= Solution
Use $t=|\beta|$ and $\gamma=|\beta_{sp}|<t$. Direct substitution gives
$$
\boxed{\begin{array}{c|c|c}
k&E_{k,1}&E_{k,2}\\\hline
0&+2t&-2t\\
\pi/(2a)&+2\gamma&-2\gamma\\
\pi/a&+2t&-2t
\end{array}}
$$
The upper branch falls to a minimum at the middle point and rises symmetrically; the lower branch rises to a maximum and falls symmetrically. These are the lower-right curves in the figure in part (a)(ii).
At $k=0$ the lower state is pure $s$ and the upper state pure $p$. At $k=\pi/a$ the assignments are reversed. At $k=\pi/(2a)$ the two states have equal orbital weights because their diagonal energies coincide and the off-diagonal coupling is nonzero.
More explicitly, for the lower band and $u=ka$, define $D=\sqrt{t^2\cos^2u+\gamma^2\sin^2u}$. Then
$$
\boxed{w_s^{(-)}=\frac12\left(1+\frac{t\cos u}{D}\right),\qquad
w_p^{(-)}=1-w_s^{(-)}.}
$$
The upper-band weights are interchanged. For zero mixing, the energies still reduce to the crossing limit, but the orbital weights at the exact crossing are not uniquely fixed by <energy>. These normalized squared amplitudes quantify the smooth change of orbital character; the Bloch-basis phase in the off-diagonal matrix element does not change them.
Back to article page
