Competing Orders and Electronic Nematicity

Magnetic, charge, pairing, and nematic orders compete or cooperate through symmetry-allowed couplings because they draw on the same fermions. A coupled free energy classifies coexistence and exclusion locally, while susceptibilities and finite-size scaling decide which tendency actually orders. Coincident crossover scales do not establish causal hierarchy.

Required background. Use Pomeranchuk and density-wave distinctions and Hubbard symmetries. Helpful background. Functional RG compares weak-coupling channels.

Coupled magnetic, nematic, and superconducting orders

Let $\mathbf M_x$ and $\mathbf M_y$ denote symmetry-related stripe magnetic fields at $(Q,0)$ and $(0,Q)$, and let $\Psi$ be a superconducting field. A minimal free energy is

$$\begin{aligned} F={}&\frac r2(M_x^2+M_y^2) +\frac u4(M_x^2+M_y^2)^2 -\frac g4(M_x^2-M_y^2)^2\\ &+r_\Psi|\Psi|^2+u_\Psi|\Psi|^4 +w|\Psi|^2(M_x^2+M_y^2)-h\phi, \qquad \phi=M_x^2-M_y^2. \end{aligned}$$

For $g>0$, fluctuations prefer one stripe orientation, so the composite Ising variable $\phi$ is nematic. It may order before $\langle\mathbf M_{x,y}\rangle$ and is then vestigial. This is distinct from a primary $q=0$ fermionic Pomeranchuk field, even though both break the same point-group symmetry.

The coupling $w>0$ makes magnetic amplitude and superconductivity compete at mean-field level; sufficiently small $w^2$ relative to the quartic self-couplings permits coexistence. Fluctuations can change this classification, so the functional is a symmetry analysis, not a microscopic mechanism.

Susceptibility and explicit strain

For fields $h_i$ conjugate to order parameters $O_i$, the susceptibility matrix

$$\chi_{ij}=\frac{\partial\langle O_i\rangle}{\partial h_j} =\beta\big(\langle O_iO_j\rangle-\langle O_i\rangle\langle O_j\rangle\big)$$

contains the covariance that must accompany comparisons. Uniaxial strain is a field $h$ conjugate to nematicity, rounding an Ising transition and selecting domains. One must extrapolate $h\to0$ after the thermodynamic limit before calling the response spontaneous.

The broad intertwined-order framework and experimental coupling issues are reviewed in Fradkin, Kivelson, and Tranquada 2015, §§ II–V.

Exercises

Why can $\langle\phi\rangle\ne0$ while $\langle\mathbf M_x\rangle=\langle\mathbf M_y\rangle=0$?

Solution

$\phi$ is a spin-rotation-invariant composite. Fluctuations can choose unequal variances $\langle M_x^2\rangle\ne\langle M_y^2\rangle$ without developing either vector expectation value. Point-group symmetry is then broken while spin rotation and translation remain unbroken.

References

• Eduardo Fradkin, Steven A. Kivelson, and John M. Tranquada, “Colloquium: Theory of Intertwined Orders in High Temperature Superconductors,” Reviews of Modern Physics 87 (2015) 457–482, doi:10.1103/RevModPhys.87.457, Open PDF.