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In the orthodox interpretation of quantum mechanics, the following three assumptions are made (please correct me if I am wrong):

  1. Every physical system is completely specified by a state $\lvert\psi\rangle \in H$
  2. The state of two systems is specified by a point in the tensor product space $H_{1} \otimes H_{2}$
  3. An isolated system evolves linearly in time: $\lvert\psi\rangle_{t} = \hat{U}\lvert\psi\rangle$

Now, it seems easy to show that these lead to a contradiction, and I can't find the flaw in the following argument, hence the question: where exactly is the error?

Suppose I measure the property of a system described by $\lvert\Psi\rangle$, by means of an interaction with an apparatus $\lvert\Phi\rangle$. Because of axiom 1) these states exist and completely specify the two systems at some initial time $t_{0}$.

By applying axiom 2), we can define the global system as a point in the tensor product space, now I assume the following:

  1. The global system at $t_{0}$ is separable: $\lvert\Phi\rangle \otimes \lvert\Psi\rangle$ and isolated
  2. The apparatus is $\textbf{faithful}$, meaning that there exists a basis for $\lvert\Psi\rangle = \sum_{i}c_{i}\lvert\psi_{i}\rangle$ and a state $\lvert\Phi_{R}\rangle$ for the apparatus such that $\hat{U}(\lvert\psi_{i}\rangle \otimes \lvert\Phi_{R}\rangle) = \lvert\tilde\psi_{i}\rangle \otimes \lvert\Phi_{i}\rangle$, where $\hat{U}$ is linear and the two sets of vectors $\lvert\Phi_{i}\rangle$, $\lvert\tilde\psi_{i}\rangle$ are both linearly independent.

If the apparatus was not faithful, two definite basis states $\lvert\psi_{i}\rangle$ and $\lvert\psi_{k}\rangle$ would get mapped to the same apparatus state $\lvert\Phi_{O}\rangle$, and there would not be any noticeable experimental difference between measuring two different spin-z states $\lvert1\rangle$ and $\lvert0\rangle$ for example.

Similarly, if the global system was not isolated, then by axiom 1) there exists a state $\lvert E\rangle$ for which the $system + apparatus + E$ state is isolated, because of thermodynamics. The following reasoning also applies in this case, just taking a larger system.

Now suppose we perform an experiment where we don't measure a definite state like $\lvert\psi_{k}\rangle$, but a general state in the superposition $\lvert\Psi\rangle = \sum_{i}c_{i}\lvert\psi_{i}\rangle$, then by linearity we have the following evolution for the experiment:

$\hat{U}(\sum_{i}c_{i}\lvert\psi_{i}\rangle \otimes \lvert\Phi_{R}\rangle) = \sum_{i}c_{i}(\lvert\tilde\psi_{i}\rangle \otimes \lvert\Phi_{i}\rangle)$

By axiom 1), this final state uniquely describes a composite system, however this is the wrong one, because at the end of the experiment the system is not in a superposition: the correct final state we observe is instead $\lvert\tilde\psi_{i}\rangle \otimes \lvert\Phi_{i}\rangle$ (with some probability), but then state uniqueness doesn't hold anymore.

Does this mean that axioms 1-3 taken together are inconsistent? Is there a flaw in this reasoning?

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    $\begingroup$ As the question title suggests, this seems to be a re-statement of the measurement problem (en.wikipedia.org/wiki/Measurement_problem). $\endgroup$
    – gandalf61
    Feb 7 at 12:00
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    $\begingroup$ It is worth noting that the final state in your last paragraph is not the state $|\Psi\rangle\otimes|\Omega\rangle$, but an entangled state which cannot be factored into a simple tensor product of a system and apparatus state $\endgroup$ Feb 7 at 12:34
  • $\begingroup$ Still, at the end of the experiment where $\Psi$ is a superposition, you start from a separable state, and then get the wrong state of the apparatus if you apply linear evolution. You know that the state is wrong because of axiom 1), where you assume state uniqueness, it seems like you can't hold on both. $\endgroup$
    – Davyz2
    Feb 7 at 14:08

1 Answer 1

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There is no inherent contradiction in the statements you line up, but it appears to be a wrong description of empirical reality. The missing piece is the Born rule, which tells you how to relate quantum states to measurements.

With the Born rule there is indeed a contradiction, since you have stated that all physical systems are completely described by quantum states which evolve unitarily, while the Born rule specifies a special class of physical processes, measurements, where the interaction of two physical systems leads to a non-unitary evolution generally known as the wave function collapse.

This seemingly ad hoc addition of measurement as a special physical process gives us an extremely successful theory for predicting empirical reality, but is considered unsatisfying to many.

Note that in the many-worlds interpretation, the result of the measurement is actually $|\Omega\rangle$, i.e. the measurement yields all outcomes at once, but for some reason your empirical perception chooses a random one according to the Born rule. This school of thought shifts the inconsistency such that the measurement apparatus is not the "unphysical" non-unitary system, but rather your perception is. Whether this is more or less satisfying is up to the individual, with the predictions of the theory remaining unchanged.

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    $\begingroup$ Excellent answer, particularly the final paragraph. One important note: Gleason's theorem explains how Born's rule arises. There aren't any known nice proofs of Gleason's theorem, and it's not very intuitive, so leaving it out of the body of the answer was the right decision. To sharpen the third paragraph, measurement is ad hoc from the perspective of axioms for Hilbert spaces, but if we add it, then we must add it in a way that admits Born's rule. $\endgroup$
    – Corbin
    Feb 7 at 20:21

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