# Spin operators in QM

Mathematically, the orbital angular momentum operators and the spin angular momentum operators are really two sides of the same coin. In group-theoretic language, we say that these operators arise from two different representations of the rotation group $SO(3)$ (to be precise, in quantum mechanics we are interested in projective representations because physically, two vectors that differ by a phase are indistinguishable. This requires a representation of the double cover of $SO(3)$, which is $SU(2)$). The group encodes information about the symmetries of the system and a representation of the group on a particular space gives us a way to realize these symmetries as operators on our space of states.

The difference between the spin operators and the angular momentum operators is really just what type of vector space they operate on. However, the group has a certain structure associated with it that carries through in its representations (This is related to the structure of the Lie-algebra of $SO(3)$, $\mathfrak{so}(3) \cong \mathfrak{su}(2)$ which I can elaborate on later if you would like). Hence, any representation of the group $SO(3)$ will have the same commutation relations. This includes extensions to multiple particle states. If we denote the Hilbert space of a single particle as $\mathcal{H}_1$ and the space of a second as $\mathcal{H}_2$, then the total space describing the two particles together is denoted $\mathcal{H}_1 \otimes \mathcal{H}_2$. This is nothing but a new vector space that we may represent $SO(3)$ on!

Thus, when we want to talk about the spin of a two-particle system, we are just talking about a different representation of $SO(3)$. There are some subtleties involved in this procedure due to the fact that the representation on the space $\mathcal{H}_1 \otimes \mathcal{H}_2$ is not irreducible. However, the Clebsh-Gordon decomposition gives us a way of decomposing this representation into a sum of reducible representations. This procedure gives the Clebsch-Gordon coefficients that arise when talking about multiple particle systems.

Coupling two non-interacting quantum systems $$\:\alpha,\beta\:$$ with angular momenta $$\:j_{\alpha},j_{\beta}\:$$ (no matter if orbital or spin) we reach to the following equation for the angular momentum $$\:j\:$$ of the composite system $$\:f\:$$

$$$$J_{\boldsymbol{n}}=\bigl( J^{\boldsymbol{\alpha}}_{\boldsymbol{n}}\boldsymbol{\otimes}\mathbb{I}_{\boldsymbol {\beta}}\bigr)+ \left(\mathbb{I}_{\boldsymbol {\alpha}} \boldsymbol{\otimes}J^{\boldsymbol{\beta}}_{\boldsymbol{n}}\right) \tag{A-01}$$$$ which, for the $$\:\mathbf{n}\:$$-components to be more clear, can be expressed as

$$$$\mathbf{n}\boldsymbol{\cdot}\mathbf{J}=\Bigl[\bigl(\mathbf{n}\boldsymbol{\cdot}\mathbf{J}^{\boldsymbol{\alpha}} \bigr) \boldsymbol{\otimes}\mathbb{I}_{\boldsymbol {\beta}}+ \mathbb{I}_{\boldsymbol{\alpha}} \boldsymbol{\otimes} \bigl(\mathbf{n}\boldsymbol{\cdot}\mathbf{J}^{\boldsymbol{\beta}} \bigr)\Bigr] \tag{A-02}$$$$ On above equations the symbol $$''\boldsymbol{\otimes}''$$ is used for the product of state vectors, spaces or operators. The vector $$\:\mathbf{n}=\left(n_{1},n_{2},n_{3}\right) \:$$ is of unit norm. The operators $$\:\mathbf{J}^{\boldsymbol{\alpha}},\, J^{\boldsymbol{\alpha}}_{\boldsymbol{n}},\,\mathbb{I}_{\boldsymbol {\alpha}}\:$$ act on the $$(2j_{\alpha}+1)$$-dimensional Hilbert space $$\: \mathsf{H}_{\alpha}\:$$ of system $$\:\alpha\:$$ and on the same footing the operators $$\:J^{\boldsymbol{\beta}}_{\boldsymbol{n}},\,\mathbb{I}_{\boldsymbol {\beta}}\:$$ act on the $$(2j_{\beta}+1)$$-dimensional Hilbert space $$\: \mathsf{H}_{\beta}\:$$ of system $$\:\beta\:$$, the symbol $$\:\mathbb{I}\:$$ being used for the identity. Finally the operators $$\:\mathbf{J},\, J_{\boldsymbol{n}}\:$$ act on the $$(2j_{\alpha}+1)\cdot (2j_{\beta}+1)$$-dimensional Hilbert space $$\: \mathsf{H}_{f}=\mathsf{H}_{\alpha}\boldsymbol{\otimes}\mathsf{H}_{\beta}\:$$ of the composite system $$\:f\:$$.

We write equation (A-02) for the three axes of a coordinate system \begin{align} J_{\boldsymbol{1}}=\Bigl( J^{\boldsymbol{\alpha}}_{\boldsymbol{1}}\boldsymbol{\otimes}\mathbb{I}_{\boldsymbol {\beta}}\Bigr)+ \left(\mathbb{I}_{\boldsymbol {\alpha}} \boldsymbol{\otimes}J^{\boldsymbol{\beta}}_{\boldsymbol{1}}\right) \tag{A-03a}\\ J_{\boldsymbol{2}}=\Bigl( J^{\boldsymbol{\alpha}}_{\boldsymbol{2}}\boldsymbol{\otimes}\mathbb{I}_{\boldsymbol {\beta}}\Bigr)+ \left(\mathbb{I}_{\boldsymbol {\alpha}} \boldsymbol{\otimes}J^{\boldsymbol{\beta}}_{\boldsymbol{2}}\right) \tag{A-03b}\\ J_{\boldsymbol{3}}=\Bigl( J^{\boldsymbol{\alpha}}_{\boldsymbol{3}}\boldsymbol{\otimes}\mathbb{I}_{\boldsymbol {\beta}}\Bigr)+ \left(\mathbb{I}_{\boldsymbol {\alpha}} \boldsymbol{\otimes}J^{\boldsymbol{\beta}}_{\boldsymbol{3}}\right) \tag{A-03c} \end{align} These three component equations can be expressed symbolically in one vector equation $$$$\mathbf{J}=\bigl(\mathbf{J}^{\boldsymbol{\alpha}} \boldsymbol{\otimes}\mathbb{I}_{\boldsymbol {\beta}}\bigr)+\left(\mathbb{I}_{\boldsymbol{\alpha}} \boldsymbol{\otimes} \mathbf{J}^{\boldsymbol{\beta}} \right) \tag{A-04}$$$$ Now we must check if this so constructed quantity $$\:\mathbf{J}=\left({J}_{1},{J}_{2},{J}_{3}\right) \:$$ of the composite system is a consistent angular momentum and the criterion for this is the validation of the equation $$$$\mathbf{J}\boldsymbol{\times}\mathbf{J}= i \, \mathbf{J} \tag{A-05}$$$$ or by components \begin{align} J_{\boldsymbol{1}}J_{\boldsymbol{2}}-J_{\boldsymbol{2}}J_{\boldsymbol{1}}= i \, J_{\boldsymbol{3}} \tag{A-06a}\\ J_{\boldsymbol{2}}J_{\boldsymbol{3}}-J_{\boldsymbol{3}}J_{\boldsymbol{2}}= i \, J_{\boldsymbol{1}} \tag{A-06b}\\ J_{\boldsymbol{3}}J_{\boldsymbol{1}}-J_{\boldsymbol{1}}J_{\boldsymbol{3}}= i \, J_{\boldsymbol{2}} \tag{A-06c} \end{align} To prove equations (A-06), let find a general expression for $$\:J_{\boldsymbol{n}}J_{\boldsymbol{k}}\:$$, where $$\:J_{\boldsymbol{n}},\,J_{\boldsymbol{k}}\:$$ the components of $$\:\mathbf{J}\:$$ parallel to the unit vectors $$\:\mathbf{n}\:$$ and $$\:\mathbf{k}\:$$ respectively. From equation (A-01) and the following multiplication rule $$$$\left(\mathrm{A}_{2} \boldsymbol{\otimes} \mathrm{B}_{2}\right)\left(\mathrm{A}_{1} \boldsymbol{\otimes} \mathrm{B}_{1}\right)= \left( \mathrm{A}_{2}\mathrm{A}_{1}\right) \boldsymbol{\otimes} \left( \mathrm{B}_{2}\mathrm{B}_{1}\right) \tag{A-07}$$$$ we have \begin{align} J_{\boldsymbol{n}}J_{\boldsymbol{k}} & = \Bigl[\Bigl( J^{\boldsymbol{\alpha}}_{\boldsymbol{n}}\boldsymbol{\otimes}\mathbb{I}_{\boldsymbol {\beta}}\Bigr)+ \Bigl(\mathbb{I}_{\boldsymbol {\alpha}} \boldsymbol{\otimes}J^{\boldsymbol{\beta}}_{\boldsymbol{n}}\Bigr)\Bigr] \left[\Bigl( J^{\boldsymbol{\alpha}}_{\boldsymbol{k}}\boldsymbol{\otimes}\mathbb{I}_{\boldsymbol {\beta}}\Bigr)+ \left(\mathbb{I}_{\boldsymbol {\alpha}} \boldsymbol{\otimes}J^{\boldsymbol{\beta}}_{\boldsymbol{k}}\right)\right] \nonumber\\ & =\Bigl( J^{\boldsymbol{\alpha}}_{\boldsymbol{n}}\boldsymbol{\otimes}\mathbb{I}_{\boldsymbol {\beta}}\Bigr)\Bigl( J^{\boldsymbol{\alpha}}_{\boldsymbol{k}}\boldsymbol{\otimes}\mathbb{I}_{\boldsymbol {\beta}}\Bigr)+\Bigl(\mathbb{I}_{\boldsymbol {\alpha}} \boldsymbol{\otimes}J^{\boldsymbol{\beta}}_{\boldsymbol{n}}\Bigr)\left(\mathbb{I}_{\boldsymbol {\alpha}} \boldsymbol{\otimes}J^{\boldsymbol{\beta}}_{\boldsymbol{k}}\right) \nonumber\\ & + \Bigl( J^{\boldsymbol{\alpha}}_{\boldsymbol{n}}\boldsymbol{\otimes}\mathbb{I}_{\boldsymbol {\beta}}\Bigr)\left(\mathbb{I}_{\boldsymbol {\alpha}} \boldsymbol{\otimes}J^{\boldsymbol{\beta}}_{\boldsymbol{k}}\right) +\Bigl(\mathbb{I}_{\boldsymbol {\alpha}} \boldsymbol{\otimes}J^{\boldsymbol{\beta}}_{\boldsymbol{n}}\Bigr)\Bigl( J^{\boldsymbol{\alpha}}_{\boldsymbol{k}}\boldsymbol{\otimes}\mathbb{I}_{\boldsymbol {\beta}}\Bigr) \tag{A-08} \end{align} so $$$$J_{\boldsymbol{n}}J_{\boldsymbol{k}} = \Bigl[\Bigl( J^{\boldsymbol{\alpha}}_{\boldsymbol{n}}J^{\boldsymbol{\alpha}}_{\boldsymbol{k}}\Bigr)\boldsymbol{\otimes}\mathbb{I}_{\boldsymbol {\beta}}\Bigr]+\Bigl[\mathbb{I}_{\boldsymbol {\alpha}}\boldsymbol{\otimes}\Bigl( J^{\boldsymbol{\beta}}_{\boldsymbol{n}}J^{\boldsymbol{\beta}}_{\boldsymbol{k}}\Bigr)\Bigr] +\Bigl( J^{\boldsymbol{\alpha}}_{\boldsymbol{n}}\boldsymbol{\otimes}J^{\boldsymbol{\beta}}_{\boldsymbol{k}}\Bigr) +\Bigl( J^{\boldsymbol{\alpha}}_{\boldsymbol{k}}\boldsymbol{\otimes}J^{\boldsymbol{\beta}}_{\boldsymbol{n}}\Bigr) \tag{A-09}$$$$ Permutation of $$\:n\:$$ and $$\:k\:$$ yields $$$$J_{\boldsymbol{k}}J_{\boldsymbol{n}} = \Bigl[\Bigl( J^{\boldsymbol{\alpha}}_{\boldsymbol{k}}J^{\boldsymbol{\alpha}}_{\boldsymbol{n}}\Bigr)\boldsymbol{\otimes}\mathbb{I}_{\boldsymbol {\beta}}\Bigr]+\Bigl[\mathbb{I}_{\boldsymbol {\alpha}}\boldsymbol{\otimes}\Bigl( J^{\boldsymbol{\beta}}_{\boldsymbol{k}}J^{\boldsymbol{\beta}}_{\boldsymbol{n}}\Bigr)\Bigr] +\Bigl( J^{\boldsymbol{\alpha}}_{\boldsymbol{k}}\boldsymbol{\otimes}J^{\boldsymbol{\beta}}_{\boldsymbol{n}}\Bigr) +\Bigl( J^{\boldsymbol{\alpha}}_{\boldsymbol{n}}\boldsymbol{\otimes}J^{\boldsymbol{\beta}}_{\boldsymbol{k}}\Bigr) \tag{A-10}$$$$ Subtracting (A-10) from (A-09)

$$$$J_{\boldsymbol{n}}J_{\boldsymbol{k}}-J_{\boldsymbol{k}}J_{\boldsymbol{n}}= \Bigl[\Bigl( J^{\boldsymbol{\alpha}}_{\boldsymbol{n}}J^{\boldsymbol{\alpha}}_{\boldsymbol{k}}-J^{\boldsymbol{\alpha}}_{\boldsymbol{k}}J^{\boldsymbol{\alpha}}_{\boldsymbol{n}}\Bigr)\boldsymbol{\otimes}\mathbb{I}_{\boldsymbol {\beta}}\Bigr]+\Bigl[\mathbb{I}_{\boldsymbol {\alpha}}\boldsymbol{\otimes}\Bigl(J^{\boldsymbol{\beta}}_{\boldsymbol{n}}J^{\boldsymbol{\beta}}_{\boldsymbol{k}}- J^{\boldsymbol{\beta}}_{\boldsymbol{k}}J^{\boldsymbol{\beta}}_{\boldsymbol{n}}\Bigr)\Bigr] \tag{A-11}$$$$ For $$\:n=1\:$$ and $$\:k=2\:$$ above equation (A-11) gives \begin{align} J_{\boldsymbol{1}}J_{\boldsymbol{2}}-J_{\boldsymbol{2}}J_{\boldsymbol{1}} & = \Bigl[\overbrace{\Bigl( J^{\boldsymbol{\alpha}}_{\boldsymbol{1}}J^{\boldsymbol{\alpha}}_{\boldsymbol{2}}-J^{\boldsymbol{\alpha}}_{\boldsymbol{2}}J^{\boldsymbol{\alpha}}_{\boldsymbol{1}}\Bigr)}^{i \,J^{\boldsymbol{\alpha}}_{\boldsymbol{3}} }\boldsymbol{\otimes}\mathbb{I}_{\boldsymbol {\beta}}\Bigr]+\Bigl[\mathbb{I}_{\boldsymbol {\alpha}}\boldsymbol{\otimes} \overbrace{\Bigl(J^{\boldsymbol{\beta}}_{\boldsymbol{1}}J^{\boldsymbol{\beta}}_{\boldsymbol{2}}- J^{\boldsymbol{\beta}}_{\boldsymbol{2}}J^{\boldsymbol{\beta}}_{\boldsymbol{1}}\Bigr)}^{i \,J^{\boldsymbol{\beta}}_{\boldsymbol{3}}}\Bigr] \nonumber\\ & = \Bigl[\Bigl(i \,J^{\boldsymbol{\alpha}}_{\boldsymbol{3}}\Bigr)\boldsymbol{\otimes}\mathbb{I}_{\boldsymbol{\beta}}\Bigr] +\Bigl[\mathbb{I}_{\boldsymbol{\alpha}}\boldsymbol{\otimes}\Bigl(i\,J^{\boldsymbol{\beta}}_{\boldsymbol{3}}\Bigr)\Bigr] \nonumber\\ & = i\,\Bigl[\Bigl(J^{\boldsymbol{\alpha}}_{\boldsymbol{3}}\boldsymbol{\otimes}\mathbb{I}_{\boldsymbol{\beta}}\Bigr) +\Bigl(\mathbb{I}_{\boldsymbol{\alpha}}\boldsymbol{\otimes}J^{\boldsymbol{\beta}}_{\boldsymbol{3}}\Bigr)\Bigr] \nonumber\\ & = i \, J_{\boldsymbol{3}} \tag{A-12} \end{align} so proving (A-06a). By cyclic permutation (A-06b) and (A-06c) are proved also.

For the treatment of the angular momentum we make use of equation (A-03c), repeated here for convenience: $$$$J_{\boldsymbol{3}}=\Bigl( J^{\boldsymbol{\alpha}}_{\boldsymbol{3}}\boldsymbol{\otimes}\mathbb{I}_{\boldsymbol {\beta}}\Bigr)+ \left(\mathbb{I}_{\boldsymbol {\alpha}} \boldsymbol{\otimes}J^{\boldsymbol{\beta}}_{\boldsymbol{3}}\right) \tag{A-03c}$$$$ This relation has the advantage that if the matrices representing the components $$\:J^{\boldsymbol{\alpha}}_{\boldsymbol{3}}\:$$ and $$\:J^{\boldsymbol{\beta}}_{\boldsymbol{3}}\:$$ of the component systems are diagonal, then the matrix representing the component $$\:J_{\boldsymbol{3}}\:$$ of the composite system is diagonal too(1). But for the full treatment of the angular momentum we need the matrix representing the quantity $$\:\mathbf{J}^{\boldsymbol{2}}=J^{\boldsymbol{2}}_{\boldsymbol{1}}+J^{\boldsymbol{2}}_{\boldsymbol{2}}+J^{\boldsymbol{2}}_{\boldsymbol{3}}\:$$ also. We'll find an expression of $$\:\mathbf{J}^{\boldsymbol{2}}\:$$ convenient for the determination of its matrix, which isn't from the beginning diagonal as $$\:J_{\boldsymbol{3}}\:$$ does. So, inserting in equation (A-09) the pair of values $$\:(n,k)=(1,1)\:$$, $$\:(n,k)=(2,2)\:$$ and $$\:(n,k)=(3,3)\:$$ we have respectively

\begin{align} J_{\boldsymbol{1}}^{\boldsymbol{2}} = \Bigl[\bigl( J^{\boldsymbol{\alpha}}_{\boldsymbol{1}}\bigr)^{\boldsymbol{2}}\boldsymbol{\otimes}\mathbb{I}_{\boldsymbol {\beta}}\Bigr]+\Bigl[\mathbb{I}_{\boldsymbol {\alpha}}\boldsymbol{\otimes}\bigl( J^{\boldsymbol{\beta}}_{\boldsymbol{1}}\bigr)^{\boldsymbol{2}}\Bigr] +2\Bigl( J^{\boldsymbol{\alpha}}_{\boldsymbol{1}}\boldsymbol{\otimes}J^{\boldsymbol{\beta}}_{\boldsymbol{1}}\Bigr) \tag{A-13a}\\ J_{\boldsymbol{2}}^{\boldsymbol{2}} = \Bigl[\bigl( J^{\boldsymbol{\alpha}}_{\boldsymbol{2}}\bigr)^{\boldsymbol{2}}\boldsymbol{\otimes}\mathbb{I}_{\boldsymbol {\beta}}\Bigr]+\Bigl[\mathbb{I}_{\boldsymbol {\alpha}}\boldsymbol{\otimes}\bigl( J^{\boldsymbol{\beta}}_{\boldsymbol{2}}\bigr)^{\boldsymbol{2}}\Bigr] +2\Bigl( J^{\boldsymbol{\alpha}}_{\boldsymbol{2}}\boldsymbol{\otimes}J^{\boldsymbol{\beta}}_{\boldsymbol{2}}\Bigr) \tag{A-13b}\\ J_{\boldsymbol{3}}^{\boldsymbol{2}} = \Bigl[\bigl( J^{\boldsymbol{\alpha}}_{\boldsymbol{3}}\bigr)^{\boldsymbol{2}}\boldsymbol{\otimes}\mathbb{I}_{\boldsymbol {\beta}}\Bigr]+\Bigl[\mathbb{I}_{\boldsymbol {\alpha}}\boldsymbol{\otimes}\bigl( J^{\boldsymbol{\beta}}_{\boldsymbol{3}}\bigr)^{\boldsymbol{2}}\Bigr] +2\Bigl( J^{\boldsymbol{\alpha}}_{\boldsymbol{3}}\boldsymbol{\otimes}J^{\boldsymbol{\beta}}_{\boldsymbol{3}}\Bigr) \tag{A-13c} \end{align} Having in mind that \begin{align} \bigl(\mathbf{J}^{\boldsymbol{\alpha}}\bigr)^{\boldsymbol{2}} & =\bigl( J^{\boldsymbol{\alpha}}_{\boldsymbol{1}}\bigr)^{\boldsymbol{2}} +\bigl( J^{\boldsymbol{\alpha}}_{\boldsymbol{2}}\bigr)^{\boldsymbol{2}}+\bigl( J^{\boldsymbol{\alpha}}_{\boldsymbol{3}}\bigr)^{\boldsymbol{2}} = j_{\alpha}(j_{\alpha}+1)\mathbb{I}_{\alpha} \tag{A-14}\\ \bigl( \mathbf{J}^{\boldsymbol{\beta}}\bigr)^{\boldsymbol{2}} &=\bigl( J^{\boldsymbol{\beta}}_{\boldsymbol{1}}\bigr)^{\boldsymbol{2}} +\bigl( J^{\boldsymbol{\beta}}_{\boldsymbol{2}}\bigr)^{\boldsymbol{2}}+\bigl( J^{\boldsymbol{\beta}}_{\boldsymbol{3}}\bigr)^{\boldsymbol{2}} = j_{\beta}(j_{\beta}+1) \mathbb{I}_{\beta} \tag{A-15}\\ \mathbb{I}_{\alpha} \boldsymbol{\otimes}\mathbb{I}_{\beta} & \equiv \mathbb{I}_{f}=\text{identity in } \mathsf{H}_{f}=\mathsf{H}_{\alpha}\boldsymbol{\otimes}\mathsf{H}_{\beta} \tag{A-16} \end{align} addition of equations (A-13) yields $$$$\mathbf{J}^{\boldsymbol{2}} =\bigl[ j_{\alpha}(j_{\alpha}+1)+ j_{\beta}(j_{\beta}+1) \bigr] \mathbb{I}_{f} +2\sum_{q=1}^{q=3}\Bigl( J^{\boldsymbol{\alpha}}_{\boldsymbol{q}}\boldsymbol{\otimes}J^{\boldsymbol{\beta}}_{\boldsymbol{q}}\Bigr) \tag{A-17}$$$$

(1) More precisely : from the definition of the product of operators and given that $$\:J^{\boldsymbol{\alpha}}_{\boldsymbol{3}}\:$$ is represented by the $$(2j_{\alpha}+1)$$-square matrix $$$$J^{\alpha}_{3} = \begin{bmatrix} j_{\alpha} & 0 & \cdots & 0 \\ 0 & j_{\alpha}-1 & \cdots & 0 \\ \vdots & \vdots & m_{\alpha} & \vdots \\ 0 & 0 & \cdots & -j_{\alpha} \end{bmatrix} \tag{foot-01}$$$$ and $$\:J^{\boldsymbol{\beta}}_{\boldsymbol{3}}\:$$ is represented by the $$(2j_{\beta}+1)$$-square matrix $$$$J^{\beta}_{3} = \begin{bmatrix} j_{\beta} & 0 & \cdots & 0 \\ 0 & j_{\beta}-1 & \cdots & 0 \\ \vdots & \vdots & m_{\beta} & \vdots \\ 0 & 0 & \cdots & -j_{\beta} \end{bmatrix} \tag{foot-02}$$$$ equation (A-03c) gives that $$\:J_{\boldsymbol{3}}\:$$ is represented by the following $$(2j_{\alpha}+1)\cdot (2j_{\beta}+1)$$-square diagonal matrix $$$$J_{\boldsymbol{3}}=\Bigl( J^{\boldsymbol{\alpha}}_{\boldsymbol{3}}\boldsymbol{\otimes}\mathbb{I}_{\boldsymbol {\beta}}\Bigr)+ \left(\mathbb{I}_{\boldsymbol {\alpha}} \boldsymbol{\otimes}J^{\boldsymbol{\beta}}_{\boldsymbol{3}}\right)= \nonumber\\$$$$ $$$$\begin{bmatrix} \begin{matrix} j_{\alpha}+ j_{\beta} & 0 & \cdots & 0 \\ 0 & j_{\alpha}+ j_{\beta}-1 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & j_{\alpha} -j_{\beta} \end{matrix} & & & \\ & \begin{matrix} j_{\alpha}-1+ j_{\beta} & 0 & \cdots & 0 \\ 0 & j_{\alpha}-1+ j_{\beta}-1 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & j_{\alpha}-1-j_{\beta} \end{matrix} & & \\ & &\ddots & \\ & & & -j_{\alpha}-j_{\beta} \end{bmatrix}$$$$ $$$$\tag{foot-03}$$$$ Example : for $$\:j_{\alpha}=\tfrac{1}{2}\:$$ and $$\:j_{\beta}=1\:$$ $$$$J^{\alpha}_{3} = \begin{bmatrix} \begin{array}{cc} +\frac{1}{2}&0\\ &\\ 0&-\frac{1}{2} \end{array} \end{bmatrix} \:,\qquad J^{\beta}_{3} = \begin{bmatrix} \begin{array}{ccc} +1&0&0\\ 0&0&0\\ 0&0&-1 \end{array} \end{bmatrix} \tag{foot-04}$$$$
so $$$$\Bigl( J^{\boldsymbol{\alpha}}_{\boldsymbol{3}}\boldsymbol{\otimes}\mathbb{I}_{\boldsymbol {\beta}}\Bigr)= \begin{bmatrix} \begin{array}{cc} +\frac{1}{2}\cdot\mathbb{I}_{\boldsymbol {\beta}}&0\cdot\mathbb{I}_{\boldsymbol {\beta}}\\ &\\ 0\cdot\mathbb{I}_{\boldsymbol {\beta}}&-\frac{1}{2}\cdot\mathbb{I}_{\boldsymbol {\beta}} \end{array} \end{bmatrix} = \begin{bmatrix} \begin{array}{cccccc} +\frac{1}{2}&0&0&0&0&0\\ 0&+\frac{1}{2}&0&0&0&0\\ 0&0&+\frac{1}{2}&0&0&0\\ 0&0&0&-\frac{1}{2}&0&0\\ 0&0&0&0&-\frac{1}{2}&0\\ 0&0&0&0&0&-\frac{1}{2} \end{array} \end{bmatrix} \tag{foot-05}$$$$ $$$$\left(\mathbb{I}_{\boldsymbol {\alpha}} \boldsymbol{\otimes}J^{\boldsymbol{\beta}}_{\boldsymbol{3}}\right)= \begin{bmatrix} \begin{array}{cc} 1\cdot J^{\boldsymbol{\beta}}_{\boldsymbol{3}}&0\cdot J^{\boldsymbol{\beta}}_{\boldsymbol{3}}\\ &\\ 0\cdot J^{\boldsymbol{\beta}}_{\boldsymbol{3}}&1\cdot J^{\boldsymbol{\beta}}_{\boldsymbol{3}} \end{array} \end{bmatrix} = \begin{bmatrix} \begin{array}{cccccc} +1&0&0&0&0&0\\ 0&0&0&0&0&0\\ 0&0&-1&0&0&0\\ 0&0&0&+1&0&0\\ 0&0&0&0&0&0\\ 0&0&0&0&0&-1 \end{array} \end{bmatrix} \tag{foot-06}$$$$ Adding (foot-05), (foot-06) we have $$$$J_{\boldsymbol{3}} =\Bigl( J^{\boldsymbol{\alpha}}_{\boldsymbol{3}}\boldsymbol{\otimes}\mathbb{I}_{\boldsymbol {\beta}}\Bigr)+ \left(\mathbb{I}_{\boldsymbol {\alpha}} \boldsymbol{\otimes}J^{\boldsymbol{\beta}}_{\boldsymbol{3}}\right)= \begin{bmatrix} \begin{array}{cccccc} +\frac{3}{2}&0&0&0&0&0\\ 0&+\frac{1}{2}&0&0&0&0\\ 0&0&-\frac{1}{2}&0&0&0\\ 0&0&0&+\frac{1}{2}&0&0\\ 0&0&0&0&-\frac{1}{2}&0\\ 0&0&0&0&0&-\frac{3}{2} \end{array} \end{bmatrix} \tag{foot-07}$$$$
which after rearrangement of rows and columns becomes $$$$\widehat{J}_{\boldsymbol{3}} = \begin{bmatrix} \begin{array}{cccccc} +\frac{3}{2}&0&0&0&0&0\\ 0&+\frac{1}{2}&0&0&0&0\\ 0&0&-\frac{1}{2}&0&0&0\\ 0&0&0&-\frac{3}{2}&0&0\\ 0&0&0&0&+\frac{1}{2}&0\\ 0&0&0&0&0&-\frac{1}{2} \end{array} \end{bmatrix} \tag{foot-08}$$$$ recognized later on as the direct sum of $$\:j_{1}=\tfrac{1}{2}\:$$ and $$\:j_{2}=\tfrac{3}{2}\:$$ $$$$\boldsymbol{2}\boldsymbol{\otimes}\boldsymbol{3}=\boldsymbol{2}\boldsymbol{\oplus}\boldsymbol{4} \tag{foot-09}$$$$ a special case of the more general expression of the product space as the direct sum of mutually orthogonal and invariant under SU(2) subspaces $$$$(2j_{\alpha}+1)\boldsymbol{\otimes}(2j_{\beta}+1)=\bigoplus_{j=\vert j_{\beta}-j_{\alpha} \vert }^{j=\left(j_{\alpha}+j_{\beta}\right)}(2j+1) \tag{foot-10}$$$$

For a more detailed treatment see my answers here : Total spin of two spin-1/2 particles.

Both orbital angular momentum and spin are related to rotations in three dimensions. Their commutation relations can be derived from the properties of the group of rotations alone, so they should be equal.

The group of rotations of three-dimensional space is known as $SO(3)$. Quantum states are vectors in a space $V$ over which this group has a (projective) representation. This means that for each rotation $R$ there is a $n\times n$ matrix $U(R)$ ($n$ is the dimension of $V$) so that every quantum state $\left|\psi\right>$ changes to $U(R)\left|\psi\right>$ when the system is rotated by $R$.

You may know that a rotation in two dimensions (the complex plane) is given by multiplication by $e^{-i\theta}$, where $\theta$ is the only parameter that characterizes a rotation of two-dimensional space: the angle of the rotation. In three dimensions, a rotation can be parametrized by the angle $\theta$ and a unit vector $\hat{u}$ stating the axis of rotation. This is equivalent to just a vector $\vec{u}=\theta\hat{u}$. Proceeding in the same way as in two dimensions a rotation can be written as $$U(\vec{u})=e^{-i\vec{u}\cdot \vec{J}}$$ where now we need three objects $J_x$, $J_y$ and $J_z$ (the components of $\vec{J}$), one to multiply each component of $\vec{u}$. They must be $n\times n$ matrices, to make $U(R)$ be also such a matrix (the exponential of matrices can be defined by its power series).

Notice that the derivative of a rotation of a 3d vector is orthogonal to the axis of rotation and to the vector itself and it is proportional to the vector, so that it should be $\hat{u}\times \vec{v}$. You can imagine $\vec{v}$ as a point on a sphere with radius $|\vec{v}|$, and $\hat{u}\times \vec{v}$ as an arrow starting at that point and pointing in the direction to which it moves when it's rotated.

On the other hand $\frac{d}{d\theta}U(\theta\hat{u})= -i\hat{u}\cdot\vec{J}$, so, writing the vectors of the three dimensional representation as $\vec{v}$ instead of $\left|\psi\right>$ we have the equation $\hat{u}\times \vec{v}=(-i\hat{u}\cdot\vec{J})v$. Now we wish to compute the commutator $[J_x,J_y]$: \begin{align} [J_x,J_y]\vec{v}=J_xJ_y\vec{v}-J_yJ_x\vec{v}= -\hat{x}\times(\hat{y}\times\vec{v}) + \hat{y}\times(\hat{x}\times\vec{v}) = -(\hat{x}\times\hat{y})\times\vec{v} = -\hat{z}\times\vec{v} = iJ_z\vec{v} \end{align} where I have used the properties of the triple cross product. We have just derived one of the commutation relations: $[J_x,J_y]=iJ_z$. The others follow in the same manner.

The spin operators $S_x$, $S_y$, $S_z$ and the orbital angular momentum operators $L_x$, $L_y$, $L_z$ are both just the generators $J_x$, $J_y$, $J_z$ of three-dimensional rotations.

The only difference between them is that the name spin (and the notation $S_i$) refers to the representations under $SO(3)$ for states of a single particle without movement in space, the "internal" rotations, whereas the name orbital angular momentum (and the symbols $L_i$) is commonly used for the representations under $SO(3)$ of states of systems that have some extension or some movement in space.

The combination of representations of rotations is again a representation of rotations, so it will still have the same generators with the same commutation relations. This is true for any combination, such as the combined spins for the electron and the proton, the combination of the orbital angular momentum and the spin of a particle or angular momenta for multi-particle systems.

You're right about derivations with ladder operators. You can use the approach that you know in every case, because it's derived from commutation relations.