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.



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*In what sense is there an infinite energy barrier? How can we see this explicitly?

*What does this have to do with spontaneous symmetry breaking?

 A: 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). 
Now let's address your questions : 


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*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. 

*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. 
