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I Phase factors of eigenstates in time-dependent Hamiltonians

  1. Nov 17, 2016 #1
    For a time-dependent Hamiltonian, the Schrodinger equation is given by

    $$i\hbar\frac{\partial}{\partial t}|\alpha;t\rangle=H(t)|\alpha;t\rangle,$$

    where the physical time-dependent state ##|\alpha;t\rangle## is given by

    $$|\alpha;t\rangle = \sum\limits_{n}c_{n}(t)e^{i\theta_{n}(t)}|n;t\rangle$$

    and

    $$\theta_{n}(t)\equiv -\frac{1}{\hbar}\int_{0}^{t}E_{n}(t')dt'.$$

    ---

    ##e^{i\theta_{n}(t)}## is the phase factor that has been pulled out from the eigenstate-expansion coefficients of ##|\alpha;t\rangle##.

    Why is ##\theta_{n}(t)## given by

    $$\theta_{n}(t)\equiv -\frac{1}{\hbar}\int_{0}^{t}E_{n}(t')dt'?$$
     
  2. jcsd
  3. Nov 17, 2016 #2

    DrClaude

    User Avatar

    Staff: Mentor

    Because the time evolution operator ##\hat{U}(t)## is a solution of the TDSE
    $$
    i \hbar \frac{d\hat{U}(t)}{dt} = \hat{H}(t) \hat{U}(t)
    $$
    hence
    $$
    \hat{U}(t) = \exp\left[-\frac{i}{\hbar} \int_0^t \hat{H}(t') dt' \right]
    $$
     
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