- #1
JRudolfo
- 4
- 7
A scalar field theory with potential $$V(\phi)=-\mu^2\phi^2+\lambda \phi^4$$ is spontaneously broken and as a consequence, for the ground state, $$\langle \phi(x) \rangle \neq 0$$.
However, the path integral, which should give ground state expectation values, looks to be zero by oddness of the integrand:
$$\langle \phi(x) \rangle=\int d\phi \, e^{i \int d^4x\, \partial^\mu \phi \partial_\mu \phi+\mu^2\phi^2-\lambda \phi^4}\phi(x)=0$$
How can this be reconciled with the fact that $$\langle \phi(x) \rangle \neq 0$$?
However, the path integral, which should give ground state expectation values, looks to be zero by oddness of the integrand:
$$\langle \phi(x) \rangle=\int d\phi \, e^{i \int d^4x\, \partial^\mu \phi \partial_\mu \phi+\mu^2\phi^2-\lambda \phi^4}\phi(x)=0$$
How can this be reconciled with the fact that $$\langle \phi(x) \rangle \neq 0$$?