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Solution $S^2\alpha = \left(S_x^2 + S_y^2 +S_z^2 \right)\alpha= S_x(S_x(\alpha)) + S_y(S_y(\alpha)) +S_z(S_z(\alpha))$ from the ''rules", where for instance $S_x$ operating on $\alpha$ gives $\frac{1}{2}\hbar\beta$, we make the three replacements to get $=S_x(\frac{1}{2}\hbar\beta) +S_y(\frac{1}{2}i\hbar\beta) +S_z(\frac{1}{2}\hbar\alpha)$ Now take ...

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One possible way: As you say $${\bf{S}^2} = S_x^2+S_y^2+S_z^2$$ Thus \begin{align} {\bf{S}^2} &= \left( \frac{\hbar}{2}\right)^2 \left[\begin{pmatrix} 0 & 1 \\ 1 & 0 \\ \end{pmatrix}\begin{pmatrix} 0 & 1 \\ 1 & 0 \\ \end{pmatrix} + \begin{pmatrix} 0 & -i \\ i & 0 \\ \end{pmatrix} \begin{pmatrix} 0 & -i \\ i & 0 \\...

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This a problem of radiation chemistry. It has been thoroughly studied in the years 1950 - 1960 - 1970 with gamma rays. When intense source of gamma rays strikes water, electrons are extracted from the water molecule. And some of these electrons are coming from the valence bond O-H. The bond is then weakened, and may even be broken. Usually the electron soon ...

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Actually, this question is not as bad as the comments may make you think. It's actually quite a sensible question, because when we solve the usual, Schrödinger's equation-based, QM problem of energy states of hydrogen atom (or any other atom), we get eigenstates of the Hamiltonian we inserted into the equation. These solutions should thus be stable for ...

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$\newcommand{\Ket}[1]{\left|#1\right>}$ $\newcommand{\Bra}[1]{\left<#1\right|}$ $\newcommand{\BraKet}[2]{{\left<#1}\left|#2\right>}$ It's an older question, but one worth answering! For the uninitiated, second quantization is a bookkeeping technique that describes many-particle systems as excited states of a field, usually taken to be the ...

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