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I am currently working on the angular momentum part (chapter 3) of Sakurai's QM 2nd. From eq. 3.159 and eq. 3.154, the book says $[J^2, J_+] = 0$ and $[J^2, J_z] = 0$, but I don't understand why $| l, m \rangle $ is not an eigenstate of $J_+$ (eq. 3.164)? Doesn't the commutation relation tell us $J^2$ and $J_+$ share the same eigenstates, so the eigenstates of $J^2$ are also the eigenstates of $J_+$?

Mathematically, if we apply two states $ | l, m \rangle $ and $ | l', m' \rangle $, $$ \langle l', m'|[J^2, J_+]| l, m \rangle = \hbar ^2(l'(l'+1) - l(l+1))\langle l', m'|J_+ | l, m \rangle = 0$$ Doesn't this mean $ | l, m \rangle $ is an eigenstate of $J_{+}$?

CWL
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1 Answers1

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It all comes down to the theorem that states that when two (finite-dimensional) linear functions commute and are both diagonalizable then they are simultaneously diagonalizable. The operators $J^2$ and $J_\pm$ do indeed commute, as you correctly stated, but they do not share a basis of eigenvectors because the $J_\pm$ do not have such a basis, as they are not diagonalizable in the first place.

nico617
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