# Geometric product of 2 vectors derived from Pauli matrices

The 3 Pauli matrices are:

${\color{blue}{\sigma_1}}$ = $\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$

${\color{blue}{\sigma_2}}$ = $\begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}$

${\color{blue}{\sigma_3}}$ = $\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$

Define 2 "vectors" to be:

$\mathbf{\vec{u}} = u_1 {\color{blue}\sigma_1} + u_3 {\color{blue}\sigma_3}= \begin{pmatrix} u_3 & u_1 \\ u_1 & -u_3 \end{pmatrix}$

$\mathbf{\vec{v}} = v_1 {\color{blue}\sigma_1} + v_3 {\color{blue}\sigma_3} = \begin{pmatrix} v_3 & v_1 \\ v_1 & -v_3 \end{pmatrix}$

It then follows that:

\begin{align} \mathbf{\vec{u}\vec{v}} &= \left( \begin{array}{rr} u_3 & u_1 \\ u_1 & -u_3 \end{array} \right) \left( \begin{array}{rr} v_3 & v_1 \\ v_1 & -v_3 \end{array} \right) \\ &= \left( \begin{array}{rr} u_1 v_1 + u_3 v_3 & u_3 v_1 - u_1 v_3 \\ u_1 v_3 - u_3 v_1 & u_1 v_1 + u_3 v_3 \end{array} \right) \\ &= (u_1 v_1 + u_3 v_3) \cdot {\color{red} I_2} + (u_3 v_1 - u_1 v_3 ) \cdot {\color{green} {\boldsymbol{\hat{\jmath}}}} \\ &= \mathbf{\vec{u}} \cdot \mathbf{\vec{v}} + \mathbf{\vec{u}} \wedge \mathbf{\vec{v}} \end{align}

Where:

${\color{green} {\boldsymbol{\hat{\jmath}}}} = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} = quaternion$

${\color{green} {\boldsymbol{\hat{\jmath}}}}$ is rotation from $\sigma_3$ to $\sigma_1$

Edit: Sorry. I got my u's and v's mixed up and got a little confused with the wedge product. The solution to the problem is that there was never any problem to begin with.

• The wedge product is $u\wedge v = \frac{1}{2}(uv-vu)$. If you use this you obtain the correct answer. Aug 29, 2018 at 21:49

$\mathbf{\vec{u}} = u_1 {\color{blue}\sigma_1} + u_2 {\color{blue}\sigma_2} + u_3 {\color{blue}{\sigma_3}}$
$\mathbf{\vec{v}} = v_1 {\color{blue}\sigma_1} + v_2 {\color{blue}\sigma_2} + v_3 {\color{blue}\sigma_3}$
\begin{align} \mathbf{\vec{u}\vec{v}} &= \left( \begin{array}{ll} u_3 & \phantom{-}u_1-u_2 i \\ u_1+u_2 i & -u_3 \end{array} \right) \left( \begin{array}{ll} v_3 & \phantom{-}v_1-v_2 i \\ v_1+v_2 i & -v_3 \end{array} \right) \\ &= \left( \begin{array}{rr} (u_3)(v_3) + (u_1-u_2 i)(v_1+v_2 i) & (u_3)(v_1-v_2 i) + (u_1-u_2 i)(-v_3) \\ (u_1+u_2 i)(v_3) + (-u_3 )(v_1+v_2 i) & (u_1+u_2 i)(v_1-v_2 i) + (-u_3 )(-v_3) \end{array} \right) \\ &= \left( \begin{array}{rr} u_1 v_1 + u_2 v_2 + v_3 u_3 + (u_1 v_2 - u_2 v_1)i & u_3 v_1 - u_1 v_3 + (u_2 v_3 - u_3 v_2)i \\ u_1 v_3 - u_3 v_1 + (u_2 v_3 - u_3 v_2)i & + u_1 v_1 + u_2 v_2 + v_3 u_3 + (u_2 v_1 - u_1 v_2)i \end{array} \right) \\ &= (u_1 v_1 + u_2 v_2 + v_3 u_3) \cdot {\color{red} I_2} + (u_1 v_2 - u_2 v_1) \cdot {\color{green} {\boldsymbol{\hat{\imath}}}} + (u_3 v_1 - u_1 v_3) \cdot {\color{green} {\boldsymbol{\hat{\jmath}}}} + (u_2 v_3 - u_3 v_2) \cdot {\color{green} {\boldsymbol{\hat{k}}}} \\ & \text{Now switching back to ordinary vector notation} \\ &= \mathbf{\vec{u}} \cdot \mathbf{\vec{v}} + \mathbf{\vec{u}} \wedge \mathbf{\vec{v}} = \left( \begin{array}{rr} (\mathbf{\vec{u}} \cdot \mathbf{\vec{v}}) & {\color{green}{(u_1 v_2 - u_2 v_1)}} & (u_1 v_3 - u_3 v_1) \\ (u_2 v_1 - u_1 v_2) & (\mathbf{\vec{u}} \cdot \mathbf{\vec{v}}) & {\color{green}{(u_2 v_3 - u_3 v_2)}} \\ {\color{green}{(u_3 v_1 - u_1 v_3)}} & (u_3 v_2 - u_2 v_3) & (\mathbf{\vec{u}} \cdot \mathbf{\vec{v}}) \end{array} \right) \end{align}