Deriving the square angular momentum in spherical coordinates

In summary, the conversation is about deriving the square of the total angular momentum using the angular momentum operator in spherical coordinates. The x, y, and z components of angular momentum are shown in the given link. The individual components are summed up to obtain the squared total angular momentum. The issue is how to deal with partial differential operators in the process. The solution involves using the chain rule of partial differentiation and spherical coordinates, without the need for binomial formulae.
  • #1
Raziel2701
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0

Homework Statement


I want to derive the square of the total angular momentum as shown here: http://en.wikipedia.org/wiki/Angular_momentum_operator#Angular_momentum_computations_in_spherical_coordinates


Homework Equations



The x,y, and z components of angular momentum are shown in the above link. I'm attempting to solve it by [tex]L^2=(L_x)^2 + (L_y)^2+(L_z)^2[/tex]

The Attempt at a Solution


I think my problem is that I do not know how to deal with those partial differential operators. How do you go about squaring that binomial term that has differential operators? I think that's what I need to know and I don't know what to google to look for it.

Thank you.
 
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  • #2
You just need the chain rule of partial differentiation and a use of spherical coordinates...No binomial formulae needed.
 

1. What is the formula for deriving the square angular momentum in spherical coordinates?

The formula for deriving the square angular momentum in spherical coordinates is L² = Lx² + Ly² + Lz², where Lx, Ly, and Lz represent the components of the angular momentum along the x, y, and z axes, respectively.

2. Why is it important to derive the square angular momentum in spherical coordinates?

Deriving the square angular momentum in spherical coordinates allows us to express the total angular momentum of a particle in terms of its components along different axes, making it easier to analyze and understand its rotational motion.

3. How is the square angular momentum in spherical coordinates related to the moment of inertia?

The square angular momentum in spherical coordinates is directly proportional to the moment of inertia, with the proportionality constant being the square of the distance from the axis of rotation to the particle. This relationship is given by the formula L² = Iω², where ω is the angular velocity.

4. Can the square angular momentum in spherical coordinates be negative?

No, the square angular momentum in spherical coordinates cannot be negative. This is because it represents the square of a physical quantity (angular momentum), which is always positive.

5. How is the square angular momentum in spherical coordinates different from the square of the magnitude of the angular momentum?

The square angular momentum in spherical coordinates takes into account the components of the angular momentum along different axes, while the square of the magnitude of the angular momentum only considers the total magnitude of the angular momentum. The former is a vector quantity, while the latter is a scalar quantity.

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