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Wave-function solution in time-reversal transformation

  1. Jan 3, 2016 #1
    In obtaining (5.362) from (5.359), we first get

    ##U_{\tau}i\hbar\frac{\partial}{\partial t}\Psi(t) = U_{\tau}H^*\Psi^*(-t)##

    In order to obtain the LHS of (5.362), ##U_{\tau}## must commute with ##\frac{\partial}{\partial t}##. But how do we know that they commute?

    Screen Shot 2016-01-04 at 6.26.09 am.png
     
  2. jcsd
  3. Jan 4, 2016 #2

    ShayanJ

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    The Hamiltonian is assumed to be time-independent, so its reasonable to expect U to be time-independent too. So we just assume it and continue. If it was a bad assumption, it would give us non-sense. Then we could go back and relax this assumption. But if it works(as it did here), good for us!
     
  4. Jan 4, 2016 #3
    But it appears I could get non-sense out of this assumption:

    Since ##U_{\tau}## commutes with ##i\hbar\frac{\partial}{\partial t}##, it also commutes with ##H## by the time-dependent Schrondinger equation. From (5.361), ##H^*=U_{\tau}^{\dagger}HU_{\tau}##. Since ##U_{\tau}## commutes with ##H##, ##H^*=U_{\tau}^{\dagger}U_{\tau}H=H##, which contradicts the assumption that ##H## is not real.
     
  5. Jan 4, 2016 #4

    ShayanJ

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    I don't see how that follows!
     
  6. Jan 4, 2016 #5
    ##UH\Psi=Ui\hbar\frac{\partial}{\partial t}\Psi=i\hbar\frac{\partial}{\partial t}U\Psi=HU\Psi##

    Since this is true for any ##\Psi##, ##UH=HU##.
     
  7. Jan 4, 2016 #6

    ShayanJ

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    Here you're assuming that ## U\Psi ## is a solution to Schrodinger's equation only because ## \Psi ## is! Can you prove this?
     
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