Section 3: (Note: Keep scrolling to the bottom for more problems/solutions)
Chapter 2, Section 3: Selected problems and solutions to NFCM, 2nd Ed.
Note: This is a new websection on my website. I expect to be updating
this section from time to time.
Click on this link to find Problem (3.3) on page 62 of NFCM.
Problem (3.3) on page 62 of NFCM.
Click on this link to find Problem (3.8) on page 63 of NFCM.
Problem (3.8) on page 63 of NFCM.
On page 63 of NFCM, we find Problem (3.8):
Let $\bB = \half B_{\ell p}\,\bsigma_\ell\wedge\bsigma_p$ (sum on repeated indices)
be a bivector and $\bb = b_k\bsigma_k$
(sum on repeated indices) be a vector and they are related by the equation
\begin{equation}
\bB = i\bb\,.\label{eq:FirstEq}
\end{equation}
Prove that $B_{ij} = \epsilon_{ijk}b_k$, where $ \epsilon_{ijk} \definedas i^\dagger\bsigma_i\wedge \bsigma_j\wedge\bsigma_k$.
Lemma:
\begin{equation}
\bsigma_\ell\wedge\bsigma_p\cdot \bsigma_i\wedge\bsigma_j =
\delta_{pi} \delta_{\ell j} - \delta_{pj} \delta_{i\ell}\,.\label{eq:deltas}
\end{equation}
This is left to the reader to prove.
Proof of main result:
Expanding (\ref{eq:FirstEq}), we have
\begin{equation}
\half B_{\ell p}\,\bsigma_\ell\wedge\bsigma_p = i b_k\bsigma_k\,.\label{eq:FirstEqExpanded}
\end{equation}
On dotting through by $ \bsigma_i\wedge\bsigma_j$ on the right and using (\ref{eq:deltas}), we have
\begin{equation}
\half B_{\ell p}\,( \delta_{pi} \delta_{\ell j} - \delta_{pj} \delta_{i\ell})
= b_k(i\bsigma_k)\cdot( \bsigma_i\wedge\bsigma_j)
= b_k\langle i\bsigma_k( \bsigma_i\wedge\bsigma_j)\rangle\,.\label{eq:FirstEqExpanded.bivector}
\end{equation}
Which becomes
\begin{equation}
\half (B_{ji} - B_{ij}) = b_k(i\bsigma_k)\cdot( \bsigma_i\wedge\bsigma_j) = b_k\langle i\bsigma_k( \bsigma_i\wedge\bsigma_j)\rangle\,.\label{eq:FirstEqExpanded.bivector.2}
\end{equation}
And this becomes
\begin{equation}
- B_{ij} = - b_k( i^\dagger\bsigma_i\wedge \bsigma_j\wedge\bsigma_k)\,.\label{eq:FirstEqExpanded.bivector.simpler}
\end{equation}
And finally, we get that
\begin{equation}
B_{ij} = \epsilon_{ijk}b_k\,.
\end{equation}
Click on this link to find Problem (3.10) on page 64 of NFCM.
Problem (3.10) on page 64 of NFCM.
NFCM Problems and Solutions