# What does intrinsic and extrinsic contribution to Hall effect mean?

In the context of transportation theory (electron and thermal conductivities), what is the physical meaning of extrinsic and intrinsic contributions to the Hall effect (i.e. transverse conductivity)? For example, in the article "Anomalous Hall effect in the Dirac electron system with a split term", they define the Hall conductivity as

$$\sigma_{xy} = \sum_n \frac{e^2}{\hbar}\int \frac{d\vec{k}}{(2\pi)^d} \Omega_{k_x,k_y}^n (\vec{k}) f(E(\vec{k}))$$ where $$\Omega$$ is Berry curvature, and $$f(E)$$ is Fermi distribution function. They call it intrinsic Hall conductivity. What exactly does "intrinsic" mean here? What could be extrinsic?

Just another confusion, I have seen usually the formula for Hall conductivity does not contain $$f(E)$$ term in it (unlike in the above formula), does $$f(E)$$ has to do something with "intrinsic"?

## 1 Answer

In this context, "intrinsic" means that the Hall conductivity comes from the Berry curvature. That is, it's a contribution intrinsic to the band structure. Disorder can produce "extrinsic" contributions to the anomalous Hall effect through so-called side-jump and skew-scattering mechanisms. If you're interested, you can see this answer of mine for some references.

The Fermi distribution function is there to account for temperature effects, and does not have anything to do with intrinsic vs. extrinsic. At $$T=0$$ it can be set to unity below the Fermi energy (as is stated in the paper you linked), and zero above it. At finite $$T$$ quantization in a filled band is no longer exact, which the $$f(E)$$ factor accounts for.

• thank you very much for so clear answer. I have two follow-up questions. (1) In this sense, can we say that conductivity calculated using Berry curvature always gives intrinsic contribution of Hall effect and not the extrinsic? (2) If we take $f(E)=1$, the remaining formula becomes $\sigma_{xy}=e^2/\hbar \sum_n C_n$, where $C_n$ is Chern number of nth band. And as $\sum_n C_n =0$, so, the intrinsic conductivity is always zero at $T=0$? – Luqman Saleem Nov 10 '20 at 11:50
• @LuqmanSaleem (1) I believe so. (2) Sorry, I was unclear. You want to set $f(E)=1$ for occupied bands - i.e. for $E<E_F$ and $f(E>E_F)=0$. That way you sum over occupied states only, and the intrinsic conductivity can be non-zero. See the equation on bottom of page 2 in the article you linked. – Anyon Nov 10 '20 at 15:15
• got it! thank you once again. – Luqman Saleem Nov 10 '20 at 15:47