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Interpretation of the functional Z (in Zee).

  1. Jun 23, 2013 #1
    In Zee's book at page 12 in both editions he finds that he can write the amplitude

    $$\langle q_f|e^{-iHT} |q_i\rangle = \int Dq(t) e^{iS} $$

    where T is the time between emission at ##q_i## and observation at ##q_f##. He then states that we often define

    $$Z = \langle 0 | e^{-iHT} |0 \rangle $$

    And he observes that by inserting a complete set of states we can write

    $$ Z = \int q_f \int q_i \langle 0 | q_f\rangle \langle q_f | e^{-iHT} |q_i\rangle \langle q_i |0 \rangle = int q_f \int q_i \psi_0^* (q_f) \psi_0 (q_i) \langle q_f | e^{-iHT} |q_i\rangle. $$

    But then a few sentences down he writes

    $$Z = \int Dq(t) e^{iS}. $$

    Is there an error in Zee here or ##Z## as defined above actually equal to the path integral given by ##\langle q_f|e^{-iHT} |q_i\rangle##? If so, how does the integration over the ground state wavefunction vanish?
     
  2. jcsd
  3. Jun 24, 2013 #2
    I don't see how are you coming to this conclusion.
     
  4. Jun 24, 2013 #3

    Avodyne

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    Zee often leaves out details like this. He is presumably now defining Dq to include the integration over the ground-state wave function. For a more thorough explanation of how this works, see Srednicki's text.
     
  5. Jun 25, 2013 #4

    vanhees71

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    If I remember right, Zee leaves out the crucial detail to introduce a little imaginary part to the Hamiltonian such that you get the vacuum-to-vacuum-transition amplitude, i.e., in the path integral contributions from all other states are exponentially damped out in the limit [itex]t_i \rightarrow -\infty[/itex] and [itex]t_f \rightarrow \infty[/itex], where [itex]t_i[/itex] and [itex]t_f[/itex] are the initial and final time in the original path integral for the time-evolution kernel.

    A very good treatment of this issue can be found, e.g., in

    Bailin, Love, Gauge Theories
     
  6. Jun 25, 2013 #5
    I did not mean "vanish" in the sence of becoming zero, but in the sense of disappearing from the expression :)
     
  7. Jun 26, 2013 #6
    I thought a substitution t→it is done so as to give a proper meaning to the exponential
    exp(iS),not to hamiltonian.
     
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