A little background knowledge. We know that the imaginary quaternions $\mathrm{Im}\mathbb{H} = \mathrm{span}\{\sigma_1, \sigma_2, \sigma_3\}$and $\mathbb{R}^3$ are lie algebra isomorphims, i.e. $$ (\mathrm{Im} \mathbb{H}, [ , ]) \simeq (\mathbb{R}^3, \times), $$ where $\sigma_\alpha$ denotes the Pauli-matrices. The identification $$ X = -i\sum\limits_{\alpha=1}^{3}X_\alpha \sigma_\alpha \in \mathrm{Im}\mathbb{H} \longleftrightarrow X = (X_1,X_2,X_3) \in \mathbb{R}^3 $$ provides us with the following matrix reprensentation $$ X = \begin{pmatrix}-iX_3&-iX_1-X_2\\-iX_1 + X_2&iX_3\end{pmatrix}. $$ Now i have a matrix
$$ U = \frac{2u}{\beta^2}\begin{pmatrix}-i(u+u^{-1})&-i\bar{a}\\-ia&i(u+u^{-1})\end{pmatrix} \in \mathrm{Im}\mathbb{H}, $$ where $u$ is a positive real-valued function, $a$ a complex-valued function and $\beta^2 = 2 + \left|a\right|^2 + u^2 + u^{-2}$. I am having a hard time to transform $U$ into a vector in $\mathbb{R}^3$ by using this identification from above. I know that $X_3 = u + u^{-1}$. But I don't know how to obtain $X_1$ and $X_2$, respectively , since we have different entries on the off-diagonals. In the end I want to calculate the norm of the desired vector in $\mathbb{R}^3$. The desired result should be $$ \left|U\right|_{\mathbb{R}^3} = \frac{2u}{\beta} $$ Any suggenstions?

Thanks in advance :)

  • $\begingroup$ You could use the $4\times 4$ real valued matrix representation and just solve it with a couple of matrix equations as the real and imaginary parts have their well defined places in the respective $2\times 2$ blocks in that matrix representation. $\endgroup$ – mathreadler May 14 '17 at 20:15
  • $\begingroup$ In the end I want to calculate the norm of the desired vector in $\mathbb{R}^3$. I should probably mention that.. $\endgroup$ – aGer May 14 '17 at 20:17
  • $\begingroup$ isn't suppose that the FOUR pauli matrices in dirac equation represent the quaternions ? $\endgroup$ – Jose Garcia May 14 '17 at 20:30
  • $\begingroup$ You mean something like this $$ - X = X_1 \begin{pmatrix}0&i\\i&0\end{pmatrix} + X_2 \begin{pmatrix}0&1\\-1&0\end{pmatrix} + X_3 \begin{pmatrix}i&0\\0&-i\end{pmatrix} $$? $\endgroup$ – aGer May 14 '17 at 20:35
  • $\begingroup$ the 4 by 4 version I like is, to take two complex numbers and make the 2 by 2 $$ \left( \begin{array}{rr} \alpha & \beta \\ - \bar{\beta} & \bar{\alpha} \end{array} \right) $$ which, I think, comes out to the second set of 4 by 4 matrices at en.wikipedia.org/wiki/Quaternion#Matrix_representations $\endgroup$ – Will Jagy May 14 '17 at 21:11

By visual comparison, $X_3 = \dfrac {2u} {\beta^2} (u + u^{-1})$. If $a = A + \Bbb i B$, then

$$- \Bbb i X_1 + X_2 = \frac {2u} {\beta^2} (- \Bbb i a) = \frac {2u} {\beta^2} [- \Bbb i (A + \Bbb i B)] = - \Bbb i A \frac {2u} {\beta^2} + B \frac {2u} {\beta^2} \ ,$$

which gives $X_1 = A \dfrac {2u} {\beta^2} = \Re (a) \dfrac {2u} {\beta^2}$ and $X_2 = B \dfrac {2u} {\beta^2} = \Im (a) \dfrac {2u} {\beta^2}$.

Therefore, the isomorphism is

$$U \mapsto \frac {2u} {\beta^2} \big( u + u^{-1}, \Re (a), \Im (a) \big) \ .$$

  • $\begingroup$ Indeed! Thank you very much. I was not quite sure how $a$ has to be written. I thought at the same representation but was not sure how to handle $\Im(a)$ and $\Re(a)$ as entries in the vecor.. but then I realized that $\Re (a)^2 + \Im (a)^2 = \left| a \right|^2$.. $\endgroup$ – aGer May 14 '17 at 23:10

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