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4

When labeling states of the hydrogen atom, one doesn't refer to the z component of the angular momentum, but rather to the total angular momentum. The total angular momentum is positive, but, as you've stated, there are two states for $J=\frac{1}{2}$ with $L=0$, and those are $J_z=\pm\frac{1}{2}$ (Or some linear combination of them) As to why this is, ...

2

The instructor hasn't thrown out $r^4$. If you do the calculations properly you will get the desired result. What the instructor is telling is that while simplifying and taking commons out for $\frac{\text{dP(r)}}{\text{dr}}=0$ you get,$$(\text r^3)(6-\frac{\text r}{\text a_0})(f(\text r))=0$$which gives, $\text r=0$ and $6\text a_0$ from the first two ...

1

Firstly, you cannot simply 'neglect' $a_0$ (as the *.pdf actually shows). In some test books the following substitution is used: $\rho=\frac{r}{a_0}$, so the $a_0$ doesn't have to be 'carried around'. As your expression for $P(\rho)$ is made of three factors, each factor can be evaluated for extrema individually. $\rho=0$ is an obvious minimum from the ...

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