Transforming Spin Matrices (Sx, Sy, Sz) to a Spherical Basis

Click For Summary

Discussion Overview

The discussion revolves around the transformation of spin matrices (specifically Sx) from a Cartesian basis to a spherical basis. Participants explore the implications of representing spin as a vector and the appropriate mathematical transformations involved, touching on concepts of rotation matrices and spherical tensors.

Discussion Character

  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant presents a transformation of the spin operator Sx into a spherical basis using a rotation matrix R, questioning the validity of their approach when the expected outcome does not match the original operator.
  • Another participant argues that the matrix representing the spin component in the direction of a unit vector should involve the entire vector operator, not just a single component Sx.
  • A different viewpoint suggests that spin cannot be accurately modeled as a vector, referring to it as a "flagpole." This implies a conceptual disagreement on the nature of spin.
  • Another participant clarifies that while the quantum state is a spinor, the vector operator S is indeed a vector operator, referencing external sources for support.
  • One participant points out that the matrix Sx is based on the eigenvectors of Sz, not Cartesian basis vectors, and questions the completeness of the rotation matrix R, suggesting it should have three independent parameters. They also discuss the distinction between vector operators and matrix elements, introducing the concept of spherical tensors that transform under rotation.

Areas of Agreement / Disagreement

Participants express differing views on the nature of spin and the appropriate mathematical treatment of spin operators. There is no consensus on the correct approach to transforming Sx into a spherical basis, and multiple competing perspectives remain unresolved.

Contextual Notes

Participants highlight limitations in the understanding of how vector operators and their matrix representations relate, as well as the specific properties of rotation matrices and spherical tensors. The discussion reflects a need for clarity on definitions and assumptions regarding the transformation process.

chi_rho
Messages
10
Reaction score
0
Say I have {S_{x}=\frac{1}{\sqrt{2}}\left(\begin{array}{ccc}<br /> 0 &amp; 1 &amp; 0\\<br /> 1 &amp; 0 &amp; 1\\<br /> 0 &amp; 1 &amp; 0\\<br /> \end{array}\right)}<br />
Right now, this spin operator is in the Cartesian basis. I want to transform it into the spherical basis. Since, {\vec{S}} acts like a vector I think that I only need to transform S_{x} like a vector:
{<br /> R=\left(\begin{array}{ccc}<br /> \sin \theta \cos \phi &amp; \sin \theta \sin \phi &amp; \cos \theta \\<br /> \cos\theta \cos \phi &amp; \cos \theta \sin \phi &amp; -\sin \theta \\<br /> -\sin \phi &amp; \cos \phi &amp; 0 \\<br /> \end{array}\right)}<br />
So my transformation looks like:
S_{spherical}=RS_{x}
I thought this would be fine, so I performed a cross check on the angles I obtained. My cross-check was determined from the fact that I think the spin operator "points" solely in the x-direction so after performing my rotation in order to make sure I'm right when {\theta=\frac{\pi}{2} \hspace{5mm} \phi=0} I should return with the original Sx in the Cartesian basis...This does not happen though. Is there something wrong with my transformation, or my reasoning behind the cross-check, or both?
 
Physics news on Phys.org
The matrix of the component of the spin in the direction ##\hat{n} = (\sin \theta \cos \phi , \sin \theta \sin \phi, \cos \theta)## is ##\hat{n} \cdot \vec{S}## where $$ \vec{S} = \left( \begin{array}{c} S_x \\ S_y \\ S_z \end{array} \right). $$ ##R## should act on the 3-vector, not just one component ##S_x##.
 
First off, the matrix ##S_x## you have there is written in basis formed by the eigenvectors of ##S_z##, it has nothing to do with the Cartesian basis vectors ##\mathbf i##, ##\mathbf j##, and ##\mathbf k##. Secondly, I don't feel familiar with matrix ##R## as you wrote above. If it's a rotation matrix then it should contain three independent parameters, but your ##R## matrix only has two.

As for your main question, I got the impression that you confused the way vector operator ##\mathbf S = S_x\mathbf i+ S_y\mathbf j+S_z\mathbf k## is expanded with the matrix element of each of the spin components. Although it's true that ##\mathbf S## transforms like an ordinary vector under rotation, but it doesn't connect to the matrix elements. There is the so-called Spherical Tensor, though, which transforms like spherical harmonics under rotation. Especially for a vector operator like ##\mathbf S##, you can simply rearrange the basis vectors ##\mathbf i##, ##\mathbf j##, and ##\mathbf k## such that ##\mathbf S## is expressed in rank-1 spherical tensors ##S_{+1}##, ##S_{0}##, and ##S_{-1}##.
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 4 ·
Replies
4
Views
5K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 5 ·
Replies
5
Views
4K
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K