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The one-point function (and all other odd correlation functions) in the $\phi^4$-theory, for example, calculated from the generating functional, always gives zero value in absence of external source i.e., $J=0$. To prove this it requires the invariance of the action under $\phi\to -\phi$.

However, if there is spontaneous symmetry breaking (SSB), the one-point function simply represent the vacuum expectation value of the field operator $\phi$ and is non-zero. But symmetry of the action continues to hold even after SSB takes place.

How do we reconcile these two apparent contradictions?

Qmechanic
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SRS
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2 Answers2

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You should work out the minimum energy state of your system (classically) to find the vacuum expectation value. I assume you're working with the standard $\phi^4$-Lagrangian $$\mathcal L=\frac{1}{2}(\partial \phi)^2-\frac{1}{2}m^2\phi^2-\frac{\lambda}{4}\phi^4 $$ which corresponds to the Hamiltonian $$\mathcal H=\frac{1}{2}\dot\phi^2+\frac{1}{2}(\nabla\phi)^2+\underbrace{\frac{1}{2}m^2\phi^2+\frac{\lambda}{4}\phi^4}_{=: V}$$ It is easy to see that the lowest energy solution for arbitrary $V(\phi)$ is always $\phi=\text{constant}$, and in this case the potential is minimized by $\phi=0$. Thus, the true vacuum of the theory is, indeed, located at $\phi=0$ (this indeed also yields the one-point function $\langle \phi\rangle$).

Now, to see the difference with spontaneous symmetry breaking, one really only needs to look at the relevant Lagrangian: It has a different potential. Usually, the potential for something similar to the abelian Higgs model is of the form $$V(\phi)=-\frac{1}{2}m^2\phi^2+\frac{\lambda}{4}\phi^4$$ which we can easily minimize to find that the lowest energy state corresponds to $$\phi^2=\frac{m^2}{\lambda}$$ so that we see that the true vacuum of theory is not located at the "origin", i.e. we find a nonzero vacuum expectation value.

Danu
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  1. OP is essentially making the suggestive argument that since (i) the action $S[\phi]$ is $\mathbb{Z}_2$-invariant, and (ii) the path integral measure ${\cal D}\phi$ (including boundary conditions) formally looks $\mathbb{Z}_2$-invariant then the 1-point function $\langle\phi\rangle_{J=0}$ should vanish, cf. e.g. my Phys.SE post here.

  2. However, the fallacy in the SSB case is, that the vacuum state $|\Omega\rangle$ breaks the $\mathbb{Z}_2$-symmetry. The main point is that the path integral $$Z~=~\int\!{\cal D}\phi ~\exp\left\{\frac{i}{\hbar}S[\phi]\right\}\tag{A}$$ is now implicitly defined relative to the vacuum state $|\Omega\rangle$, so that the path integral measure ${\cal D}\phi$ is actually not $\mathbb{Z}_2$-invariant.

  3. See also e.g. this related Phys.SE post.

Qmechanic
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