# Difference between Charge Density Wave (CDW) and Superconductivity?

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?

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.
• @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. – Everett You Jun 15 '17 at 4:40