quantum field theory - 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 $Z(0,0)$?

Answer

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 $(1)$.