The following was written down as a solution to a problem,[tex](adsbygoogle = window.adsbygoogle || []).push({});

\begin{eqnarray}

P(\alpha_n) & = & \frac{1}{25} \left[ 9| \langle \phi_n \mid \psi_1 \rangle |^2 + 16 | \langle \phi_n \mid \psi_2 \rangle |^2 + 12 i \langle \phi_n \mid \psi_1 \rangle \langle \phi_n \mid \psi_2 \rangle^* - 12 i \langle \phi_n \mid \psi_2 \rangle \langle \phi_n \mid \psi_1 \rangle^* \right]\\

& = & \frac{1}{25} \left( 9| \langle \phi_n \mid \psi_1 \rangle |^2 + 16 | \langle \phi_n \mid \psi_2 \rangle |^2 + 2 \Re \left[ 12 i \langle \phi_n \mid \psi_1 \rangle \langle \phi_n \mid \psi_2 \rangle^* \right] \right)

\end{eqnarray}

[/tex]How do you get from the first line to the second line? How does [itex]12 i \langle \phi_n \mid \psi_1 \rangle \langle \phi_n \mid \psi_2 \rangle^* = - 12 i \langle \phi_n \mid \psi_2 \rangle \langle \phi_n \mid \psi_1 \rangle^*[/itex] ?

Is this solution wrong?

Here, [itex]\mid \psi_1 \rangle[/itex] and [itex]\mid \psi_2 \rangle [/itex] are two orthonormal states, while [itex]\mid \phi_n \rangle[/itex] is a normalized state, if that makes any difference.

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