Answer :
To show that a single nucleon state, which is an eigenstate of the operators [tex]\( J^2, L^2, S^2, \)[/tex] and [tex]\( J_z \)[/tex], is also an eigenstate of the operator [tex]\( \mathbf{L} \cdot \mathbf{S} \)[/tex] with the eigenvalue
[tex]\[ \frac{1}{2}[j(j+1) - l(l+1) - s(s+1)] \hbar^2, \][/tex]
we need to use some fundamental properties of angular momentum in quantum mechanics. Let's follow a step-by-step approach:
1. Review the Definitions:
- [tex]\( \mathbf{J} = \mathbf{L} + \mathbf{S} \)[/tex]: Total angular momentum operator, where [tex]\( \mathbf{L} \)[/tex] is the orbital angular momentum and [tex]\( \mathbf{S} \)[/tex] is the spin angular momentum.
- [tex]\(J^2\)[/tex], [tex]\(L^2\)[/tex], and [tex]\(S^2\)[/tex] are the magnitudes squared of total, orbital, and spin angular momenta, respectively.
2. Eigenvalue Equations:
- For [tex]\(J^2\)[/tex]: [tex]\( J^2 |j, l, s \rangle = \hbar^2 j(j+1) |j, l, s \rangle \)[/tex]
- For [tex]\(L^2\)[/tex]: [tex]\( L^2 |j, l, s \rangle = \hbar^2 l(l+1) |j, l, s \rangle \)[/tex]
- For [tex]\(S^2\)[/tex]: [tex]\( S^2 |j, l, s \rangle = \hbar^2 s(s+1) |j, l, s \rangle \)[/tex]
3. Express [tex]\( J^2 \)[/tex] in Terms of [tex]\( L \)[/tex] and [tex]\( S \)[/tex]:
[tex]\[ J^2 = (\mathbf{L} + \mathbf{S}) \cdot (\mathbf{L} + \mathbf{S}) \][/tex]
Expanding this, we get:
[tex]\[ J^2 = L^2 + S^2 + 2 \mathbf{L} \cdot \mathbf{S} \][/tex]
4. Rearrange the Expression:
[tex]\[ 2 \mathbf{L} \cdot \mathbf{S} = J^2 - L^2 - S^2 \][/tex]
5. Apply Eigenvalue Equations:
When applying to the state [tex]\( |j, l, s \rangle \)[/tex]:
[tex]\[ 2 \mathbf{L} \cdot \mathbf{S} |j, l, s \rangle = (\hbar^2 j(j+1) - \hbar^2 l(l+1) - \hbar^2 s(s+1)) |j, l, s \rangle \][/tex]
6. Simplify:
[tex]\[ \mathbf{L} \cdot \mathbf{S} |j, l, s \rangle = \frac{1}{2} [ \hbar^2 j(j+1) - \hbar^2 l(l+1) - \hbar^2 s(s+1)] |j, l, s \rangle \][/tex]
7. Extract the Eigenvalue:
Thus, the operator [tex]\( \mathbf{L} \cdot \mathbf{S} \)[/tex] has the eigenvalue
[tex]\[ \frac{1}{2} [j(j+1) - l(l+1) - s(s+1)] \hbar^2 \][/tex]
for the state [tex]\( |j, l, s \rangle \)[/tex].
Therefore, we have shown that the single nucleon state is indeed an eigenstate of [tex]\( \mathbf{L} \cdot \mathbf{S} \)[/tex] with the desired eigenvalue
[tex]\[ \frac{1}{2}[j(j+1) - l(l+1) - s(s+1)] \hbar^2. \][/tex]
[tex]\[ \frac{1}{2}[j(j+1) - l(l+1) - s(s+1)] \hbar^2, \][/tex]
we need to use some fundamental properties of angular momentum in quantum mechanics. Let's follow a step-by-step approach:
1. Review the Definitions:
- [tex]\( \mathbf{J} = \mathbf{L} + \mathbf{S} \)[/tex]: Total angular momentum operator, where [tex]\( \mathbf{L} \)[/tex] is the orbital angular momentum and [tex]\( \mathbf{S} \)[/tex] is the spin angular momentum.
- [tex]\(J^2\)[/tex], [tex]\(L^2\)[/tex], and [tex]\(S^2\)[/tex] are the magnitudes squared of total, orbital, and spin angular momenta, respectively.
2. Eigenvalue Equations:
- For [tex]\(J^2\)[/tex]: [tex]\( J^2 |j, l, s \rangle = \hbar^2 j(j+1) |j, l, s \rangle \)[/tex]
- For [tex]\(L^2\)[/tex]: [tex]\( L^2 |j, l, s \rangle = \hbar^2 l(l+1) |j, l, s \rangle \)[/tex]
- For [tex]\(S^2\)[/tex]: [tex]\( S^2 |j, l, s \rangle = \hbar^2 s(s+1) |j, l, s \rangle \)[/tex]
3. Express [tex]\( J^2 \)[/tex] in Terms of [tex]\( L \)[/tex] and [tex]\( S \)[/tex]:
[tex]\[ J^2 = (\mathbf{L} + \mathbf{S}) \cdot (\mathbf{L} + \mathbf{S}) \][/tex]
Expanding this, we get:
[tex]\[ J^2 = L^2 + S^2 + 2 \mathbf{L} \cdot \mathbf{S} \][/tex]
4. Rearrange the Expression:
[tex]\[ 2 \mathbf{L} \cdot \mathbf{S} = J^2 - L^2 - S^2 \][/tex]
5. Apply Eigenvalue Equations:
When applying to the state [tex]\( |j, l, s \rangle \)[/tex]:
[tex]\[ 2 \mathbf{L} \cdot \mathbf{S} |j, l, s \rangle = (\hbar^2 j(j+1) - \hbar^2 l(l+1) - \hbar^2 s(s+1)) |j, l, s \rangle \][/tex]
6. Simplify:
[tex]\[ \mathbf{L} \cdot \mathbf{S} |j, l, s \rangle = \frac{1}{2} [ \hbar^2 j(j+1) - \hbar^2 l(l+1) - \hbar^2 s(s+1)] |j, l, s \rangle \][/tex]
7. Extract the Eigenvalue:
Thus, the operator [tex]\( \mathbf{L} \cdot \mathbf{S} \)[/tex] has the eigenvalue
[tex]\[ \frac{1}{2} [j(j+1) - l(l+1) - s(s+1)] \hbar^2 \][/tex]
for the state [tex]\( |j, l, s \rangle \)[/tex].
Therefore, we have shown that the single nucleon state is indeed an eigenstate of [tex]\( \mathbf{L} \cdot \mathbf{S} \)[/tex] with the desired eigenvalue
[tex]\[ \frac{1}{2}[j(j+1) - l(l+1) - s(s+1)] \hbar^2. \][/tex]