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Why is $\textbf{J}$ called angular momentum operator? Can anyone explain why the expectation value of J is angular momentum?

Here is how $J$ is defined: The rotation operator $$ U(\alpha)=\exp(-i {\bf \alpha} \cdot \mathbf{J}) $$ Rotates the system around the direction of $\alpha$, through an angle $|\alpha |$.

Sorry that this is not quite research level, but I cannot find some intuitive explanation about this elsewhere.

  • This question would be more suitable for Physics SE, and Wikipedia already has an answer:"The most general and fundamental definition of angular momentum is as the generator of rotations". It is essentially the same in classical mechanics, just the generation is described in more elementary terms with cross product or quaternions. – Conifold Jan 29 '18 at 00:09
  • As pointed out you should ask the question in physics SE, but it is because you can check that it obeys $[J_k, J_p]=\varepsilon_{kpi} J_i$. This property is also obeyed by the "orbital" angular momentum defined by $\mathbf{L}=\mathbf{\hat{x}}\times\mathbf{\hat{p}}$ in analogy with classical mechanics, and it turns out that commutation relation, loosely speaking, is the most general way to define something that behaves like an angular momentum – user2723984 Feb 15 '18 at 01:52

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