# Spin operators on 2 spin half particles

I have been given the operator: $$S_{12}=3(\sigma_1.e)(\sigma_2.e)-\sigma_1.\sigma_2,$$ where $e$ is a unit vector connecting the 2 particles and $\sigma_i$ is the pauli vector operator acting on particle $i$.

Show for a system is a singlet state that: $$S_{12}|\Psi\rangle=0.$$

so, $$|\Psi\rangle=\frac{1}{\sqrt{2} }(|+\rangle\otimes|-\rangle-|-\rangle\otimes|+\rangle).$$

What I assume happens is:

$$S_{12}|+\rangle\otimes|-\rangle=3(\sigma_1.e)|+\rangle\otimes(\sigma_2.e)|-\rangle-\sum_i \sigma_{1i}|+\rangle\otimes\sigma_{2i}|-\rangle$$

However this on the singlet state doesn't get $0$

How does this $S_{12}$ operator work?

• Is it $S_{12} = 3 (\sigma_1\cdot e)(\sigma_2\cdot e) +\sigma_1 \cdot \sigma_2$ or $S_{12} = 3 (\sigma_1\cdot e)(\sigma_2\cdot e) - \sigma_1 \cdot \sigma_2$? – secavara Feb 1 '18 at 15:14
• The tensor operator defined with negative sign as pointed out by secavara – SAKhan Feb 1 '18 at 16:06
• Honestly, now you have the sign right and also you have written how the operator works accurately on $|+-\rangle$. If you do the math carefully you'll get the result. Remember picking $e$ to be a normalized vector. You can pick $e=\{\sin \theta \cos \phi, \sin \theta \sin \phi,\cos \theta\}$, for instance, or even just $e=\{0,0,1\}$ to begin with. You can also use a computational too. – secavara Feb 1 '18 at 19:19
• I only managed to get it to work if $e=0$ which would imply that the spin aligns along this unit vector. – Toby Peterken Feb 1 '18 at 22:00
• Yes, you have $e \cdot \sigma=\begin{bmatrix} \cos \theta & \mathrm{e}^{-i\phi} \sin \theta \\ \mathrm{e}^{i\phi} \sin \theta & -\cos \theta \end{bmatrix}$ and therefore $e \cdot \sigma |+\rangle$ gives the result you have there. – secavara Feb 1 '18 at 22:24

The space of the 2-particle system is the 4-dimensional Hilbert space $\:\mathbb{C}^{4}=\mathbb{C}^{2}\boldsymbol{\otimes}\mathbb{C}^{2}$, the product space of the two 1-particle system spaces $\:\mathbb{C}^{2}$.

If for the particles we use the symbols $\:\alpha,\beta\:$ instead of $\:1,2\:$ then the single state $\:\Psi\:$ mentioned in the question is a complex 4-dimensional vector and more exactly $$\Psi \sim \begin{bmatrix} 1\\ 0 \end{bmatrix}_{\!\alpha} \!\!\!\otimes\! \begin{bmatrix} 0\\ 1 \end{bmatrix}_{\!\beta} \!\!\!-\!\! \begin{bmatrix} 0\\ 1 \end{bmatrix}_{\!\alpha} \!\!\!\otimes\! \begin{bmatrix} 1\\ 0 \end{bmatrix}_{\!\beta} = \begin{bmatrix} 0\\ 1\\ 0\\ 0 \end{bmatrix}_{\!\alpha\beta} \!\!\!\!\!-\!\! \begin{bmatrix} 0\\ 0\\ 1\\ 0 \end{bmatrix}_{\!\alpha\beta} \!\!\!= \begin{bmatrix} \hphantom{\!-\!}0\:\:\\ \hphantom{\!-\!}1\:\:\\ \!-\!1\:\:\\ \hphantom{\!-\!}0\:\: \end{bmatrix}_{\!\alpha\beta} \tag{01}$$ andthe operator $\:S_{\alpha\beta} \equiv S_{12}\:$ is sum of operator products as follows : $$S_{\alpha\beta}=3\left(\boldsymbol{\sigma}^{\alpha}\boldsymbol{\cdot}\mathbf{e}\vphantom{\boldsymbol{\sigma}^{\beta}}\right)\boldsymbol{\otimes}\left(\boldsymbol{\sigma}^{\beta}\boldsymbol{\cdot}\mathbf{e}\right)-\boldsymbol{\sigma}^{\alpha}\boldsymbol{\otimes}\boldsymbol{\sigma}^{\beta} \tag{02}$$ where $$\boldsymbol{\sigma}^{\alpha}\boldsymbol{\otimes}\boldsymbol{\sigma}^{\beta}\equiv \sigma^{\alpha}_{1}\otimes\sigma^{\beta}_{1}+\sigma^{\alpha}_{2}\otimes\sigma^{\beta}_{2}+\sigma^{\alpha}_{3}\otimes\sigma^{\beta}_{3} \tag{03}$$ Without loss of generality1 we may suppose that $$\mathbf{e}= \begin{bmatrix} 1\\ 0\\ 0 \end{bmatrix} \tag{04}$$ and (02) gives $$S_{\alpha\beta}=2\sigma^{\alpha}_{1}\otimes\sigma^{\beta}_{1}-\sigma^{\alpha}_{2}\otimes\sigma^{\beta}_{2}-\sigma^{\alpha}_{3}\otimes\sigma^{\beta}_{3} \tag{05}$$ Now, \begin{align} \sigma^{\alpha}_{1}\otimes\sigma^{\beta}_{1} & = \begin{bmatrix} 0&\hphantom{-}1\\ 1&\hphantom{-}0 \end{bmatrix} \otimes \begin{bmatrix} 0&\hphantom{-}1\\ 1&\hphantom{-}0 \end{bmatrix} = \begin{bmatrix} \hphantom{-}0&\hphantom{-}0&\hphantom{-}0&\hphantom{-}\color{blue}{\bf 1}\hphantom{-}\\ \hphantom{-}0&\hphantom{-}0&\hphantom{-}\color{blue}{\bf 1}&\hphantom{-}0\hphantom{-}\\ \hphantom{-}0&\hphantom{-}\color{blue}{\bf 1}&\hphantom{-}0&\hphantom{-}0\hphantom{-}\\ \hphantom{-}\color{blue}{\bf 1}&\hphantom{-}0&\hphantom{-}0&\hphantom{-}0\hphantom{-} \end{bmatrix} \tag{06.1}\\ \sigma^{\alpha}_{2}\otimes\sigma^{\beta}_{2} & = \begin{bmatrix} 0&-i\\ i&\hphantom{-}0 \end{bmatrix} \otimes \begin{bmatrix} 0&-i\\ i&\hphantom{-}0 \end{bmatrix} = \begin{bmatrix} \hphantom{-}0&\hphantom{-}0&\hphantom{-}0&-\color{blue}{\bf 1}\hphantom{-}\\ \hphantom{-}0&\hphantom{-}0&\hphantom{-}\color{blue}{\bf 1}&\hphantom{-}0\hphantom{-}\\ \hphantom{-}0&\hphantom{-}\color{blue}{\bf 1}&\hphantom{-}0&\hphantom{-}0\hphantom{-}\\ -\color{blue}{\bf 1}&\hphantom{-}0&\hphantom{-}0&\hphantom{-}0\hphantom{-} \end{bmatrix} \tag{06.2}\\ \sigma^{\alpha}_{3}\otimes\sigma^{\beta}_{3} & = \begin{bmatrix} 1&\hphantom{-}0\\ 0&-1 \end{bmatrix} \otimes \begin{bmatrix} 1&\hphantom{-}0\\ 0&-1 \end{bmatrix} = \begin{bmatrix} \hphantom{-}\color{blue}{\bf 1}&\hphantom{-}0&\hphantom{-}0&\hphantom{-}0\hphantom{-}\\ \hphantom{-}0&-\color{blue}{\bf 1}&\hphantom{-}0&\hphantom{-}0\hphantom{-}\\ \hphantom{-}0&\hphantom{-}0&-\color{blue}{\bf 1}&\hphantom{-}0\hphantom{-}\\ \hphantom{-}0&\hphantom{-}0&\hphantom{-}0&\hphantom{-}\color{blue}{\bf 1}\hphantom{-} \end{bmatrix} \tag{06.3} \end{align} so $$S_{\alpha\beta}=2 \begin{bmatrix} \hphantom{-}0&\hphantom{-}0&\hphantom{-}0&\hphantom{-}1\hphantom{-}\\ \hphantom{-}0&\hphantom{-}0&\hphantom{-}1&\hphantom{-}0\hphantom{-}\\ \hphantom{-}0&\hphantom{-}1&\hphantom{-}0&\hphantom{-}0\hphantom{-}\\ \hphantom{-}1&\hphantom{-}0&\hphantom{-}0&\hphantom{-}0\hphantom{-} \end{bmatrix} \!-\! \begin{bmatrix} \hphantom{-}0&\hphantom{-}0&\hphantom{-}0&-1\hphantom{-}\\ \hphantom{-}0&\hphantom{-}0&\hphantom{-}1&\hphantom{-}0\hphantom{-}\\ \hphantom{-}0&\hphantom{-}1&\hphantom{-}0&\hphantom{-}0\hphantom{-}\\ -1&\hphantom{-}0&\hphantom{-}0&\hphantom{-}0\hphantom{-} \end{bmatrix} \!-\! \begin{bmatrix} \hphantom{-}1&\hphantom{-}0&\hphantom{-}0&\hphantom{-}0\hphantom{-}\\ \hphantom{-}0&-1&\hphantom{-}0&\hphantom{-}0\hphantom{-}\\ \hphantom{-}0&\hphantom{-}0&-1&\hphantom{-}0\hphantom{-}\\ \hphantom{-}0&\hphantom{-}0&\hphantom{-}0&\hphantom{-}1\hphantom{-} \end{bmatrix} \tag{07}$$ or $$S_{\alpha\beta}= \begin{bmatrix} -1&\hphantom{-}0&\hphantom{-}0&\hphantom{-}3\hphantom{-}\\ \hphantom{-}0&\hphantom{-}1&\hphantom{-}1&\hphantom{-}0\hphantom{-}\\ \hphantom{-}0&\hphantom{-}1&\hphantom{-}1&\hphantom{-}0\hphantom{-}\\ \hphantom{-}3&\hphantom{-}0&\hphantom{-}0&-1\hphantom{-} \end{bmatrix} \tag{08}$$ and finally $$S_{\alpha\beta}\Psi\sim \begin{bmatrix} -1&\hphantom{-}0&\hphantom{-}0&\hphantom{-}3\hphantom{-}\\ \hphantom{-}0&\hphantom{-}1&\hphantom{-}1&\hphantom{-}0\hphantom{-}\\ \hphantom{-}0&\hphantom{-}1&\hphantom{-}1&\hphantom{-}0\hphantom{-}\\ \hphantom{-}3&\hphantom{-}0&\hphantom{-}0&-1\hphantom{-} \end{bmatrix} \begin{bmatrix} \hphantom{-}0\hphantom{-}\\ \hphantom{-}1\hphantom{-}\\ -1\hphantom{-}\\ \hphantom{-}0\hphantom{-} \end{bmatrix} = \begin{bmatrix} \hphantom{-}0\hphantom{-}\\ \hphantom{-}0\hphantom{-}\\ \hphantom{-}0\hphantom{-}\\ \hphantom{-}0\hphantom{-} \end{bmatrix} \tag{09}$$

EDIT

$^{1}$ To be sure that equation (09) is valid in general let
$$\mathbf{e}= \begin{bmatrix} e_{1}\\ e_{2}\\ e_{3} \end{bmatrix} \in \mathbb{R}^{3},\quad \Vert\mathbf{e}\Vert^{2}=e_{1}^{2}+e_{2}^{2}+e_{3}^{2}=1 \tag{ed-01}$$ From equations (06) $$\boldsymbol{\sigma}^{\alpha}\boldsymbol{\otimes}\boldsymbol{\sigma}^{\beta}\equiv \sigma^{\alpha}_{1}\otimes\sigma^{\beta}_{1}+\sigma^{\alpha}_{2}\otimes\sigma^{\beta}_{2}+\sigma^{\alpha}_{3}\otimes\sigma^{\beta}_{3}= \begin{bmatrix} \hphantom{-}1&\hphantom{-}0&\hphantom{-}0&\hphantom{-}0\hphantom{-}\\ \hphantom{-}0&-1&\hphantom{-}2&\hphantom{-}0\hphantom{-}\\ \hphantom{-}0&\hphantom{-}2&-1&\hphantom{-}0\hphantom{-}\\ \hphantom{-}0&\hphantom{-}0&\hphantom{-}0&\hphantom{-}1\hphantom{-} \end{bmatrix} \tag{ed-02}$$ Now \begin{align} \left(\boldsymbol{\sigma}^{\alpha}\boldsymbol{\cdot}\mathbf{e}\vphantom{\boldsymbol{\sigma}^{\beta}}\right)\boldsymbol{\otimes}\left(\boldsymbol{\sigma}^{\beta}\boldsymbol{\cdot}\mathbf{e}\right) & = \begin{bmatrix} e_3 & \left(e_1-i e_2\right) \vphantom{\dfrac12} \\ \left(e_1+i e_2\right) & -e_3 \vphantom{\dfrac12} \end{bmatrix} \boldsymbol{\otimes} \begin{bmatrix} e_3 & \left(e_1-i e_2\right) \vphantom{\dfrac12} \\ \left(e_1+i e_2\right) & -e_3 \vphantom{\dfrac12} \end{bmatrix} \nonumber\\ & = \begin{bmatrix} \hphantom{\left(e_1\:\,\, i e_2\right)}e_3 \begin{bmatrix} e_3 & \left(e_1-i e_2\right) \\ \left(e_1+i e_2\right) & -e_3 \vphantom{\dfrac12} \end{bmatrix} & \left(e_1-i e_2\right)\vphantom{\dfrac12} \begin{bmatrix} e_3 & \left(e_1-i e_2\right) \\ \left(e_1+i e_2\right) & -e_3 \vphantom{\dfrac12} \end{bmatrix} \\ \left(e_1+i e_2\right)\vphantom{\dfrac12} \begin{bmatrix} e_3 & \left(e_1-i e_2\right) \\ \left(e_1+i e_2\right) & -e_3 \vphantom{\dfrac12} \end{bmatrix} & \hphantom{\left(e_1\!e_2\right)}-e_3 \begin{bmatrix} e_3 & \left(e_1-i e_2\right) \\ \left(e_1+i e_2\right) & -e_3 \vphantom{\dfrac12} \end{bmatrix} \vphantom{\dfrac12} \end{bmatrix} \nonumber\\ &= \begin{bmatrix} e_3^{2} & \left(e_3e_1-i e_3e_2\right) & \left(e_1e_3-i e_2e_3\right) & \left(e_1-i e_2\right)^2 \vphantom{\dfrac12} \\ \left(e_3e_1+i e_3e_2\right) & -e_3^{2}& \left(e_1^{2}+e_2^{2}\right) & -\left(e_1e_3-i e_2e_3\right)\vphantom{\dfrac12}\\ \left(e_1e_3+i e_2e_3\right) & \left(e_1^{2}+e_2^{2}\right) & -e_3^{2} & -\left(e_3e_1-i e_3e_2\right) \vphantom{\dfrac12}\\ \left(e_1+i e_2\right)^2 & -\left(e_1e_3+i e_2e_3\right) & -\left(e_3e_1+i e_3e_2\right) & e_3^{2} \vphantom{\dfrac12} \end{bmatrix} \tag{ed-03} \end{align} that is $$\left(\boldsymbol{\sigma}^{\alpha}\boldsymbol{\cdot}\mathbf{e}\vphantom{\boldsymbol{\sigma}^{\beta}}\right)\boldsymbol{\otimes}\left(\boldsymbol{\sigma}^{\beta}\boldsymbol{\cdot}\mathbf{e}\right) = \begin{bmatrix} e_3^{2} & \left(e_3e_1-i e_3e_2\right) & \left(e_1e_3-i e_2e_3\right) & \left(e_1-i e_2\right)^2 \vphantom{\dfrac12} \\ \left(e_3e_1+i e_3e_2\right) & -e_3^{2}& \left(e_1^{2}+e_2^{2}\right) & -\left(e_1e_3-i e_2e_3\right)\vphantom{\dfrac12}\\ \left(e_1e_3+i e_2e_3\right) & \left(e_1^{2}+e_2^{2}\right) & -e_3^{2} & -\left(e_3e_1-i e_3e_2\right) \vphantom{\dfrac12}\\ \left(e_1+i e_2\right)^2 & -\left(e_1e_3+i e_2e_3\right) & -\left(e_3e_1+i e_3e_2\right) & e_3^{2} \vphantom{\dfrac12} \end{bmatrix} \tag{ed-04}$$ So $$S_{\alpha\beta}= \begin{bmatrix} \left(3e_3^{2}-1\right) & 3\left(e_3e_1-i e_3e_2\right) & 3\left(e_1e_3-i e_2e_3\right) & 3\left(e_1-i e_2\right)^2 \vphantom{\dfrac12} \\ 3\left(e_3e_1+i e_3e_2\right) & -\left(3e_3^{2}-1\right)& 3\left(e_1^{2}+e_2^{2}\right)-2 & -3\left(e_1e_3-i e_2e_3\right)\vphantom{\dfrac12}\\ 3\left(e_1e_3+i e_2e_3\right) & 3\left(e_1^{2}+e_2^{2}\right)-2 & -\left(3e_3^{2}-1\right) & -3\left(e_3e_1-i e_3e_2\right) \vphantom{\dfrac12}\\ 3\left(e_1+i e_2\right)^2 & -3\left(e_1e_3+i e_2e_3\right) & -3\left(e_3e_1+i e_3e_2\right) & \left(3e_3^{2}-1\right) \vphantom{\dfrac12} \end{bmatrix} \tag{ed-05}$$ or $$S_{\alpha\beta}= \begin{bmatrix} \left(3e_3^{2}-1\right) & 3\left(e_3e_1-i e_3e_2\right) & 3\left(e_1e_3-i e_2e_3\right) & 3\left(e_1-i e_2\right)^2 \vphantom{\dfrac12} \\ 3\left(e_3e_1+i e_3e_2\right) & -\left(3e_3^{2}-1\right)& -\left(3e_3^{2}-1\right) & -3\left(e_1e_3-i e_2e_3\right)\vphantom{\dfrac12}\\ 3\left(e_1e_3+i e_2e_3\right) & -\left(3e_3^{2}-1\right) & -\left(3e_3^{2}-1\right) & -3\left(e_3e_1-i e_3e_2\right) \vphantom{\dfrac12}\\ 3\left(e_1+i e_2\right)^2 & -3\left(e_1e_3+i e_2e_3\right) & -3\left(e_3e_1+i e_3e_2\right) & \left(3e_3^{2}-1\right) \vphantom{\dfrac12} \end{bmatrix} \tag{ed-06}$$ and again in general $$S_{\alpha\beta}\Psi\sim \begin{bmatrix} \hphantom{-}0\hphantom{-}\\ \hphantom{-}0\hphantom{-}\\ \hphantom{-}0\hphantom{-}\\ \hphantom{-}0\hphantom{-} \end{bmatrix} \tag{ed-07}$$ Note that from equations (01) and (ed-02) we have $$\left[\!\!\left[\boldsymbol{\sigma}^{\alpha}\boldsymbol{\otimes}\boldsymbol{\sigma}^{\beta}\right]\!\!\right]\Psi= \begin{bmatrix} \hphantom{-}1&\hphantom{-}0&\hphantom{-}0&\hphantom{-}0\hphantom{-}\\ \hphantom{-}0&-1&\hphantom{-}2&\hphantom{-}0\hphantom{-}\\ \hphantom{-}0&\hphantom{-}2&-1&\hphantom{-}0\hphantom{-}\\ \hphantom{-}0&\hphantom{-}0&\hphantom{-}0&\hphantom{-}1\hphantom{-} \end{bmatrix} \begin{bmatrix} \hphantom{\!-\!}0\:\:\\ \hphantom{\!-\!}1\:\:\\ \!-\!1\:\:\\ \hphantom{\!-\!}0\:\: \end{bmatrix} = \begin{bmatrix} \hphantom{\!-\!}0\:\:\\ \!-\!3\:\:\\ \hphantom{\!-\!}3\:\:\\ \hphantom{\!-\!}0\:\: \end{bmatrix} =-3\Psi \tag{ed-08}$$ while from equations (01) and (ed-04) we have $$\left[\!\!\left[\left(\boldsymbol{\sigma}^{\alpha}\boldsymbol{\cdot}\mathbf{e}\vphantom{\boldsymbol{\sigma}^{\beta}}\right)\boldsymbol{\otimes}\left(\boldsymbol{\sigma}^{\beta}\boldsymbol{\cdot}\mathbf{e}\right)\right]\!\!\right]\Psi= \begin{bmatrix} \hphantom{\!-\!}0\:\:\\ \!-\!\left(e_1^{2}+e_2^{2}+e_3^{2}\right)\:\:\\ \hphantom{\!-\!}\left(e_1^{2}+e_2^{2}+e_3^{2}\right)\:\:\\ \hphantom{\!-\!}0\:\: \end{bmatrix} = \begin{bmatrix} \hphantom{\!-\!}0\:\:\\ \!-\!1\:\:\\ \hphantom{\!-\!}1\:\:\\ \hphantom{\!-\!}0\:\: \end{bmatrix} =-\Psi \tag{ed-09}$$ that is the lhs expression $\:\left[\!\!\left[\left(\boldsymbol{\sigma}^{\alpha}\boldsymbol{\cdot}\mathbf{e}\vphantom{\boldsymbol{\sigma}^{\beta}}\right)\boldsymbol{\otimes}\left(\boldsymbol{\sigma}^{\beta}\boldsymbol{\cdot}\mathbf{e}\right)\right]\!\!\right]\Psi\:$ is independent of the choice of the unit vector $\:\mathbf{e}$.

• but there is loss of generality assuming $e=(1,0,0)$ as it is now perpendicular to the direction if spin – Toby Peterken Feb 2 '18 at 8:32
• @Toby Peterken : Have you appreciated what the scalar products of the 3-vectors $\hat e$ and $\vec \sigma$ mean, and how they transform under a 3-rotation? secavara solved the problem for you in excruciating, painful, detail. Why don't you just rotate $\hat{e}\to \hat{z}$, plug in, and check? – Cosmas Zachos Feb 4 '18 at 14:23