quantum mechanics - Total spin of two spin-$1/2$ particles

On my book I read:

$S_{z-tot}\chi_+(1)\chi_+(2)=[S_{1z}+S_{2z}]\chi_+(1)\chi_+(2)=[S_{1z}\chi_+(1)]\chi_+(2)+[S_2\chi_+(2)]\chi_+(1)=...$

Now, I have two questions:

• What's $\chi_+(1)\chi_+(2)$ ? I know that $\chi_+=(1,0)$ but I really don't understand that writing (what is a product between vectors?!). Is it maybe just a way to indicate a vector in $C^4$? Or instead it is a matrix or other?


• What's $S_{z-tot}$? Is it 2x2 matrix or a, 4x4 matrix or other?

I also don't understand why the book distinguishes $S_{1z}$ and $S_{2z}$, aren't them the same 2x2 matrix defined for the single electron?

Thanks for your attenction and please answer in a simple way (I'm a begginer in these subjects).

Answer

You must study about product states, product space of two (linear) spaces, product of linear transformations etc (product symbol $\;'\otimes\;'$) $$\begin{equation} \chi_+(1)\chi_+(2) \equiv \chi_+(1) \otimes\chi_+(2) \tag{01} \end{equation}$$ $$\begin{equation} S_{z-tot}= S_{1z}+S_{2z}\equiv \left(S_{1z} \otimes I_2\right)+ \left(I_1 \otimes S_{2z}\right) \tag{02} \end{equation}$$

$$\begin{align} &S_{z-tot}\chi_+(1)\chi_+(2)=[S_{1z}+S_{2z}]\chi_+(1)\chi_+(2) \nonumber\\ &\equiv \left[\left(S_{1z} \otimes I_2\right)+ \left(I_1 \otimes S_{2z}\right)\right]\left[\chi_+(1) \otimes\chi_+(2)\right] \nonumber\\ &=\left(S_{1z} \otimes I_2\right)\left[\chi_+(1) \otimes\chi_+(2)\right]+\left(I_1 \otimes S_{2z}\right)\left[\chi_+(1) \otimes\chi_+(2)\right] \nonumber\\ &=\left[S_{1z}\chi_+(1)\right] \otimes\chi_+(2)+\chi_+(1) \otimes\left[S_{2z}\chi_+(2)\right] \tag{03} \end{align}$$

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^{A representation : $$\begin{equation} \chi_+(1)= \begin{bmatrix} \xi_1\\ \xi_2 \end{bmatrix}\;,\; \chi_+(2)= \begin{bmatrix} \eta_1\\ \eta_2 \end{bmatrix} \quad \Longrightarrow \quad \chi_+(1) \otimes\chi_+(2) = \begin{bmatrix} \xi_1 \eta_1\\ \xi_1 \eta_2\\ \xi_2 \eta_1\\ \xi_2 \eta_2 \end{bmatrix} \tag{04} \end{equation}$$ Now $$\begin{align} & S_{1z}= \begin{bmatrix} a_{11} & a_{12}\\ a_{21} & a_{22} \end{bmatrix}\;,\; I_2= \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} \nonumber\\ &\quad \Rightarrow \quad S_{1z} \otimes I_2= \begin{bmatrix} a_{11} & a_{12}\\ a_{21} & a_{22} \end{bmatrix} \otimes \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} = \begin{bmatrix} a_{11}\cdot\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} & a_{12}\cdot\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}\\ &\\ a_{21}\cdot\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} & a_{22}\cdot\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} \end{bmatrix} \nonumber\\ &\quad \Rightarrow \quad S_{1z} \otimes I_2= \begin{bmatrix} a_{11} & 0 & a_{12} & 0\\ 0 & a_{11} & 0 & a_{12} \\ a_{21} & 0 & a_{22} & 0\\ 0 & a_{21} & 0 & a_{22} \end{bmatrix} \tag{05} \end{align}$$ and $$\begin{align} & I_1= \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}\;,\; S_{2z}= \begin{bmatrix} b_{11} & b_{12}\\ b_{21} & b_{22} \end{bmatrix} \nonumber\\ &\quad \Rightarrow \quad I_1 \otimes S_{2z}= \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} \otimes \begin{bmatrix} b_{11} & b_{12}\\ b_{21} & b_{22} \end{bmatrix} = \begin{bmatrix} 1\cdot \begin{bmatrix} b_{11} & b_{12}\\ b_{21} & b_{22} \end{bmatrix}&0\cdot\begin{bmatrix} b_{11} & b_{12}\\ b_{21} & b_{22} \end{bmatrix}\\ &\\ 0\cdot\begin{bmatrix} b_{11} & b_{12}\\ b_{21} & b_{22} \end{bmatrix}& 1\cdot\begin{bmatrix} b_{11} & b_{12}\\ b_{21} & b_{22} \end{bmatrix} \end{bmatrix} \nonumber\\ &\quad \Rightarrow \quad I_1 \otimes S_{2z}= \begin{bmatrix} b_{11} & b_{12} & 0 & 0\\ b_{21} & b_{22} & 0 & 0 \\ 0 & 0 & b_{11} & b_{12}\\ 0 & 0 & b_{21} & b_{22} \end{bmatrix} \tag{06} \end{align}$$ From equations (05) and (06) $$\begin{equation} S_{z-tot}=\left(S_{1z} \otimes I_2\right)+ \left(I_1 \otimes S_{2z}\right)= \begin{bmatrix} \left(a_{11}+b_{11}\right) & b_{12} & a_{12} & 0\\ b_{21} & \left(a_{11}+b_{22}\right) & 0 & a_{12} \\ a_{21} & 0 & \left(a_{22}+b_{11}\right) & b_{12}\\ 0 & a_{21} & b_{21} & \left(a_{22}+b_{22}\right) \end{bmatrix} \tag{07} \end{equation}$$ 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
$$\begin{equation} S'_{z-tot}= \begin{bmatrix} \begin{array}{c|cccc} 0 & 0 & 0 & 0\\ \hline 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & \!\!\!\!-\!1 \end{array} \end{bmatrix} = \begin{bmatrix} \begin{array}{c|c} S_{z}^{(j=0)} & 0_{1\times 3}\\ \hline 0_{3\times1} & S_{z}^{(j=1)} \end{array} \end{bmatrix} \tag{10} \end{equation}$$ because, as could be proved(1), the product 4-dimensional Hilbert space is the direct sum of two orthogonal spaces : the 1-dimensional space of the angular momentum $\;j=0\;$ and the 3-dimensional space of the angular momentum $\;j=1\;$ : $$\begin{equation} \boldsymbol{2}\boldsymbol{\otimes}\boldsymbol{2}=\boldsymbol{1}\boldsymbol{\oplus}\boldsymbol{3} \tag{11} \end{equation}$$ In general for two independent angular momenta $\;j_{\alpha}\;$ and $\;j_{\beta}\;$, living in the $\;\left(2j_{\alpha}+1\right)-$ dimensional and $\;\left(2j_{\beta}+1\right)-$ dimensional spaces $\;\mathsf{H}_{\boldsymbol{\alpha}}\;$ and $\;\mathsf{H}_{\boldsymbol{\beta}}\;$ respectively, their coupling is achieved by constructing the $\;\left(2j_{\alpha}+1\right)\cdot\left(2j_{\beta}+1\right)-$ dimensional product space $\;\mathsf{H}_{\boldsymbol{f}}\;$
$$\begin{equation} \mathsf{H}_{\boldsymbol{f}}\equiv \mathsf{H}_{\boldsymbol{\alpha}}\boldsymbol{\otimes}\mathsf{H}_{\boldsymbol{\beta}} \tag{12} \end{equation}$$ Then the product space $\:\mathsf{H}_{\boldsymbol{f}}\:$ is expressed as the direct sum of $\:n\:$ mutually orthogonal subspaces $\:\mathsf{H}_{\boldsymbol{\rho}}\: (\rho=1,2,\cdots,n-1,n)$ $$\begin{equation} \mathsf{H}_{\boldsymbol{f}}\equiv \mathsf{H}_{\boldsymbol{\alpha}}\boldsymbol{\otimes}\mathsf{H}_{\boldsymbol{\beta}} = \mathsf{H}_{\boldsymbol{1}}\boldsymbol{\oplus}\mathsf{H}_{\boldsymbol{2}} \boldsymbol{\oplus} \cdots \boldsymbol{\oplus} \mathsf{H}_{\boldsymbol{n}}=\bigoplus_{{\boldsymbol{\rho}}={\boldsymbol{1}}}^{{\boldsymbol{\rho}}={\boldsymbol{n}}} \mathsf{H}_{\boldsymbol{\rho}} \tag{13} \end{equation}$$ where the subspace $\:\mathsf{H}_{\boldsymbol{\rho}}\:$ corresponds to angular momentum $\;j_{\rho}\;$ and has dimension $$\begin{equation} \dim \left(\mathsf{H}_{\boldsymbol{\rho}}\right) =2\cdot j_{\rho}+1 \tag{14} \end{equation}$$ with $$\begin{align} j_{\rho} & = \vert j_{\beta}-j_{\alpha} \vert +\rho - 1\: , \quad \rho=1,2,\cdots,n-1,n \tag{15a}\\ n & =2\cdot\min (j_{\alpha}, j_{\beta})+1 \tag{15b} \end{align}$$ Equation (13) is expressed also in terms of the dimensions of spaces and subspaces as : $$\begin{equation} (2j_{\alpha}+1)\boldsymbol{\otimes} (2j_{\beta}+1)=\bigoplus_{\rho=1}^{\rho=n}(2j_{\rho}+1) \tag{16} \end{equation}$$ Equation (11) is a special case of equation (16) : $$\begin{equation} j_{\alpha}=\tfrac{1}{2} \:,\:j_{\beta}=\tfrac{1}{2} \: \quad \Longrightarrow \quad \: j_{1}=0 \:,\: j_{2}=1 \tag{17} \end{equation}$$}

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^{(1) the square of total angular momentum $\mathbf{S}^2$ expressed in the basis of its common with $\:S_{z-tot}\:$ eigenvectors has the following diagonal form : $$\begin{equation} \mathbf{S'}^2= \begin{bmatrix} \begin{array}{c|cccc} 0 & 0 & 0 & 0\\ \hline 0 & 2 & 0 & 0 \\ 0 & 0 & 2 & 0\\ 0 & 0 & 0 & 2 \end{array} \end{bmatrix} = \begin{bmatrix} \begin{array}{c|c} \left(\mathbf{S'}^2\right)^{(j=0)} & 0_{1\times 3}\\ \hline 0_{3\times1} & \left(\mathbf{S'}^2\right)^{(j=1)} \end{array} \end{bmatrix} \tag{10'} \end{equation}$$ since for $$\begin{equation} S_{1x}=\tfrac{1}{2} \begin{bmatrix} 0 & 1\\ 1 & 0 \end{bmatrix}\;,\; S_{2x}=\tfrac{1}{2} \begin{bmatrix} 0 & 1\\ 1 & 0 \end{bmatrix} \tag{18} \end{equation}$$ $$\begin{equation} S_{1y}=\tfrac{1}{2} \begin{bmatrix} 0 &\!\!\! -\!i\\ i & 0 \end{bmatrix}\;,\; S_{2y}=\tfrac{1}{2} \begin{bmatrix} 0 &\!\!\! -\!i\\ i & 0 \end{bmatrix} \tag{19} \end{equation}$$ we have $$\begin{equation} S_{x-tot}=\left(S_{1x} \otimes I_2\right)+ \left(I_1 \otimes S_{2x}\right) =\tfrac{1}{2} \begin{bmatrix} 0 & 1 & 1 & 0\\ 1 & 0 & 0 & 1 \\ 1 & 0 & 0 & 1\\ 0 & 1 & 1 & 0 \end{bmatrix} \tag{20} \end{equation}$$ $$\begin{equation} S_{y-tot}=\left(S_{1y} \otimes I_2\right)+ \left(I_1 \otimes S_{2y}\right)=\tfrac{1}{2} \begin{bmatrix} 0 & \!\!\! -\!i & \!\!\! -\!i & 0\\ i & 0 & 0 & \!\!\! -\!i \\ i & 0 & 0 & \!\!\! -\!i \\ 0 & i & i & 0 \end{bmatrix} \tag{21} \end{equation}$$ and consequently $$\begin{align} S^{2}_{x-tot} & =\tfrac{1}{4} \begin{bmatrix} 0 & 1 & 1 & 0\\ 1 & 0 & 0 & 1 \\ 1 & 0 & 0 & 1\\ 0 & 1 & 1 & 0 \end{bmatrix}^{2} =\tfrac{1}{2} \begin{bmatrix} 1 & 0 & 0 & 1\\ 0 & 1 & 1 & 0 \\ 0 & 1 & 1 & 0\\ 1 & 0 & 0 & 1 \end{bmatrix} \tag{22x}\\ S^{2}_{y-tot} & =\tfrac{1}{4} \begin{bmatrix} 0 & \!\!\! -\!i & \!\!\! -\!i & 0\\ i & 0 & 0 & \!\!\! -\!i \\ i & 0 & 0 & \!\!\! -\!i \\ 0 & i & i & 0 \end{bmatrix}^2 =\tfrac{1}{2} \begin{bmatrix} 1 & 0 & 0 & \!\!\! -\!1 \\ 0 & 1 & 1 & 0\\ 0 & 1 & 1 & 0\\ \!\!\! -\!1 & 0 & 0 & 1 \end{bmatrix} \tag{22y}\\ S^{2}_{z-tot} & =\quad \!\! \begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 &\!\!\!-\!1 \end{bmatrix}^{2} =\quad \!\! \begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 1 \end{bmatrix} \tag{22z} \end{align}$$ From $$\begin{equation} \mathbf{S}^{2}_{tot}=S^{2}_{x-tot}+S^{2}_{y-tot}+S^{2}_{z-tot} \tag{23} \end{equation}$$ we have finally $$\begin{equation} \mathbf{S}^{2}_{tot}= \begin{bmatrix} 2 & 0 & 0 & 0\\ 0 & 1 & 1 & 0 \\ 0 & 1 & 1 & 0\\ 0 & 0 & 0 & 2 \end{bmatrix} \tag{24} \end{equation}$$ For its eigenvalues $\lambda$ $$\begin{equation} \det\left(\mathbf{S}^{2}_{tot}-\lambda I_{4}\right)= \begin{vmatrix} 2-\lambda & 0 & 0 & 0\\ 0 & 1-\lambda & 1 & 0 \\ 0 & 1 & 1-\lambda & 0\\ 0 & 0 & 0 & 2-\lambda \end{vmatrix} =-\lambda \left(2-\lambda \right)^{3} \tag{25} \end{equation}$$ So the eigenvalues of $\;\mathbf{S}^{2}_{tot}\;$ are: the eigenvalue $\lambda_{1}=0=j_{1}\left(j_{1}+1\right)$ with multiplicity 1 and the eigenvalue $\lambda_{2}=2=j_{2}\left(j_{2}+1\right)$ with multiplicity 3.}

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