# Inner products

1. May 14, 2013

### omoplata

The following was written down as a solution to a problem,$$\begin{eqnarray} P(\alpha_n) & = & \frac{1}{25} \left[ 9| \langle \phi_n \mid \psi_1 \rangle |^2 + 16 | \langle \phi_n \mid \psi_2 \rangle |^2 + 12 i \langle \phi_n \mid \psi_1 \rangle \langle \phi_n \mid \psi_2 \rangle^* - 12 i \langle \phi_n \mid \psi_2 \rangle \langle \phi_n \mid \psi_1 \rangle^* \right]\\ & = & \frac{1}{25} \left( 9| \langle \phi_n \mid \psi_1 \rangle |^2 + 16 | \langle \phi_n \mid \psi_2 \rangle |^2 + 2 \Re \left[ 12 i \langle \phi_n \mid \psi_1 \rangle \langle \phi_n \mid \psi_2 \rangle^* \right] \right) \end{eqnarray}$$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.

2. May 14, 2013

### CompuChip

The crucial observation is that

$$i { \langle \phi_n \mid \psi_1 \rangle } { \langle \phi_n \mid \psi_2 \rangle^* } = \left( -i { \langle \phi_n \mid \psi_1 \rangle^* } { \langle \phi_n \mid \psi_2 \rangle } \right)^*$$

(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, $z + z* = 2 \Re[z]$.

3. May 14, 2013

### omoplata

OK. Got it. Thanks.