# Physical meaning of partition function in QFT

When we have the generating functional $Z$ for a scalar field

$$Z(J,J^{\dagger}) = \int{D\phi^{\dagger}D\phi \; \exp\left[{\int L+\phi^{\dagger}J(x)+J^{\dagger}(x)}\phi\right]},$$

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

$$Z(0,0) = \langle 0,t_f|0,t_i \rangle.$$

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 $Z(0,0)$?

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 $(1)$.

• The correspondence of $Z(0) = 1$ and $H\lvert \Omega \rangle = 0$ is not so easy. $Z(0) = \langle \Omega\vert\Omega \rangle$, how do you conclude from $\langle \Omega \vert \Omega \rangle = 1$ that we "effectively set $H\lvert \Omega \rangle = 0$"? Apr 13, 2016 at 13:19
• @ACuriousMind See e.g., Pesking&Schroeder, page 98, eqs. 4.55 and 4.56. Apr 13, 2016 at 13:24
• Hm, well, I guess that's also true. Apr 13, 2016 at 14:08
• @Numrok (1/2) yes: the state $|0\rangle$ is the "vacuum state", i.e., the state with no particles at all. In perturbation theory, this ket is time dependent, which means that we should write $|0,t\rangle$, but it is still the vacuum. In QM a product $\langle\varphi,t_1|\phi,t_2\rangle$ represents the amplitude to go from state $\varphi$ at a time $t_1$ to a state $\phi$ at a time $t_2$, where $\varphi,\phi$ are any two states. In the equation (2) both states are the vacuum (i.e., the system is empty). Apr 13, 2016 at 17:29
• @Numrok (2/2) Therefore, (2) represents the amplitude to go from a state with no particles to a state with no particles. Naturally, the amplitude for this is one: if you have nothing at some point, you can't suddenly have something some time later, because of conservation of energy and momentum. But it doesn't only mean that "on average we have nothing": we don't have nothing at all, not even on average! ($|0\rangle$ is an "exact" or "perfect" vacuum, not just an approximate one) Apr 13, 2016 at 17:29

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 $\ln(Z) \sim \langle 0\vert T \int H \vert 0 \rangle$.