So, functional integral is defined to be (with $\lvert\Omega\rangle$ is the vacuum state):

$$\frac{\langle\Omega\rvert X \lvert\Omega\rangle}{\langle\Omega\vert\Omega\rangle} = \frac{\int \mathcal{D} \Phi e^{iS} X}{\int \mathcal{D} \Phi e^{iS} }.$$

The $X$ part is just the insertions one wants to calculate.

However, in a theory with degeneracy vacuum states, such as spontaneous symmetry breaking theory, it's not very clear how this formalism can be used. The question is, how is the functional integral formulated in such a theory?


2 Answers 2


Note: In the case of gauge symmetry, the degeneracy of the vacuum under gauge transformations leads to topologically inequivalent vacua characterized by the winding number of the gauge fields, in which case the lagrangian in the path integral has a term which indeed depends on which (theta) vacuum you choose. However here we will consider a vacuum degeneracy due to a global symmetry.

Proof that defining equation for path integral is independent upon choice of vacuum (it is assumed that you agree that the in and out vacuum are the same, because otherwise the matrix element vanishes), for simplicity let's consider a one parameter continuous symmetry:

First remember that for a general S-matrix element \begin{align*} \left\langle \beta^+ \right| T\{ \ldots \} \left| \alpha^- \right\rangle &= \int \prod_{\tau,\vec{x},m}d \phi_m (\vec{x},\tau)\{ \ldots\} \text{exp }\left(i\int_{-\infty}^{+\infty}d\tau L[\phi(\vec{x},\tau),\dot{\phi}(\vec{x},\tau)]\right) \\ &\qquad\qquad \times\left\langle \beta^+ \right| \left.\phi(+\infty);+\infty\right\rangle \left\langle \phi(-\infty);-\infty\right|\left. \alpha^- \right\rangle \end{align*}

The $\pm$ superscript is for out and in state respectively. We parametrize the vacuum by $\theta$ $$ e^{Q\theta}\left| \Omega,0\right\rangle = \left| \Omega,\theta\right\rangle $$ Where $Q(\pi,\phi)$ is the generator of this continuous global symmetry. Fist, chose the in and out states to be the $\theta=0$ vacuum. Then you can easily prove that $$ \left\langle \Omega^{\pm},0\right| \left.\phi(\pm\infty);\pm\infty\right\rangle \propto \text{exp} \left( -\frac{1}{2}\int d^3 xd ^3 y\mathcal{E}(x,y)\phi(x)\phi(y) \right) $$ Where $\mathcal{E}(x,y) = \mathcal{E}(x-y)$ is the Fourier transform of the free energy as a function of the momentum.

Then if we chose the in and out states to be the vacuum with small $\theta>0$ \begin{align*} \left\langle \Omega^{\pm},\theta\right| \left.\phi(\pm\infty);\pm\infty\right\rangle &\propto \left(1 + \theta Q\left[\phi,\frac{\delta}{\delta \phi}\right] \right)\text{exp} \left( -\frac{1}{2}\int d^3 xd ^3 y\mathcal{E}(x,y)\phi(x)\phi(y) \right) \\ &\approx \text{exp} \left( -\frac{1 + \theta}{2}\int d^3 xd ^3 y\mathcal{E}(x,y)\phi(x)\phi(y) \right) \end{align*} And we deduce that in both cases $\theta = 0$ and $\theta $ slightly larger, the term $$ \left\langle \Omega^+, \theta \right| \left.\phi(+\infty);+\infty\right\rangle \left\langle \phi(-\infty);-\infty\right|\left. \Omega^- ,\theta \right\rangle $$ Will have the only purpose of providing the $i\epsilon$ term, and there for $\theta$ is irrelevant.


You don't have to write the generating functional in the ground state. More generally $$ \langle\mathcal{O}\rangle=\frac{\text{tr}(\rho \mathcal{O})}{\text{tr}\;\rho} $$ where the trace is over all states and $\rho$ is the density matrix. This can be written as a path integral in general for any $\rho$, so I don't see a particular formal difficulty in defining the functional integral in a symmetry breaking case.

If you are in a certain vacuum you can write the path integral in terms of fields that are perturbations from that vacuum and the remaining gauge symmetry can be dealt with using the standard Fadeev-Popov procedure for gauge invariant theories.

  • $\begingroup$ Thank you. So, just do perturbation theory around the vacuum of interest, then applied the same procedure for the functional integration, right? Yes, I know this trick. But, the point is, functional integral is supposed to be a way to go to the nonperturbative definition of the QFT, so I really want to avoid using any kind of expansion around the minimum. But still ... $\endgroup$
    – user109798
    Commented Mar 24, 2015 at 18:45
  • $\begingroup$ One can always think of the functional integration as a way to read-off correlation of n-points insertion in the theory. And I guess with this information alone, it's sufficient to study the theory (well, assume that one has infinite calculational power). I do agree with you that the term "ground state" isn't needed. $\endgroup$
    – user109798
    Commented Mar 24, 2015 at 18:46
  • $\begingroup$ The path integral indeed is a nonperturbative definition of QFT, but that does not mean that we, puny humans, are able to solve it nonperturbatively is well. If we could, the path integral would (among others) contain a sum over all vacua, restoring the broken symmetry. $\endgroup$ Commented Mar 27, 2015 at 9:18
  • $\begingroup$ @DavidVercauteren can you elaborate in an answer? it sounds interesting $\endgroup$
    – innisfree
    Commented Apr 24, 2015 at 13:26
  • $\begingroup$ Which part do you want elaboration on? $\endgroup$ Commented May 5, 2015 at 8:44

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