That's the entire problem.
  - mablap 10y ago
    And that state is a completely made up mathematical artifact. The wavefunction is not measurable, but its absolute value squared is. This does not mean it's not "real". I understand what you mean, but you need to provide us with another better answer that supports all observations before saying it's "completely made up". And yes, the Schrödinger equation predicts exactly how a state evolves, even after a measurement. It does not tell you what the outcome of a measurement is - you give it the outcome, and it gives you the time evolution. There is a problem, and it lies in what happens during measurement - or what is the exact nature of measurement. But given any state, the SE tells you how it evolves.