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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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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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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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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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