Why is there an infinite potential barrier between degenerate vacua in QFT?

I recently read "Fifty Years of Yang-Mills Theory and my Contribution to it" by R. Jackiw, and stumbled upon an intriguing sentence:

Usually in quantum field theory tunneling is suppressed by infinite energy barriers between degenerate vacua – this leads to spontaneous symmetry breaking.

1. In what sense is there an infinite energy barrier? How can we see this explicitly?
2. What does this have to do with spontaneous symmetry breaking?
• There is a more or less intuitive discussion of this in section 2.11 of arxiv.org/abs/1703.05448 – AccidentalFourierTransform Mar 27 '17 at 13:21
• @AccidentalFourierTransform Thanks! I'll have a look at it – jak Mar 27 '17 at 13:27
• @AccidentalFourierTransform If I'm not missing sth crucial, I think his argument only holds for an infinite volume system, which our universe is not. In addition, from his argument it seems very well possible that transitions take place. The only problem seems to be that it can not happen everywhere in an infinite volume system, because that would need, well an infinite amount of energy. For a finite volume system the energy would be finite. However, these lecture notes were incredibly helpful for some other problems I was currently struggling with. – jak Mar 27 '17 at 14:32

Suppose you have a quantum field theory with one scalar field $\phi$, and the equation of motion for this field tells you that the lowest energy configuration is obtained when $\phi$ is constant throughout spacetime, with $\phi^2 = \phi_0^2$ (here $\phi_0$ is a constant determined by the parameters of the Lagrangian, for instance).

1. There are two classical vacua, characterized by $\langle \phi \rangle = \pm \phi_0$. If you want to tunnel from one to the other, the energy barrier will be some constant integrated on all of space. Because of the infinite volume of space, this energy barrier is infinite, independently of how small the energy density is.
2. As a consequence, the quantum states which are superpositions of the two classical vacua are completely suppressed. This means that there are two quantum vacuum, which are identical to the classical ones. Hence the $\mathbb{Z}_2$ symmetry of the equation $\phi^2 = \phi_0^2$ is broken in any given vacuum by $\langle \phi \rangle = \pm \phi_0$. This is spontaneous breaking of the $\mathbb{Z}_2$ symmetry.
In summary, tunneling in QFT in strictly more than one dimension (as opposed to quantum mechanics, that you can see as QFT in $0+1$ dimension) costs an infinite amount of energy because of the integration over space. This suppresses superposition states, and entails spontaneous symmetry breaking.
• For your other remark, yes you can have a bubble where the scalar takes the other VEV, but in that case the bubble is surrounded by a domain wall, which has some tension. In other words, this configuration does not have minimal energy, because at the boundary of the bubble the field $\phi$ is varying. Because of the tension of the wall, the bubble will ultimately disappear. – Antoine Mar 27 '17 at 16:29