Physical meaning of partition function in QFT When we have the generating functional $Z$ for a scalar field
\begin{equation}
Z(J,J^{\dagger}) = \int{D\phi^{\dagger}D\phi \; \exp\left[{\int L+\phi^{\dagger}J(x)+J^{\dagger}(x)}\phi\right]},
\end{equation}
the partition function is $Z(0,0)$. We know that the derivatives of the generating functional give the propagator for the system, and it is often said that $Z(0,0)$ relates to the vacuum energy, and it is formally given by
\begin{equation}
Z(0,0) = \langle 0,t_f|0,t_i \rangle.
\end{equation}
How does this matrix element represent the vacuum energy of the system? Is it to do with the size of the fluctuations between the times $t_i$ and $t_f$? Or what is another interpretation of $Z(0,0)$?
 A: In terms of Feynman diagrams, the partition is represented by the sum over so-called vacuum bubbles - diagrams with no external legs. In formulae and in terms of the interaction picture and the free vacuum $\lvert 0 \rangle$ and the interacting vacuum $\lvert \Omega \rangle$, we have that
$$ \lvert \Omega\rangle = \lim_{T\to\infty(1-\mathrm{i}\epsilon)} \left(\mathrm{e}^{-\mathrm{i}E_\Omega T}\langle\Omega \vert 0\rangle\right)^{-1}\mathrm{e}^{-\mathrm{i}H T}\lvert 0 \rangle$$
and hence
$$ Z = \langle \Omega \vert \Omega\rangle = \lim_{T\to\infty(1-\mathrm{i}\epsilon)} \lvert \langle \Omega\vert 0 \rangle\rvert^2\mathrm{e}^{\mathrm{i}E_\Omega 2T}$$
Now, if you write $Z$ as $\mathrm{e}^{\sum_i V_i}$ where $V_i$ is the contribution of the vacuum bubbles of order $i$, you see that, schematically, $\sum_i V_i \propto E_\Omega T$, so the partition function is the exponential of the vacuum energy.
Heuristically, it should not be surprising that the logarithm of the partition function is the vacuum energy, since $Z \sim \langle 0 \rvert\mathrm{e}^{-\mathrm{i}\int H} \vert 0 \rangle$ so $\ln(Z) \sim \langle 0\vert T \int H \vert 0 \rangle$.
A: The partition function $Z[J]$, both in QM and in CM, is underdetermined: any multiple of $Z[J]$ gives rise to the same dynamics. This means that $Z[0]$ is arbitrary, and is usually set to one:
$$
Z[0]\equiv 1 \tag{1}
$$
effectively getting rid of vacuum diagrams, that is, we set $H|\Omega\rangle=0$. In other words: the energy of the vacuum is not measurable and can be set to any number we want. We can only measure differences in energies (except in GR), which means that a constant offset of energies is irrelevant.
The matrix element
$$
\langle 0,t_f|0,t_i\rangle \tag{2}
$$
can be interpreted as the amplitude of ending up with a vacuum state at the time $t_f$ if you start with vacuum at a time $t_i$. Or put it another way, it is the amplitude to get nothing if you initially have nothing. This number is, naturally, one:
$$
\langle 0,t_f|0,t_i\rangle\equiv 1 \tag{3}
$$
in agreement with $(1)$.
