Why does an in-plane field suppress the 2D metallic state?
In plain words
A magnetic field along the sheet, which aligns electron spins but does not bend their paths, raises the resistance many times and in silicon transistors removes the metal-like behavior. Why spin alignment has such a large effect, and why some ultra-clean samples stay metallic even with all spins aligned, is not understood.
Precise statement
In Si-MOSFETs near $n_c$ an in-plane field $B_{\mathrm{parallel}}$ raises $\rho$ by one to two orders of magnitude and saturates at $B_c(n)$, where the electrons become fully spin polarized; above $B_c$ the metallic $\mathrm{d}\rho/\mathrm{d}T > 0$ is lost in Si-MOSFETs, while in ultra-clean SiGe/Si/SiGe wells a metallic state and a metal-insulator transition survive in the fully polarized (spinless, two-valley) system (Melnikov et al. 2020). Explain the magnitude of the positive magnetoresistance and its saturation at B_c, explain why the spinless two-valley system stays metallic while the Si-MOSFET does not, and give the ratio $n_c(B > B_c)/n_c(0)$.
What would settle it
A theory that predicts $\rho(B_{\mathrm{parallel}}, T, n)$ including $B_{c}(n)$ with parameters fixed independently, plus transport at $B > B_{c}$ down to $T << 0.01 E_{F}$ in both Si-MOSFETs and SiGe/Si/SiGe wells.
Status in the literature
A metallic state of fully spin-polarized electrons was reported in ultra-clean SiGe/Si/SiGe wells (Melnikov et al., Physical Review B 101, 045302, 2020); no theory accounts for the material dependence.