# 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?

## 1 Answer

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
• Could it be that after SSB, the limits or boundary conditions (in the path integral) are not symmetric causing the odd correlation functions to be nonzero? @Danu – SRS Dec 21 '17 at 9:54
• I think that that's one way of thinking about what I said above---the asymptotic (vacuum) state has no symmetry. – Danu Dec 21 '17 at 9:58