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Heisenberg's equation of motion

  1. Mar 1, 2008 #1
    The equation of motion for an observeable A is given by [itex]\dot{A} = \frac{1}{i \hbar} [A,H][/itex].

    If we change representation, via some unitary transformation [itex] \widetilde{A} \mapsto U^\dag A U[/itex] is the corresponding equation of motion now

    [itex]\dot{\widetilde{A}} = \frac{1}{i \hbar} [\widetilde{A},U^\dag H U][/itex]
    [itex]\dot{\widetilde{A}} = \frac{1}{i \hbar} [\widetilde{A},H][/itex]?

    I'm asking because I want to write the time derivative of the Dirac representation of the position operator in the Foldy-Wouthusyen representation.
    Last edited: Mar 1, 2008
  2. jcsd
  3. Mar 1, 2008 #2


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    If you know how to derive Heisenberg eq of Motion, then you should have no problem to find the answer.
  4. Mar 1, 2008 #3


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    They're the same, the first equation of motion for the operator UAUt gives the second EOM for A.
  5. Mar 1, 2008 #4
    Are you saying that the transformed operator satisfies the first equation but not the second?
  6. Mar 2, 2008 #5


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    If the generator of the unitary transform U depends on t -- like going from Schrodinger picture to the Interaction Picture -- then noospace, you have left out a term. Standard stuff, can be found in most QM or QFT texts.
    Reilly Atkinson
  7. Mar 2, 2008 #6


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    see Messiah QM vol 2.
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