In $1$ dimension, we know that lattice breaks continuous translational symmetry into discrete translational symmetry, which generates $1$ Goldstone boson, i.e. $1$ longitudinal phonons.

In $d$ dimensions, if there are only $1$ type of atoms, then there are $1$ longitudinal phonon and $d-1$ transverse phonons. However, in $d$ dimensions, the continous symmetries of $d$ dimensional Euclidean group are broken, and in principle we should have $d+\frac{d(d-1)}{2}=d(d+1)/2$ Goldstone bosons. What are the other Goldstone Bosons?


3 Answers 3


In systems that break continuous translation symmetry (regular crystals,lamellar solids, smectics etc.) one also necessarily breaks rotational invariance (but note that this is not true the other way around, allowing liquid crystals to exist). As you rightly pointed out the simple goldstone mode count does not seem to work in such ordered phases.

The reason is essentially because the orientational degrees of freedom get slaved to the translational phonon modes, which results in the orientational modes no longer being soft (they get gapped) and hence not Goldstone modes. This is analogous to the Anderson-Higgs mechanism but is different in that we only have global symmetries being spontaneously broken and no gauge fields involved in forming a crystal.

To see it explicitly, we can take for example a regular 3d crystal (with reciprocal lattice vectors ${\bf G}_n$). Then the relevant non-vanishing order parameter is going to be the fourier component of the mass density $$ \rho({\bf r})=\rho_0+\sum_{{\bf G}_n}\left[\psi_{n}e^{i{\bf G}_n\cdot{\bf r}}+\mathrm{c.c}\right] $$ where $\rho_0$ is the mean density and $\psi_n$ are the complex fourier amplitudes ($\langle|\psi_n|\rangle\neq0$ in the crystal). As the energy is invariant under uniform translations we require ${\bf r}\rightarrow{\bf r}+{\bf u}\implies\psi_n\rightarrow\psi_n e^{-i{\bf G}_n\cdot{\bf u}}$.

Now a global rigid rotation is also a symmetry of the energy: ${\bf G}_n\rightarrow{\bf G}_n+\delta\theta\times{\bf G}_n$. This leads to a corresponding displacement of lattice points as ${\bf u}=\delta\theta\times{\bf r}$.

As ${\bf u}$ is the translational goldstone mode and $\delta\theta$ is the rotational one, we immediately see that the orientational zero modes are slaved to the translation phonon modes ($\bf u$), which in this particular case is specified by $$ \delta\theta_i=\dfrac{1}{4}\epsilon_{ijk}(\partial_j u_k-\partial_k u_j) $$ which is the antisymmetric part of the strain tensor. As $u_i$ is a goldstone mode, its two point correlator will have a pole at zero wave-vector (momentum). This can also be obtained using linear classical elasticity for the phonons. Therefore heuristically (being cavalier about indices) as $\delta\theta\sim\partial u$ and $\langle|u(q)|^2\rangle\sim1/q^2$, we have $\langle|\delta\theta(q)|^2\rangle=$const. as $q\rightarrow0$ showing that the orientaional modes are no longer soft but are instead gapped (even though the rotational symmetry was spontaneously broken). This point was already noted by Mermin (Phys. Rev. 176, 250 (1968)) as it leads to true long-ranged orientational order even in 2d crystals.

You can also look at the book "Principles of Condensed Matter Physics" by Chaikin and Lubensky, where this point is discussed in multiple cases (including for the smectic-A phase, in which case a generalization of this point leads to the spectacular analogy between the SmA-Nematic transition to that of a superconductor).

  • 2
    $\begingroup$ Thanks, may you point out explicitly which chapter in Chaikin's textbook covers this part? $\endgroup$
    – 346699
    Jan 19, 2017 at 20:44

For the case of translational symmetry, the gapless point of the phonon mode is understood as a uniform translation of the lattice, which of course costs zero energy. However, by saying that the phonon is gapless, we practically compute the energy $E(k)$ of a phonon of lattice momentum $k$, and take the limit $$\lim_{k\to 0}E(k)=0.$$ The limit of lattice momentum $k\to 0$ is approachable because in the limit of crystal size $L\to \infty$, $k$ becomes quasi-continuous.

Now let's look at the case of rotational symmetry. First, for a uniform rotation of the whole lattice, again we know it costs no energy, which should serve as the gapless point of the Goldstone mode. However, we cannot define a "lattice angular momentum", which is a "generator" of the discrete rotation group, and which becomes quasi-continuous in the large size limit.

So to summarize, it is easy to confirm that there is a zero mode corresponding to rotational symmetry breaking (uniform rotation of the crystal). However, unlike for phonons, there is no quasi-continuous quantum number (the would-be lattice angular momentum) we can write its "dispersion" in terms of. Thus this mode is less talked about in literatures.

However, consider the following example:

enter image description here

This system is not a lattice, but nevertheless has a discrete rotational symmetry. In this case one can define a quasi-continuous analog of angular momentum, and the spectrum of the Goldstone mode corresponding to rotation symmetry breaking can be easily obtained.

This system also breaks translational symmetry. However, one cannot define a phonon mode that has linear dispersion as a function of momentum, although there does exist a Goldstone corresponding to translational symmetry breaking.


The Wikipedia article on "Goldstone boson" says

In general, the phonon is effectively the Nambu–Goldstone boson for spontaneously broken Galilean/Lorentz symmetry. However, in contrast to the case of internal symmetry breaking, when spacetime symmetries are broken, the order parameter need not be a scalar field, but may be a tensor field, and the corresponding independent massless modes may now be fewer than the number of spontaneously broken generators, because the Goldstone modes may now be linearly dependent among themselves: e.g., the Goldstone modes for some generators might be expressed as gradients of Goldstone modes for other broken generators.

I won't pretend to fully understand that, but apparently the usual rule - that in a relativistic system, each spontaneously broken symmetry gets an independent Goldstone mode - only applies for internal symmetries, not for spatial ones.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.