Deriving B.31 in PDF on Squared-Angular Momentum

  • Context: Graduate 
  • Thread starter Thread starter TheCanadian
  • Start date Start date
  • Tags Tags
    Derivation
Click For Summary

Discussion Overview

The discussion centers on the derivation of the squared-angular momentum operator as presented in a specific PDF. Participants are particularly focused on understanding the derivation of equation B.31 and the role of spherical unit vectors in this context.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant questions whether B.31 is simply the dot product of B.24 with itself and expresses confusion about the presence of sine terms in the numerator and denominator.
  • Another participant asserts that while it is a dot product, the non-fixed nature of spherical unit vectors means that their differentiation does not yield zero, providing a mathematical expression to illustrate this point.
  • A later reply seeks clarification on the value of the derivative of the spherical unit vector with respect to phi, acknowledging the complexity introduced by non-fixed unit vectors.
  • Another participant asks how to express the spherical unit vectors in terms of Cartesian unit vectors, suggesting a need to explore the coefficients as functions of spherical coordinates.

Areas of Agreement / Disagreement

Participants appear to agree on the importance of the non-fixed nature of spherical unit vectors in the derivation, but there is no consensus on the specific expressions or simplifications related to these vectors.

Contextual Notes

There are unresolved questions regarding the exact expressions for the derivatives of spherical unit vectors and their representation in Cartesian coordinates, which may affect the understanding of the derivation.

TheCanadian
Messages
361
Reaction score
13
I was looking at this PDF (http://www.ifi.unicamp.br/~maplima/fi001/2012/aula20b.pdf) showing the derivation of the squared-angular momentum operator. Everything seems okay although I am just slightly lost in how exactly B.31 was derived. Isn't B.31 equal to the dot product of B.24 with itself? Where exactly does the sin in the numerator and denominator in B.31 come from when the theta component of B.24 is simply the partial derivative with respect to theta with no other terms? Shouldn't the theta components dotted with each simply yield the second order partial derivative with respect to theta when considering only this component?
 
Physics news on Phys.org
It's indeed just the dot product, but you cannot neglect the fact that the spherical unit vectors are not fixed, therefore differentiating them against some variable will not necessarily equal zero. For example
$$
\frac{\hat{\theta}}{\sin \theta} \frac{\partial}{\partial \phi} \cdot \frac{\hat{\theta}}{\sin \theta} \frac{\partial}{\partial \phi} = \frac{\hat{\theta}\cdot \hat{\theta}}{\sin^2 \theta} \frac{\partial^2}{\partial \phi^2} + \frac{\hat{\theta}}{\sin^2 \theta} \cdot \frac{\partial \hat{\theta}}{\partial \phi} \frac{\partial }{\partial \phi}
$$
 
blue_leaf77 said:
It's indeed just the dot product, but you cannot neglect the fact that the spherical unit vectors are not fixed, therefore differentiating them against some variable will not necessarily equal zero. For example
$$
\frac{\hat{\theta}}{\sin \theta} \frac{\partial}{\partial \phi} \cdot \frac{\hat{\theta}}{\sin \theta} \frac{\partial}{\partial \phi} = \frac{\hat{\theta}\cdot \hat{\theta}}{\sin^2 \theta} \frac{\partial^2}{\partial \phi^2} + \frac{\hat{\theta}}{\sin^2 \theta} \cdot \frac{\partial \hat{\theta}}{\partial \phi} \frac{\partial }{\partial \phi}
$$

Interesting. What exactly is the value of ## \frac{\partial \hat{\theta}}{\partial \phi} ##? I don't seem to know of an expression that would help simplify this, although I seem to understand what you are saying now. Thank you for pointing out that these unit vectors are not fixed...so easy to forget.
 
TheCanadian said:
What exactly is the value of ∂^θ∂ϕ∂θ^∂ϕ \frac{\partial \hat{\theta}}{\partial \phi} ?
Do you know how to express the spherical unit vectors in terms of Cartesian unit vectors? ##\hat{\theta}## is something like
$$
\hat{\theta} = A_{\theta x} \hat{x} + A_{\theta y} \hat{y} + A_{\theta z} \hat{z}.
$$
Find out how the ##A##'s look like as a function of ##r##, ##\theta##, and ##\phi##. The Cartesian unit vectors have fixed direction, thus they won't be acted upon by any differentiation.
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 9 ·
Replies
9
Views
1K
  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 0 ·
Replies
0
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 20 ·
Replies
20
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K