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?


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!

| improve this answer | |
  • $\begingroup$ 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 Aug 18 '15 at 15:12
  • $\begingroup$ What else did you mean by multiplying by sign(theta) ? $\endgroup$ – Alexei Boulbitch Aug 19 '15 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 Aug 19 '15 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. $\endgroup$ – Alexei Boulbitch Aug 19 '15 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 Aug 19 '15 at 8:42

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.