0
$\begingroup$

I have written the following code

\[Psi] = {\[Alpha] Cos[B t/2] + 
I \[Beta] Sin[B t/2] E^(-I \[Phi]), \[Alpha] I Sin[B t/2] E^(
 I \[Phi]) + \[Beta] Cos[B t/2]};
\[Psi]1 =  Assuming[{B, t, \[Phi] } \[Element] Reals , Conjugate[\[Psi]]]
R = KroneckerProduct[\[Psi], \[Psi]1]
v = Eigenvectors[R]
Normalize /@ v // FullSimplify

but the matrix [Psi]1 takes the complex conjugate of the whole element and not just that of i, a and b, which, in turn, makes the normalized eigenvectors somewhat complicated to work with. What can i change to my code? The ideal would be to get

$$ \psi1=\begin{pmatrix} \alpha^{*}\cos(Bt/2)-i\beta^{*}\sin(Bt/2)e^{i\varphi} & -ia^{*}\sin(Bt/2)e^{-i\varphi}+\beta^{*}\cos(Bt/2) \end{pmatrix} $$

$\endgroup$
5
  • $\begingroup$ Not clear what you are asking for, can you can explain better. $\endgroup$
    – yarchik
    Apr 26 at 15:19
  • $\begingroup$ My question is if there is any way to have mathametica take the conjugate of i, a and b and not of the whole matrix element of [Psi] $\endgroup$
    – Assassinos
    Apr 26 at 15:33
  • $\begingroup$ Also, is the code i have written correct? I want B, t and [\Phi] to be real. $\endgroup$
    – Assassinos
    Apr 26 at 15:37
  • $\begingroup$ Note that by default all variables and parameters are complex. Do you mean that they should all be real except for $a$ and $b$ (and of course for $i$, but that is not a variable) ? Also, I cannot see $a$ and $b$ in your code. $\endgroup$
    – A.G.
    Apr 26 at 16:05
  • $\begingroup$ Yes, all should be real except for [Alpha] ,[Beta]. My bad, i wrote a and b instead of the previous ones. $\endgroup$
    – Assassinos
    Apr 26 at 17:55
1
$\begingroup$

ComplexExpand assumes that all variables are real except those mentioned as second argument.

Try:

ψ = {α Cos[B t/2] + 
    I β Sin[B t/2] E^(-I ϕ), α I Sin[
      B t/2] E^(I ϕ) + β Cos[B t/2]};

ComplexExpand[Conjugate[ψ], {α, b }] // Simplify

This gives:

enter image description here

$\endgroup$

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.