testing

Maybe we can sense that from rotation operator. Since $latex e^{iL_{z}\varphi/\hbar} $ can rotate a function with angle $latex \varphi $ and rotate $latex 2\pi $ should remain original value, $latex e^{2\pi i L_{z}/\hbar}\,=\, e^{m2\pi i}\,=\, 1 $ if $latex L_{z} \Psi=m\hbar\Psi$