energy - Doesn't the "Mexican hat" potential give a misleading impression that the barrier height between two vacua is finite?

For $\mathbb{Z}_2$ symmetry breaking in a Classical Field Theory described by a potential $$V(\phi)=\lambda(\phi^2-v^2)^2,\tag{1}$$ there is a finite energy barrier of height $\epsilon=\lambda v^4$ that exists between two minima $\phi=\pm v$ of $V(\phi)$ separated by a maximum at $\phi=0$. This can be seen clearly from the picture of a double-well potential. This finite potential barrier when multiplied with the infinite volume of space gives rise to infinite (potential) energy barrier. To be explicit, the infinite energy barrier comes from $$[V(0)-V(\pm v)]\int d^3x=\epsilon\int d^3x\to \infty!\tag{2}$$

But for $U(1)$ symmetry breaking, the potential energy barrier cannot be seen between two neighbouring vacua from the picture of Mexican hat potential. According to (1), this would mean by the potential barrier between two minima can be finite! But probably changing the vacuum costs infinite energy. If so, does it mean that the "Mexican hat" potential give a misleading impression? Why does the figure incapable of showing the actual barrier?

Answer

Good question. You are correct that, in principle, it is possible to rigidly, uniformly, and adiabatically rotate the entire field configuration by an identical angle in such a way that you always stay exactly inside the degenerate ground-state manifold. This gives the misleading impression that you can tunnel from one ground state to another with zero energy cost, as you suggest in your question.

But in practice, this isn't really realistically physically possible. That's because of the local nature of the thermal or quantum fluctuations. For concreteness, imagine the $XY$ model $$H = J \sum_{\langle i, j\rangle} \cos(\theta_i - \theta_j)$$ on a hypercubic lattice in $D \geq 3$ spacetime dimensions, in which the global rotational symmetry gets simultaneously broken. You can't realistically set up a situation in which every spin in the entire lattice rotates from pointing toward angle $\theta_i$ to angle $\theta_f$ exactly simultaneously, in one giant nonlocal perturbation. (For one thing, such a "perturbation" would require an infinite amount of time to set up, since the "planning message" about what angle to rotate by would need to propagate arbitrarily far before the process actually begins.) In practice, all you can do is first rotate a small little local cluster of spins from $\theta_i$ to $\theta_j$, then rotate that cluster's neighboring spins, then the next neighboring spins, etc. That's because by assumption, in a local QFT (or lattice system), the microscopic couplings are purely local, so nonlocal "update moves" (in Monte Carlo language) are not allowed.

Eventually, the effect of the initial perturbation will have propagated out far enough that a huge region of spacetime will have rotated its angle, so that to a good approximation you're locally in a new ground state. But the point is that as you do this process, at any given instant there will always be a boundary between the old and new ground states (kind of like a domain wall, but less sharply localized). That boundary will need to move and hit every spin in order for the entire system to rotate. So even if the rotation angle is infinitesimal and each individual spin's rotation only costs energy $\epsilon J$, the total energy of the entire tunneling process will still get multiplied by the volume of spacetime and will be a huge $\epsilon J V$.

Thought of another way: you can create a Goldstone mode with arbitrarily low energy by delocalizing it enough, but in practice it's impossible to create a Goldstone mode with exactly zero energy. The difference is subtle, but conceptually important.