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

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.

• I guess you're asking whether the state moves around in the valley, or if it stays fixed ("remains broken" in the same orientation), and if it stays fixed, what keeps it there? ... I'm not sure, so I just leave this as a comment. But I think you can have a different value of the field (position in the valley) at every point in space, and it doesn't matter which one it is, because you can't measure the field directly. What matters is the form of the potential - the position of the minimum gives you the higgs VEV, the curvature (oscillations up the walls) creates the gauge boson masses.
– jdm
Jan 5 '14 at 12:17
• What I was trying to say was, I think the symmetry breaking only implies that you are anywhere in the circle, not that you are in a certain fixed point. I don't know how you move around in the valley, or if there is a physical meaning to the "angular position" but I'd like to know, too.
– jdm
Jan 6 '14 at 11:09
• I think this is a good question, I'm not sure what the confusion is. The question is, after spontaneous symmetry breaking does the system stay in a particular vacuum or does it move around. Furthermore, does this have any implications. Jan 27 '14 at 12:48
• the vacua are disjoint, $\langle \theta | \phi \rangle = \delta(\phi - \theta)$, so once you're in a vacuum, you stay there. also, superselection rules are relevant, i think, because they forbid the preparation of a superposition of vacua, eg $|0\rangle = |\phi\rangle + |\theta\rangle$, so you can't preserve the symmetry with eg $|0\rangle = \int d\theta |\theta\rangle$ Jan 21 '15 at 9:14
• As for the question itself, it's as unclear to me as to the other commenters. I suspect that the issue is your understanding of what a Goldstone boson is. In the proof of Goldstone's theorem, where exactly did you get the impression that the Goldstone bosons really induce "motion" along the vacuum manifold? It's a heuristic picture sometimes given, but since the different vacua have zero overlap, such perturbative excitations cannot mediate between different vacua. Jan 2 '17 at 14:00

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.)

• minor comment: I don't think that moving from $\theta=0$ to $\theta=\delta$ requires an energy density proportional to $\delta$ since the the energy density is $\theta$-independent. It's all the gradients $\partial\theta$ that matter for Goldstone bosons. One could perhaps imagine a very adiabatic process where all $\theta$-derivatives are very small and send them to zero faster than the volume is increased, I am not sure... But I agree that the crucial aspect suppressing the amplitude is the infinite volume, mathematically reflected in the vanishing amplitudes in th soft limit. Aug 28 '17 at 6:57
• @TwoBs Yeah, I guess it depends on exactly how you do it. If you move tiny little domains one at a time, the energy cost for each little domain would go like (domain wall surface area) * ($\delta$ / domain wall thickness), so $\delta$ times some geometric factor. One could also flip a large region in a very controlled uniform way to keep keep all spatial gradients small, but of course such a fine-tuned, nonlocal operation isn't very realistic. You may be right about a very adiabatic process being able to make the energy cost arbitrarily small. Aug 28 '17 at 7:38
– SRS
Oct 27 '18 at 15:27

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.

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.

• Yes. After the symmetry (say, U(1)) is spontaneously broken the system randomly chooses one particular solution, our of infinite such solutions. It is also true that, after the symmetry is broken, the solutions no longer exhibit the symmetry of the Lagrangian and the action of an U(1) group element maps one solution to another. But that is not what my confusion was...
– SRS
Jan 5 '14 at 9:38
• Great, then I have no idea what your question could possibly be. Jan 6 '14 at 7:25
• What do you mean when you essentially say that the degenerate vacua are isomorphic to each other? I know about isomorphisms between two groups. @Lubos Motl
– SRS
Apr 7 '17 at 9:56
• Dear @SRS - isomorphism is defined not only for groups but also for rings, fields, Lie algebras, whatever - and for objects. The general point is that there exists a map - the isomorphism - from one to the other that is simple, onto, and that respects all the structures on the mathematical or physical object. So you can map the Hamiltonians, operators, they have the operators that are just translations of each other etc. But they're not "equal", you can't really identify them element by element in the same set/space. They just have the same shape - iso morph ;-). Apr 8 '17 at 15:34

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.