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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 $Z(0,0)$?

Orca
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2 Answers2

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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 $(1)$.

AccidentalFourierTransform
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    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 $H\lvert \Omega \rangle = 0$"? – ACuriousMind Apr 13 '16 at 13:19
  • @ACuriousMind See e.g., Pesking&Schroeder, page 98, eqs. 4.55 and 4.56. – AccidentalFourierTransform Apr 13 '16 at 13:24
  • I know these formulae (you might notice the preceding formula appears in my answer :P). It is not straightforward to deduce from them that "$H\lvert \Omega \rangle = 0$ because there is the non-convergent limit in $T$. And when you get rid of $T$, you are left with 4.56, and you have to explain that $Z$ really is that sum over vacuum bubbles. I guess what I'm saying is that your answer makes the correspondence sound much easier than it is. – ACuriousMind Apr 13 '16 at 13:53
  • @ACuriousMind "non-convergent" you say? since when do we care about convergence issues in QFT? [I'm obviously joking: its true that I might have oversimplified the issue, but strictly speaking neither my formulas nor yours are mathematically justified. There are no vacuum diagrams in rigorous treatments of perturbation theory: everything is normal ordered from the beginning, and $E_\Omega=0$] – AccidentalFourierTransform Apr 13 '16 at 14:07
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    Hm, well, I guess that's also true. – ACuriousMind Apr 13 '16 at 14:08
  • @AccidentalFourierTransform Can $\langle \Omega | \Omega \rangle$ be interpreted as "there is no particle creation". Or as "on average there is no particle creation"? – Wolpertinger Apr 13 '16 at 17:15
  • @Numrok In QM states are usually normalised: $\langle \Omega|\Omega\rangle=1$. Perhaps you are taking both $\Omega$ at different times: $\langle\Omega,t_1|\Omega,t_2\rangle$? – AccidentalFourierTransform Apr 13 '16 at 17:19
  • @AccidentalFourierTransform yes, I meant the different time one. Or the matrix element in equation (2) of your answer. Does that mean there is no particle creation? – Wolpertinger Apr 13 '16 at 17:23
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    @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). – AccidentalFourierTransform Apr 13 '16 at 17:29
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    @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) – AccidentalFourierTransform Apr 13 '16 at 17:29
  • This is an old question, but I think I have to disagree. For instance it is known that the partition function of conformal field theories on a sphere has a universal contribution that cannot be scaled away by local counterterms, and this contribution is related to the entanglement entropy. – Ryan Thorngren Mar 08 '18 at 20:08
  • Could you please explain how choosing $Z[0]\equiv 1$ gets rid of vacuum diagrams? – HelloGoodbye Nov 22 '21 at 05:56
  • @HelloGoodbye sure, see here: https://web.physics.ucsb.edu/~mark/ms-qft-DRAFT.pdf#page=76 – AccidentalFourierTransform Nov 23 '21 at 15:43
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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 $\ln(Z) \sim \langle 0\vert T \int H \vert 0 \rangle$.

ACuriousMind
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