- #1
lion8172
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Homework Statement
I have a question relating to Griffith's treatment of spin. Griffiths shows, using the commutation relations for angular momentum, that L^2 has eigenvalues l(l+1), where l=0, 1/2, 1, 3/2, ... He also shows that the operator
[tex] -\hbar^2 \left( \frac{1}{\sin(\theta)} \frac{\partial}{\partial \theta} \left( \sin(\theta) \frac{\partial}{\partial \theta} \right) + \frac{1}{\sin^2(\theta)} \frac{\partial^2}{\partial \phi^2} \right) [/tex]
only admits eigenvalues with integral values of l, by directly solving the differential equation. At one point, he states that the above expression is equal to L^2. My question is as follows: Is it correct to state that the half-integral eigenvalues are automatically excluded when L^2 is parametrized in terms of angles, and is it true to state that L^2 is not, in general, equal to the expression above? Spin and angular momentum satisfy the same commutation relations, so would it be correct to state that "angular momentum" is a kind of "spin" which is generated by an operator of the form [tex] (\hbar/i) (\vec{r} \times \nabla) [/tex]?