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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,
[tex] k_F=\left(2\pi n\right)^{1/2} [/tex]
where [tex] n [/tex] is the electron surface density.
Assuming this 2D electron system can be considered as 2D tightbinding square lattice whose eigenergy can be written as,
[tex]E(k_x,k_y)=t[(\cos(k_xa))+\cos(k_ya) [/tex]
Then the electron surface density is [tex] \frac{N_e}{S}[/tex], where [tex]N_e[/tex] is the
total number of electron and [tex]S=a^2*m*n[/tex] is the total surface area (m,n is the lattice number along x and y direction respectively).
Then [tex] 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} [/tex]
Since [tex]N_e=m\cdot n[/tex], then
[tex] 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}[/tex]
where we have neglected spin freedom.
That is to say [tex] k_F\cdot a=\sqrt{2\pi}\approx 2.5 [/tex]
my question is why the following article (the paragraph in the third page, before section III, see attachment) said that
[tex] k_s a_s=\frac{\pi}{3} [/tex] corresponds to half filling?
OF course, in this article, the two dimensional square lattice has been simplified as an effective onedimensional lattice due to the conservation of the [tex]k_y [/tex], and [tex] k_s [/tex] is the [tex]k_x[/tex].
The wavefunction along the y directionc an be expressed as [tex]e^{ik_y y}[/tex]， and the effective one dimensional Hamiltonian can be written as
[tex] \sum\limits _i\bigg( t[i\rangle \langle i+1+h.c.] +2t\cos(k_s a_s)i\rangle \langle i\bigg) [/tex]
https://journals.aps.org/prb/abstract/10.1103/PhysRevB.76.155433
[tex] k_F=\left(2\pi n\right)^{1/2} [/tex]
where [tex] n [/tex] is the electron surface density.
Assuming this 2D electron system can be considered as 2D tightbinding square lattice whose eigenergy can be written as,
[tex]E(k_x,k_y)=t[(\cos(k_xa))+\cos(k_ya) [/tex]
Then the electron surface density is [tex] \frac{N_e}{S}[/tex], where [tex]N_e[/tex] is the
total number of electron and [tex]S=a^2*m*n[/tex] is the total surface area (m,n is the lattice number along x and y direction respectively).
Then [tex] 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} [/tex]
Since [tex]N_e=m\cdot n[/tex], then
[tex] 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}[/tex]
where we have neglected spin freedom.
That is to say [tex] k_F\cdot a=\sqrt{2\pi}\approx 2.5 [/tex]
my question is why the following article (the paragraph in the third page, before section III, see attachment) said that
[tex] k_s a_s=\frac{\pi}{3} [/tex] corresponds to half filling?
OF course, in this article, the two dimensional square lattice has been simplified as an effective onedimensional lattice due to the conservation of the [tex]k_y [/tex], and [tex] k_s [/tex] is the [tex]k_x[/tex].
The wavefunction along the y directionc an be expressed as [tex]e^{ik_y y}[/tex]， and the effective one dimensional Hamiltonian can be written as
[tex] \sum\limits _i\bigg( t[i\rangle \langle i+1+h.c.] +2t\cos(k_s a_s)i\rangle \langle i\bigg) [/tex]
https://journals.aps.org/prb/abstract/10.1103/PhysRevB.76.155433
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