If we look at the simple matrix

{evals, evecs} = 
Eigensystem[{{\[CapitalDelta] - J/2 Cos[\[Theta]], 
    I*J/2*Sin[\[Theta]]}, {-I*J/2 Sin[\[Theta]], \[CapitalDelta] + 
    J/2 Cos[\[Theta]]}}]

and compute its normalized eigenvectors and simplify

(Normalize /@ evecs // Transpose // 
Assuming[-Pi < \[Theta] < Pi, FullSimplify[TrigToExp@#]] &) /. 
    {(1 + Abs[Cot[\[Theta]/2]]^2)^(-1/2) -> Abs[Sin[\[Theta]/2]]} 

we obtain the eigenvector matrix (each column one eigenvector):

$\left(\begin{array}{cc}-i\,\left|\sin(\frac \theta 2)\right|\,\cot(\frac\theta 2) & i\sin (\frac \theta 2)\\ \left|\sin(\frac\theta 2)\right| & \cos(\frac\theta 2)\end{array}\right)$

Now the left eigenvector looks rather complicated, however, since we can choose the phase freely and still have a normalized eigenvector, we could multiply the left eigenvector by the phase:


Which would greately simplify the eigenvectors to:

$\left(\begin{array}{cc}-i\cos(\frac\theta 2) & i\sin (\frac \theta 2)\\ \sin(\frac\theta 2) & \cos(\frac\theta 2)\end{array}\right)$

This was all done by hand now. Is there a way that Mathematica can help, i.e. can this result be obtained somehow automatically or semiautomatically?


1 Answer 1


Try this:

   (Normalize /@ evecs // Transpose // 
   Assuming[0 < \[Theta] < Pi, 
     FullSimplify[TrigToExp@#]] &) /. {(1 + 
      Abs[Cot[\[Theta]/2]]^2)^(-1/2) -> Abs[Sin[\[Theta]/2]]}

(*  {{-I Cos[\[Theta]/2], I Sin[\[Theta]/2]}, {Sin[\[Theta]/2], 
  Cos[\[Theta]/2]}}   *)

Edit: if theta<0,

(Normalize /@ evecs // Transpose // 
FullSimplify[TrigToExp@# /. \[Theta] -> -x, 
  0 < x < Pi] &) /. {(1 + Abs[Cot[\[Theta]/2]]^2)^(-1/2) -> 
Abs[Sin[\[Theta]/2]]} /. x -> -\[Theta]

yielding this:

 (*  {{I Cos[\[Theta]/2], I Sin[\[Theta]/2]}, {-Sin[\[Theta]/2], 
  Cos[\[Theta]/2]}}  *)

Have fun!

  • $\begingroup$ This yields the correct result. However, your assumption is of course a bit restrictive. Who knows, maybe for $\theta<0$ this does not work at all? ($\theta$ in my case has nothing to do with spherical coordinates and is not restricted in my case to [0,Pi). $\endgroup$
    – NOhs
    Commented Aug 18, 2015 at 15:12
  • $\begingroup$ What else did you mean by multiplying by sign(theta) ? $\endgroup$ Commented Aug 19, 2015 at 8:03
  • $\begingroup$ Well if $\vec{v}$ is an eigenvector then so is $c\cdot\vec{v}$. Now for every $\theta$ I try to find a $c$ so that in the end the resulting entries in the eigenvector as a function of $\theta$ are simple (smooth etc). $\endgroup$
    – NOhs
    Commented Aug 19, 2015 at 8:07
  • $\begingroup$ Well I only see that c v is zero at theta<0and is nonzero at theta>=0. Since a zero eigenvector is of no use, we are effectively left with the latter case. $\endgroup$ Commented Aug 19, 2015 at 8:13
  • $\begingroup$ Do I miss something? $c\cdot\vec v$ is not zero for my choice of $c$ for any $\theta$ imho. $\endgroup$
    – NOhs
    Commented Aug 19, 2015 at 8:42

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