How can the expression for orthonormal states be simplified?

[omoplata]

The following was written down as a solution to a problem,<br /> \begin{eqnarray}<br /> P(\alpha_n) &amp; = &amp; \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]\\<br /> &amp; = &amp; \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)<br /> \end{eqnarray}<br />How do you get from the first line to the second line? How does 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^* ?

Is this solution wrong?

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

 

[CompuChip]

The crucial observation is that

i { \langle \phi_n \mid \psi_1 \rangle } { \langle \phi_n \mid \psi_2 \rangle^* } = <br /> \left( -i { \langle \phi_n \mid \psi_1 \rangle^* } { \langle \phi_n \mid \psi_2 \rangle } \right)^*

(because (abc)^* = a^* b^* c^* and (a^*)^* = a, and because the brakets are complex numbers which commute).

Then it's easy to verify that for any complex number z, z + z* = 2 \Re[z].

 

[omoplata]

OK. Got it. Thanks.