I'm assuming it is a jth state with m value as total angular momentum?
$\begingroup$
$\endgroup$
2
-
$\begingroup$ Wikipedia discusses quantization of angular momentum e.g. here and here. $\endgroup$– Qmechanic ♦Jan 28, 2013 at 1:45
-
$\begingroup$ Also here, given that these are essentially the same indices we index spherical harmonics with. $\endgroup$– user10851Jan 28, 2013 at 2:13
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
2
The states $|j,m\rangle$ are simultaneous eigenstates of the total angular momentum squared operator $\mathbf J^2$ and the $z$-component of the total angular momentum operator $J_z$. The letter $j$ is related to the eigenvalue of the operator $\mathbf J^2$, while the letter $m$ gives the eigenvalue of the operator $J_z$. Specifically $$ \mathbf J^2|j,m\rangle =\hbar^2 j(j+1)|j,m\rangle, \qquad J_z|j,m\rangle = \hbar \,m|j,m\rangle $$ Given these considerations, $j$ is called the total angular momentum quantum number.
Hope that helps! Let me know if you'd like more details.
Cheers!
answered Jan 28, 2013 at 1:53
-
$\begingroup$ But does J here include spin or not? If it doesn't, isn't it notationally more common to denote the total ang. momentum by $\ell$ and not $j$? I thought $j$ was for the total ang. momentum including spin... $\endgroup$– daaxixJan 29, 2013 at 18:30
-
$\begingroup$ @daaxix The letter J is generically used for all forms of angular momentum in quantum. What $J_i$ denotes depends on the context. As long as the components of $\mathbf J$ satisfy the angular momentum commutation relations $[J_i, J_j] = i\hbar\epsilon_{ijk}J_k$, one can show that the analysis of its eigenvalues and eigenvectors leads to states labeled in this way. So $J_i$ could denote the components of orbital angular momentum, or spin, or any combination of both, and the notation will carry through. $\endgroup$ Jan 29, 2013 at 18:40