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Inner products

  1. May 14, 2013 #1
    The following was written down as a solution to a problem,[tex]
    \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.
     
  2. jcsd
  3. May 14, 2013 #2

    CompuChip

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    Science Advisor
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    The crucial observation is that

    [tex]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)^*[/tex]

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

    Then it's easy to verify that for any complex number z, [itex]z + z* = 2 \Re[z][/itex].
     
  4. May 14, 2013 #3
    OK. Got it. Thanks.
     
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