# Interpretation of the functional Z (in Zee).

1. Jun 23, 2013

### 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?

2. Jun 24, 2013

### andrien

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

3. Jun 24, 2013

### 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.

4. Jun 25, 2013

### 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

5. Jun 25, 2013

### center o bass

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

6. Jun 26, 2013

### andrien

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