Interpretation of the functional Z (in Zee).

[center o bass]

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?

 

[andrien]

If so, how does the integration over the ground state wavefunction vanish?

I don't see how are you coming to this conclusion.

 

[Avodyne]

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.

 

[vanhees71]

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 t_i \rightarrow -\infty and t_f \rightarrow \infty, where t_i and t_f 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

 

[center o bass]

I don't see how are you coming to this conclusion.

I did not mean "vanish" in the sense of becoming zero, but in the sense of disappearing from the expression :)

 

[andrien]

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 thought a substitution t→it is done so as to give a proper meaning to the exponential
exp^(iS),not to hamiltonian.