So­lu­tion an­guc-b


Can you in­vert the state­ment about zero an­gu­lar mo­men­tum and say: if a par­ti­cle can be found at all an­gu­lar po­si­tions com­pared to the ori­gin with equal prob­a­bil­ity, it will have zero an­gu­lar mo­men­tum?


No. To be at zero an­gu­lar mo­men­tum, not just the prob­a­bil­ity $\vert\Psi\vert^2$, but $\Psi$ it­self must be in­de­pen­dent of the spher­i­cal co­or­di­nate an­gles $\theta$ and $\phi$. As an ar­bi­trary ex­am­ple, $\Psi$ $\vphantom0\raisebox{1.5pt}{$=$}$ $R(r)e^{{\rm i}\phi\sin\theta}$ would have a prob­a­bil­ity of find­ing the par­ti­cle in­de­pen­dent of $\theta$ and $\phi$, but not zero an­gu­lar mo­men­tum.