I am trying to solve:

$\frac{d U_t}{dt} = Tr(G^{\dagger}U_t)G - Tr(U_t^{\dagger}G)U_t G^{\dagger} U_t$

Where $U_t \in SU(4)$ and $G \in SU(4)$ is given and constant.

Is it possible to solve this equation? A solution not in coordinates would be most helpful. I've tried several series methods previously.

  • $\begingroup$ for $G$ hermitian and traceless I see $U=I$ as a solution. $\endgroup$ – James S. Cook Aug 14 '15 at 21:20
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    $\begingroup$ If $n \ne 4$,how can this work? So you must mean $n = 4$, right? Cheers! $\endgroup$ – Robert Lewis Aug 14 '15 at 21:30
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    $\begingroup$ Are you sure you have $U_tG^\dagger U_t$ instead of $U_t^\dagger G^\dagger U_t$ or $U_tG^\dagger U_t^\dagger$ on the right hand side? If so, putting $V_t=G^\dagger U_t\in SU(4)$ will simplify the equation a little bit: $\dot{V_t} = tr(V_t)I - tr(V_t^\dagger)V_t^2$. $\endgroup$ – user1551 Aug 15 '15 at 11:37
  • $\begingroup$ Stating the obvious: thus there are infinitely many constant solutions, because any constant and real $V\in SO(4)$ with $V^2=I$ would satisfy the equation. $\endgroup$ – user1551 Aug 15 '15 at 15:09
  • $\begingroup$ Yes, I'm sure it is as stated. Awkward as this may be, nature did it apparently. $\endgroup$ – Benjamin Aug 15 '15 at 16:37

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