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===Relation to dot and cross product=== Pauli vectors elegantly map these commutation and anticommutation relations to corresponding vector products. Adding the commutator to the anticommutator gives :<math> \begin{align} \left[ \sigma_j, \sigma_k\right] + \{\sigma_j, \sigma_k\} &= (\sigma_j \sigma_k - \sigma_k \sigma_j ) + (\sigma_j \sigma_k + \sigma_k \sigma_j) \\ 2i\varepsilon_{j k \ell}\,\sigma_\ell + 2 \delta_{j k}I &= 2\sigma_j \sigma_k \end{align} </math> so that, {{Equation box 1 |indent =: |equation = <math>~~ \sigma_j \sigma_k = \delta_{j k}I + i\varepsilon_{j k \ell}\,\sigma_\ell ~ .~</math> |cellpadding= 6 |border |border colour = #0073CF |bgcolor=#F9FFF7 }} [[tensor contraction|Contracting]] each side of the equation with components of two {{math|3}}-vectors {{math|''a{{sub|p}}''}} and {{math|''b{{sub|q}}''}} (which commute with the Pauli matrices, i.e., {{math|1=''a{{sub|p}}Ο{{sub|q}}'' = ''Ο{{sub|q}}a{{sub|p}}'')}} for each matrix {{math|''Ο{{sub|q}}''}} and vector component {{math|''a{{sub|p}}''}} (and likewise with {{math|''b{{sub|q}}''}}) yields :<math>~~ \begin{align} a_j b_k \sigma_j \sigma_k & = a_j b_k \left(i\varepsilon_{jk\ell}\,\sigma_\ell + \delta_{jk}I\right) \\ a_j \sigma_j b_k \sigma_k & = i\varepsilon_{jk\ell}\,a_j b_k \sigma_\ell + a_j b_k \delta_{jk}I \end{align}.~</math> Finally, translating the index notation for the [[dot product]] and [[cross product#Index notation for tensors|cross product]] results in {{NumBlk||{{Equation box 1 |indent =: |equation = <math>~~\Bigl(\vec{a} \cdot \vec{\sigma}\Bigr)\Bigl(\vec{b} \cdot \vec{\sigma}\Bigr) = \Bigl(\vec{a} \cdot \vec{b}\Bigr) \, I + i \Bigl(\vec{a} \times \vec{b}\Bigr) \cdot \vec{\sigma}~~</math> |cellpadding= 6 |border |border colour = #0073CF |bgcolor=#F9FFF7 }} |{{EquationRef|1}} }} If {{mvar|i}} is identified with the pseudoscalar {{math|''Ο{{sub|x}}Ο{{sub|y}}Ο{{sub|z}}''}} then the right hand side becomes <math> a \cdot b + a \wedge b </math>, which is also the definition for the product of two vectors in geometric algebra. If we define the spin operator as {{math|1='''''J''''' = {{sfrac|''Δ§''|2}}'''''Ο'''''}}, then {{math|1='''''J'''''}} satisfies the commutation relation:<math display="block">\mathbf{J} \times \mathbf{J} = i\hbar \mathbf{J}</math>Or equivalently, the Pauli vector satisfies:<math display="block">\frac{\vec{\sigma}}{2} \times \frac{\vec{\sigma}}{2} = i\frac{\vec{\sigma}}{2}</math>
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