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

In summary: Then you can use ##\mathbf S = S_{+1}+\mathbf iS_{0}+ S_{-1}\mathbf j## to connect ##\mathbf S## with the matrix elements.In summary, the conversation discusses transforming a spin operator from the Cartesian basis to the spherical basis. The transformation involves using a rotation matrix and checking the angles obtained from the rotation. However, there is confusion about how the spin operator should be modeled and the relation between the vector operator and the matrix elements. It is suggested to use spherical tensors to connect the vector operator with the matrix elements.
  • #1
chi_rho
10
0
Say I have [tex]{S_{x}=\frac{1}{\sqrt{2}}\left(\begin{array}{ccc}
0 & 1 & 0\\
1 & 0 & 1\\
0 & 1 & 0\\
\end{array}\right)}
[/tex]
Right now, this spin operator is in the Cartesian basis. I want to transform it into the spherical basis. Since, [tex]{\vec{S}}[/tex] acts like a vector I think that I only need to transform [tex]S_{x}[/tex] like a vector:
[tex]{
R=\left(\begin{array}{ccc}
\sin \theta \cos \phi & \sin \theta \sin \phi & \cos \theta \\
\cos\theta \cos \phi & \cos \theta \sin \phi & -\sin \theta \\
-\sin \phi & \cos \phi & 0 \\
\end{array}\right)}
[/tex]
So my transformation looks like:
[tex]S_{spherical}=RS_{x}[/tex]
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 [tex]{\theta=\frac{\pi}{2} \hspace{5mm} \phi=0}[/tex] 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?
 
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  • #2
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##.
 
  • #5
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}##.
 

1. How do you transform spin matrices to a spherical basis?

To transform spin matrices (Sx, Sy, Sz) to a spherical basis, you can use the following equations:
Sx = (1/√2)(S+ + S-)
Sy = (1/√2i)(S+ - S-)
Sz = Sz
where S+ and S- are the ladder operators defined as S+ = Sx + iSy and S- = Sx - iSy.

2. What is the purpose of transforming spin matrices to a spherical basis?

The purpose of transforming spin matrices to a spherical basis is to simplify the mathematical representation of spin angular momentum in quantum mechanics. The spherical basis allows for easier visualization and interpretation of spin states, and also simplifies the calculations involved in spin-related measurements and experiments.

3. Can you explain the spherical basis in more detail?

The spherical basis is a set of basis vectors that describe the orientation of a particle's spin angular momentum in three-dimensional space. It consists of three basis vectors: |s, ms = +1>, |s, ms = 0>, and |s, ms = -1>, where s is the spin quantum number and ms is the z-component of the spin angular momentum. These basis vectors are related to the spin matrices through the transformation equations mentioned in the first question.

4. Are there any other ways to represent spin angular momentum besides the spherical basis?

Yes, there are other ways to represent spin angular momentum, such as the Cartesian basis and the Pauli basis. In the Cartesian basis, the spin matrices are represented as Sx, Sy, and Sz, while in the Pauli basis, they are represented as σx, σy, and σz. However, the spherical basis is the most commonly used representation for spin angular momentum in quantum mechanics.

5. What are some applications of transforming spin matrices to a spherical basis?

The transformation of spin matrices to a spherical basis has many practical applications in quantum mechanics, such as in the study of atomic and molecular physics, nuclear physics, and particle physics. It is also used in technologies such as magnetic resonance imaging (MRI) and nuclear magnetic resonance (NMR) spectroscopy to analyze the spin states of particles. Additionally, the spherical basis is essential in understanding and predicting the behavior of spin-based quantum systems, which have potential applications in quantum computing and information processing.

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