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Propagators for time-dependent Hamiltonians

  1. Aug 8, 2013 #1
    Suppose I know

    [tex]
    H \psi(x) = \left( -\frac{1}{2m} \Delta_x + V(x) \right) \psi(x) = E\psi(x).
    [/tex]

    Then

    [tex]
    \psi(x,t) = e^{-iEt}\psi(x)
    [/tex]

    solves the time-dependent Schrodinger equation

    [tex]
    \left( i \frac{\partial}{\partial t} + \frac{1}{2m} \Delta_x - V(x) \right)\psi(x,t) = 0.
    [/tex]

    I've done some computations, and it looks like

    [tex]
    \Psi(x,t) = e^{-imvx}e^{-imv^2t/2}\psi(x+vt)
    [/tex]

    is a solution to the time-dependent Schrodinger equation

    [tex]
    \left( i \frac{\partial}{\partial t} + \frac{1}{2m} \Delta_x - V(x+vt) \right)\Psi = 0.
    [/tex]

    I have a couple of questions about this:
    1. What is going on here physically? That is, what are those two phase factors telling me?
    2. What does this mean the propagator is? When [itex]H[/itex] is time-independent, [itex]U(t) = e^{-iEt}[/itex]...but what is it in the time-dependent case? Is there a neat little formulation of it?
     
    Last edited: Aug 8, 2013
  2. jcsd
  3. Aug 8, 2013 #2

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    With the ansatz
    [tex]\psi(t,\vec{x})=\exp(-\mathrm{i} E t) \phi_{E,\alpha}(\vec{x}),[/tex]
    you only get the energy-eigensolutions (stationary states) for a time-independent Hamiltonian,
    [tex]\hat{H} \phi_{E,\alpha}(\vec{x})=E \phi_{E,\alpha}(\vec{x}).[/tex]
    Here [itex]\alpha[/itex] stands for all necessary additional observables, compatible with energy, to label the possible degeneracy of the energy eigenstates.

    The most general solution of the time-dependent Schrödinger equation is then given as a superposition of energy eigensolutions
    [tex]\psi(t,\vec{x})=\sum_{E,\alpha} \exp(-\mathrm{i} E t) \phi_{E,\alpha}(\vec{x}).[/tex]

    This is all for time-independent Hamiltonians. For time-dependent Hamiltonians, the problem is a bit more complicated. Here you get a formal solution by using the time-evolution operator for the state in the Schrödinger picture,
    [tex]\hat{C}(t,t_0)=\mathcal{T}_c \exp[-\mathrm{i} \int_{t_0}^t \mathrm{d} t' \hat{H}(t)],[/tex]
    where [itex]\mathcal{T}_c[/itex] is the time-ordering operator that orders products of time-dependent operators such that the time arguments are ordered from right to left.

    The exponential must be formaly expanded in terms of a power series to make sense of this time-ordering symbol. The nth-order contribution is
    [tex]\hat{C}_n(t,t_0)=\frac{(-\mathrm{i})^n}{n!} \int_{t_0}^t \mathrm{d} t_1 \cdots \int_{t_0}^t \mathrm{d} t_n \mathcal{T}_c \hat{H}(t_1) \cdots \hat{H}(t_n).[/tex]
    Then
    [tex]\hat{C}(t,t_0)=\sum_{n=0}^{\infty} \hat{C}_n(t,t_0).[/tex]

    The propagator of the Schrödinger equation is then given by
    [tex]U(t,\vec{x};t_0,\vec{x}')=\langle \vec{x}|\hat{C}(t,t_0)|\vec{x}' \rangle,[/tex]
    and the solution of the time-dependent Schrödinger equation reads
    [tex]\psi(t,\vec{x})=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x'} U(t,\vec{x};t_0,\vec{x}') \psi(t_0,\vec{x}').[/tex]
     
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