# One-point function and vacuum expectation value in $\phi^4$-theory

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?

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.
• But if the action is invariant under $\phi\to-\phi$, all odd correlation functions must vanish. The symmetry of the action never spoils even after SSB. @Danu – SRS Dec 19 '17 at 19:32
• @SRS I think the point is that the odd correlation functions w.r.t. the $\phi=0$-state do vanish, but not with respect to the vacuum state. There is no symmetry when expanding around the vacuum (the usual $\phi=\phi_0+\rho$). – Danu Dec 20 '17 at 21:43