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Direct calculation of the propagator

  1. Oct 24, 2013 #1
    Hello all,

    I have a fundamental problem with the concept of the propagator regarding to time dependent potential. If the propagator is based on the expansion of the time evolution operator, how can it act on initial wave function in the presence of time dependent potential? You cannot use the evolution operator with these kind of potentials. I can accept the solution of Feynman path integrals, because in this frame, the time is split to many short intervals in which the time evolution operator is valid. However, can you still calculate the propagator by direct calculation?

    Thanks, The Samurai.
  2. jcsd
  3. Oct 25, 2013 #2


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    Actually, you can form an evolution operator even in the case of time dependent potential. The only difference is that if there's explicit time-dependence in the Hamiltonian, the evolution operator is a function of both initial and final times, ##U(t_{1},t_{2})##, instead of only the time difference, ##U(t_{2}-t_{1})##.
  4. Oct 25, 2013 #3
    Hi Hilbert, thank you for your response. I'm still not sure if the propagator is still valid though. When the propagator is calculated directly, we employ ##\langle x|e^{-iH(t_{1}-t_{2})/\hbar}|x'\rangle##. Can we calculate it by ##\langle x|U(t_{1},t_{2})|x'\rangle##?
  5. Oct 25, 2013 #4


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    No we don't :-).

    Rather we solve
    [tex]\mathrm{i} \hbar \partial_t \hat{U}(t,t')=\hat{H}(t) \hat{U}(t,t'), \quad \hat{U}(t,t)=1[/tex]
    (in the Schrödinger picture of time evolution).

    The formal solution is
    [tex]\hat{U}(t,t') = \mathcal{T}_c \exp \left [-\frac{\mathrm{i}}{\hbar}\int_{t'}^{t} \mathrm{d} t'' \hat{H}(t'') \right ],[/tex]
    where [itex]\mathcal{T}_c[/itex] is the causal time-ordering symbol. For details, see my QFT manuscript (Sect. 1.3)


    Note that there the general interaction picture of time-evolution is discussed!
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