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If we look at the representations of the $su(2)$ algebra we can construct a representation for every possible spin as follows $$|0,0\rangle\,\,\, \text{(spin 0)}$$ $$|-1/2,1/2\rangle,|1/2,1/2\rangle\,\,\, \text{(spin 1/2)}\,\,\, \text{(spin 1/2)}$$ $$|-1,1\rangle,|0,1\rangle,|1,1\rangle\,\,\, \text{(spin 1)}$$ $$\vdots$$

With the following spin operators to be $$0\,\,\, \text{(spin 0)}$$ $$\frac{1}{2}\sigma_x,\frac{1}{2}\sigma_y,\frac{1}{2}\sigma_z\,\,\, \text{(spin 1/2)}$$ $$ S_x = \frac{1}{\sqrt{2}}\begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \\ \end{pmatrix},S_y = \frac{1}{\sqrt{2}}\begin{pmatrix} 0 & -i & 0 \\ i & 0 & -i \\ 0 & i & 0 \\ \end{pmatrix},S_z = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -1 \\ \end{pmatrix} \,\,\, \text{(spin 1)}$$ $$\vdots$$

For the spin 1/2 representation this is the only way of expressing spin 1/2. However for the spin 1 they can be described by vectors with the following angular momentum matrices. $$ J_x = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & i \\ 0 & -i & 0 \\ \end{pmatrix},J_y = \begin{pmatrix} 0 & 0 & i \\ 0 & 0 & 0 \\ -i & 0 & 0 \\ \end{pmatrix},J_z = \begin{pmatrix} 0 & -i & 0 \\ i & 0 & 0 \\ 0 & 0 & 0 \\ \end{pmatrix} \,\,\, \text{(spin 1)}$$ Where the elements of the matrices are described by spacetime elements. What is the mathematical relationship between the two representations of spin 1 .i.e how do we go from the spin space to space-time space and vice versa? And what is the relationship of the angular momentum matrices in the space-time space and spin space.

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    $\begingroup$ MSE handles your specific spin 1 basis change . $\endgroup$ Aug 20, 2023 at 17:01

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The basis states are different and related by a change of basis. This is "obvious" because in one basis $S_z$ is diagonal but $J_z$ is not in the other. Going to a basis where $J_z$ is diagonal should convert your second set of matrices to your first.

To be explicit, the first set uses the spherical basis, where $$ \hat e_{\pm 1}=\frac{1}{\sqrt{2}} (\hat e_x\pm i \hat e_y)\, ,\qquad \hat e_0 = \hat e_z $$ in terms of Cartesian basis vectors. (There may be some signs somewhere but that's basically right.) In your second basis, you are using directly the Cartesian basis.

Notice in particular that, in the case of your $J_x$ and $J_y$, they are not hermitian in the Cartesian basis. When dealing with rotations in Cartesian 3d space, the factor of $i$ is often taken out and the matrices for $J_x,J_y$ and $J_z$ are real antisymmetric matrices, as you can find in many class mech. textbooks.

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