# Calculation of the quantized Hall coefficient in the Integral Quantum Hall Effect

I have been reading about the QHE over the past couple of days. I am facing difficulty understanding a calculation in this review.

In this paper, equation 16 gives the energy levels of a 2d electron gas as $$E_k =\frac{\hbar^2}{2m}[k_x^2+k_y^2]$$ which I have written as $$E_k=\alpha k^2$$ The author then says (eqn 17) that the number of states with energy less than $E$ is $$N(E)=\frac{S_k}{(2\pi)^2}=\frac{\pi k^2}{(2 \pi)^2}$$ $S_k$ is the area in k space. I don't know what the $k$ in this equation is, and if it is the same $k$ that I have taken.

I am not being able to understand how they got this equation. I integrated the equation for $E_k$ from $-\sqrt{\frac{E}{\alpha}}$ to $\sqrt{\frac{E}{\alpha}}$. But I am getting it as $\frac{2}{3}\sqrt{\alpha}k^3$.

How do I get that equation? Why is that equation true in the first place? What is the $(2\pi)^2$ in the denominator for?

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It seems to me that your question is not about the calculation of the quantized Hall coefficient, but the calculation of numer density of 2D electron gas. I would suggest to modify the title of this question. –  Everett You Apr 20 '13 at 19:08
You should calculate the number of $k$s inside the circle $k<k_f$. If the two dimensional medium is finite with sizes $L_x$ , $L_y$ the density of points will be $\frac{(2\pi)^2}{L_xL_y}$. The number of $k$s inside the circle is $\pi k_f^2/(\frac{(2\pi)^2}{L_xL_y})$. Then the number of $k$s per unit area is that $N(E)$ above.