- #1
cooev769
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So I understand the commutation laws etc, but one thing I can't get my head around is the fact that L^2 commutes with Lx,y,z but L does not.
I mean if you found L^2 couldn't you just take the square root of it and hence know the total angular momentum. It seems completely ridiculous that you could know the square of the total angular momentum is 100, but not know that the angular momentum is therefore 10. Either that, or the textbook explains it horrifically and you can know the total angular momentum squared which is a scalar and hence you can know L scalar but L generally has direction, so they use L^2 to be explicit that it's a scalar.
Please elaborate. Cheers.
I mean if you found L^2 couldn't you just take the square root of it and hence know the total angular momentum. It seems completely ridiculous that you could know the square of the total angular momentum is 100, but not know that the angular momentum is therefore 10. Either that, or the textbook explains it horrifically and you can know the total angular momentum squared which is a scalar and hence you can know L scalar but L generally has direction, so they use L^2 to be explicit that it's a scalar.
Please elaborate. Cheers.