= Solution
Choose orthonormal localized orbitals $|\mathbf R\rangle$. The on-site energy is $E_0=\langle\mathbf R|H|\mathbf R\rangle$, and the sign convention in the question defines $t(\mathbf T)=-\langle\mathbf R|H|\mathbf R+\mathbf T\rangle$. <Bloch state> diagonalizes this translationally invariant hopping problem. For real, inversion-symmetric nearest-neighbor amplitudes, $t(\pm a\hat x)=t_x$ and $t(\pm b\hat y)=t_y$, so the exponential pairs yield
$$
\boxed{E(\mathbf k)=E_0-2t_x\cos(k_xa)-2t_y\cos(k_yb).}
$$
The first zone is the rectangle $|k_x|\leq\pi/a$, $|k_y|\leq\pi/b$, wider in $k_x$ because $a<b$. Use $\Gamma=(0,0)$, $X=(\pi/a,0)$, $Y=(0,\pi/b)$ and $M=(\pi/a,\pi/b)$.
For positive $t_y=t$ and $t_x=2t$, the four corner energies and the requested path are
$$
\boxed{E_\Gamma=E_0-6t,\quad E_M=E_0+6t,\quad
E_X=E_0+2t,\quad E_Y=E_0-2t;\qquad
\Gamma\to M\to X\to\Gamma\to Y\to M\to\Gamma.}
$$
Along a path parametrized in reduced coordinates $x=k_xa$, $y=k_yb$, the full curve is simply $E-E_0=-4t\cos x-2t\cos y$. In particular the two appearances of $\Gamma$ and the two diagonal traversals must have consistent endpoint energies.<Solid State Physics by David Tong>[§2.1.1, localized orbitals and hopping sign convention; chapter 2, Bloch-band construction]{ref} A reversed hopping sign changes which point is the band minimum; the usual positive-hopping convention is stated explicitly here.
\Image[/past-exam-of-the-condensed-matter-physics-major-option-course-of-the-university-of-oxford/2015-rectangular-tight-binding-band-and-fermi-lines.png]
{title=Rectangular tight-binding zone, the specified band path and half-filled Fermi lines}
{description=Original evaluation for tx equal to twice ty. Reduced coordinates identify the symmetry points; the physical reciprocal cell is wider along kx when a is less than b. The Fermi lines are open in the repeated-zone picture.}
Back to article page
