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Dynamics in the heizenberg picture

  1. Feb 17, 2012 #1
    The time-derivative of an operator [itex] A(t) = U^\dagger a(t)U [/itex] in the heizenberg picture is given by:

    [itex]\frac{dA(t)}{dt} = \frac{i}{\hbar} [H,A(t)] + U^\dagger(\frac{da(t)}{dt})U [/itex]

    Now, I know that under some conditions, we can write:

    [itex]\frac{dA(t)}{dt} = \frac{i}{\hbar} U^\dagger[H,a(t)]U + U^\dagger(\frac{da(t)}{dt})U [/itex]

    My question is- what are those conditions?

    Last edited: Feb 17, 2012
  2. jcsd
  3. Feb 17, 2012 #2


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    Under the condition that the Hamiltonian is not dependent on time.
  4. Feb 19, 2012 #3
    is there a more specific rule?
    take for example:

    [itex] H = a^\dagger e^{-i\omega t} + a e^{i\omega t} [/itex]

    (here [itex]a[/itex] is the annihilation operator).

    This Hamiltonian is of course time-dependant,
    but, non-the-less, it is also true that:

    [itex]\frac{dA(t)}{dt} = \frac{i}{\hbar} U^\dagger[H,a]U[/itex]

    (where [itex]A(t)[/itex] is the annihilation operator in heizenberg's picture)
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