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Dirac's Quantum Mechanics - the definition of the time evolution operator

  1. Aug 7, 2012 #1
    Dirac's "Quantum Mechanics" - the definition of the time evolution operator

    I'm reading Dirac's "Principles Of Quantum Mechanics" to learn more about the formal side of the subject.

    I have a question about the way he defines the time evolution operator in the book. Either there's a mistake or I'm missing something.

    In chapter 27 he says (eqn 1) that [itex]\hat{T}[/itex] is defined such that:

    [itex]|P(t)> = \hat{T} |P(0)>[/itex]

    Where |P(0)> is a ket at time t=0 , and |P(t)> - at time t
    Or equivalently |P(0)> is a ket in the Heisenberg picture, and |P(t)> - in the Schrodinger picture.

    So this implies that:

    [itex]<P(t)| = <P(0)| \hat{T}^{\dagger} [/itex]

    And then in chapter 32, eqn 45 implies that:

    [itex]<P(t)| = <P(0)| \hat{T} [/itex]

    And I understand, that we can define it both ways, since it's a unitary operator. But we should stick to one way of defining it, and I'm sure Dirac does. So what it is here, that I'm not understanding properly?
     
    Last edited: Aug 7, 2012
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  3. Aug 8, 2012 #2

    strangerep

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    Re: Dirac's "Quantum Mechanics" - the definition of the time evolution operator

    I think this is essentially the same misunderstanding about selfadjoint operators which I clarified in your other thread.
     
  4. Aug 8, 2012 #3
    Re: Dirac's "Quantum Mechanics" - the definition of the time evolution operator

    Thanks again,

    But here [itex]\hat{T}[/itex] isn't self-adjoint. In fact it's unitary.
     
  5. Aug 8, 2012 #4

    dextercioby

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    Re: Dirac's "Quantum Mechanics" - the definition of the time evolution operator

    Yes, the T's are not identical, one is the adjoint of the other.
     
  6. Aug 8, 2012 #5
    Re: Dirac's "Quantum Mechanics" - the definition of the time evolution operator

    Thanks,

    Yes. I'm sorry, but I still don't understand how this connects to my question.
     
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