Propagator for free particle in momentum space

Propagator for free particle in momentum space

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3 Responses to Propagator for free particle in momentum space

  1. Pedro Dardengo Mesquita says:

    About the solution of the problem on the book QFT for the gifted amateur that asks you to derive the propagator for a free particle in momentum space.
    As a first comment I would like suggest to explicitly write the time dependence of the creation and annihilation operators, since they are given in this case as the standard time dependence in interaction picture. (hbar = 1)
    a(t) = e^(iHt) a e^(-iHt)
    a^dagger = e^(-iHt) a^dagger e^(iHt)

    Also it would be interesting to specify the operator e^(-iHt) acting on the ground state of this non interacting system |0>. Since there is no interaction the Hamiltonian is quadratic on the creation and annihilation operators. This makes the ground state |0> an eigenvalue of the operator e^(-iHt). So:
    e^(-iHt)|0> = e^0|0> = |0>
    The same happens in the other side of the expression when e^(iHt) acts on <0| .
    This suggestion are merely to facilitate the comprehension of how the problem was solved.

    In the last few lines you say you are not sure if the operator e^(-i H*(tx-ty)) can be split. They can be just like you did, since the Baker-Campbell-Hausdorff formula ensures this is valid the commutators
    [H*t_x,H*t_y] = 0
    e^(-iH (t_x – t_y)) = e^(-iH*t_x) * e^(iH*t_y)

    Thank you for posting the solutions to these problems, they are helping me a lot.

    Like

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