Functional integral in spontaneous symmetry breaking

Question

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?

Answer 1

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.

Answer 2

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.