# How to convert quantum anomalous Hall conductivity values from (Ohm.cm)-1 to e2/hc?

I am currently working on anomalous Hall conductivity calculations using both Wanniertools and wannier90. The output values from these calculations are provided in units of (Ohm$$\cdot$$cm)-1. I am interested in converting these output values into the more familiar unit of e2/hc. Could you kindly provide guidance on the most straightforward and simple method for performing this conversion? Your expertise in this matter would be greatly appreciated and would help me immensely in my research. Thank you in advance for your assistance, and I look forward to your response.

• +1 but e2 or $e^2$? Sep 26, 2023 at 22:17
• Are you sure there is a $c$ in there? Hall conductance is typically expressed in units of $e^2/h$. Sep 28, 2023 at 17:49

# The unit conversions

The SI base units for Ohms are:

$$\Omega = \frac{\textrm{kg}\cdot \textrm{m}^2}{\textrm{s}^3\cdot \textrm{A}^2}\tag{1}$$

Since 1 Joule is:

$$\tag{2} \textrm{J} = \frac{\textrm{kg}\cdot \textrm{m}^2}{\textrm{s}^2},$$

and 1 Coulomb is:

$$\tag{3} \textrm{C} = \textrm{A}\cdot \textrm{s},$$

we have:

$$\tag{4} \Omega = \frac{\textrm{J}\cdot \textrm{s}}{\textrm{C}^2}.$$

The final units of the output from your software, are as follows:

$$\tag{5} \frac{1}{\Omega \cdot \textrm{cm}} = \frac{\textrm{C}^2}{\textrm{J}\cdot \textrm{s}\cdot \textrm{cm}}.$$

You then said:

"I am interested in converting these output values into the more familiar unit of e2/hc."

So let's first see what the units are for that quantity:

$$\tag{6} \left[ \frac{e^2}{hc}\right] = \frac{\textrm{C}^2}{\textrm{J}\cdot \textrm{m}}.$$

Even if we convert cm to m in Eq. (5):

$$\tag{7} \frac{1}{\Omega \cdot \textrm{cm}} = \frac{\textrm{C}^2}{\textrm{J}\cdot \textrm{s}\cdot \textrm{cm}}\times \frac{100~\textrm{cm}}{\textrm{m}} = \frac{100 ~ \textrm{C}^2}{\textrm{J}\cdot \textrm{s}\cdot \textrm{m}},$$

we still have incompatible units because what the software is giving you has "seconds" in the denominator and what you want does not.

Perhaps what you want is not $$e^2/hc$$ but $$e^2/(\epsilon_0hc)$$, but even then, and even if you were to convert cm$$^{-1}$$ to Joules, you would not get compatible units.

# The final recommendation

I recommend to stick with the units that the software is using when printing the output, until you know better what unit system you would like your publication to use. Almost always, the authors of the software know very much what they're doing, and they will print the output in a unit that is acceptable for their own publications.