# For a continuous symmetry breakdown, why is a superposition of all vacua impossible?

I understand that for a discrete symmetry breakdown where the wave functional for your field has two vacua, say $|+\rangle$ and $|-\rangle$, the reason why the true groundstate can't be a superposition of these two is because they are separated by a potential barrier and, assuming the field extends over an infinite volume, the total energy barrier is infinite. So tunneling is impossible.

But let's say you have a continuous symmetry and a Mexican hat potential. Then you have infinitely many vacua, denoted by $|\theta\rangle$ where $0\le\theta<2\pi$. If I pick two of these states, say $|0\rangle$ and $|\varepsilon\rangle$ where $\varepsilon$ is infinitesimal, the potential barrier between them is zero: the two states have the same energy and are next to each other, so tunneling should be permitted. And then once you have tunneled from $|0\rangle$ to $|\varepsilon\rangle$, you can tunnel to $|2\varepsilon\rangle$ then $|3\varepsilon\rangle$ and so on all the way back to $|0\rangle$. You will then have a superposition of all such states, and the superposition would respect the symmetry.

Except that doesn't happen. Where's the flaw in the reasoning?

• The vacua are disjoint. It takes infinite energy to tunnel from one vacuum to the next for a infinite system. Goldstone modes will not take you from one vacuum to another. – SRS Nov 16 '16 at 10:17
• – Bruce Greetham Nov 16 '16 at 10:58
• @Bruce The answer to that question doesn't fully solve my confusion. Peter says that any matrix element between two vacua vanishes, and I know this needs to be the case, but I don't why. For discrete symmetries, there's the intuitive notion that you have a potential barrier multiplied by an infinite volume, but I don't know how the same idea would apply for continuous symmetry, since one vacuum can be right next to the other, without barrier. – Donjon Nov 16 '16 at 12:41

## 1 Answer

It's true that rotating the value of the field at a single point by an infinitesimal amount only costs infinitesimally small energy $\epsilon$. But to take the field into a new ground state, you need to rotate the field value at every point. So the total energy cost is $E = \epsilon V$, where the local energy cost $\epsilon$ is infinitesimal but the total spacetime volume $V$ is infinite. Whether $E$ ends up infinitesimal, finite, or infinite depends on the order of limits - is $\epsilon$ "smaller than $V$ is infinite?" Your simple intuitive argument isn't precise enough to answer that question - you need to do a calculation. And if you do so, it turns out that the total energy cost $E$ to tunnel between different vacua is infinite, but you do get Goldstone mode excitations which locally (though not globally) take you into a different ground state, and these Goldstone modes can have arbitrarily low energy (although strictly speaking not zero energy, because an exactly zero energy Goldstone mode isn't an excitation at all, and just leaves you in the original ground state).

• "but you do get Goldstone mode excitations which locally (though not globally) take you into a different ground state" I intuitively understand this. But what does it mean mathematically? Different vacua are disjoint but still Goldstone modes locally connect them in some manner. Which Hilbert space does a Goldstone mode belong to? @tparker – SRS Oct 26 '18 at 18:57