Propagator in Path Integrals

Question

I am reading Section I.2 in Zee's QFT in a Nutshell. The amplitude for a particle to start at position $I$ and end at $F$ is (eq. (I.2.6)): $$ \langle q_f|e^{-iHT}|q_I\rangle=\int Dq(t)\ e^{i\int_0^T dt\ L(\dot q, q)}\tag{6} $$

In Section I.3 he then says it's better to consider an initial and final state, which are both taken as the vacuum state (eq. (I.3.3)): $$ Z\equiv\langle0|e^{-iHT}|0\rangle=\int Dq(t)\ e^{iS(q)}\tag{3} $$

When this integral is evaluated it gives (eq. (I.3.18)): $$ Z(J)=\mathcal Ce^{-(i/2)\iint d^4xd^4y\ J(x)D(x-y)J(y)}\equiv\mathcal Ce^{iW(J)}\tag{18} $$

$D(x - y)$ is the propagator, ie. the amplitude for the particle to travel from $y$ to $x$. But isn't this just equal to (6)? If so, how does it end up in the exponent of the solution to the path integral when considering vacuum states and a source term J(x)?

I am probably missing something obvious, any help would be great.

Answer 1

Conceptually, the biggest difference between $\langle q_F | e^{-iHT} |q_I\rangle$ and $D_F(x-y)$ is certainly that the former describes the amplitude of a real particle moving from $q_I$ to $q_F$ whereas the latter describes a virtual particle(s) propagating from $x$ to $y$. The difference between real and virtual particles is that the former fulfill $p^2=m^2$, whereas the latter don't fulfill it. If in both cases it could be spoken of propagation -- of which I am not completely sure of -- this propagation is not the same. So both expressions are not the same.

There are certainly other differences. I think, in the first case it is one particle described by 1-particle non-relativistic Quantum mechanics, whereas in the second case it is actually the field that corresponds to a particle (or particles) that propagates which is described by a multi-particle relativistic Quantum field theory. In case of a complex field a multi-particle propagation embraces not only particles, but also their anti-particles. Even in case of a real field, where anti-particle and particles are the same --- actually there is propagation between $x$ and $y$ forward and backwards that is to be included into the probability amplitude.

If the propagator $D(x-y)$ is not meant to be Feynman-propagator, then anti-particles might not be participating in the propagation, still it would be not the same because of the mentioned point at the beginning, $D(x-y)$ describes virtual particles, whereas the path integral amplitude for 1 particle describes real particles.

This means, both expressions, on first sight very similar, are actually very different.