- #1
center o bass
- 560
- 2
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?
$$\langle q_f|e^{-iHT} |q_i\rangle = \int Dq(t) e^{iS} $$
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?