How to treat spin orbit operator directly

  • Thread starter hokhani
  • Start date
  • #1
443
7

Main Question or Discussion Point

For an electron the spin operator [itex]S_z[/itex]is represented by a [itex]2×2[/itex] matrix, with spin up and down as its bases. Consider the angular momentum operator [itex]L_z[/itex] with [itex]l=1[/itex] which is a [itex]3×3[/itex] matrix. How can we treat the [itex]L_z S_z[/itex] operator directly in matrix form?
 

Answers and Replies

  • #2
38
3
If with ##L_z S_z## you mean the total angular momentum ##J_z## than you have that ##J=L+S##, so ##j=l+s=\frac{3}{2}##. The eigenvalues of ##J_z## are ##m_j=-\frac{3}{2},-\frac{1}{2},\frac{1}{2},\frac{3}{2}## (##-j \le m_j \le j##). By choosing a basis where ##J^2## and ##J_z## are diagonal, your matrix operator of ##J_z## is:

##J_z = \hbar
\begin{pmatrix}
\frac{3}{2} & 0 & 0 & 0 \\
0 & \frac{1}{2} & 0 & 0 \\
0 & 0 & -\frac{1}{2} & 0 \\
0 & 0 & 0 & -\frac{3}{2}
\end{pmatrix}
##

P.S.: i'm studying this right now, so maybe i could probably make some mistakes
 
  • #3
vanhees71
Science Advisor
Insights Author
Gold Member
2019 Award
14,406
5,986
In non-relativistic physics (I guess that's all about non-relativistic quantum theory) spin and orbital angular momentum commute, and total angular momentum is given by
$$\vec{J}=\vec{L}+\vec{S}.$$
Then according to representation theory of the rotation group, if the orbital angular angular momentum has definite ##l## and spin ##s##, then the total angular momentum has ##j \in \{|l-s|,|l-s|+1,\ldots,l+s \}##, i.e., in your case you have ##j \in \{1/2,3/2 \}##.

Indeed the total-angular momentum space is ##(2l+1)(2s+1)##, i.e., in your case ##3 \cdot 2=6## dimensional. The possible values for ##j## imply the same dimension, i.e., ##2+4=6##.

For the eigenvalues of ##j_z=l_z+s_z##. You should look this up in a good textbook. Look for Clebsch-Gordan coefficients.
 
  • #4
443
7
Thanks, but my question is how to multiply the two matrixes directly without solving the problem this way.
 
  • #5
88
45
  • #6
Khashishi
Science Advisor
2,815
493
The usual way to get ##L S## is to use the identities
##J^2 = (L + S)^2 = L^2 + S^2 + 2LS##
##S## is a constant, so the result depends on ##L^2## and ##J^2##. With ##l=1## fixed, you can write out the basis states of ##j## in terms of the basis states of ##s##.
 

Related Threads on How to treat spin orbit operator directly

  • Last Post
Replies
1
Views
3K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
7
Views
1K
  • Last Post
Replies
1
Views
1K
Replies
4
Views
620
Replies
2
Views
1K
  • Last Post
Replies
4
Views
7K
Replies
1
Views
1K
  • Last Post
Replies
8
Views
2K
Top