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I read that the generators of the Lie algebra in this representation are $$ J_k= \begin{pmatrix} \frac{1}{2}\sigma_k & 0 \\ 0 & \frac{1}{2}\sigma_k \end{pmatrix} $$ (Rotations) and $$ K_k= \begin{pmatrix} -\frac{i}{2}\sigma_k & 0 \\ 0 & \frac{i}{2}\sigma_k \end{pmatrix} $$ (Boosts) Since $u\in SU(2)$ is parametrized by $$\exp(-\theta^k\sigma_k)=\begin{pmatrix} e^{i\theta^1}\cos(\theta^2) & -e^{-i\theta^3}\sin(\theta^2) \\ e^{i\theta^3}\sin(\theta^2) & e^{-i\theta^1}\cos(\theta^2) \end{pmatrix}$$ It would seem $$ exp(ia^kJ_k)=\begin{pmatrix} exp(\frac{ia^k}{2}\sigma_k) & 0 \\ 0 & exp(\frac{ia^k}{2}\sigma_k) \end{pmatrix}$$ $$=\begin{pmatrix} e^{\frac{-a^1}{2}}cosh(\frac{a^2}{2}) & ie^{\frac{a^3}{2}}sinh(\frac{a^2}{2}) & 0 & 0 \\ -ie^{\frac{-a^3}{2}}sinh(\frac{a^2}{2}) & e^{\frac{a^1}{2}}cosh(\frac{a^2}{2}) & 0 & 0 \\ 0 & 0 & e^{\frac{a^1}{2}}cosh(\frac{a^2}{2}) & -ie^{-\frac{a^3}{2}}sinh(\frac{a^2}{2}) \\ 0 & 0 & ie^{\frac{a^3}{2}}sinh(\frac{a^2}{2}) & e^{-\frac{a^1}{2}}cosh(\frac{a^2}{2}) \\ \end{pmatrix}$$ Which I find odd since a boost would be of the same format but with trigonometric instead of hyperbolic functions-which seems backwards since rotations are parametrized with 3 spherical angles and boosts with 2 spherical angles and 1 hyperbolic angle. Did I mess up my algebra or is there an explanation for this that I'm missing?

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  • $\begingroup$ The last step is wrong. Your matrix expression is not correct. $\endgroup$ – OkThen Nov 17 '16 at 20:42
  • $\begingroup$ @OkThen I suspected this, if $-\theta^k=\frac{ia^k}{2}$ then $\theta^k=-\frac{ia^k}{2}$ so I forgot the factor of one half-I'll edit this in a sec but this doesn't explain the presence of hyperbolic functions in the rotations. $\endgroup$ – Eben Cowley Nov 17 '16 at 20:53
  • $\begingroup$ Your mistake is in passing from $\exp(ia\sigma/2)$ to the 2x2 block matrix in the final equation. I don't know what mistake you made, but Wikipedia has the correct expression for $\exp(ia\sigma/2)$ en.wikipedia.org/wiki/Pauli_matrices $\endgroup$ – Luke Pritchett Nov 17 '16 at 21:01
  • $\begingroup$ There are no hyperbolic functions. The exponential of $i \sigma$ is unitary. $\endgroup$ – OkThen Nov 17 '16 at 21:07
  • $\begingroup$ @LukePritchett Ah I see where I made a silly assumption now $\endgroup$ – Eben Cowley Nov 17 '16 at 21:11

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