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Neglecting any effect of spin,it is a well known fact,that orbital angular momentum of a system in ground state is zero.(For potential $V=V(r)$ .)

The angle-angular momentum uncertainty,gives $\Delta{L}\Delta{\phi}\ge\hbar/2$ , where $\Delta$L=$\sqrt{<L^2>-<L>^2}$.Now,<$L^2$>=$l(l+1)=0$(Since,$l=0$ for ground state).Therefore the uncertainty relation to be valid, expectation of Angular momentum <$L$>should be non-zero,and...non-real too!Why is this happening?

Qmechanic
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Manas Dogra
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    Intuitively, the distribution of the ground state is perfectly symmetric, so it makes sense to put "$\Delta \phi = \infty$". Mathematically, $\Delta \phi$ is not well-defined in the first place; there is no self-adjoint angle operator. This question has been answered here: https://physics.stackexchange.com/questions/338044/why-doesnt-the-uncertainty-principle-contradict-the-existence-of-definite-angul – Noiralef Apr 06 '20 at 09:08

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