The quantity $Z[J]$ (which is the generating functional for all Green functions) physically represents the probability amplitude for a system to remain in the vacuum state. Can we find a similar physical meaning of the quantity $$W[J]=\frac{\hbar}{i}\ln Z[J]$$ (which is the generating functional for all connected Green functions)?


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Recall that, by definition, $$ Z[J]=e^{-iW[J]}=\langle\Omega|e^{-iHT}|\Omega\rangle $$ where $H$ is the Hamiltonian of your system, and $\Omega$ is the vacuum state. Therefore, $W[J]$ can be interpreted as the energy of the vacuum $E_\Omega$ in the presence of a source $J$, where the origin is chosen at $E=0$ for $J=0$. In other words, $W[J]$ is how much more energy does the vacuum have when we turn on an external source $J$.

More precisely, and as per $Z[J]=e^{-iW[J]}$, you can think of $W$ as the Helmholtz free energy of the system.

For more details, see e.g. ref.1., §11.3, or ref.2., §6.1.2 (and Appendix 18). See also this PSE post.


  1. Peskin & Schroesder - An introduction To Quantum Field Theory.

  2. Zinn-Justin - Quantum Field theory and Critical Phenomena.

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    $\begingroup$ At a mathematical level, I understand what you said. Thanks. But even now, it is physically very abstract. I wonder if there is a way to understand this physically (in terms of a real quantum mechanical system coupled to a real source field)? Please forgive my English. $\endgroup$ Commented Jan 19, 2020 at 13:36
  • $\begingroup$ @mithusengupta123 Sorry, but I haven't said anything mathematical, or at least I tried not to. Physically, $W[J]$ is the energy of the vacuum when you introduce a source $J$ in the system. I don't think one can by any more concrete than this... $\endgroup$ Commented Jan 19, 2020 at 13:39
  • $\begingroup$ I see. I thought I can think of the source as something like an external electric field or something and the $W[J]$ could be the energy in the presence and absence of that field. Thanks though. :-) $\endgroup$ Commented Jan 19, 2020 at 13:43
  • $\begingroup$ @mithusengupta123 That is absolutely right: the source $J$ could be an external electromagnetic field, or the current density of a fermion, or any other object that can couple to the system. $\endgroup$ Commented Jan 19, 2020 at 13:50

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