# Why does the breaking of a 'continuous' symmetry lead to gapless excitations but not that of a discrete symmetry?

Goldstone theorem, especially in the context of condensed matter physics, can be stated as:

Whenever there is spontaneous breakdown of a continuous global symmetry, the spectrum of the theory contains gapless excitations. However, this is not true for breakdown of discrete symmetries.

For example, the spin wave of Heisenberg ferromagnet is a classic example of a gapless excitation resulting from the spontaneous breakdown of the spin rotational symmetry of the Heisenberg model. There exists rigorous, formal proofs of this theorem.

Question But physically, can we understand why this happens (or doesn't happen)?

It's true that in the Heisenberg model $$H=-J\sum_{\langle i,j\rangle}{\bf s}_i\cdot {\bf s}_j\tag{1}$$ the spins $${\bf s}_i$$ can rotate in $${\rm 3D}$$ while the Ising spins can only point up or down. It's tempting to think, if the spins can rotate continuously, we can excite the system with arbitrarily small energy by gradually tilting the spins infinitesimally w.r.t its immediate neighbors. This is not allowed if the spins are only allowed to flip (e.g., Ising model) and therefore, it requires a finite energy cost to excite the system! In the first case, we thus expect gapless excitation and gapped in the second. But this argument (based only on the freedom of movement of spins) is half-baked!

Continuous movement in $${\rm 3D}$$ doesn't seem to ensure emergence of gapless excitations. The Hamiltonian must also have a continuous symmetry. For example, if we think of a generalized Heisenberg model $$H=-\sum_{}(Js_{ix}s_{jx}+J's_{iy}s_{jy}+J''s_{iz}s_{jz}),\tag{2}$$ the spins can still rotate continuously in $${\rm 3D}$$, but the Hamiltonian of Eq.$$(2)$$ has no rotational symmetry unless the constants $$J,J',J''$$ are all equal (in which case $$(2)$$ reduces to form $$(1)$$)! Therefore, Goldstone's theorem cannot be applied here and there is no certainty that the spectrum will have gapless excitations.

So the question is why is it necessary to have a continuous symmetry in the first place to have gapless excitations? I am not looking for a formal proof.

DISCLAIMER: the heuristic argument provided below largely pertains to classical systems. I make no pretense of rigor, nor of simple generalization to quantum systems.

It's tempting to think, if the spins can rotate continuously, we can excite the system with arbitrarily small energy by gradually tilting the spins infinitesimally w.r.t its immediate neighbors.

Let me start by reviewing the basic physical argument for Goldstone modes. If a continuous global symmetry is broken in the ground state, then the ground state picks out one particular broken symmetry state. By slowly varying the system over space by the continuous symmetry, the system always appears to be in the ground state locally. Globally, there can be as small a "twist" as you want, so the energy must be gapless.

Now, suppose your system has continuous degrees of freedom, but does not have a continuous symmetry. Then by slowly varying the spins, there is no way the system could appear to be locally in the ground state everywhere -- indeed, if it did look like this, then you would certainly have a continuous symmetry! Instead, your system will have larger and larger energy densities the more you "twist".

• Can you work this out? I tried to write an answer along those likes, but when I did, the terms of the "small" twists did not add up to something small (essentially, each twist contributes a sin(1/N) term, and there are N of those, so this is O(1)). So it seems you need to put some extra information in to make sure that the small twists only contribute O(1/N^2). (To put it more bluntly: Your statement "Globally, there can be as small a "twist" as you want, so the energy must be gapless." is ad hoc completely unjustified, as you are summing many of those small things!) Commented Feb 18, 2021 at 18:17
• Just because many people iterate an supposedly intuitive explanation doesn't make it correct. The multiple of 2pi/L is not 1D specific. My point is more general: Your argument says "each energy contribution is as small as you want". I am saying "But you are adding more and more of them." It is entirely unclear from this "argument" whether this growing number of smaller and smaller numbers will add up to zero, 1, or infinity. As it stands, the intuition could be just as well plain wrong. You should at least try to estimate how this small energy scales with the amount of twist! Commented Feb 18, 2021 at 22:25
• There is, by the way, also the possibility that the overall explanation is indeed plain wrong (and not just an argument missing): From the way spin wave excitations are constructed, it does in fact not look very much like a system which has been slowly twisted (indeed, you apply one spin raising operator at some position, which does not seem to change the average state in any region!) Commented Feb 18, 2021 at 22:30
• Of course, your point is well-taken and I certainly agree. Perhaps I can assuage your initial concern, now that I've thought about it a bit. Consider a 1D Heisenberg or XY model -- I'll use an XY model, $H = -J \sum_i \cos(\theta_{i+1} - \theta_i)$, for simplicity. The lowest energy Goldstone mode has the small "twist" $\delta \theta = 2\pi / N$ from spin to spin. Then the energy of this mode is $JN (1 - \cos(\delta \theta)) \sim \pi J / N$ as $N \rightarrow \infty$ (I'm not sure where you're $\sin(1/N)$ came from, now that I think about it).
– Zack
Commented Feb 18, 2021 at 23:23
• Stackexchange is warning me not to have an extended discussion in the comments, so I will end it here, although I'd be happy to discuss elsewhere. The last thing I'll say is this: of course I agree that my answer is not extremely careful, and also that not all commonly-held dogma is correct in physics. But that does not mean that these sorts of heuristic arguments are not useful: clearly, just by looking at a few examples, the picture I described is at least true of Goldstone modes in classical systems, and is at least a plausible picture for "why" Goldstone modes are gapless.
– Zack
Commented Feb 18, 2021 at 23:30

TL;DR:

If your Hamiltonian has a continuous symmetry, you can take the ground state and rotate it by an arbitrarily small amount, and it is still a ground state. If you patch together two blocks which differ by a small rotation, you only pay a price at the boundary between the regions, and this penalty can become arbitrarily small as you make the rotation smaller and smaller.

You can then gradually change this small rotation throughout your system, and the price you pay is only the change in rotation, which can be made arbitrarily small. (Symmetry breaking ensures that this state is different from the ground state; otherwise, you might just construct a state which is equal to the ground state.)

However (fortunately? unfortunately?), this is only half the story, since you have to add up these small energy penalties over a large number of positions (since you need to complete a full $$2\pi$$ rotation globally), and there is no guarantee that the sum of many small things will be small.

To show that this sum is indeed small, you also have to use that the ground state has an extremality property (namely that it is the ground state), which implies that the dependence of the energy on the rotation angle must be quadratic in the angle (a linear term would increase the energy in one direction, but it would necessarily decrease the energy in the other direction), while the number of terms is linear in the inverse angle, so the sum goes to zero for a large system (and thus small angles) at least as one over the system size.

What is different is the symmetry is discrete? For a discrete symmetry, a small rotation of the ground state gives a state which is no longer a ground state, so when you patch together two regions which differ by a small rotation, you pay a price proportional to the volume of one of the regions. On the other hand, if you patch together two regions related by a discrete symmetry transformation, the price you pay at the boundary is constant (as the angle is large). Thus, the excitations you can construct this way have a constant energy.

Consider a 1D Hamiltonian with a continuous symmetry $$U_g$$, $$U_g^{\otimes N}H(U_g^\dagger)^{\otimes N} = H\ .$$ Then, if $$\vert\psi\rangle$$ is a ground state of $$H$$, so is $$U_g^{\otimes N}\vert\psi\rangle$$.

Now split the chain into two parts A and B, and apply $$U_g$$ to the sites in A, and $$U_h$$ to the sites in B. Then, the energy for the Hamiltonian terms inside A and B will still be the same. However, the terms at the boundary will have some higher energy. Assuming a two-body Hamiltonian $$H=\sum h$$, and denoting the two-site reduced density matrix by $$\rho$$, this energy will be $$e'=\mathrm{tr}[h(U_g\otimes U_h)\rho(U_g\otimes U_h)^\dagger] = \mathrm{tr}[h(I\otimes U_{g^{-1}h})\rho(I\otimes U_{g^{-1}h})^\dagger]\ ,$$ where I have used the symmetry of the Hamiltonian.

Now if the group is continuous, you can choose $$g^{-1}h$$ such that $$\|U_{g^{-1}h}-I\|\le \epsilon\ .$$ This is what is special about the continuous symmetry, and what fails for a discrete symmetry!

Then, you will find that the change in energy at each boundary is $$\Delta e = e' - \mathrm{tr}[h\rho] \le 2 \|h\|\epsilon\ .\tag{1}$$

You might now think that we are done, since we can now proceed to combine this by splitting the system into $$L$$ blocks which you rotate by $$2\pi/L$$ each. Then, $$\epsilon \propto 1/L\to 0$$. However, the problem is that you are adding $$L$$ such terms, so from (1), you only find that the energy is $$O(\epsilon L)=O(1)$$, which comes as no surprise.

So to make sure this energy is actually vanishing as $$L\to\infty$$, we have to make sure that in fact $$\epsilon\propto 1/L^2$$ or smaller. How can this be the case?

Clearly, this requires that the first-order term in (1) when expanding $$U_{g^{-1}h} = 1+\epsilon G+O(\epsilon)$$ vanishes; this term is nothing but $$R:=\epsilon\,\mathrm{tr}[h(I\otimes G)\rho-h\rho(I\otimes G)]\ .$$

It is easy to see that $$R\stackrel{!}{=}0$$ does not follow from the symmetry of the Hamiltonian. Instead, it must be a property of the ground state, or the ground state and Hamiltonian together!

Indeed, imagine $$\Delta e = \epsilon K + O(\epsilon^2)$$ for some non-zero constant $$K$$. Then, summed up over the whole system, the total energy difference is $$\Delta E = \epsilon L K + O(L\epsilon^2)\ .$$ Indeed, if $$\epsilon>0$$, this gives a term or order $$1$$. However, we can also change the sign of $$\epsilon$$, such that $$\epsilon L K<0$$ -- in that case, we get an ansatz for a state such that $$\Delta E < 0\ !$$ Clearly, this is impossible since $$\Delta E$$ is the energy difference to the ground state!

Thus, the fact that we are in a ground state rules out a term linear in $$\epsilon$$, and thus, the energy per site/boundary is $$O(\epsilon^2)=O(1/L^2)$$, and thus the total energy difference is $$\Delta E = O(1/L^2)\ ,$$ where $$L$$ can be taken to be the system size (if we twist at each site by an angle proportional to the position).

• I should also acknowlegde that Zack's classical example gave me the intuition why it is 1/N^2 - also in the classical case a sin(x-y) coupling would be linear in the difference, but then the FM state would not be the ground state! Commented Feb 19, 2021 at 20:00
• Is there a typo in the line where you say $\Vert U_{g^{-1}h} \Vert < \epsilon$? If $U$ is unitary, then $\Vert U\Vert = 1$. Presumably you mean $\Vert U - \mathbb I\Vert$? Commented Feb 19, 2021 at 20:26
• An interesting follow-up question is how (whether?) this is related to the usual spin wave ansatz which is obtained by applying an operator $\sum e^{ikx} S^+_x$ -- this does not seem obvious. Commented Feb 19, 2021 at 21:00
• Glad that your answer worked out, the result is very pretty. The missing piece about the necessity of being in the ground state is a good catch. It's also nice to know that the commonly held classical intuition applies to the quantum case as well (although I would have been very surprised if it didn't).
– Zack
Commented Feb 20, 2021 at 21:29
• Thanks. And conversely, the classical case requires (unsurprisingly) the same extra condition of being a ground state to actually work. I'd still be very curious to understand how this links to a spin-wave ansatz, which seems rather different. -- In any case, I had the whole answer written more than a week ago and then I got stuck on the last part - why the 1/N part shoud vanish. In retrospect, it is obvious one has to use this condition. It was really the classical case which made me realize it! Commented Feb 20, 2021 at 22:56

Short answer: There are other gapless excitations than Nambu-Goldstone modes.

No, certainly not, but it's close. You're skipping over the part about the energy cost. In your proposed generalized Heisenberg model, $$J\neq J'\neq J''$$ would generically imply the existence of one or more discrete ground states, separated from one another by higher energy configurations. Starting from the ground state, it would require a finite amount of energy to excite the system, which means that it's gapped.
If all of the $$J$$'s are very close to one another, then you have an approximate (not exact) symmetry, which would correspond to low-energy (but not exactly massless) excitations. This is the case e.g. for pions, which can be thought of as the pseudo-Goldstone modes corresponding to the approximate chiral-flavor symmetries which are broken by the different masses of the quarks, and which therefore have far smaller masses than most hadronic particles.