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

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

  • $\begingroup$ Not clear what you are asking for, can you can explain better. $\endgroup$
    – yarchik
    Apr 26, 2021 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, 2021 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, 2021 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, 2021 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, 2021 at 17:55

1 Answer 1


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


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


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