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

Table of contents
- i a
  - Solution i
- ii a
  - Solution ii
- iii a
  - Solution iii

i

Solution

In the orthogonal nearest-neighbour tight-binding model,

α_{s} = ⟨ s_{j} ∣ H ∣ s_{j} ⟩, β_{s} = ⟨ s_{j} ∣ H ∣ s_{j + 1} ⟩ .

The on-site integral

α_{s}

sets the band centre, while the hopping integral

β_{s}

describes coupling between neighbouring orbitals. With a consistent phase convention, the bonding combination at the zone centre has

β_{s} < 0

.

For p orbitals directed along the chain, use

E_{p} (k) = α_{p} + 2 β_{p} cos (ka), β_{p} > 0.

Here every positive

p

lobe points in the same spatial direction. Neighbouring equal-sign coefficients then bring opposite-sign facing lobes together and make the zone-centre combination antibonding; the alternating zone-boundary combination is bonding. Thus the usual axial

p

hopping has the opposite sign to

s

hopping. Orbital phase and the sign of one-dimensional hopping must be treated consistently; replacing only

α_{s}

by

α_{p}

while silently keeping the same physical hopping sign would give the wrong band ordering.

ii

Solution

Write

t_{s} = ∣ β_{s} ∣

and

t_{p} = ∣ β_{p} ∣

, so

E_{s} = α_{s} - 2 t_{s} cos (ka), E_{p} = α_{p} + 2 t_{p} cos (ka) .

The labelled key points of the two dispersion relations are

k 0 π / (2 a) π / a E_{s} α_{s} - 2 t_{s} α_{s} α_{s} + 2 t_{s} E_{p} α_{p} + 2 t_{p} α_{p} α_{p} - 2 t_{p}

The

s

band rises from centre to boundary and the axial

p

band falls. The separated-band assumption requires

α_{p} - α_{s} > 2 (t_{s} + t_{p})

. The upper-left panel shows a representative pair, with the other panels supplying the requested comparisons in part (b).

Figure 1.
Original one-dimensional s and axial-p band sketches, comparing separated bands, mixing, and the equal-on-site-energy special case
. Details.

iii

Solution

The energy-band widths are the difference between their extrema:

W_{s} = 4 t_{s} = 4∣ β_{s} ∣, W_{p} = 4 t_{p} = 4∣ β_{p} ∣.

For the separated bands, the top of the

s

band and bottom of the

p

band are both at

k = π / a

. Hence

E_{g}^{m i n} = α_{p} - α_{s} - 2 (t_{s} + t_{p}) .

It is a direct band gap at the zone boundary, because both limiting states have the same wavevector. This result uses the opposite signs of the

s

and axial-

p

hoppings.

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