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Aug 19, 2021 at 7:47 comment added ric.san I'm sorry, but... what does change of basis has to do with $n$?
Jul 18, 2021 at 23:04 comment added ZeroTheHero Yes but only $\ell$ - not $n$ - enters in angular momentum coupling, i.e. the change of basis does not depend on $n$.
Jul 18, 2021 at 20:11 comment added ric.san Ok. That's really strange though. The atomic states are given by three quantum numbers $n , l$ and $m$, and $l$ goes from $0$ to $n-1$...
Jul 18, 2021 at 19:49 comment added ZeroTheHero No. For fixed $j_1,j_2$ you have $\sum_{m_1m_2} \vert j_1m_1;j_2m_2\rangle \langle j_1m_1;j_2m_2\vert =\mathbb{I}$ and, for fixed $j_1m_1,j_2m_2$, you have $\sum_{j} \vert j m\rangle\langle j m \vert =\mathbb{I}$ where $m=m_1+m_2$.
Jul 18, 2021 at 18:16 comment added ric.san @ZeroTheHero does that mean I cannot say if the set $\displaystyle \sum_{j1_,j_2,m_1,m_2} |j_1,j_2,m_1,m_2⟩⟨j_1,j_2,m_1,m_2|$ is complete or not, even if it is for each fixed value of $j_1$ and $j_2$?
Jul 18, 2021 at 18:13 vote accept ric.san
Jul 18, 2021 at 0:00 comment added ZeroTheHero Your last equation is incorrect: there is no sum on $j_1,j_2$ since $\mathbb{I}=\sum_{jm}\vert jm\rangle\langle jm\vert$. In the context of this equation $j_1,j_2$ are fixed.
Jul 17, 2021 at 22:17 answer added Cosmas Zachos timeline score: 4
Jul 15, 2021 at 19:19 comment added Davide Morgante What do you know about group theory? You know the difference between reducible and irriducible representations? Do you know what a Casimir operator is?
Jul 15, 2021 at 18:22 history edited ric.san CC BY-SA 4.0
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Jul 15, 2021 at 17:44 history asked ric.san CC BY-SA 4.0