# In QFT, why are the vacuum partition function and the zero-temperature imaginary-time partition function the same?

When doing thermal field theory, one can start with the definition of the (thermal) partition function $$Z = Tr[e^{-\beta H}]$$, and inserting a number of completness-relations, we can arrive at (I am using a scalar field for illustration)

$$Z[\beta] = \int D\varphi \, \exp[- \int_0^\beta d\tau \int d\vec x \, \mathcal{L}_E(\varphi(\tau, \vec x))]$$ Here, $$\mathcal L_E$$ is the Euclidian Lagrangian, defined by $$\mathcal L_E(\tau, \vec x) = - \mathcal L(t = -i\tau, \vec x)$$. Thus, the thermal partition function and the vacuum partition function $$Z' = \int D\varphi \, \exp[i\int_0^\infty dt \int d\vec x \, \mathcal{L}(\varphi(t, \vec x))]$$ is related by a simple procedure: rotate the time integral from the real line to the imaginary line, then change variable of integration to $$\tau$$, before restricting this new variable to $$\tau \in [0, \beta]$$. The thermal partition function gives free energy as a function of temperature, $$F(\beta) = - \frac{1}{\beta} \ln(Z[\beta])$$ My understanding is that in the low temperature limit, $$\beta \rightarrow \infty$$, the vacuum partition function $$Z'$$ also gives the free energy through $$\beta F = i \ln(Z')$$ (Peskin & Schröder, 9.3 seems to suggest so, but i do not understand the explanation) Is this true? And if yes, how does it hold? I do not see why the partition function should be unaltered after changing the integration for $$\mathbb{R}$$ to $$i\mathbb{R}^+$$. If not, what is the relation between $$Z$$ and $$Z'$$.

It's only true in the limit as $$\beta$$ increases. Your $$Z'$$ contains no dependence on $$\beta$$, so it can not give you the full function of $$F(\beta)$$.
$$Z'=\lim_{\beta\rightarrow \infty}\langle \Omega| e^{-iH\beta}|\Omega\rangle=\lim_{\beta\rightarrow \infty}e^{-iE_0\beta}$$, where $$\Omega$$ is the vacuum and $$E_0$$ is the vacuum energy (which is infrared divergent since the volume of space is infinite). This expression for the vacuum to vacuum transition amplitude is where Peskin and Schroeder (for instance) started from in deriving the path integral in 6.2.
The partition function $$Z= \sum_n e^{-E_n\beta}$$, where the sum over $$n$$ runs over all energy eigenstates. It is true that $$\lim_{\beta\rightarrow \infty}Z\sim e^{-E_0\beta}$$ since all states with higher energy are much smaller. So it is true that $$\lim_{\beta\rightarrow \infty} \beta F(\beta)= i\log Z'.$$
Whether you are considering the Euclidean or Minkowski path integral, on the quantum mechanical side this is the same system with the same energy eigenvalues $$E_n$$. We can even think of this as a discrete spectrum if we have both a UV and IR cutoff. All of the subtlety about this lies when we define our system through a Euclidean path integral and are wondering whether it corresponds to some quantum mechanics. But that is not what we are doing here, we are going from a presumably well-defined quantum mechanics to two different path integral descriptions.
Think of the Wick rotation as something that happens from $$e^{-itH}\rightarrow e^{-\tau H}$$ where it is seen as something trivial. Both $$\text{Tr}\, e^{-itH}$$ and $$\text{Tr}\,e^{-\tau H}$$ may then be turned into a valid path integral through the well-defined series of steps given in your textbook. It is confusing to think of Wick rotation after the chain of steps leading to the path integral has already been performed, but quite clear when it is thought of in terms of exponentials of Hamiltonians.
• Yes, i understand that it is only in the limit $\beta \rightarrow \infty$, maybe I was not clear on that. I think your answer is really helpfull, I will look into it. However, it does not explain why we can do wick-rotation without altering the partition function. Oct 5, 2021 at 18:36