Is Griffith's Treatment of Spin in Quantum Mechanics Misinterpreted?

[lion8172]

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
$$-\hbar^2 \left( \frac{1}{\sin(\theta)} \frac{\partial}{\partial \theta} \left( \sin(\theta) \frac{\partial}{\partial \theta} \right) + \frac{1}{\sin^2(\theta)} \frac{\partial^2}{\partial \phi^2} \right)$$
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 $$(\hbar/i) (\vec{r} \times \nabla)$$?

Homework Equations

The Attempt at a Solution

 

[nrqed]

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
$$-\hbar^2 \left( \frac{1}{\sin(\theta)} \frac{\partial}{\partial \theta} \left( \sin(\theta) \frac{\partial}{\partial \theta} \right) + \frac{1}{\sin^2(\theta)} \frac{\partial^2}{\partial \phi^2} \right)$$
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 $$(\hbar/i) (\vec{r} \times \nabla)$$?

Homework Equations

The Attempt at a Solution

Yes, what you say is basically correct. The language people use is slightly different, however. What you call angular momentum (represented as a differential operator acting in space) should be called "orbital angular momentum". What you call spin for the more general concept is usually called "angular momentum".

SO spin and orbital angular momenta are two types of "angular momentum". Spin is also called "intrinsic" angular momentum and is represented by matrices acting in an internal space. Orbital angular momentum is represented by differential operators acting in ordinary space.