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Here is my problem:

I understand what is meant by 'observable' but don't have a formal definition at hand. How do I 'show' it?

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closed as off-topic by Emilio Pisanty, BMS, Qmechanic May 27 '15 at 16:33

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  • $\begingroup$ You show this with experiments, just like you show with experiments that a force does what Newton said it does. $\endgroup$ – CuriousOne May 27 '15 at 16:12
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In quantum mechanics, an observable is basically an hermitian operator. You can see a definition of it in chapter 4 of Le Bellac's Quantum Physics.

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  • $\begingroup$ Thank you for the reply. So to show that an operator is an observable I just show that it is self-adjoint? I thought this was obvious, and his wording in the question almost implies that it needs more work than that. I'm sure you're right, though. Thanks, again. $\endgroup$ – Ben Gerry May 27 '15 at 15:06
  • $\begingroup$ Just to be clear: in real life just because an operator is self adjoint does not mean you can actually observe it in practice. The real criterion in that case is whether or not you can construct a system where that operator couples to a measurement apparatus strongly enough. $\endgroup$ – DanielSank May 27 '15 at 16:34

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