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The answer to the OP's question is that this is an order of magnitude estimation and the person doing the estimation used values that were known to be closer to the correct values to make the order of magnitude estimation come out closer to the true answer. The majority of my post shows that there is a simple choice for "r" and "p", for which you could say ...


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$\newcommand{\d}[1]{\,\mathrm{d}#1}\newcommand{\pdv}[2]{\frac{\partial #1}{\partial #2}}\newcommand{\p}{\psi_{100}}\newcommand{\pdvt}[2]{\frac{\partial^2 #1}{\partial #2^2}}$The hydrogen ground state is the following: $$\psi_{100}=Y_{00}\frac{2}{a_0^{3/2}}e^{-r/a_0}$$ The expectation value of the position operator on this state is the following: ...


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Angular momentum is a conserved quantity (in a closed system) and this is true also for the angular momentum that is carried by the electromagnetic (EM) field. This conservation is a manifestation of rotational symmetry and the azimuthal part of the EM field emitted must be single valued. In other words, when rotating the EM field in the azimuthal ($\phi$) ...


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L I Komarov and T S Romanova 1985 J. Phys. B: At. Mol. Phys. 18 859 The algebraic method of solution of the Dirac equation for a particle in a Coulomb potential Abstract:An equation is constructed in two-dimensional complex space, in the set of solutions of which solutions of the Dirac equation for a particle in a Coulomb potential are present. These ...


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The two questions are slightly different. Each individual measurement of $L^2$ or $L_z$ will return an eigenvalue. In this case, you have only one possible measurement for $L^2$ (corresponding to $l=1$), but you have two possible measurements for $L_z$; 2/3 of the time you'll get $m=1$, and 1/3 of the time you'll get $m=0$. The expectation value, on the ...


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The 'quantum' angular momentum, in the classical limit, does reduce to the 'classical' angular momentum. In this sense, they are the same thing. Where the classical angular momentum occurs in the classical Hamiltonian, we replace that with the quantum angular momentum in the quantum Hamiltonian. Of course, 'QM being QM', brings with it some new aspects. In ...


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It is a misconception to say that In QM [...] we know that the the electron does not radiate EM-Waves because it is not actually circling around the nucleus. It is sometimes here and there. In QM the notion of "circling round the nucleus" does indeed fail to make sense, but this is not why electrons don't radiate EM waves. Instead, an atom in its ...



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