Griffith's Treatment of Spin: Exploring Angular Momentum and Spin

In summary, Griffiths' treatment of spin shows that L^2 has eigenvalues l(l+1) and that the operator for orbital angular momentum only allows integral values of l. Spin and orbital angular momentum are both types of angular momentum, with spin being represented by matrices and orbital angular momentum being represented by differential operators.
  • #1
lion8172
29
0

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]?

Homework Equations





The Attempt at a Solution

 
Physics news on Phys.org
  • #2
lion8172 said:

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]?

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.
 

1. What is Griffith's Treatment of Spin?

Griffith's Treatment of Spin is a model used in quantum mechanics to describe the properties and behavior of particles with intrinsic angular momentum, known as spin. This model was developed by physicist David J. Griffiths and is based on the mathematical framework of quantum mechanics.

2. How does Griffith's Treatment of Spin explain angular momentum?

Griffith's Treatment of Spin explains angular momentum by assigning a spin quantum number to each particle, which describes the amount of angular momentum it possesses. This model also introduces the concept of spin states, which describe the orientation of a particle's spin in space. The spin states are quantized, meaning they can only take on discrete values, and they follow specific rules for how they can interact and change.

3. What is the significance of Griffith's Treatment of Spin in quantum mechanics?

Griffith's Treatment of Spin is significant in quantum mechanics because it explains certain experimental results that cannot be explained by classical physics. It also provides a mathematical framework for understanding the properties and behavior of particles with spin, such as electrons and protons. This model has been successfully applied in various fields of physics, including quantum chemistry, solid state physics, and particle physics.

4. How does Griffith's Treatment of Spin differ from other models of spin?

Griffith's Treatment of Spin differs from other models of spin, such as the Pauli spin matrix model, in that it treats spin as an intrinsic property of particles rather than a result of motion. This model also introduces the concept of spin operators, which are used to calculate the spin of a particle in different directions. Additionally, Griffith's Treatment of Spin is based on the Dirac equation, which is a relativistic equation, while other models are based on non-relativistic equations.

5. What are some real-world applications of Griffith's Treatment of Spin?

Griffith's Treatment of Spin has various real-world applications, such as in nuclear magnetic resonance (NMR) imaging, which is used in medical diagnostics. It is also used in electron spin resonance (ESR) spectroscopy to study the properties of materials. This model has also been applied in the development of quantum computing, which utilizes the spin states of particles as information carriers. Additionally, Griffith's Treatment of Spin has contributed to our understanding of the behavior of subatomic particles and the structure of matter.

Similar threads

  • Advanced Physics Homework Help
Replies
9
Views
884
  • Advanced Physics Homework Help
Replies
2
Views
914
  • Advanced Physics Homework Help
Replies
9
Views
827
  • Advanced Physics Homework Help
Replies
17
Views
1K
  • Advanced Physics Homework Help
Replies
21
Views
1K
Replies
1
Views
1K
  • Advanced Physics Homework Help
Replies
1
Views
1K
  • Advanced Physics Homework Help
Replies
1
Views
720
Replies
25
Views
2K
  • Advanced Physics Homework Help
Replies
1
Views
1K
Back
Top