0
$\begingroup$

Consider

\begin{align} &\hat{e}_{\pm}=(1,\pm i,0) &\hat{z}=(0,0,1) \end{align}

Then $(\hat{e}_{\pm})^*\cdot (\hat{e}_{\mp})=0$ and $(\hat{e}_{\pm})^*\cdot \hat{z}=0$, so they are mutually orthogonal.

The cross product is

$$ \hat{e}_{\pm}\times \hat{z}=\pm i \hat{e}_{\pm}$$

So the result of the cross product is not orthogonal to one of the two vectors in the cross product. That means that the right hand rule is not valid for complex vectors?

$\endgroup$
  • $\begingroup$ How would you right hand rule this? $\endgroup$ – Kaynex Mar 8 '17 at 13:09
  • $\begingroup$ I just mean that the result of the cross product is not orthogonal to one of the two vector in the product. $\endgroup$ – Alessandro Zunino Mar 8 '17 at 13:37
  • 1
    $\begingroup$ I guess this question might help you. $\endgroup$ – Alex Vong Mar 8 '17 at 14:02

Your Answer

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

Browse other questions tagged or ask your own question.