# Half filling? Free electron gas? 2D tight binding lattice?

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## Main Question or Discussion Point

It is well known that the 2D free electron gas fermi momentum can be expressed as follows,

$$k_F=\left(2\pi n\right)^{1/2}$$
where $$n$$ is the electron surface density.
Assuming this 2D electron system can be considered as 2-D tight-binding square lattice whose eigenergy can be written as,
$$E(k_x,k_y)=t[(\cos(k_xa))+\cos(k_ya)$$
Then the electron surface density is $$\frac{N_e}{S}$$, where $$N_e$$ is the
total number of electron and $$S=a^2*m*n$$ is the total surface area (m,n is the lattice number along x and y direction respectively).

Then $$k_F=\left(2\pi n\right)^{1/2} =\left(2\pi \frac{N_e}{S}\right)^{1/2}= \left(2\pi \frac{N_e}{m\cdot n\cdot a^2}\right)^{1/2}$$
Since $$N_e=m\cdot n$$, then
$$k_F= \left(2\pi \frac{N_e}{m\cdot n\cdot a^2}\right)^{1/2} =\left(2\pi \frac{1}{a^2}\right)^{1/2}$$
where we have neglected spin freedom.
That is to say $$k_F\cdot a=\sqrt{2\pi}\approx 2.5$$
my question is why the following article (the paragraph in the third page, before section III, see attachment) said that
$$k_s a_s=\frac{\pi}{3}$$ corresponds to half filling?

OF course, in this article, the two dimensional square lattice has been simplified as an effective one-dimensional lattice due to the conservation of the $$k_y$$, and $$k_s$$ is the $$k_x$$.
The wavefunction along the y directionc an be expressed as $$e^{ik_y y}$$， and the effective one -dimensional Hamiltonian can be written as
$$\sum\limits _i\bigg( t[|i\rangle \langle i+1|+h.c.] +2t\cos(k_s a_s)|i\rangle \langle i|\bigg)$$

https://journals.aps.org/prb/abstract/10.1103/PhysRevB.76.155433

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