Exploring Quantum Mechanics - even more about cross product.

Believe it or not, there are even more property in the cross product. This time our goal is the BAC-CAB identity

$\vec{a} \times (\vec{b} \times \vec{c}) = \vec{b}(\vec{a} \cdot \vec{c}) - \vec{c}(\vec{a} \cdot \vec{b})$.

To prove this, we need this result first, the so-called contraction identity.

$\epsilon_{ijk}\epsilon_{ist} = \delta_{js}\delta_{kt} - \delta_{jt}\delta_{ks}$.

To be honest, I don't really understand how does one come up with the identity, but it is relatively easy to verify that it is correct, and under what condition (that $i$ does not equal any of $j, k, s, t$

With that, we express the right hand side as follow:

$\begin{eqnarray*} \vec{a} \times (\vec{b} \times \vec{c}) &=& \vec{a} \times (\epsilon_{pqr}b_pc_qe_r) \\ &=& \epsilon_{srt}a_s\epsilon_{pqr}b_pc_qe_t \\ &=& \epsilon_{srt}\epsilon_{pqr}a_sb_pc_qe_t \\ &=& \epsilon_{rts}\epsilon_{rpq}a_sb_pc_qe_t \\ &=& (\delta_{tp}\delta_{sq} - \delta_{tq}\delta_{sp})a_sb_pc_qe_t \\ &=& \delta_{tp}\delta_{sq}a_sb_pc_qe_t - \delta_{tq}\delta_{sp}a_sb_pc_qe_t \\ &=& a_sb_tc_se_t - a_sb_sc_te_t \\ &=& \vec{b}(\vec{a} \cdot \vec{c}) - \vec{c}(\vec{a} \cdot \vec{b}) \end{eqnarray*}$