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I am trying to derive the spin matrices $S_x$ and $S_y$ in the z-basis for spin $s>1/2$ using the direct product (Kronecker product) method. For simplicity, let's focus on the case $s=1$.

I understand how to derive $S_z$. As show in https://physics.stackexchange.com/a/342156/50484,

if for example \begin{equation} S_{1z}=\tfrac{1}{2} \begin{bmatrix} 1 & 0\\ 0 &\!\!\! -\!1 \end{bmatrix}\;,\; S_{2z}=\tfrac{1}{2} \begin{bmatrix} 1 & 0\\ 0 &\!\!\! -\!1 \end{bmatrix} \tag{08} \end{equation} then \begin{equation} S_{z-tot}=\left(S_{1z} \otimes I_2\right)+ \left(I_1 \otimes S_{2z}\right)= \begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 &\!\!\! -\!1 \end{bmatrix} \tag{09} \end{equation} The matrix in (09) is already diagonal with eigenvalues 1,0,0,-1. Rearranging rows and columns we have
\begin{equation} S'_{z-tot}= \begin{bmatrix} \begin{array}{c|cccc} 0 & 0 & 0 & 0\\ \hline 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & \!\!\!\!-\!1 \end{array} \end{bmatrix} = \begin{bmatrix} \begin{array}{c|c} S_{z}^{(j=0)} & 0_{1\times 3}\\ \hline 0_{3\times1} & S_{z}^{(j=1)} \end{array} \end{bmatrix} \tag{10} \end{equation}

and so

$S_{z,s=1}= \begin{bmatrix} 1 & 0 & 0\\ 0 & 0 & 0 \\ 0 & 0 & -1\\ \end{bmatrix}$.

It is precisely the steps starting from (09) and getting to $S_{z,s=1}$ that I am wondering about, but for $S_{x,s=1}$ and $S_{y,s=1}$. Specifically, I can get

\begin{equation} S_{x-tot}=\left(S_{1x} \otimes I_2\right)+ \left(I_1 \otimes S_{2x}\right) =\tfrac{1}{2} \begin{bmatrix} 0 & 1 & 1 & 0\\ 1 & 0 & 0 & 1 \\ 1 & 0 & 0 & 1\\ 0 & 1 & 1 & 0 \end{bmatrix} \tag{20} \end{equation} \begin{equation} S_{y-tot}=\left(S_{1y} \otimes I_2\right)+ \left(I_1 \otimes S_{2y}\right)=\tfrac{1}{2} \begin{bmatrix} 0 & \!\!\! -\!i & \!\!\! -\!i & 0\\ i & 0 & 0 & \!\!\! -\!i \\ i & 0 & 0 & \!\!\! -\!i \\ 0 & i & i & 0 \end{bmatrix} \tag{21} \end{equation}

but from these I do not know how to get

$S_{x,s=1}= \frac{1}{\sqrt 2} \begin{bmatrix} 0 & 1 & 0\\ 1 & 0 & 1 \\ 0 & 1 & 0\\ \end{bmatrix}$

and

$S_{y,s=1}= \frac{1}{\sqrt 2} \begin{bmatrix} 0 & -i & 0\\ i & 0 & -i \\ 0 & i & 0\\ \end{bmatrix}$.

It seems $S_{z-tot}$ is just abnormally easy to work with because it is a diagonal matrix.

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  • $\begingroup$ The tensor product of two spin-1/2's contain both spin-0 and spin-1, so you need to disentangle the two parts, by e.g. projecting to the $\mathbf{S}^2=2$ subspace. $\endgroup$
    – Meng Cheng
    Feb 22 at 2:39

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The problem with your approach is there are two $m=0$ states and the $S=1$ states are linear combination of those. In other words, you have hastily identified the $(1,1)$ as entirely in the $S=0$, and the $(3,3)$ as entirely in the $S=1$ subspace. In fact, it's a linear combination of those basis states which are in the relevant subspace.

Making the change of basis will solve your problem. You can make this change of basis "by hand" by searching for the linear combination of your first and third basis states that is killed by $S_+$, identifying this linear combination with the $S=0$ states.

Another method is to take the $m=1$ state (it is unique), and use $S_-=S_x-iS_y$ to lower to a single state, which must be the $S=1,m=0$ state.

A more elegant method uses Clebsch-Gordan coefficients, and another more computationally intensive method is to diagonalize $L^2=L_x^2+L_y^2+L_z^2$ to transform to the correct $S$ subspace.


Edit: For your single-particle states you are using the basis $\{\vert +\rangle,\vert -\rangle\}$ since the diagonal form of $S_z$ produces the eigenvalues $\pm \frac{1}{2}$ on the basis vectors $\vert+\rangle\mapsto (1,0)^\top$ and $\vert -\rangle \mapsto (0,1)^\top$.

A basis for your tensor product is thus $\{\vert +\rangle\vert +\rangle,\vert +\rangle\vert -\rangle,\vert -\rangle\vert +\rangle,\vert -\rangle\vert -\rangle$ with eigenvalues of your $S_{z-tot}$ given by $1,0,0,-1$ respectively.

The states $\vert +\rangle\vert +\rangle$ and $\vert -\rangle\vert -\rangle$ must live in the $S=1$ and must be $\vert 1,1\rangle$ and $\vert 1,-1\rangle$ by inspection. On the other hand the states $\vert +\rangle\vert -\rangle$ and $\vert +\rangle\vert -\rangle$ do not live in an irreducible subspace. This is immediately seen by lowering from $\vert +\rangle\vert +\rangle$.

Indeed, the $S=1,m=0$ state is the combination $$ \vert 1,0\rangle = C^{1,0}_{1/2,1/2;1/2,-1/2}\vert +\rangle \vert -\rangle +C^{1,0}_{1/2,-1/2;1/2,1/2}\vert -\rangle \vert +\rangle $$ where $C^{1,0}_{1/2,1/2;1/2,-1/2}$ is a Clebsch-Gordan coefficient. Likwise $$ \vert 0,0\rangle = C^{0,0}_{1/2,1/2;1/2,-1/2}\vert +\rangle \vert -\rangle +C^{0,0}_{1/2,-1/2;1/2,1/2}\vert -\rangle \vert +\rangle\, . $$ The general change of basis transformation $$ \vert S,m_s\rangle =\sum_{m_1m_2} C^{Sm}_{1/2,m_1;1/2,m_2}\vert m_1\rangle \vert m_2\rangle $$ defines a matrix $T$ with elements $C^{Sm}_{m_1m_2}$ that will bring your $4\times 4$ matrices to the correct block diagonal form.

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  • $\begingroup$ I'm not familiar with the notation (1,1) and (3,3) in this context. Also, What are the first and third states you're referring to? I haven't written any states, only the operators. Finally, are you saying a similarity transformation of (20) and of (21) will give a matrix direct sum of the j=0 and j=1 matrices? $\endgroup$ Feb 22 at 4:19
  • $\begingroup$ they are just the location of your entries in your matrix. Since your $S_z$ is diagonal with eigenvalues $\pm 1/2$, your basis states are ordered as $\vert m=1/2\rangle$ and $\vert m=-1/2\rangle$. And yes the same similarity transformation will bring (20) and (21) to block diagonal form. $\endgroup$ Feb 22 at 5:01
  • $\begingroup$ Please look up examples of combining two spin-1/2 states. There are plenty of such questions on this site. $\endgroup$ Feb 22 at 5:06
  • $\begingroup$ Thank you for you time @ZeroTheHero. I don't have a problem doing that work myself for the spin 1/2 case as it is straightforward, but this was only an example. I am looking for the form of a solution that would apply to all spin cases, not a method that would work in all cases. This is what attracted me to the direct product stuff in the first place. For example, is there a general form/formula for the similarity matrix that converts (20) and (21) into block diagonal form for all spins, since that seems to be what I want? Perhaps I should ask this as a separate question. $\endgroup$ Feb 22 at 21:28
  • $\begingroup$ Yea there is a general form. The change of basis is given by the Clebsch-Gordan coefficients, which form the unitary matrix required for the similarity transformation. $\endgroup$ Feb 22 at 23:36

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