Anomalous Hall Effect Saturation Field with Berry Curvature

Question

Whenever you see anomalous Hall effect resistivity/conductivity vs. external magnetic field curves, there is always some low-field part with a greater slope that changes to a lesser slope at some characteristic field. In magnets, this low-field region is usually attributed to magnetic domains aligning to the direction of the external field, and the point where the slope changes is called the "saturation magnetization"--this is where all the domains are aligned.
What causes this characteristic field in non-magnetic systems that show the AHE (topological semimetals)?

I.e., What causes the "shoulder" to be at ~1.5 T in this plot of the AHE in ZrTe5--a Dirac semimetal--since it is not magnetic and are therefore no magnetic domains?

Alex · Answer

In this case, the slope is not just "lesser", it is zero at high fields. This is because the high field linear resistance has been subtracted off of the total Hall resistivity to isolate the anomalous component (i.e. this plot does not show the true Hall resistivity that would be measured directly).
There is evidence that this anomalous contribution is Berry curvature induced (see reference by Sun et. al. below), in which case the vanishing of the anomalous contribution would be indicative of a topological phase transition (Chern insulator -> trivial insulator). The Hall resistivity above 1.5 T becomes linear with negative slope, as the charge carriers of $text{Zr}text{Te}_5$ are holes.
Reference:

Sun, Z., Cao, Z., Cui, J. et al. Large Zeeman splitting induced
anomalous Hall effect in ZrTe5. npj Quantum Mater. 5, 36 (2020).
https://doi.org/10.1038/s41535-020-0239-z

For completeness, the source article citation for this image is:

Liang, T., Lin, J., Gibson, Q. et al. Anomalous Hall effect in ZrTe5.
Nature Phys 14, 451–455 (2018).
https://doi.org/10.1038/s41567-018-0078-z

Anomalous Hall Effect Saturation Field with Berry Curvature

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