# How might I show that an operator is, by definition, an 'observable'? [closed]

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?

• You show this with experiments, just like you show with experiments that a force does what Newton said it does. – CuriousOne May 27 '15 at 16:12