If a Goldstone boson is an excitation moving between degenerate vacua, how do symmetries remain broken?

Question

In spontaneous symmetry breaking, moving around the circular valley of the Mexican hat potential doesn’t change the potential energy. These angular excitations are called Goldstone bosons. But doesn't the change of angle implies that the system moves from one vacuum to another because different points on the circular valley represent degenerate vacua?

If Goldstone excitations are like this, how does the symmetry remain broken? Goldstone excitation, by definition (because they represent variations in the coordinate on the circular valley), will then take the system from one vacuum to another. However, this doesn't happen.

Answer 1

This is a good question. I believe you are applying a mistaken analogy with an example that is often used to first introduce the notion of SSB: a single nonrelavistic particle in a double well with a potential barrier separating the two minima. As long as the barrier between the minima is finite, the particle can tunnel through it and inhabit both minima, so the reflectional symmetry is unbroken. But when the barrier height is formally taken to infinity, the particle gets "stuck" in one minimum or the other, breaking the symmetry. You are thinking that in a Mexican-hat-type potential, there is no potential barrier, so the system should be free to tunnel to all the minima, restoring the symmetry.

But an infinitely high potential barrier is a rather artificial notion. It's really a schematic representation of a more realistic situation in which, rather than a single particle, you have a spatially extended field (or a many-body system on a huge lattice) defined over a large volume $V$. Then the two minima in the one-particle picture really represent two distinct field configurations with equal total energy. Since going from one configuration to the other requires changing the value of the field at every single point in space (or spacetime), it requires a huge energy proportional to the total system volume $V$. So the "barrier height" in the one-particle picture really corresponds to the volume $V$ of the system in the field-theory picture. Only in the formal infinite-volume ("thermodynamic") limit is the symmetry truly broken.

Now moving the field configuration from, say, $\theta(x) \equiv 0$ to $\theta(x) \equiv \delta$ for some tiny angle $\delta$ only requires a tiny energy density proportional to $\delta$ (in the appropriate units), but the total energy $\delta \times V$ can still be very large. It's an order-of-limits subtlety: for any shift $\delta$ in the field value, no matter how small, we can image a system so large (roughly speaking, much larger than $1/\delta$) that shifting the entire system over by that amount requires an arbitrarily large amount of energy. So you still get an "infinitely high potential barrier to tunnel over" in the one-particle picture, even though the field's potential energy density $V(\varphi)$ has no barrier at all.

(To make things more explicitly quantum-mechanical, consider the quantum Heisenberg model on a lattice. If $|\psi\rangle$ and $|\psi'\rangle$ are two individual spin-$1/2$s rotated on the Bloch sphere by a small angle $\delta$, then the inner product $\langle \psi | \psi' \rangle = \cos \delta \approx 1 - (1/2) \delta^2$ is quite large, so a spin-$1/2$ could easily tunnel between the two states. But if we consider two huge systems of $N \gg 1$ aligned spins $|\Psi\rangle = \otimes_{i = 1}^N |\psi\rangle_i$ and $|\Psi'\rangle = \otimes_{i = 1}^N |\psi'\rangle_i$, then the tunneling amplitude $\langle \Psi | \Psi' \rangle = \cos(\delta)^n$ between the two systems is tiny, so it's very difficult to flip one state into the other.)

(To tie all this into TwoBs's answer: in the formal infinite-volume limit, the field's Fourier series representation becomes a continuous Fourier transform indexed by a continuous parameter $k$, and we can talk about Taylor expanding the energy density dispersion relation $\epsilon(k)$ about $k = 0$. For a Goldstone mode we have $\epsilon \to 0$ as $k \to 0$, but this is just an energy density - the total energy $E(k) = V \times \epsilon(k)$ is still huge, so the "energy barrier" is still very high. A Goldstone mode would need to be infinitely spatially extended in order to truly destroy the long-range order and restore the symmetry, and such an infinitely-large Goldstone mode is still too energetically costly to create.)

Answer 2

I am not sure I fully understand the question, but I will give it a try.

I think the answer to your question is in the Adler's zero condition. Indeed, the Goldstone bosons (GB)would represent a new minimum only in the zero momentum limit(otherwise their kinetic energy raises the total energy up, and it creates spacetime gradients that aren't there for the vacuum), which is precisely the limit for which GB's amplitudes vanish. Hence, no non-trivial transition is really taking place.

Extra Edits after extra thoughts My feeling that my answer above makes sense and is correct is reinforced by thinking about how you would actually move from one vacuum to another, namely by acting with the exponential of the (would-be*) charge operator $Q=\int d^3x \, J^0=\lim_{p\rightarrow 0} \hat{J}^0(p)$. Except that for a spontaneously broken continuous global symmetry the current starts linearly in the GB field, $J^\mu=-f\partial_\mu\pi+\ldots$, so that $Q|0\rangle$ doesn't move you into another vacuum but rather into a coherent state of zero momentum GBs, starting with a single one-particle soft GB: $Q|0\rangle=\lim_{p_\rightarrow 0}|\pi(p)\rangle+\ldots$, for which again the amplitudes vanish. This is nicely explained (and in fact the soft theorems are even derived from) in section 4.1 of this beautiful paper https://arxiv.org/abs/0808.1446 .

*bonus comment: one could be worried about the Fabri-Picasso theorem (see e.g. https://en.wikipedia.org/wiki/Goldstone_boson) that tells us that the charge, strictly speaking, doesn't really exists (although its commutator always does). But this statement is an overkill as it is simply the statement that the one-particle states of definite momenta, such as $|\pi(p\rightarrow 0)\rangle$ that is generated by the charge, have infinite norm, i.e. $\langle\pi(p)|\pi(k)\rangle=(2\pi)^3 2|k| \delta^3(\vec{k}-\vec{p})$. Incidentally, this IR divergence of the norm of the state for $\vec{k}\rightarrow \vec{p}=\vec{0}$ is prortional to the volume $V\rightarrow \infty$, making contact with the answer by @tparker . The moral of this story is: the charge doesn't exist, strictly speaking, but just because one is creating one-particles states of soft momenta. It makes perfect sense to consider soft limits taking the limit carefully, again as in the reference cited above.

Answer 3

The proposition that the symmetry G is spontaneously broken means that by acting by G on the vacuum configuration, we obtain an isomorphic but different configuration. For the symmetry to be unbroken, the transformations in G would have to map the vacuum configuration onto the same one, not just isomorphic one.

If you reflect the letter R along the vertical axis, you will get Я. This "ya" is isomorphic but it is different, so R isn't left-right-symmetric; the symmetry is broken; it is not ever possible for a symmetry to produce an object that even looks different (isn't isomorphic). It's always isomorphic; the question is whether it is identical. The letter H is mapped to H again so H is left-right-symmetric.

The Goldstone bosons' being nontrivial excitations proves that the action by G is nontrivial so the vacuum is not symmetric under G.

Answer 4

Not sure I understand the question. The symmetry is broken because you're in the valley. As you say correctly, moving around in the valley doesn't change anything about that.