0
$\begingroup$

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 :)

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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

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

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.