Timeline for Classifying irreducible subspaces for angular momentum according to symmetrization
Current License: CC BY-SA 4.0
5 events
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| Apr 12, 2023 at 14:28 | comment | added | Gattu Mytraya | I was thinking along the lines of something like this. Is it always true that only the symmetric sum can produce the highest allowed state? Also, can the hook length formula account for the "accidental" zeros of CG coefficients? | |
| Apr 12, 2023 at 2:57 | comment | added | JEB | like combing $l=1$ orbital angular momentum and $s=\frac 1 2$? I would make the $l=1$ from 2 $s=\frac 1 2$'s and proceed normally with 2 x 2 x 2 = 4 + 2 + 2, but keep only irreps that have the two spin half's symmetric, you'd be left with the 4 and one of the 2's, which makes sense: a vector boson plus a spinor has a $j=\frac 3 2$ and a $j=\frac 1 2$ total angular momentum eigenstate. I would write is "3 x 2 = 4 + 2", but I am not sure if you can get that from the hook length formula... | |
| Apr 11, 2023 at 17:21 | comment | added | Gattu Mytraya | How would you do this when taking the tensor product between states from different representations? | |
| May 4, 2021 at 15:48 | comment | added | Jagerber48 | Thank you very much for this great tutorial! I've seen in a few places the relationship between irreps of angular momentum spaces and young tableaux but I've never seen a clear explanation until this one! I'm still not totally clear on the Young symmetrizer. I understand how you constructed the Young symmetrizer $S$ for a particular Young Tableaux. Why, in this case, does it make sense to apply the symmetrizer to the state $|\uparrow \uparrow \downarrow \rangle$ and not some other state? | |
| May 4, 2021 at 5:17 | history | answered | JEB | CC BY-SA 4.0 |