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Schlosshauer (in Decoherence and the Quantum-to-Classical Transition) defines ideal quantum measurement as a von-Neumann measurement in which

(1) the apparatus states correspond 1-to-1 to given system states (so in the measurement scheme we would have $(\sum |s_i \rangle)|a_r \rangle \to \sum |s_i \rangle|a_i \rangle$ where $|a_r \rangle$ is the ready state of the apparatus and $s$ denotes the system of interest.

(2) the measurement process does not change the system states.

My question is with respect to (1). Do we need also to require that the apparatus states are orthogonal (else we don't truly have 1-to-1 correspondence)?

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Yes orthogonality is required here, but it's basically a given because with the number of degrees of freedom in a macroscopic state, the probability of two macroscopic states having significant overlap is essentially zero.

Another way of arguing orthogonality here is that you can physically see with your eyes that the macroscopic states after a measurement have particles in different positions, i.e. if you imagine a macroscopic pointer it will be pointing in a different place for different outcomes. The overlap of the wave functions of those different pointer positions is zero since they are localized to where you can see the pointer actually is. That's just one element of it though, in reality not just the pointer itself is different but many other particles along the way that helped to correlate the measured particle with the pointer. These could in principle also be measured and would have to be significantly different by design (otherwise you couldn't tell the difference between different measurement outcomes).

Still it's good to keep in mind when considering these things that orthogonality must be part of the picture. So it is good you noticed.

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  • $\begingroup$ Thank you for your answer; it certainly helps in elucidating what's physically going in terms of why orthogonality is demanded. I want to confirm though -- when people in the measurement field use the term "ideal", definitionally do they also mean that the apparatus states are orthogonal? $\endgroup$
    – EE18
    Commented Mar 12, 2023 at 0:55
  • $\begingroup$ Yes. I think many physicists think they are orthogonal anyway, others might say they are so close to orthogonal that you couldn't tell the difference, but if ideal measurements are being talked about it's safe to say the apparatus states should be treated as orthogonal. $\endgroup$
    – don't train ai on me
    Commented Mar 12, 2023 at 1:10
  • $\begingroup$ Thank you; I have accepted your answer. Given that you seem to be quite expert in this field of measurement and if you get the chance, can you please take a look at this (physics.stackexchange.com/questions/686213/…) question? I had an identical question and tried to supply an answer but am not certain. $\endgroup$
    – EE18
    Commented Mar 12, 2023 at 1:20
  • $\begingroup$ @EE18 Sure thing, I've written a bit more to supplement your answer $\endgroup$
    – don't train ai on me
    Commented Mar 12, 2023 at 4:29

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