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Time dependence of scalar product

  1. Apr 27, 2010 #1
    How do I show that the scalar product is time independent?

    I have: [tex]\frac{d}{dt}\int\Psi^{*}_{1}(x,t)\Psi_{2}(x,t)dx = 0[/tex]

    And have proceeded to take the derivatives inside the integral and using the time dependent Schrodinger eq. ending up with:

    [tex]\frac{i}{\hbar}\int\left(\Psi^{*}_{1}\widehat{H}\Psi_{2}-\Psi_{2}\widehat{H}\Psi^{*}_{1}\right)dx[/tex]

    Using the Hermitian properties of [tex]\widehat{H}[/tex] I then got to:

    [tex]\frac{i}{\hbar}\int\left(\Psi^{*}_{1}\widehat{H}\Psi_{2} - \Psi_{1}\widehat{H}\Psi^{*}_{2}\right)dx[/tex]

    Either these two terms equivalent meaning they cancel and I have my result or I've made a drastic error somewhere. Can anyone help me out?
     
  2. jcsd
  3. Apr 27, 2010 #2

    Meir Achuz

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    Your last step is wrong. The hermitian conjugate of [tex]\Psi^*_1 H|\Psi_2[/tex] is [tex]\Psi^*_1 H|\Psi_2[/tex].
    The stars on Psi don't change.
     
    Last edited: Apr 27, 2010
  4. Apr 27, 2010 #3
    [tex]
    \frac{i}{\hbar}\int\left(\Psi^{*}_{1}\widehat{H}\Psi_{2}-\Psi_{2}\widehat{H}\Psi^{*}_{1}\right)dx
    [/tex]

    should read

    [tex]
    \frac{i}{\hbar}\int\left(\Psi^{*}_{1}(\widehat{H}\Psi_{2})-(\widehat{H}\Psi^{*}_{1})\Psi_{2}\right)dx
    [/tex]

    hermiticy:

    [tex]
    \int \psi _1 ^* (\hat{O} \psi _2 ) \, dx = \int (\hat{O}\psi _1 )^* \psi _2 \, dx
    [/tex]
    gives
    [tex]
    - \frac{i}{\hbar}\int\left( (\widehat{H}\Psi_{1})^{*}\Psi_{2}-(\widehat{H}\Psi^{*}_{1})\Psi_{2}\right)dx
    [/tex]
     
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