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I was learning about Qubit measurement and the basics of Quantum computing. The instructor forced a lot on the following statement : Assume an isolated system S. Let an observer O interact with the system S to measure some property, state of Observer O will change as opposed to the state of S i.e. $|S\rangle (|O_1\rangle+|O_2\rangle) $

Almost every other resource mentions the opposite, i.e. upon observation, state of the system is changed or Wavefunction of the system is collapsed when observed. (At first, I thought they are equivalent just like different observers in moving frames as discussed in relativity. But I still wanted to confirm)

Q1: Are these statements equivalent or do they have different meanings (shouldn't the above equation be $ |O\rangle (|S_1\rangle+|S_2\rangle) $ )? And if they do have different meanings, why did the instructor use the former?

Later, the instructor attempted to connect the given statement with the Many-Worlds interpretation which raised another question in a different context.

Q2: If Many-worlds interpretation is true, why do we always observe only the outcomes with the highest probability? According to Many-worlds, there exists some universe where results with lower probability would have been observed. Why is it always some different universe and not ours? (Is it again because the probability of observing those results is very, very small and the number of parallel universes is infinite? Does this question has some physics-related answer at all?)

Any insights will be helpful. Thanks in advance.

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  • $\begingroup$ "Observer" is a classical object. $\endgroup$
    – Roger V.
    Commented May 14, 2021 at 7:13
  • $\begingroup$ So that means both the statements are not equal? $\endgroup$
    – Samarth
    Commented May 14, 2021 at 7:30
  • $\begingroup$ The two statements seem to me somewhat incomplete as formulated. The state of observer changes, and the state of system also changes. If the state of observer didn't change we would measure nothing. But observer is classical - it is not described by a wave function. $\endgroup$
    – Roger V.
    Commented May 14, 2021 at 7:34
  • $\begingroup$ I'm sorry if that's so. He just said that and went on. Anyways, saying that change in the state of only one entity (observer or system) would be wrong? That does sound more logical. Thanks! $\endgroup$
    – Samarth
    Commented May 14, 2021 at 7:38
  • $\begingroup$ @ChiralAnomaly Observer is postulated to be a classical object, which destroyes the superposition (causes "wave function collapse"). Trying to use a theory self-consistently to describe its postulate is circular logic. But yes, once we believe in QM, we say that observer is a macroscopic, thermodynamic, etc. $\endgroup$
    – Roger V.
    Commented May 14, 2021 at 14:59

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According to the MWI we can also observe outcomes that are less probable. They just are not seen as much as the most probable outcome. I think your teacher has read much of Bohr, who says that measurements alter or knowledge of physical systems. Which means the state of the observer is changed after measurement. This is of course the case but the physical system on which a measurement is made has changed too after the measurement.

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    $\begingroup$ If I understand correctly, what you are saying is that measurement is a process like : $$|O\rangle (|S_1\rangle + |S_2\rangle) \longrightarrow |O_1\rangle |S_1\rangle + |O_2\rangle |S_2\rangle$$ $\endgroup$
    – SolubleFish
    Commented May 14, 2021 at 7:58
  • $\begingroup$ @SolubleFish There is only one state for the observer. Like you write it seems as there are two observer states. $\endgroup$
    – Deschele Schilder
    Commented May 14, 2021 at 8:03
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    $\begingroup$ After measurement, the measurement apparatus can be in two state depending on which outcome was observed, no ? $\endgroup$
    – SolubleFish
    Commented May 14, 2021 at 8:05
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    $\begingroup$ But the state of the observer also changes, right? (by any chance, can you suggest some good reference material? That'll be really helpful) $\endgroup$
    – Samarth
    Commented May 14, 2021 at 13:12
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    $\begingroup$ Because the observer made an observation and his state should be altered as well? $\endgroup$
    – Samarth
    Commented May 14, 2021 at 13:15

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