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I am asking this question to all the Quantum Computing or Quantum Mechanics practitioners.

I have studied that when you measure a state, then you will get an eigenvalue for sure corresponding to whatever measurement operator you have. So suppose if I toss a coin, and upon measurement get a head, then is the value of head be the eigenvalue corresponding to the measurement operator?

If so, then what is the exact measurement operator? Measure coin state?

Sorry if this question feels dumb to you, I am just starting to study the subject and I am reading the quantum postulates. I was thinking and got stuck here.

Thank you for your help.

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Yes, one of the eigenvalues represents the outcome "heads".

To model a measurement in the operator formalism, you must assign distinct real numbers to the possible outcomes. It's a requirement of the operator formalism, not of quantum mechanics itself.

If the measurement outcomes don't naturally map to real numbers, you can assign whatever numbers you want. If there are just two outcomes, then $0$ and $1$ is often a good choice. If you choose other numbers, you'll get a numerically different measurement operator, but it represents the same physical measurement.

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  • $\begingroup$ Thank you @benrg, got your point, and understood it. $\endgroup$
    – Deb Prakash Chatterjee
    May 1 at 7:15

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