Past exam of the chemistry A course of the University of Cambridge / 2022 / Paper 2 / A9 / b / iii

Use $t=|\beta |$ and $\gamma =|{\beta }_{sp}|<t$. Direct substitution givesThe 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 $D=\sqrt{t^2{\cos }^{2}u+{\gamma }^2{\sin }^{2}u}$. ThenThe 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.