Past exam of the chemistry course of the year 3 of the natural sciences course of the University of Cambridge / 2022 / Paper 4 / 59 / b / iv

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- Solution iv

Solution

The midpoint-odd perturbation couples the ground state only to even-index states. Direct integration gives

⟨ 1∣ x - a /2∣2 j ⟩ = - \frac{16 aj}{π ^{2} ( 4 j ^{2} - 1 ) ^{2}}, E_{2 j}^{(0)} - E_{1}^{(0)} = \frac{h ^{2}}{8 m a ^{2}} (4 j^{2} - 1) .

Insertion in the second-order expression yields

E_{1}^{(2)} = - \frac{2048 m a ^{4}}{h ^{2} π ^{4}} \sum_{j = 1}^{\infty} \frac{j ^{2}}{( 4 j ^{2} - 1 ) ^{5}} .

It is negative because all coupled states lie above the ground state. This is the coefficient of

λ^{2}

; the physical energy shift includes the additional factor

q^{2} ε^{2}

.

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