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I don't understand why the stated equation is not equal to the expectation value of $m^2$ calculation

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  • $\begingroup$ Are the matrices $M$ promised to have eigenvalues/eigenvectors? $\endgroup$
    – DaftWullie
    Nov 2, 2022 at 11:28
  • $\begingroup$ Yes, that is specified in the text. $\endgroup$ Nov 2, 2022 at 13:49
  • $\begingroup$ Are you sure? It's not in my older version of the book (and shouldn't be) $\endgroup$
    – DaftWullie
    Nov 2, 2022 at 14:01
  • $\begingroup$ Hopefully that explains my misunderstanding. Here is the text: Postulate 3: Quantum measurements are described by a collection {Mm} of measurement operators. These are operators acting on the state space of the system being measured. The index m refers to the measurement outcomes that may occur in the experiment. If the state of the quantum system is |ψ immediately before the measurement then the probability that result m occurs is given by ... and the equation follows $\endgroup$ Nov 2, 2022 at 14:06
  • $\begingroup$ I see that in the following text, they do a simple example where M^2=M, but that does not answer my question $\endgroup$ Nov 2, 2022 at 14:25

1 Answer 1

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You seem to have changed a collection of operators $\{M_m\}$ into a single operator $M=\sum_mmM_m$ (this is actually an observable).

The whole point of the collection is that you handle each of the cases $m$ separately - you do a measurement and your measurement apparatus gives you one value $m$ (corresponding to a specific $M_m$) and you calculate the consequences of that.

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  • $\begingroup$ I'm not sure where you got that sum from. I didn't sum anything. Neither does (2.92). Rather, I am treating the "collection of operators" as a collection with a single element. In that case, (2.92) appears to be in error. $\endgroup$ Nov 2, 2022 at 15:16
  • $\begingroup$ The sum is what I've had to introduce to recreate your calculation in the notation of N&C. Note that if you think you're doing the calculation with a single operator, that single operator is something that has to square to identity because it must satisfy the completeness relation. Then all the $m_i^2=1$. $\endgroup$
    – DaftWullie
    Nov 2, 2022 at 15:23
  • $\begingroup$ Take a look at Box 2.5 on page 91. About halfway down you see the statement PiPj=dijPi. So, as you said, they really are assuming that m^2=I. But nowhere in the text (that I can see prior to section 2.2.3) do they mention this important restriction. That's pretty messed up. $\endgroup$ Nov 2, 2022 at 15:37
  • $\begingroup$ It's stated in equation 2.94 as part of the postulate. $\endgroup$
    – DaftWullie
    Nov 3, 2022 at 7:22

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