It's true that in general, A and J do not commute. In particular, all we assumed is that A is a vector operator. This means, the following equation must hold: (1) [ A_i , J_j ] = i epsilon_ijk A_k If the components of A fulfill eq. (1), then A is a vector operator. Now, the term we are interested in is A_i J_i. However when you set the indices i and j equal in eq. (1), then: (2) [ A_i , J_i ] = i epsilon_iik A_k = 0 (because two indices are the same) This means that A_i J_i = J_i A_i. If A has some additional commutator relations with J (independent of it being a vector operator), then things might get trickier.
How can you make such topic sound so easy? You are a legend!
Thanks for the kind words! :)
Thank you for putting the time into making extremely helpful physics videos! They are extremely interesting. Thank you!
What a nice comment, thank you very much for your feedback! :)
*Loved it. Thanks for this*
Thanks Quahntasy!
In Sakurai, I found the theorem with the scalar product of A and J the other way around. But they don't commute?!
It's true that in general, A and J do not commute. In particular, all we assumed is that A is a vector operator. This means, the following equation must hold:
(1) [ A_i , J_j ] = i epsilon_ijk A_k
If the components of A fulfill eq. (1), then A is a vector operator. Now, the term we are interested in is A_i J_i. However when you set the indices i and j equal in eq. (1), then:
(2) [ A_i , J_i ] = i epsilon_iik A_k = 0 (because two indices are the same)
This means that A_i J_i = J_i A_i. If A has some additional commutator relations with J (independent of it being a vector operator), then things might get trickier.
Well you sure have a very affective community needless to say
🙈
Btw what's wrong with the comments lol? All of them seem to be in Love with you LOL
I guess that has to be the power of the projection theorem 🥰
@@PrettyMuchPhysics lol