- #1
Jezza
- 37
- 0
There are two types of angular momentum: orbital and spin. If we define their operators as pseudo-vectors [itex]\vec{L}[/itex] and [itex]\vec{S}[/itex], then we can also define the total angular momentum operator [itex]\vec{J} = \vec{L}+\vec{S}[/itex].
Standard commutation relations will show that we can have simultaneous well defined values for [itex]J^2[/itex] and [itex]J_z[/itex] and etc. for [itex]\vec{L}[/itex] & [itex]\vec{S}[/itex]. i.e. We can have well defined total angular momentum and one component of it (usually z) for each type. The eigenvalues of these operators are then [itex]\hbar^2 j(j+1)[/itex] and [itex]\hbar m_j[/itex] respectively when we consider a simultaneous eigenstate [itex]|j,m_j>[/itex] of [itex]J^2[/itex] and [itex]J_z[/itex] only, and etc. for [itex]l[/itex] and [itex]s[/itex]
My question is really about how these different types combine.
Using these standard commutation relations:
[tex]
[J_i, J_j] = i \sum_k \epsilon_{ijk} J_k \hspace{10mm} [J_i, L_j] = i \sum_k \epsilon_{ijk} L_k \hspace{10mm} [J_i, S_j] = i \sum_k \epsilon_{ijk} S_k
[/tex]
It's very easy to show that we can have a simultaneous eigenstate [itex]|m_j,m_l,m_s>[/itex] of [itex]J_z, L_z, S_z[/itex] respectively, and thus the relation between the eigenvalues is
[tex]m_j=m_l+m_s[/tex]
We also have the commutation relations:
[tex]
[J^2, L^2] = 0 \hspace{10mm} [J^2, S^2] = 0 \hspace{10mm} [L^2, S^2] = 0
[/tex]
So we can have a simultaneous eigenstate [itex]|jls>[/itex] of [itex]J^2, L^2, S^2[/itex]. My question is then what is the relationship between [itex]j, l[/itex] and [itex]s[/itex]? So far as I can see, it is not straightforwards because:
[tex]
J^2 = L^2+S^2+2\vec{L}\cdot\vec{S} = L^2+S^2+2\sum_iL_iS_i \\
j(j+1) = l(l+1) + s(s+1) + \frac{2}{\hbar^2} \sum_i <jls|L_iS_i|jls>
[/tex]
Which presents a problem, since [itex][L_i, J^2], [S_i,J^2] \neq 0[/itex] so the state [itex]|jls>[/itex] cannot be an eigenstate of any of [itex]S_i, L_i[/itex] and so the relationship between the 3 numbers is not well defined.How can we have a state in which [itex]j,l,s[/itex] are well defined and yet their relationship is not well defined?
Standard commutation relations will show that we can have simultaneous well defined values for [itex]J^2[/itex] and [itex]J_z[/itex] and etc. for [itex]\vec{L}[/itex] & [itex]\vec{S}[/itex]. i.e. We can have well defined total angular momentum and one component of it (usually z) for each type. The eigenvalues of these operators are then [itex]\hbar^2 j(j+1)[/itex] and [itex]\hbar m_j[/itex] respectively when we consider a simultaneous eigenstate [itex]|j,m_j>[/itex] of [itex]J^2[/itex] and [itex]J_z[/itex] only, and etc. for [itex]l[/itex] and [itex]s[/itex]
My question is really about how these different types combine.
Using these standard commutation relations:
[tex]
[J_i, J_j] = i \sum_k \epsilon_{ijk} J_k \hspace{10mm} [J_i, L_j] = i \sum_k \epsilon_{ijk} L_k \hspace{10mm} [J_i, S_j] = i \sum_k \epsilon_{ijk} S_k
[/tex]
It's very easy to show that we can have a simultaneous eigenstate [itex]|m_j,m_l,m_s>[/itex] of [itex]J_z, L_z, S_z[/itex] respectively, and thus the relation between the eigenvalues is
[tex]m_j=m_l+m_s[/tex]
We also have the commutation relations:
[tex]
[J^2, L^2] = 0 \hspace{10mm} [J^2, S^2] = 0 \hspace{10mm} [L^2, S^2] = 0
[/tex]
So we can have a simultaneous eigenstate [itex]|jls>[/itex] of [itex]J^2, L^2, S^2[/itex]. My question is then what is the relationship between [itex]j, l[/itex] and [itex]s[/itex]? So far as I can see, it is not straightforwards because:
[tex]
J^2 = L^2+S^2+2\vec{L}\cdot\vec{S} = L^2+S^2+2\sum_iL_iS_i \\
j(j+1) = l(l+1) + s(s+1) + \frac{2}{\hbar^2} \sum_i <jls|L_iS_i|jls>
[/tex]
Which presents a problem, since [itex][L_i, J^2], [S_i,J^2] \neq 0[/itex] so the state [itex]|jls>[/itex] cannot be an eigenstate of any of [itex]S_i, L_i[/itex] and so the relationship between the 3 numbers is not well defined.How can we have a state in which [itex]j,l,s[/itex] are well defined and yet their relationship is not well defined?