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I am struggling to see the difference between these two mechanisms.

If they are both electron-phonon mediated and both distort the lattice then why don't Cooper pairs form at the CDW transition temperature?

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A brief answer is that charge density wave (CDW) and superconductivity (SC) are two different symmetry breaking phases that break different symmetries. The CDW breaks the lattice translation and particle-hole symmetry but preserves the charge conservation symmetry. The SC breaks the charge conservation symmetry but preserves the lattice translation and particle-hole symmetry. It is not true that superconductivity necessarily distorts the lattice. Although in some materials SC may coexist with CDW, yet they are still distinct orders.

However, from a higher point of view, CDW and SC order can be unified into an O(3) vector $$\boldsymbol{n}=\psi_{\boldsymbol{k}+\boldsymbol{Q}}^\dagger\ \boldsymbol{\sigma}\psi_{\boldsymbol{k}}$$ where $\psi_{\boldsymbol{k}}=(c_{\boldsymbol{k}\uparrow},c_{-\boldsymbol{k}\downarrow}^\dagger)^\intercal$ is the Nambu spinor. The $n_1+\mathrm{i}n_2=\Delta$ forms the SC order parameter and $n_3$ is the CDW order parameter, they can be rotated to each other by the SU(2) transformation of the Nambu spinor $$\psi_{\boldsymbol{k}}\to \exp(\mathrm{i}\boldsymbol{\theta}\cdot\boldsymbol{\sigma})\psi_{\boldsymbol{k}},$$ which is also called the particle-hole SU(2) transformation. In this sense, the relation between CDW and SC is like that between Ising order and XY order in the spin system. Therefore, if the Hamiltonian indeed has the particle-hole SU(2) symmetry (i.e. there is no anisotropy between CDW and SC), then, yes, there is no difference between CDW and SC. The CDW transition and the SC transition becomes the same transition, happening at the same temperature and belongs to the same universality class. One such example is the negative-$U$ Hubbard model at half-filling $$H=-t\sum_{\langle ij\rangle,\sigma}(c_{i\sigma}^\dagger c_{j\sigma}+\text{h.c.})+U\sum_{i}n_{i\uparrow}n_{i\downarrow},$$ with $U<0$ providing the attractive interaction which is necessary to for the formation of both the CDW and the SC order. In this model, the CDW and SC orders are unified.

But in reality, the particle-hole SU(2) symmetry is always broken by the chemical potential term $-\mu\sum_{i,\sigma}c_{i\sigma}^\dagger c_{i\sigma}$. If there is no obvious reason why the chemical potential must be pinned to the half-filling point, then the degeneracy between CDW and SC is lifted. The CDW and the SC transition becomes two different transitions at different temperatures. Because the Fermi surface is always perfectly nested for SC, so the SC instability is generally stronger than CDW (at least in the negative-$U$ Hubbard model away from half-filling), thus we will enter the SC phase before entering the CDW phase. But more generally, even the interaction term itself can break the particle-hole SU(2) symmetry, for example, the phonon-mediated electronic interaction does not need to be the same in the CDW channel as in the SC channel. If there is no symmetry relating the CDW and the SC orders, then there is no reason why these two transitions should happen at the same point.

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  • $\begingroup$ Can you explain why the CDW break the particle-hole symmetry? $\endgroup$ – Chuan Chen Jun 15 '17 at 4:07
  • $\begingroup$ @ChuanChen Because the CDW order parameter changes sign under the particle-hole transformation $c\to c^\dagger$, so the CDW order breaks the particle-hole symmetry. $\endgroup$ – Everett You Jun 15 '17 at 4:40
  • $\begingroup$ This answer leaves a lot to be desired. What about a discussion on the differences in phenomenology between the two orders? The microscopic discussion of CDWs never involves electron-hole pairs, and dissipation-free flow is never achieved. Why do the different symmetries necessitate a different interaction with phonons and different physics etc. $\endgroup$ – KF Gauss Jun 15 '17 at 6:23
  • $\begingroup$ @ChuanChen, physically speaking a charge density wave means the charge density oscillates in space. This mean electrons will accumulate in the negative regions and holes in the positive regions, hence the obvious asymmetry $\endgroup$ – KF Gauss Jun 15 '17 at 6:27
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    $\begingroup$ @Scientized You need to distinguish the difference between lattice distortion as a virtual process and the lattice distortion as a ground state order. The formal is a quantum effect, just like a particle tunneling a barrier, it is just a theoretical picture to illustrate electron-phonon interaction, there is no static and persistent lattice distortion that can be observed by experiments. CDW is a ground state order. The lattice distortion is static and real, and there are experimental consequences, such as diffraction peaks in X-ray and neutron scattering, band folding in ARPES ... $\endgroup$ – Everett You Jun 16 '17 at 20:16

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