# Is a phason a Goldstone mode?

Suppose we have a lattice system whose ground state is an incommensurate charge-density wave (CDW). Strictly speaking, this ground state does not have Goldstone modes because the only symmetry that is spontaneously broken is the discrete translational symmetry of the lattice. But the possible symmetry-broken states vary continuously (they can be continuously parameterized by the arbitrary phase shift by which the CDW is translated relative to some reference configuration), which is why phasons (distortions of the CDW with arbitrarily long wavelengths and low energies) are gapless. This seems like pretty much exactly what we'd expect from a Goldstone mode. Should I think of a phason as a Goldstone mode? Is the existence of a continuous degenerate ground-state manifold more important for "Goldstone-like" behavior than the existence of a continuous spontaneously broken symmetry?

• Just out of curiosity since I know nothing about quasicrystals: how does one ensure exact continuous degeneracy of the ground states without a spontaneously broken continuous symmetry? – Tomáš Brauner Aug 5 '16 at 20:58
• A naive question: can the CDW transition be thought of precisely as spontaneously breaking of this continuous phase shift "symmetry" (in addition to the discrete lattice symmetry)? And if not, why? – Rococo Aug 5 '16 at 21:27
• @TomášBrauner I think it's quite mathematically nontrivial to prove. As I understand it, the rough idea is to think of every site on the lattice as being pinned in position by a restoring force with potential energy $k\, \delta x^2$. If you translate a periodic lattice with unit spacing $a$ a slight distance $\delta x \ll a$ from where it "wants" to be, the springs all stretch and you get an extensively scaling energy $N\, k\, \delta x^2$. But if you do the same thing to a quasicrystal, one can show that since there's no unit lattice spacing, it turns out that the new position of many ... – tparker Aug 5 '16 at 23:35
• @TomášBrauner ... sites will be extremely close to the equilibrium positions of different sites. So instead of all stretching by the same length $\delta x$, many of the springs will "snap" to a new pinning site and reduce their energy, so the total energy of the translation will scale subextensively, and from there you somehow argue that any such translation is energetically costless? Out of my depths here. – tparker Aug 5 '16 at 23:37
• @TomášBrauner It's much easier to prove for an incommensurate charge-density wave: you can often just find an explicit band dispersion relation $\epsilon(k)$ and show that it has two degenerate minima at $\pm k_0$, where $a k_0 \neq \pi/n$ is incommensurate. Then by choosing linear combinations of the $\pm k_0$ states with the appropriate phase offset, you can translate the CDW by whatever phase you want. – tparker Aug 5 '16 at 23:46

## 3 Answers

This question nagged me for most of Friday. It seems obvious that it is a Goldstone mode. You can translate the ICDW and the energy does not chage. However it is not clear what continuous symmetry remains since the lattice has already broken translation symmetry. To get to the bottom of the issue we should focus on the relevant Hamiltonian which is electron+phonon $$H=\sum_k\epsilon_k c_k^\dagger c_k+\sum_q\hbar\omega_q b_q^\dagger b_q+\sum_{k,q}g(k)c_{k+q}^\dagger c_kb_q+h.c.$$ where $c_k^\dagger$ and $b_q^\dagger$ are electron and phonon creation operators respectively and $h.c.$ denotes Hermitian conjugate of the interaction term. This Hamiltonian is invariant under the continuous transformation $$c_k\to c_k e^{ika\varphi} \\ b_q\to b_q e^{iqa\varphi}$$ for any choice of $\varphi$, and $a$ is the lattice constant. To see this is the relevant phase for CDW consider a simple 1D system with Peierls transition. There the CDW causes phonons to condense and the complex order parameter $\Delta$ is $$\Delta=|\Delta| e^{i\varphi}=g(2k_f)\langle b_{2k_f}+b_{-2k_f}^\dagger\rangle.$$ The phase of $\Delta$ is chosen by spontaneous symmetry breaking with $\varphi$ parameterizing the continuous symmetry so there is a goldstone mode. For completeness the charge density is $$\rho_0+\delta\rho\cos(2k_f x+\varphi).$$ The CDW order parameter I got from this reference.

Upon initial inspection the continuous symmetry here appears to be ordinary translation invariance ($\psi_k\to \psi_k e^{ikr}$). This cannot be correct as translation symmetry was already broken in forming the lattice and phonons. The continuous symmetry that $H$ posses is a $U(1)$ symmetry that is a remnant of the full translation symmetry $\mathbb{R}=\mathbb{Z}\times U(1)$. The $U(1)$ component of $\mathbb{R}$ is a translation symmetry with translation only defined within a unit cell of the lattice. Translation by multiple unit cells comes from the $\mathbb{Z}$ factor of $\mathbb{R}$.

• Isn't this continuous symmetry equivalent to continuous translation invariance? If you think of the electrons and phonons as being described by some local fields, then in the Fourier decomposition the annihilation operators usually appear in the combination $c_ke^{ikx}$ and the same for $b_q$. Your phase transformation is then equivalent to $x\to x+a\varphi$, no? – Tomáš Brauner Aug 7 '16 at 18:40
• It is not translation invariance. The wavenumbers in the electron+phonon Hamiltonian are restricted to the first Brillouin zone. This was not obvious to me when I first read your comment so thank you for pointing it out. – James Rowland Aug 8 '16 at 17:19
• Aha, of course, the system is defined on a discrete lattice! Can one still define a local density of the conserved charge for this symmetry? A Goldstone mode is much more than zero gap. It should couple to the conserved current (charge), which in turn restricts its interactions. – Tomáš Brauner Aug 9 '16 at 14:36
• This is probably a very dumb question, but is the interaction term actually invariant under your transformation? Doesn't $c^\dagger c b + c^\dagger c b^\dagger$ transform to $c^\dagger c b e^{i q a \varphi} + c^\dagger c b^\dagger e^{-i q a \varphi}$? Also, as @TomášBrauner asks, what is the conserved current to which the U(1) symmetry couples? – tparker Aug 23 '16 at 19:12
• You have to keep track of the $q$ indices $c_{k+q}^\dagger c_k b_q\to (c_{k+q}^\dagger e^{-i(k+q)a\varphi})(c_{k} e^{ika\varphi})(b_q e^{iqa\varphi}).$ The other term is satisfied because it is Hermitian conjugate. Since the $U(1)$ is a remnant of translation invariance the conserved current is momentum. Note that the conserved current associated with phonons is crystal momentum. – James Rowland Aug 24 '16 at 3:04

References

I have found multiple references in scientific literature referring to phasons as a Goldstone mode. Below are some examples with links:

• "...the soft, amplitudon and phason (Goldstone) modes..." from Phase Transitions in Liquid Crystals edited by Arthur N. Chester (google book format open to the correct page here)
• " The Goldstone modes evolving from the magnetic satellites consist of transverse spin-wave modes and longitudinal phason modes..." from the abstract of Goldstone Modes and Low-Frequency Dynamics of Incommensurate Chromium Alloys by R.S. Fishman and S.H. Liu (the abstract can be found here)
• "...the strongest mode is the phason (Goldstone) mode..." from Probability Measures on Semigroups: Convolution Products, Random Walks and Random Matrices by Goran Hognas and Arunava Mukherjea (google book format open to the correct page here)
• "...the phason (Goldstone) mode..." from Relaxation Phenomena: Liquid Crystals, Magnetic Systems, Polymers, High-Tc Superconductors, Metallic Glasses edited by Wolfgang Haase (google book format open to the correct page here)

This google search contains more references. This makes it seem like (as tparker stated in the comments) it is "morally" though not strictly a Goldstone mode. This now begs the question, what is the actual relationship between phason/Goldstone modes.

Relationship

After looking around for a more in-depth description of the relationship between phasons and Goldstone modes, I found this physics.SE question. Below the one answer, the second comment says

In my understanding, the Goldstone mode always corresponds to the fluctuation of the "phase" while the fluctuation of "amplitude" may be called Higgs mode. So when we talk about the gapless Goldstone mode in SDW, it should relates to the fluctuation of spin directions rather than spin length. Thus I think there is only "Phason" excitations in SDW, but conventionally we call the excitations "magnons" or spin-wave (classical partner). I hope this comment may be helpful to you.

This paper also has some relevant sections on phasons and Goldstone modes. This book has some information that might be helpful, but unfortunately I cannot find a free copy online and the google book sample I've linked too doesn't include some of the relevant pages.

I also found a quote from the book Liquid Crystals in the Nineties and Beyond (edited by S. Kumar; google book format open to the correct page here) - "The Goldstone mode, which is a phason mode with the wave vector in the center of the dispersion..." which makes it seem like a Goldstone mode is a phason mode, not the other way around.

I found a paper that might also help: Phason dynamics in nonlinear photonic quasicrystals by Barak Freedman, Ron Lifshitz, Jason Fleischer, and Mordechai Segev (the pdf can be downloaded here). The section that seems to be relevant to your Hamiltonian question is on the last page, in the last paragraph...". As such, the observed phason behaviour is representative of a more general hamiltonian dynamics commonly found in non-equilibrium pattern-forming systems". There are, of course, other relevant sections.

Hope this helps! I will continue to update this as I find more information.

• @tparker, let me know if I need to add anything else to my answer; I'd like to improve it wherever possible. – heather Aug 4 '16 at 18:39
• Well, strictly speaking a Goldstone mode must correspond to a spontaneously broken continuous symmetry, and a CDW does not break any continuous symmetries, so a phason is not literally a Goldstone mode according to the usual definition. The fact that so many authors have called it a Goldstone mode seems to reinforce my suspicion that it's "morally" (though not strictly) a Goldstone mode, but I'd like to see an actually discussion of this issue, not just a passing reference. – tparker Aug 5 '16 at 6:16
• @tparker, I just found this question that has some interesting comments in the answers; they might help you. I updated my answer with the relevant comment. I'll keep looking for more information. – heather Aug 5 '16 at 12:59
• Some good legwork there heather. – John Duffield Aug 5 '16 at 13:32
• These references are helpful, but I'd still like to see a discussion that explicitly points out that phasons don't correspond to any spontaneously broken symmetry of the original Hamiltonian. – tparker Aug 5 '16 at 14:18

Is the existence of a continuous degenerate ground-state manifold more important for "Goldstone-like" behavior than the existence of a continuous spontaneously broken symmetry?

A couple of recent papers by Takahaski and Nitta address this question: https://arxiv.org/pdf/1404.7696v3.pdf

https://arxiv.org/pdf/1410.2391v2.pdf

The authors use Bogoliubov theory to show that when the ground-state exhibits emergent symmetries--i.e. ones which have generators which do not commute with the Hamiltonian--we still anticipate gapless modes arising from the generators of the continuous ground-state degeneracy, called quasi-Goldstone-modes. Essentially, one may identify zero-modes (eigenvectors of $H_{k=0}$ with zero eigenvalue) associated with symmetry generators of the ground-state. However I think this result relies on Bogoliubov theory and is therefore most relevant to Bose-Einstein condensates.