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I have declared this matrix in my notebook

aa = (1/2)*(1 + Cos[θ]); 
bb = Sin[θ]/Sqrt[2]; 
CCC = (1/2)*(1 - Cos[θ]); 

q2 = {{aa^2, aa*bb*E^((-I)*B*t*γ + ((b*t*γ)/(4*m*σ))^2*(4*I*h*m*t*σ^2) + ι/2), 
     aa*CCC*E^(-2*I*B*t*γ + 2*ι)}, 
    {bb*aa*E^(I*B*t*γ - ((b*t*γ)/(4*m*σ))^2*(4*I*h*m*t*σ^2) + ι/2), bb^2, 
     bb*CCC*E^((-I)*B*t*γ - ((b*t*γ)/(4*m*σ))^2*(4*I*h*m*t*σ^2) + ι/2)}, 
    {CCC*aa*E^((Plus[2])*I*B*t*γ + 2*ι), 
     CCC*bb*E^((Plus[I])*B*t*γ + ((b*t*γ)/(4*m*σ))^2*(4*I*h*m*t*σ^2) + ι/2), CCC^2}}; 

Which is Hermitian because it is a square matrix, is equal to its conjugate transposition and because its elements of the main diagonal are real (although they are trigonometric functions), then, why does mathematics give me imaginary eigenvalues ​​when these should be real?

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  • $\begingroup$ HermitianMatrixQ@q2 tells me that your matrix is not Hermitian. This is because you haven't told Mathematica to treat all symbols as real numbers. $\endgroup$ – Henrik Schumacher Apr 4 '18 at 18:43
  • $\begingroup$ Thanks @HenrikSchumacher, and how can I do this?? $\endgroup$ – Julio Abraham Mendoza Fierro Apr 4 '18 at 18:53
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    $\begingroup$ Hm. In this case, probably the simplest way would be Eigenvalues[q2] // ComplexExpand. $\endgroup$ – Henrik Schumacher Apr 4 '18 at 19:07

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