# Ground state symmetry breaking in Bose-Hubbard model with spin-orbit coupling

The Hamiltonian for 2D Bose-Hubbard model with spin-orbit coupling on a square lattice is written as

$H = -t\sum_{\langle ij \rangle}\Psi_i^{\dagger}\Psi_j^{\vphantom{\dagger}} + \frac{U}{2}\sum_{i\sigma}n_{i\sigma}(n_{i\sigma}-1) + U_{\uparrow \downarrow} \sum_{i}n_{i\uparrow}n_{i\downarrow}- \mu\sum_{i\sigma}n_{i\sigma} + i\lambda\sum_{\langle ij \rangle}\Psi^{\dagger}_i \vec{e}_z \cdot (\vec{% \sigma} \times \vec{d}_{ij} ) \Psi_j^{\vphantom{\dagger}}+ H.c.,$ where the last term describes Rashba spin-orbit coupling (SOC).

In the absence of SOC ($\lambda=0$ ), we know the mean-filed superfluid ground state of the Hamiltonian has uniform phase. So the spin-dependent momentum distribution $\langle \rho _{\uparrow ,\downarrow }(\vec{k})\rangle=N^{-2}\sum_{i,j}\langle b_{i\uparrow }^{\dagger }b_{j\downarrow }^{% \vphantom{\dagger}}\rangle e^{i\vec{k}\cdot \left( \vec{R}_{i}-\vec{R}% _{j}\right) }$ has inversion symmetry in k-space.

In the presence of SOC ($\lambda \neq 0$ ), the mean-filed superfluid ground state has non-uniform phase, $\langle b_{i\uparrow }^{\dagger }b_{j\downarrow }\rangle$ is not real any more. Is the spin-dependent momentum distribution still have inversion symmetry in k-sapce? Or say can SOC break the symmetry in k-space?