Total spin of two spin-1/2 particles

Question

Total spin of two spin-1/2 particles

Landau

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).

Marc C

You have to distinguish between

S1z $S_{1z}$ and

S2z $S_{2z}$ because every operator acts on a different space, so they are different matrices if expressed in the same base (Frobenius answer). Maybe it is more clear in ket notation?

Sz−total|m,m′⟩=(S1z+S2z)|m,m′⟩=S1z|m,m′⟩+S2z|m,m′⟩=ℏ(m+m′)|m,m′⟩ $S_{z-total} |m,m' \rangle =(S_{1z}+S_{2z})|m,m' \rangle =S_{1z}|m,m' \rangle +S_{2z}|m,m' \rangle =\hbar (m+m')|m,m'\rangle$

Respuestas (2)

Total spin of two spin-1/2 particles

You have to distinguish between $S_{1z}$ and $S_{2z}$ because every operator acts on a different space, so they are different matrices if expressed in the same base (Frobenius answer). Maybe it is more clear in ket notation? $S_{z-total} |m,m' \rangle =(S_{1z}+S_{2z})|m,m' \rangle =S_{1z}|m,m' \rangle +S_{2z}|m,m' \rangle =\hbar (m+m')|m,m'\rangle$

Frobenius · Answer 1

You must study about product states, product space of two (linear) spaces, product of linear transformations etc (product symbol $\;'\otimes\;'$ )

χ + (1) χ + (2) \equiv χ + (1) \otimes χ + (2) (01)

$\begin{equation} \chi_+(1)\chi_+(2) \equiv \chi_+(1) \otimes\chi_+(2) \tag{01} \end{equation}$

S z - t o t = S 1 z + S 2 z \equiv (S 1 z \otimes I 2) + (I 1 \otimes S 2 z) (02)

$\begin{equation} S_{z-tot}= S_{1z}+S_{2z}\equiv \left(S_{1z} \otimes I_2\right)+ \left(I_1 \otimes S_{2z}\right) \tag{02} \end{equation}$

S z - t o t χ + (1) χ + (2) = [S 1 z + S 2 z] χ + (1) χ + (2) \equiv [(S 1 z \otimes I 2) + (I 1 \otimes S 2 z)] [χ + (1) \otimes χ + (2)] = (S 1 z \otimes I 2) [χ + (1) \otimes χ + (2)] + (I 1 \otimes S 2 z) [χ + (1) \otimes χ + (2)] = [S 1 z χ + (1)] \otimes χ + (2) + χ + (1) \otimes [S 2 z χ + (2)] (03)

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

^{A representation :}

χ + (1) = [ξ 1 ξ 2], χ + (2) = [η 1 η 2] ⟹ χ + (1) \otimes χ + (2) = ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ξ 1 η 1 ξ 1 η 2 ξ 2 η 1 ξ 2 η 2 ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ (04)

$\begin{equation} \chi_+(1)= \begin{bmatrix} \xi_1\\ \xi_2 \end{bmatrix}\;,\; \chi_+(2)= \begin{bmatrix} \eta_1\\ \eta_2 \end{bmatrix} \quad \Longrightarrow \quad \chi_+(1) \otimes\chi_+(2) = \begin{bmatrix} \xi_1 \eta_1\\ \xi_1 \eta_2\\ \xi_2 \eta_1\\ \xi_2 \eta_2 \end{bmatrix} \tag{04} \end{equation}$ Now

S 1 z = [a 11 a 21 a 12 a 22], I 2 = [1001] \Rightarrow S 1 z \otimes I 2 = [a 11 a 21 a 12 a 22] \otimes [1001] = ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ a 11 \cdot [1001] a 21 \cdot [1001] a 12 \cdot [1001] a 22 \cdot [1001] ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ \Rightarrow S 1 z \otimes I 2 = ⎡ ⎣ ⎢ ⎢ ⎢ a 11 0 a 21 0 0 a 11 0 a 21 a 12 0 a 22 0 0 a 12 0 a 22 ⎤ ⎦ ⎥ ⎥ ⎥ (05)

$\begin{align} & S_{1z}= \begin{bmatrix} a_{11} & a_{12}\\ a_{21} & a_{22} \end{bmatrix}\;,\; I_2= \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} \nonumber\\ &\quad \Rightarrow \quad S_{1z} \otimes I_2= \begin{bmatrix} a_{11} & a_{12}\\ a_{21} & a_{22} \end{bmatrix} \otimes \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} = \begin{bmatrix} a_{11}\cdot\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} & a_{12}\cdot\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}\\ &\\ a_{21}\cdot\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} & a_{22}\cdot\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} \end{bmatrix} \nonumber\\ &\quad \Rightarrow \quad S_{1z} \otimes I_2= \begin{bmatrix} a_{11} & 0 & a_{12} & 0\\ 0 & a_{11} & 0 & a_{12} \\ a_{21} & 0 & a_{22} & 0\\ 0 & a_{21} & 0 & a_{22} \end{bmatrix} \tag{05} \end{align}$ and

I 1 = [1001], S 2 z = [b 11 b 21 b 12 b 22] \Rightarrow I 1 \otimes S 2 z = [1001] \otimes [b 11 b 21 b 12 b 22] = ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ 1 \cdot [b 11 b 21 b 12 b 22] 0 \cdot [b 11 b 21 b 12 b 22] 0 \cdot [b 11 b 21 b 12 b 22] 1 \cdot [b 11 b 21 b 12 b 22] ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ \Rightarrow I 1 \otimes S 2 z = ⎡ ⎣ ⎢ ⎢ ⎢ b 11 b 21 00 b 12 b 22 00 00 b 11 b 21 00 b 12 b 22 ⎤ ⎦ ⎥ ⎥ ⎥ (06)

$\begin{align} & I_1= \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}\;,\; S_{2z}= \begin{bmatrix} b_{11} & b_{12}\\ b_{21} & b_{22} \end{bmatrix} \nonumber\\ &\quad \Rightarrow \quad I_1 \otimes S_{2z}= \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} \otimes \begin{bmatrix} b_{11} & b_{12}\\ b_{21} & b_{22} \end{bmatrix} = \begin{bmatrix} 1\cdot \begin{bmatrix} b_{11} & b_{12}\\ b_{21} & b_{22} \end{bmatrix}&0\cdot\begin{bmatrix} b_{11} & b_{12}\\ b_{21} & b_{22} \end{bmatrix}\\ &\\ 0\cdot\begin{bmatrix} b_{11} & b_{12}\\ b_{21} & b_{22} \end{bmatrix}& 1\cdot\begin{bmatrix} b_{11} & b_{12}\\ b_{21} & b_{22} \end{bmatrix} \end{bmatrix} \nonumber\\ &\quad \Rightarrow \quad I_1 \otimes S_{2z}= \begin{bmatrix} b_{11} & b_{12} & 0 & 0\\ b_{21} & b_{22} & 0 & 0 \\ 0 & 0 & b_{11} & b_{12}\\ 0 & 0 & b_{21} & b_{22} \end{bmatrix} \tag{06} \end{align}$ From equations (05) and (06)

S z - t o t = (S 1 z \otimes I 2) + (I 1 \otimes S 2 z) = ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ (a 11 + b 11) b 21 a 21 0 b 12 (a 11 + b 22) 0 a 21 a 12 0 (a 22 + b 11) b 21 0 a 12 b 12 (a 22 + b 22) ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ (07)

$\begin{equation} S_{z-tot}=\left(S_{1z} \otimes I_2\right)+ \left(I_1 \otimes S_{2z}\right)= \begin{bmatrix} \left(a_{11}+b_{11}\right) & b_{12} & a_{12} & 0\\ b_{21} & \left(a_{11}+b_{22}\right) & 0 & a_{12} \\ a_{21} & 0 & \left(a_{22}+b_{11}\right) & b_{12}\\ 0 & a_{21} & b_{21} & \left(a_{22}+b_{22}\right) \end{bmatrix} \tag{07} \end{equation}$ If for example

S 1 z = 1 2 [10 0 - 1], S 2 z = 1 2 [10 0 - 1] (08)

$\begin{equation} S_{1z}=\tfrac{1}{2} \begin{bmatrix} 1 & 0\\ 0 &\!\!\! -\!1 \end{bmatrix}\;,\; S_{2z}=\tfrac{1}{2} \begin{bmatrix} 1 & 0\\ 0 &\!\!\! -\!1 \end{bmatrix} \tag{08} \end{equation}$ then

S z - t o t = (S 1 z \otimes I 2) + (I 1 \otimes S 2 z) = ⎡ ⎣ ⎢ ⎢ ⎢ 100000000000 000 - 1 ⎤ ⎦ ⎥ ⎥ ⎥ (09)

$\begin{equation} S_{z-tot}=\left(S_{1z} \otimes I_2\right)+ \left(I_1 \otimes S_{2z}\right)= \begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 &\!\!\! -\!1 \end{bmatrix} \tag{09} \end{equation}$ The matrix in (09) is already diagonal with eigenvalues 1,0,0,-1. Rearranging rows and columns we have

S' z - t o t = ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ 000001000000 000 - 1 ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ = ⎡ ⎣ S (j = 0) z 0 3 \times 1 0 1 \times 3 S (j = 1) z ⎤ ⎦ (10)

$\begin{equation} S'_{z-tot}= \begin{bmatrix} \begin{array}{c|cccc} 0 & 0 & 0 & 0\\ \hline 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & \!\!\!\!-\!1 \end{array} \end{bmatrix} = \begin{bmatrix} \begin{array}{c|c} S_{z}^{(j=0)} & 0_{1\times 3}\\ \hline 0_{3\times1} & S_{z}^{(j=1)} \end{array} \end{bmatrix} \tag{10} \end{equation}$ 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 $\;j=0\;$ and the 3-dimensional space of the angular momentum

j=1 $\;j=1\;$ :

2 \otimes 2 = 1 \oplus 3 (11)

$\begin{equation} \boldsymbol{2}\boldsymbol{\otimes}\boldsymbol{2}=\boldsymbol{1}\boldsymbol{\oplus}\boldsymbol{3} \tag{11} \end{equation}$ In general for two independent angular momenta

jα $\;j_{\alpha}\;$ and

jβ $\;j_{\beta}\;$ , living in the

(2jα+1)− $\;\left(2j_{\alpha}+1\right)-$ dimensional and

(2jβ+1)− $\;\left(2j_{\beta}+1\right)-$ dimensional spaces

Hα $\;\mathsf{H}_{\boldsymbol{\alpha}}\;$ and

Hβ $\;\mathsf{H}_{\boldsymbol{\beta}}\;$ respectively, their coupling is achieved by constructing the

(2jα+1)⋅(2jβ+1)− $\;\left(2j_{\alpha}+1\right)\cdot\left(2j_{\beta}+1\right)-$ dimensional product space

Hf $\;\mathsf{H}_{\boldsymbol{f}}\;$

H f \equiv H α \otimes H β (12)

$\begin{equation} \mathsf{H}_{\boldsymbol{f}}\equiv \mathsf{H}_{\boldsymbol{\alpha}}\boldsymbol{\otimes}\mathsf{H}_{\boldsymbol{\beta}} \tag{12} \end{equation}$ Then the product space

Hf $\:\mathsf{H}_{\boldsymbol{f}}\:$ is expressed as the direct sum of

n $\:n\:$ mutually orthogonal subspaces

Hρ(ρ=1,2,⋯,n−1,n) $\:\mathsf{H}_{\boldsymbol{\rho}}\: (\rho=1,2,\cdots,n-1,n)$

H f \equiv H α \otimes H β = H 1 \oplus H 2 \oplus \dots \oplus H n = ⨁ ρ = 1 ρ = n H ρ (13)

$\begin{equation} \mathsf{H}_{\boldsymbol{f}}\equiv \mathsf{H}_{\boldsymbol{\alpha}}\boldsymbol{\otimes}\mathsf{H}_{\boldsymbol{\beta}} = \mathsf{H}_{\boldsymbol{1}}\boldsymbol{\oplus}\mathsf{H}_{\boldsymbol{2}} \boldsymbol{\oplus} \cdots \boldsymbol{\oplus} \mathsf{H}_{\boldsymbol{n}}=\bigoplus_{{\boldsymbol{\rho}}={\boldsymbol{1}}}^{{\boldsymbol{\rho}}={\boldsymbol{n}}} \mathsf{H}_{\boldsymbol{\rho}} \tag{13} \end{equation}$ where the subspace

Hρ $\:\mathsf{H}_{\boldsymbol{\rho}}\:$ corresponds to angular momentum

jρ $\;j_{\rho}\;$ and has dimension

dim (H ρ) = 2 \cdot j ρ + 1 (14)

$\begin{equation} \dim \left(\mathsf{H}_{\boldsymbol{\rho}}\right) =2\cdot j_{\rho}+1 \tag{14} \end{equation}$ with

j ρ n = | j β - j α | + ρ - 1, ρ = 1, 2, \dots, n - 1, n = 2 \cdot min (j α, j β) + 1 (15a) (15b)

$\begin{align} j_{\rho} & = \vert j_{\beta}-j_{\alpha} \vert +\rho - 1\: , \quad \rho=1,2,\cdots,n-1,n \tag{15a}\\ n & =2\cdot\min (j_{\alpha}, j_{\beta})+1 \tag{15b} \end{align}$ Equation (13) is expressed also in terms of the dimensions of spaces and subspaces as :

(2 j α + 1) \otimes (2 j β + 1) = ⨁ ρ = 1 ρ = n (2 j ρ + 1) (16)

$\begin{equation} (2j_{\alpha}+1)\boldsymbol{\otimes} (2j_{\beta}+1)=\bigoplus_{\rho=1}^{\rho=n}(2j_{\rho}+1) \tag{16} \end{equation}$ Equation (11) is a special case of equation (16) :

j α = 1 2, j β = 1 2 ⟹ j 1 = 0, j 2 = 1 (17)

$\begin{equation} j_{\alpha}=\tfrac{1}{2} \:,\:j_{\beta}=\tfrac{1}{2} \: \quad \Longrightarrow \quad \: j_{1}=0 \:,\: j_{2}=1 \tag{17} \end{equation}$

^{^{(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 :}}

S' 2 = ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ 0000020000200002 ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ = ⎡ ⎣ ⎢ ⎢ ⎢ (S' 2) (j = 0) 0 3 \times 1 0 1 \times 3 (S' 2) (j = 1) ⎤ ⎦ ⎥ ⎥ ⎥ (10')

$\begin{equation} \mathbf{S'}^2= \begin{bmatrix} \begin{array}{c|cccc} 0 & 0 & 0 & 0\\ \hline 0 & 2 & 0 & 0 \\ 0 & 0 & 2 & 0\\ 0 & 0 & 0 & 2 \end{array} \end{bmatrix} = \begin{bmatrix} \begin{array}{c|c} \left(\mathbf{S'}^2\right)^{(j=0)} & 0_{1\times 3}\\ \hline 0_{3\times1} & \left(\mathbf{S'}^2\right)^{(j=1)} \end{array} \end{bmatrix} \tag{10'} \end{equation}$ since for

S 1 x = 1 2 [0110], S 2 x = 1 2 [0110] (18)

$\begin{equation} S_{1x}=\tfrac{1}{2} \begin{bmatrix} 0 & 1\\ 1 & 0 \end{bmatrix}\;,\; S_{2x}=\tfrac{1}{2} \begin{bmatrix} 0 & 1\\ 1 & 0 \end{bmatrix} \tag{18} \end{equation}$

S 1 y = 1 2 [0 i - i 0], S 2 y = 1 2 [0 i - i 0] (19)

$\begin{equation} S_{1y}=\tfrac{1}{2} \begin{bmatrix} 0 &\!\!\! -\!i\\ i & 0 \end{bmatrix}\;,\; S_{2y}=\tfrac{1}{2} \begin{bmatrix} 0 &\!\!\! -\!i\\ i & 0 \end{bmatrix} \tag{19} \end{equation}$ we have

S x - t o t = (S 1 x \otimes I 2) + (I 1 \otimes S 2 x) = 1 2 ⎡ ⎣ ⎢ ⎢ ⎢ 0110100110010110 ⎤ ⎦ ⎥ ⎥ ⎥ (20)

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

S y - t o t = (S 1 y \otimes I 2) + (I 1 \otimes S 2 y) = 1 2 ⎡ ⎣ ⎢ ⎢ ⎢ 0 i i 0 - i 00 i - i 00 i 0 - i - i 0 ⎤ ⎦ ⎥ ⎥ ⎥ (21)

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

S 2 x - t o t S 2 y - t o t S 2 z - t o t = 1 4 ⎡ ⎣ ⎢ ⎢ ⎢ 0110100110010110 ⎤ ⎦ ⎥ ⎥ ⎥ 2 = 1 2 ⎡ ⎣ ⎢ ⎢ ⎢ 1001011001101001 ⎤ ⎦ ⎥ ⎥ ⎥ = 1 4 ⎡ ⎣ ⎢ ⎢ ⎢ 0 i i 0 - i 00 i - i 00 i 0 - i - i 0 ⎤ ⎦ ⎥ ⎥ ⎥ 2 = 1 2 ⎡ ⎣ ⎢ ⎢ ⎢ 100 - 1 01100110 - 1 001 ⎤ ⎦ ⎥ ⎥ ⎥ = ⎡ ⎣ ⎢ ⎢ ⎢ 100000000000 000 - 1 ⎤ ⎦ ⎥ ⎥ ⎥ 2 = ⎡ ⎣ ⎢ ⎢ ⎢ 1000000000000001 ⎤ ⎦ ⎥ ⎥ ⎥ (22x) (22y) (22z)

$\begin{align} S^{2}_{x-tot} & =\tfrac{1}{4} \begin{bmatrix} 0 & 1 & 1 & 0\\ 1 & 0 & 0 & 1 \\ 1 & 0 & 0 & 1\\ 0 & 1 & 1 & 0 \end{bmatrix}^{2} =\tfrac{1}{2} \begin{bmatrix} 1 & 0 & 0 & 1\\ 0 & 1 & 1 & 0 \\ 0 & 1 & 1 & 0\\ 1 & 0 & 0 & 1 \end{bmatrix} \tag{22x}\\ S^{2}_{y-tot} & =\tfrac{1}{4} \begin{bmatrix} 0 & \!\!\! -\!i & \!\!\! -\!i & 0\\ i & 0 & 0 & \!\!\! -\!i \\ i & 0 & 0 & \!\!\! -\!i \\ 0 & i & i & 0 \end{bmatrix}^2 =\tfrac{1}{2} \begin{bmatrix} 1 & 0 & 0 & \!\!\! -\!1 \\ 0 & 1 & 1 & 0\\ 0 & 1 & 1 & 0\\ \!\!\! -\!1 & 0 & 0 & 1 \end{bmatrix} \tag{22y}\\ S^{2}_{z-tot} & =\quad \!\! \begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 &\!\!\!-\!1 \end{bmatrix}^{2} =\quad \!\! \begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 1 \end{bmatrix} \tag{22z} \end{align}$ From

S 2 t o t = S 2 x - t o t + S 2 y - t o t + S 2 z - t o t (23)

$\begin{equation} \mathbf{S}^{2}_{tot}=S^{2}_{x-tot}+S^{2}_{y-tot}+S^{2}_{z-tot} \tag{23} \end{equation}$ we have finally

S 2 t o t = ⎡ ⎣ ⎢ ⎢ ⎢ 2000011001100002 ⎤ ⎦ ⎥ ⎥ ⎥ (24)

$\begin{equation} \mathbf{S}^{2}_{tot}= \begin{bmatrix} 2 & 0 & 0 & 0\\ 0 & 1 & 1 & 0 \\ 0 & 1 & 1 & 0\\ 0 & 0 & 0 & 2 \end{bmatrix} \tag{24} \end{equation}$ For its eigenvalues

λ $\lambda$

det (S 2 t o t - λ I 4) = ∣ ∣ ∣ ∣ ∣ ∣ 2 - λ 000 0 1 - λ 10 01 1 - λ 0 000 2 - λ ∣ ∣ ∣ ∣ ∣ ∣ = - λ (2 - λ) 3 (25)

$\begin{equation} \det\left(\mathbf{S}^{2}_{tot}-\lambda I_{4}\right)= \begin{vmatrix} 2-\lambda & 0 & 0 & 0\\ 0 & 1-\lambda & 1 & 0 \\ 0 & 1 & 1-\lambda & 0\\ 0 & 0 & 0 & 2-\lambda \end{vmatrix} =-\lambda \left(2-\lambda \right)^{3} \tag{25} \end{equation}$ So the eigenvalues of

S2tot $\;\mathbf{S}^{2}_{tot}\;$ are: the eigenvalue

λ1=0=j1(j1+1) $\lambda_{1}=0=j_{1}\left(j_{1}+1\right)$ with multiplicity 1 and the eigenvalue

λ2=2=j2(j2+1) $\lambda_{2}=2=j_{2}\left(j_{2}+1\right)$ with multiplicity 3.

That is a very good answer (and a lot of work - and I haven't even looked at parts 2 and 3 yet). When is the book coming out?
The book part was very much tongue-in-cheek. I've already upvoted your answer but I will now upvote the question for bringing out this answer. Thank you.
I have to ask: I write a lot of LaTeX myself, and writing this much would be very difficult for me. Do you have a special LaTeX workflow that speeds it up? There's no way you write it in the browser, right?
@knzhou In many answers of mine, like the ones here, I had already written in LaTeX (TexLive 2015 or 2016 or 2017 using Texmaker) the main equations and text. But to write the code with Texmaker-TexLive for new answers was very difficult and time consuming. Of course it's impossible to write it in the browser because it would be annoying to have your answer perpetually on the front page.
@knzhou So, I was lucky to find an online editor : Overleaf (first v1 and later v2). You edit and the compilation is direct and online showing you your probable mistakes. Completing your answer you copy-paste LaTeX code in PSE. See images here : Overleaf v2

Frobenius · Answer 2

F O U R T H___ A N S W E R

^{(upvote or downvote my 1rst answer only. My 2nd,3rd,4th and 5th answers are addenda to it)}

(continued from T H I R D___ A N S W E R )

SECTION C : Angular Momentum Coupling and Product Transformations

Let again the two systems $\alpha$ and $\beta$ with angular momentum $j_{\alpha}$ and $j_{\beta}$ respectively, independent between each other and living in the real space $\;\mathbb{R}^{3}$ . Suppose also that both $j_{\alpha}$ and $j_{\beta}$ are integers corresponding to orbital angular momentum.

Now, let an infinitesimal rotation by angle $\;\delta \theta\;$ around an axis with unit vector $\;\boldsymbol{n}=\left(n_{1},n_{2},n_{3}\right)\;$ . Since the $\;\boldsymbol{n}-$ component of orbital angular momentum is the generator of rotations around this axis, the infinitesimal changes of states $\;\boldsymbol{\xi}\in\mathsf{H}_{\boldsymbol{\alpha}}\;$ and $\;\boldsymbol{\eta}\in\mathsf{H}_{\boldsymbol{\beta}}\;$ due to this infinitesimal rotation are

δ ξ δ η = i δ θ J α n ξ = i δ θ J β n η (63a) (63b)

$\begin{align} \delta \boldsymbol{\xi} & = i\, \delta\theta \,J^{\boldsymbol{\alpha}}_{\boldsymbol{n}}\, \boldsymbol{\xi} \tag{63a}\\ \delta \boldsymbol{\eta} & = i\, \delta\theta \, J^{\boldsymbol{\beta}}_{\boldsymbol{n}}\, \boldsymbol{\eta} \tag{63b} \end{align}$ where for the

n− $\;\boldsymbol{n}-$ components of the vector operators we have

J α n J β n = n \cdot J α = n 1 J α 1 + n 2 J α 2 + n 3 J α 3 = n \cdot J β = n 1 J β 1 + n 2 J β 2 + n 3 J β 3 (64a) (64b)

$\begin{align} J^{\boldsymbol{\alpha}}_{\boldsymbol{n}} & = \boldsymbol{n}\boldsymbol{\cdot}\mathbf{J^{\boldsymbol{\alpha}}}=n_{1}J^{\boldsymbol{\alpha}}_{1}+n_{2}J^{\boldsymbol{\alpha}}_{2}+n_{3}J^{\boldsymbol{\alpha}}_{3} \tag{64a}\\ J^{\boldsymbol{\beta}}_{\boldsymbol{n}} & = \boldsymbol{n}\boldsymbol{\cdot}\mathbf{J^{\boldsymbol{\beta}}}=n_{1}J^{\boldsymbol{\beta}}_{1}+n_{2}J^{\boldsymbol{\beta}}_{2}+n_{3}J^{\boldsymbol{\beta}}_{3} \tag{64b} \end{align}$ If we intend to construct a consistent angular momentum operator

J $\;\mathbf{J}\;$ of the coupled system

f $\;f\;$ in the product space

Hf=Hα⊗Hβ $\:\mathsf{H}_{\boldsymbol{f}}=\mathsf{H}_{\boldsymbol{\alpha}}\boldsymbol{\otimes}\mathsf{H}_{\boldsymbol{\beta}}\:$ , then the infinitesimal change of the product state

(ξ⊗η) $\left(\boldsymbol{\xi} \boldsymbol{\otimes} \boldsymbol{\eta}\right)$ must be

δ (ξ \otimes η) = i δ θ J n (ξ \otimes η) (65)

$\begin{equation} \delta \left(\boldsymbol{\xi} \boldsymbol{\otimes} \boldsymbol{\eta}\right) = i\, \delta\theta \,J_{\b<script>_addload(function(){_setupIW('com');_csi('en','es','https://physics.stackexchange.com/original/5c11ae20b52c4a25f829bb94');});$

Total spin of two spin-1/2 particles

Landau

Marc C

Respuestas (2)

Frobenius

NickD

NickD

knzhou

Frobenius

Frobenius

Frobenius

¿Los giros tienen direcciones espaciales?

Momento angular de espín total de un sistema de dos partículas de espín-1/2

¿Por qué los sistemas de dos electrones generalmente se describen en base a triplete-singlete?

Probabilidad de estado de espín de electrones [cerrado]

Confusión con respecto a la adición de espín y acoplamiento LSLSLS

¿Cómo asociar un espacio de Hilbert con un sistema QM?

Interpretación geométrica de la base rotada de estados hamiltonianos y colectivos de Dicke

Cuadrado de las matrices de Pauli y la matriz identidad

De la representación de espín del vector de estado a otra representación

Rotaciones de estado de espín y rotaciones de espinores