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Amplitude's time dependence in Heisenberg representation

  1. Apr 6, 2013 #1

    A = \langle q_f(t) \mid q_i(t) \rangle = \langle q_{f,H} \mid e^{iH(t_0-t)} e^{-iH(t-t_0)} \mid q_{i,H} \rangle = \langle q_{f,H} \mid q_{i,H} \rangle


    This means that A is time-independent, and depends only on the reference point ##t_0##. How is it possibly? From Schoedinger picture it does depend on time! Thanks.
  2. jcsd
  3. Apr 6, 2013 #2


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    You have to be careful about the meaning of your vectors. From what you write, I conclude that [itex]|q(t) \rangle[/itex] are the (generalized) eigenvectors of some operator representing an observable in the Heisenberg picture. Then, from the Heisenberg equation of motion
    [tex]\mathrm{d}_t \hat{O}=\frac{1}{\mathrm{i}}[\hat{O},\hat{H}].[/tex]
    Assuming we have time independent Hamiltonian and that implies
    [tex]\hat{O}(t)=\exp(\mathrm{i} t \hat{H}) \hat{O}(0) \exp(-\mathrm{i} t \hat{H}).[/tex] For sake of simplicity, I've set [itex]t_0=0[/itex].

    Now by definition [itex]|q(t) \rangle[/itex] is the eigenvector of [itex]\hat{O}(t)[/itex] with the fixed eigenvalue [itex]q[/itex]:
    [tex]\forall t: \quad \hat{O}(t) |q(t) \rangle=q |q(t) \rangle.[/tex]
    This implies that
    [tex]|q(t) \rangle=\exp(\mathrm{i} t \hat{H}) |q(0) \rangle.[/tex]
    So up to a sign this implies your formula.

    Of course, because the time evolution is unitary, the scalar product is time independent
    [tex]\langle q(t)|q'(t) \rangle=\langle q(0) | q'(0) \rangle=\delta(q,q'),[/tex]
    where the [itex]\delta[/itex] stands for a Kronecker [itex]\delta[/itex] (discrete eigenvalues) or a Dirac [itex]\delta[/itex] distribution (continuous eigenvalues).
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